IB Chemistry SL Review

IB Chemistry SL Review

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IB Chemistry/Stoichiometry < IB Chemistry The latest reviewed version was checked on 26 March 2012. There is 1 pending change awaiting review. Jump to: navigation, search

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1 Topic 1 - Stoichiometry o 1.1 1.1 Mole concept & Avogadro's constant  1.1.1 1.1.1: Apply the mole concept to substances.  1.1.2 1.1.2: Determine the number of particles and the amount of substance (in moles). o 1.2 1.2 Formulae  1.2.1 1.2.1: Define the term molar mass (M) and calculate the mass of one mole of a species.  1.2.2 1.2.2: Atomic mass, Molecular Mass, Formula Mass  1.2.3 1.2.3: Define the terms relative molecular mass (Mr) and relative atomic mass (Ar)  1.2.4 1.2.4: The difference between moles and mass  1.2.5 1.2.5: The difference between the molecular formula and the empirical formula o 1.3 1.3 Chemical Equations  1.3.1 Anatomy of an Equation  1.3.2 Reacting Species  1.3.3 Coefficients  1.3.4 Other Information  1.3.5 Examples  1.3.6 Simple Formulas o 1.4 1.4 Mass relationships in chemical reactions o 1.5 1.5 Solutions 2 TOPIC 1: QUANTITATIVE CHEMISTRY (12.5 HOURS) o 2.1 Formula and Valency of Ions  2.1.1 SOME COMMON CATIONS  2.1.2 Some Common Anions 3 1.1 MOLE CONCEPT AND AVOGADRO’S CONSTANT (2H) 1.2 FORMULAS (3H) o 3.1 Relative molecular mass (Mr) and relative atomic mass (Ar). o 3.2 Calculating the mass of one mole of a species from its formula. o 3.3 Applying the mole concept to substances. o 3.4 Solving problems involving amount, mass and molar mass. 2

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o 3.5 Empirical formula and molecular formula. 4 1.3 CHEMICAL EQUATIONS (1H) o 4.1 Balancing chemical equations o 4.2 The mole ratios of any species in a chemical equation. o 4.3 The state symbols (s), (l), (g) and (aq). 5 1.4 MASS AND GASEOUS VOLUME RELATIONSHIPS IN CHEMICAL REACTIONS (3H) o 5.1 Calculating theoretical yields from chemical equations. o 5.2 The limiting reactant and the reactant in excess: Theoretical, experimental and percentage yield. o 5.3 Avogadro’s law of volumes of gases. o 5.4 Graphs relating to the ideal gas equation. o 5.5 Molar volume of a gas o 5.6 The relationship between temperature, pressure and volume for a fixed mass of an ideal gas. 6 1.5 SOLUTIONS (2H) o 6.1 Solute, solvent, solution and concentration

[edit] Topic 1 - Stoichiometry [edit] 1.1 Mole concept & Avogadro's constant [edit] 1.1.1: Apply the mole concept to substances. A mole is equivalent to 6.022 x 1023 (Avogadro's constant) units. Chemists refer to a mole of something much as we refer to a dozen eggs; it is a convenient unit for counting. The periodic table provides molar masses, i.e. the number of grams of an element equivalent to one mole of atoms of that specific element. This can be extrapolated to molecules of known molecular formula. [edit] 1.1.2: Determine the number of particles and the amount of substance (in moles). Number of moles = mass / molar mass (Usually found on periodic table). The coefficients in chemical equations give the molar ratios of reactants and products i.e. 2A + 3B → C. There is 2/3 as much A as B, and 3 times more B than C involved in the reaction. Assuming the reaction goes to completion, there must be 3/2 times as much B as A for neither to remain. If this ratio is not followed, one will be a limiting reactant, and so the reaction will have some of the other reactant left over when it completes.

[edit] 1.2 Formulae [edit] 1.2.1: Define the term molar mass (M) and calculate the mass of one mole of a species. 3

The molar mass (M) is the mass of one mole's worth of a substance. To find the mass, multiply the amount of moles by the molar mass. [edit] 1.2.2: Atomic mass, Molecular Mass, Formula Mass The molar mass can be found for the periodic table, and will give the mass for 1 mol of the species (or rather the average accounting for different isotopes and their relative abundance). [edit] 1.2.3: Define the terms relative molecular mass (Mr) and relative atomic mass (Ar) Mr is the ratio between the molar masses of two species. Ar is the ratio of the number of atoms between two species. These two ratios will be equal. there should be more metal compare to copper [edit] 1.2.4: The difference between moles and mass The number of moles refers to the amount of the substance, every mol being 6.02 x 10 23 individual elements. Mass is the property which results in 'weight' in the presence of gravity. Given a molar mass, M a mass m and a number of mols n then n = m / M. [edit] 1.2.5: The difference between the molecular formula and the empirical formula An 'empirical formula' is the formula describing the different atoms present in a molecules, and their ratios, but not the actual number present. For example, AxByCz could be an empirical formula if x, y, and z are in lowest common terms. The molar mass can then be used to calculate the actual numbers of each atom present per molecule. The empirical formula can be determined by percentage composition, or anything else which gives the ratios of atoms present. A 'molecular formula', on the other hand, has the actual number of atoms present in each molecule. It will be an integer multiple of the empirical formula. For example, A2xB2yC2z. Example: If you are given the empirical formula with an equation CH3, and a mass of 30.08 g mol1 . What is the molecular formula of this compound? We will compare the molecular mass of the molecular formula, 30.08 g mol -1, to that of the empirical formula, found to be 15.04 g mol-1. By dividing, we find the ratio to be 2, meaning that the molecular formula must be 2 times as large as the empirical formula. The molecular formula is therefore C2H6.

[edit] 1.3 Chemical Equations Chemical equations are a convenient, standardised system for describing chemical reactions. They contain the following information.  

The type of reactants consumed and products formed The relative amounts of reactants and products 4

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The electrical charges on ions The physical state of each species (e.g. solid, liquid, gas) The reaction conditions (e.g. temperature, catalysts)

The final two points are optional and sometimes omitted. [edit] Anatomy of an Equation

Hydrogen gas and chlorine gas will react vigorously to produce hydrogen chloride gas. The equation above illustrates this reaction. The reactants, hydrogen and chlorine, are written on the left and the products (hydrogen chloride) on the right. The large number 2 in front of HCl indicates that two molecules of HCl are produced for each 1 molecule of hydrogen and chlorine gas consumed. The 2 in subscript below H indicates that there are two hydrogen atoms in each molecule of hydrogen gas. Finally, the (g) symbols subscript to each species indicates that they are gases. [edit] Reacting Species Species in a chemical reaction is a general term used to mean atoms, molecules or ions. A species can contain more than one chemical element (HCl, for example, contains hydrogen and chlorine). Each species in a chemical equation is written:

E is the chemical symbol for the element, x is the number of atoms of that element in the species and y is the charge (if it is an ion). For example, ethanol would be written

because each molecule contains 2 carbon, 6

hydrogen and 1 oxygen atom. A magnesium ion would be written positive charge. Finally, an ammonium ion would be written contains 1 nitrogen and 4 hydrogen atoms and has a charge of 1+.

because it has a double because each molecule

[edit] Coefficients The numbers in front of each species have a very important meaning - they indicate the relative amounts of the atoms that react. The number infront of each species is called a coefficient. In the above equation, for example, one H2 molecule reacts with one Cl2 molecule to produce two molecules of HCl. This can also be interpreted as moles (i.e. 1 mol H2 and 1 mol Cl2 produces 2 mol HCl). [edit] Other Information 5

Occasionally, other information about a chemical reaction will be supplied in an equation (such as temperature or other reaction conditions). This information is often written above the reaction arrow. We will ignore this for now, as it only complicates matters (and it's hard to draw in TeX :). 1.3.1 : The mole ratio of two species in a chemical equation is the ratio of their coefficients...ie aX + bY → cZ : The ratio of X/Y is a/b, Y/Z = b/c etc. Chemical equations are useful because they give the relative amounts of the substances that react in a chemical equation. For example, from the chemical equation for the formation of ammonia, we can see that one mole of nitrogen gas will combine with three moles of hydrogen gas to form two moles of ammonia gas.

1.3.2 : Balancing equations...change only the coefficients, not the subscripts to make sure all atoms, and charge is conserved (half equations can be balanced by addition of electrons to either side...2 half equations can be added by making the number of electrons equal in each, then vertically adding.) In some cases, however, we may not know the relative amounts of each substance that reacts. Fortunately, we can always find the correct coefficients of an equation (the relative amounts of each reactant and product) by applying the law of conservation of matter. Because matter can neither be created nor destroyed, the total number of each atom on one side of the equation must be the same as the total on the other. This process of finding the coefficients is known as balancing the equation. For example, assume in the above equation that we do not know how many moles of ammonia gas will be produced:

From the left side of this equation, we see that there are 2 atoms of nitrogen gas in the molecule N2 (2 atoms per molecule x 1 molecule), and 6 atoms of hydrogen gas in the 3 H2 molecules (2 atoms per molecule x 3 molecules). Because of the law of conservation of matter, there must also be 2 atoms nitrogen gas and 6 atoms of hydrogen gas on the right side. Since each molecule of the the resultant ammonia gas (NH3) contains 1 atom of nitrogen and 3 atoms of hydrogen, 2 molecules are needed to obtain 2 atoms of nitrogen and 6 atoms of hydrogen. In a similar manner, you can use the law of conservation of matter to solve equations containing a greater number of unknown coefficients (the relative amounts of each reactant and product), or even subscripts (the number of each element in a molecule) on either side of the equation:

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1.3.3 : State symbols -- (s)-Solid , (l)-liquid, (g)-gas, (aq)-aqueous solution...i.e. something dissolved in water. Should be included in all chemical reactions (but won't be penalized).

E is the chemical symbol for the element and (state) is the physical state.

The symbols in parentheses (in subscript below each species) indicate the physical state of each reactant or product.     

(s) means solid (l) means liquid (g) means gas (aq) means aqueous solution (i.e. dissolved in water) (CH3CH2OH) means dissolved in ethanol - any solvent can be noted this way.

[edit] Examples

Above is the equation for burning methane gas (CH4) in the presence of oxygen (O2) to form carbon dioxide and water: CO2 and H2O respectively.

This is a precipitation reaction in which dissolved lead cations and iodide anions combine to form a solid yellow precipitate of lead iodide (an ionic solid). [edit] Simple Formulas Moles to Particles: number of moles x (6.02 x 1023/1 mole)= number of Particles ex: 2.50 mol Zn x (6.02*1023 atoms of Zn/1 mol Zn)= 1.51 x 1024 atoms of Zn Particles to Moles: number of Particles * (1 mole/6.02 x 1023)= number of moles

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ex: 1.204 x 1024 atoms Ag * (1 mol/6.02 x 1023 atoms Ag)= 2 moles of Ag

Moles to Mass: number of moles x (molar mass/ 1 mole)= number of grams (or other unit of measurement) ex: 3.57 mol Al x (26.982 g Al/ 1 mol Al)= 96.3 g

Mass to Moles: number of grams * (1 mole/ molar mass)= number of moles ex: 25.5 g Ag * (1 mol Ag/ 107.868 g Ag)= 0.236 mol  o 

to find molar mass

ex: Na2CO3 = 2(mass of Na)+ mass of C + 3(mass of O) =2(22.990)+12.011+3(15.999) =45.98+12.001+47.997 =105.99 g Na2CO3

[edit] 1.4 Mass relationships in chemical reactions 1.4.1 : Mass is conserved throughout reactions. This fact allows masses to be calculated based on other masses in the reaction eg burning Mg in air to produce MgO and so to find the mass or Mg present in the original sample (ie purity)...can be extended to concentrations...ie titration. 1.4.2 : When a reaction contains several reactants, some may be in excess...is more is present that can be used in the reaction. The first reactant to run out is the limiting reagent (or reactant). Knowing the number of mols of the limiting reagent allows all other species to be calculated, and so the yield, and remaining quantities of other reactants.

[edit] 1.5 Solutions 1.5.1   

Solvent - the fluid you're dissolving in e.g. water. Solute - something which has been dissolved to form a solution e.g. an ionic compound such as sodium hydroxide. Solution - a mixture of solvent and solute where the solute has dissolved into the solvent to form an indistiguishable liquid. 8

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Concentration - the amount of solute per volume of solution. Measured in moles per dm3 (i.e. moles per liter) or grams per liter.

1.5.2 : Apply the equation concentration = moles/volume...rather obvious from the units of concentration, but remember to convert everything into the same units. 1.5.3 : Use chemical equations to relate the amount of one species to the amounts of others.

[edit] TOPIC 1: QUANTITATIVE CHEMISTRY (12.5 HOURS) Notes below are based on the specification for exams in 2009

[edit] Formula and Valency of Ions [edit] SOME COMMON CATIONS Ammonium NH4+ Barium Ba+2 Hydrogen H+ Calcium Ca+2 Lithium Li+ Copper Cu+2 Potassium K+ Lead Pb+2 Sodium Na+ Magnesium Mg+2 Silver Ag+ Mercury Hg+2 Nickel Ni+2 Aluminium Al+3 Zinc Zn+2 N.B. Many metals, including some of those above, show more than one valency. The valency is then always given with the name of the compound. It is written in Roman Numerals after the name of the metal. e.g. Iron may be Fe+2 or Fe+3 Iron II Chloride is Fe+2(Cl-)2 Iron III Chloride is Fe+3(Cl-)3 Other examples which you may meet are Chromium II or Chromium III Cr+2 or Cr+3 Manganese II or Manganese IV Mn+2 or Mn+4 9

Tin II or Tin IV Sn+2 or Sn+4 Lead II or Lead IV Pb+2 or Pb+4 [edit] Some Common Anions Aluminate Bromide Chlorate (V) Chloride Cyanide Ethanoate (acetate) Ethoxide Fluoride Hydride HydrogenCarbonate HydrogenSulphate Hydroxide HypoChlorite Iodide Methanoate Nitrate Nitrite Manganate (VII) (PerManganate) ThioCyanate

Al(OH)4- (aqueous) or Zn(OH)4-2 (aqueous) or TetraHydroxoZincate AlO2- (crystalline) ZnO2-2 (crystalline) BrChromate (VI) CrO4-2 ClO3DiChromate (VI) Cr2O7-2 Ethanedioate ClC2O4-2 (oxalate) CNOxide O-2 CH3CO2Peroxide O2-2 C2H5OSulphate SO4-2 FSulphide S-2 HSulphite SO3-2 HCO3ThioSulphate S2O3-2 HSO4HexaCyanoFerrate III [Fe(CN)6]-3 OHCarbonate CO3-2 ClOINitride N-3 HCO2Phosphate PO4-3 NO3HexaCyanoFerrate II [Fe(CN)6]-4 NO2MnO4SCN-

Rules of thumb: Anions consisting of just one element have names ending -ide. 

Anions consisting of an element plus oxygen have names ending in -ate.

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Notable exception: OH- is hydroxide, not 'hydrate'.

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If an element has two anions whose names should end in -ate, the one with fewer oxygens ends in -ite.

EXERCISE 1 Work out the formula for the following ionic chemicals 10

1. Zinc Chloride ZnCl2 2. Sodium Oxide Na2O 3. Aluminium Fluoride AlF3 4. Iron III Chloride FeCl3 5. Aluminium Oxide Al2O3 6. Zinc Hydroxide Zn(OH)2 7. Copper (II) Nitrate Cu(NO3)2 8. Barium Sulphate BaSO4 9. Sodium Carbonate Na2SO4 10. Calcium Hydrogencarbonate Ca(HCO3)2 11. Lead Iodide PbI2 12. Nickel (II) Sulphate NiSO4 13. Ammonium Carbonate (NH4)2CO3 14. Manganese IV Oxide MnO2 15. Sodium Bromide NaBr 16. Magnesium Ethanoate Mg(CH3COO)2 17. Silver Nitrate AgNO3 18. Mercury II Chloride HgCl2 19. Magnesium Nitride Mg3N2 20. Potassium Phosphate K3PO4

What are the names of the following chemicals? 1. CuBr2 Copper II Bromide 2. K2O Potassium Oxide 3. K2Cr2O7 Potassium DiChromate (VI) 4. KNO3 Potasium Nitrate 5. AlCl3 Aluminium Chloride 6. FeCl2 Iron II Chloride 7. FeS Iron II Sulphide 8. Na2S2O3Sodium ThioSulphate 9. Al2(SO4)3 Aluminium Sulphate 10. Mg(OH)2 Magnesium Hydroxide 11. Pb(CH3CO2)2 Lead (II) Ethanoate 12. Na2Zn(OH)4 Sodium Zincate 13. Pb(NO3)2 Lead (II) Nitrate 14. KMnO4 Potassium Manganate (VII) 15. PbO2 Lead (IV) Oxide 16. Ni(NO3)2 Nickel Nitrate 17. Na2SO3 Sodium Sulphite/Sodium Sulphate (IV) 18. Ca(HCO3)2 Calcium HydrogenCarbonate 19. CuO Copper II Oxide 20. MgCO3 Magnesium Carbonate 11

[edit] 1.1 MOLE CONCEPT AND AVOGADRO’S CONSTANT (2H) 1.2 FORMULAS (3H) [edit] Relative molecular mass (Mr) and relative atomic mass (Ar). Definition: The relative isotopic mass of 12C = 12 All other relative isotopic masses are measured relative to this standard and are not whole numbers e.g. 13

C 13.003 19F 18.998 35Cl 34.97 37Cl 36.97

The IB requires you to use whole numbers as an approximation to the relative isotopic mass of every isotope. The relative atomic mass (Ar) is the average value of an element’s relative isotopic masses e.g. Ar (C) = 12 * 0.9889 + 13.0033 * 0.011 = 12.0098 What is Ar (Cl) if 75.77 % is 35Cl and the rest is 37Cl? Relative atomic masses are in your data booklet, written beneath each element’s symbol in Table 5. e.g. The relative molecular mass (Mr) is the sum of a compound’s relative atomic masses e.g. Mr (CO) = 12.01 + 16.00 What is Mr (CO2)? What is Mr (Al2(CO3)3)?

[edit] Calculating the mass of one mole of a species from its formula. Another way of saying the same thing is to use the term ‘Molar mass’ The molar mass (M) of carbon-12 atoms is 12 g mol-1 NB Molar mass has units (it is not ‘relative’) You must specify if it refers to atoms, isotopes, molecules, etc. EXERCISE 2: Calculate the Relative Molecular Mass, Mr, of the following 12

1. Sodium Chloride 2. Copper Sulphate 3. Lead Nitrate 4. Magnesium Oxide 5. Potassium Hydroxide 6. Sulphuric Acid 7. Carbon Dioxide 8. Calcium Carbonate 9. Aluminium Sulphate 10. Sulphur Trioxide 11. Ammonium Nitrate 12. Zinc Chloride 13. Iron II Bromide 14. Iron III Iodide 15. Oxygen 16. Aluminium Zincate 17. Nitric Acid 18. Dilead II lead IV Oxide Pb3O4. 19. Magnesium Sulphate - 7 – water MgSO4.7H2O 20. Ammonium Iron II Sulphate - 6 – water (NH4)2SO4.FeSO4.6H2O Calculate M for :1. Copper Carbonate 2. Potassium Dichromate 13

3. Magnesium Hydroxide 4. Sodium Phosphate 5. Calcium Hydrogencarbonate 6. Potassium Oxalate 7. Sodium Cyanide 8. Nitrogen Dioxide 9. Ethanol 10. Lead Chromate 11. Potassium Permanganate 12. Tin IV Chloride 13. Ethane 14. Lithium Carbonate 15. Ammonia 16. Butane 17. Sodium Chromate 18. Hydrochloric Acid

[edit] Applying the mole concept to substances. Definition: A mole of particles contains the same number of particles as there are atoms in 12 g of carbon-12. It turns out that there are 6.02 x 1023 particles per mole. 6.02 x 1023 mol-1 is Avogadro’s constant

[edit] Solving problems involving amount, mass and molar mass. Maths (learn this thoroughly): 14

Mass = Amount x Molar mass or m = nM or Grams = Moles x Grams per mole EXERCISE 3 1. What is the mass of 0.5 moles of calcium carbonate (CaCO3)? 2. What is the amount of 27 g of water? 3. What is the relative molecular mass of fullerene if 2/3 mole has a mass of 480 g? 4. What mass of magnesium contains the same number of atoms as there are in 4 grams of calcium? 5. Calculate the mass of iron which contains the same number of atoms as there are molecules in 16 grams of oxygen. 6. What is the mass of copper sulphate (CuSO4 ) which contains the same number of copper atoms as there are atoms in 6.5 grams of zinc? 7. How much nitrogen contains the same number of atoms as there are sulphur atoms in 3.96 grams of ammonium sulphate ([NH4]2SO4)? 8. What mass of manganese contains twice the number of atoms that there in 0.6 grams of carbon?

[edit] Empirical formula and molecular formula. Empirical formula: the simplest expression of the proportions of the elements in a compound. The empirical formula can be obtained from the proportions of masses of each element. Molecular formula: the numbers of atoms of each element in a molecule of a compound. Sometimes the two formulae are the same: H2O, for example, is both the molecular and empirical formula. Hydrogen peroxide has an empirical formula of HO but its molar mass is 34 g mol -1. Its molecular formula is H2O2 15

EXERCISE 4 1. Analysis of two compounds showed that they contained the following proportions of elements: A: Carbon 26.1% Oxygen 69.6% Hydrogen 4.3% B: Carbon 52.2% Oxygen 34.8% Hydrogen 13.0% Calculate the empirical formula of each compound 2. If the empirical formula of methane is CH4, and its relative molecular mass is 16, what is the molecular formula? 3. If the empirical formula of ethane is CH3, and its relative molecular mass is 30, what is the molecular formula? 4. What is the mass of carbon dioxide that is produced when 50 g of calcium carbonate decomposes? CaCO3 → CaO + CO2 5. If 12 g of magnesium react with oxygen to produce 20 g of magnesium oxide: a. What is the formula of magnesium oxide? b. What is the equation for the reaction? 6. Determine the empirical formula and the molecular formula of the following compounds from the given information. a. 0.36 mol S are combined with 0.72 mol oxygen atoms. Mr of this compound is measured to be 64. b. 0.193 mol Fe combined with 0.2895 mol O and its molar mass is 160 g mol-1 c. 0.3285 mol Pb combined with 0.438 mol O. Mr = 685 d. 0.309 mol carbon combined with 0.618 mol H. Molar mass is 42 g mol-1 e. 0.379 mol H combined with 0.379 mol O. Mr = 34 7. A compound was shown by qualitative analysis to contain nitrogen and hydrogen as the only elements. It was found by quantitative analysis that 0.02 mol of nitrogen are combined with 0.04 mol of hydrogen. Its relative molecular mass is approximately 30. Determine its empirical formula and its molecular formula. 8. The relative molecular mass of a compound of boron and hydrogen was measured as just under 28. Every 0.25 mol of boron are combined with 0.75 mol hydrogen atoms. Find its empirical and its molecular formula. 9. Calculate the empirical formulae of the following substances whose compositions are given below: a. 5g of an oxide of magnesium which contains 3g magnesium b. 7g nitrogen combined with 16g oxygen c. 0.6g magnesium combined with 1.2g oxygen and 0.3g of carbon d. 3.5g nitrogen combined with 2g oxygen e. 4.8 g oxygen combined with 3.2 g copper and 1.4 g nitrogen f. 28.2 g of an oxide of potassium which contains 23.4 g potassium g. 0.08 mol uranium combined with 3.413 g oxygen h. 1 mol zinc combined with 32 g oxygen and 2 g hydrogen i. A chloride of lead in which 1.302 g of the compound contains 0.530 g chlorine j. Aluminium bromide which contains 2.4 g bromine in every 2.67 g of bromide 10. Determine the molecular formula of the following from the given information: a. 8 g carbon are combined with 2 g hydrogen and the molar mass is 30 g mol-1. b. 1.56 g of a hydrocarbon 16

contains 1.44 g carbon and its relative molecular mass is 78 c. 0.2 mol nitrogen atoms are combined with every 6.4 g oxygen and the molar mass of this compound is 92 g mol-1. e. 3.55 g of an oxide of phosphorus contain 1.55 g phosphorus. Its relative molecular mass is estimated to be a little below 300. Find the formula for each of the following substances whose percentage composition is given 11. 40% calcium, 12% carbon & 48% oxygen 12. 40% copper, 40% oxygen & sulphur 13. the oxide of copper which contains 88.9% Cu by mass 14. the hydrocarbon which contains 85.7% carbon and has a relative molecular mass of 96 15. the carbohydrate which has a molar mass of 180 g mol-1 and contains 53.33% oxygen and 6.67% hydrogen.

[edit] 1.3 CHEMICAL EQUATIONS (1H) [edit] Balancing chemical equations Once you know the formulae of the chemicals in a reaction, you need to make a balanced chemical equation. You cannot change the formulae of the chemicals. You can only add coefficients to indicate how many molecules of each chemical react. Example 1 C10H16 + Cl2 → C + HCl This is unbalanced. Let’s balance the carbons first: 1 C10H16 + Cl2 → 10 C + HCl Now let’s try the hydrogens: 1 C10H16 + Cl2 → 10 C + 16 HCl Finally, time for the chlorines: C10H16 + 8 Cl2 → 10 C + 16 HCl 17

Example 2 Si2H3 + O2 → SiO2 + H2O This is unbalanced. Let’s balance the silicon first: 1 Si2H3 + O2 → 2 SiO2 + H2O Now let’s try the hydrogens: 2 Si2H3 + O2 → 4 SiO2 + 3 H2O Finally, time for the oxygens: 4 Si2H3 + 11 O2 → 8 SiO2 + 6 H2O Here is a trick you can use to guide you in steps 2 and 3: 1 Si2H3 + O2 → 2 SiO2 + 3/2 H2O Example 3 Ca10F2(PO4)6 + H2SO4 → Ca(H2PO4)2 + CaSO4 + HF I don’t like the calcium. It appears in more than two compounds here. Next in line is fluorine: 1 Ca10F2(PO4)6 + H2SO4 → Ca(H2PO4)2 + CaSO4 + 2 HF The phosphate and sulphate can be treated as units: 1 Ca10F2(PO4)6 + H2SO4 → 3 Ca(H2PO4)2 + CaSO4 + 2 HF 1 Ca10F2(PO4)6 + 1 H2SO4 → 3 Ca(H2PO4)2 + 1 CaSO4 + 2 HF Note that although we have coefficients for every substance, the coefficients 1, 3 and 2 are not balanced with the coefficients 1 and 1 – so I write them in two different colours. The hydrogens look tricky too, but it’s either them or the calcium now: 1 Ca10F2(PO4)6 + 7 H2SO4 → 3 Ca(H2PO4)2 + 7 CaSO4 + 2 HF Better do the calcium now… Ca10F2(PO4)6 + 7 H2SO4 → 3 Ca(H2PO4)2 + 7 CaSO4 + 2 HF

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EXERCISE 5 Balance these equations: 1. Cu + O2 → CuO 2. Fe + O2 → Fe3O4 3. Mg + HCl → MgCl2 + H2 4. Al + HCl → AlCl3 + H2 5. P + O2 → P2O5 6. NaHCO3 → Na2CO3 + CO2 + H2O 7. KClO3 → KCl + O2 8. NaNO3 → NaNO2 + O2 9. Cu(NO3)2 → CuO + NO2 + O2 10. PbO + C → Pb + CO2 11. C4H8 + O2 → CO2 + H2O 12. C4H10 + O2 → CO2 + H2O 13. Na2CO3 + HCl → NaCl + H2O + CO2 14. KOH + H2SO4 → K2SO 4 + H2O 15. (NH4)2SO4 + NaOH → NH3 + Na2SO4 + H2O 16. Mg + H2O → MgO + H2 17. Na + H2O → NaOH + H2 18. Pb3O 4 + H2 → Pb + H2O 19. Fe2O3 + CO → Fe + CO2 20. Fe(OH)3 → Fe2O 3 + H2O 21. CuO + H2 → Cu + H2O 22. Zn + HCl → ZnCl2 + H2 23. KClO3 → KCl + O2 24. S8 + F2 → SF6 25. Fe + O2 → Fe2O3 26. C2H6 + O2 → CO2 + H2O

[edit] The mole ratios of any species in a chemical equation. EXERCISE 6 1. What is the amount (unit : mole) of hydrochloric acid needed to react with 1 mol Mg and how much magnesium chloride is formed? 2. What is the amount of chlorine which will react with 4 mol Fe? How much iron III chloride is formed? 3. How much carbon dioxide can be produced by reacting 0.1 mol CuCO3 with hydrochloric acid? What is the minimum amount of acid required? 4. What is the amount of sodium hydrogencarbonate which should be heated to produce 0.2 mol Na2CO3? 5. What amount of barium sulphate can be precipitated from a solution containing 0.01 mol barium chloride? What amount of potassium sulphate will be required to react with this barium chloride? What would be the effect of adding twice the amount of potassium sulphate?

[edit] The state symbols (s), (l), (g) and (aq). EXERCISE 7 Can you add state symbols to the equations above? Aqueous solutions and Ionic equations Many substances dissolve in water to form aqueous solutions. Ionic substances are usually soluble, separating into their component ions: Al2(SO4)3 (s)  2 Al+3(aq) + 3 SO4-2(aq) The mobility of the ions in solution means that these solutions will conduct electricity (though the solid substance will not). Covalent substances do not usually dissolve in water. If they do dissolve, the molecules remain intact: C6H12O6(s)  C6H12O6(aq) When no more solute can be dissolved in a solvent, the solution is said to be saturated. Ionic equations describe only the ions that react, omitting other (‘spectator’) ions. AgNO3 (aq) + NaCl(aq) ® AgCl(s) + NaNO3 (aq) Each aqueous compound can be treated as a pair of ions: Ag+(aq) + NO-3 (aq) + Na+(aq) + Cl-(aq) ® AgCl(s) + Na+(aq) + NO-3 (aq) The sodium and nitrate ions do not take part in the reaction Ag+(aq) + Cl-(aq) ® AgCl(s) This treatment simplifies the equation and makes it more generally applicable. E.g. What if silver fluoride and 19

potassium chloride solutions were mixed? When an acid and an alkali react, ionic equations can be very useful: Sodium hydroxide (NaOH) and hydrochloric acid (HCl) react to form sodium chloride and water Magnesium hydroxide (Mg(OH)2) and sulphuric acid (H2SO4) react to form magnesium sulphate (MgSO4) and water Treating Na+, Mg2+, Cl- and SO42- as spectator ions, what is the common ionic equation when a hydroxide reacts with an acid? EXERCISE 8 Write balanced equations for the following reactions. Put in the physical states in the appropriate manner and write the ionic equation as well where this is appropriate (numbers 1, 2, 5, 11, & 15). Occasionally, oxygen appears as reactant but is not in any of the named products. In these cases, assume it is converted into water. 1. Iron filings reacting with dilute hydrochloric acid to give hydrogen plus iron II chloride solution. 2. Zinc oxide powder dissolves in dilute hydrochloric acid to give zinc chloride and water. 3. In the Contact process, sulphur dioxide and oxygen combine to form sulphur trioxide in the presence of vanadium V oxide catalyst and at a temperature of 450°C. 4. When crystalline potassium nitrate [nitrate (V)] is heated, oxygen is given off and a residue of potassium nitrite [nitrate (III)] remains. 5. When chlorine is bubbled into aqueous potassium iodide, a ppt. of iodine is formed along with a solution of potassium chloride. 6. The effect of heat on dilead II lead IV oxide (Pb3O4) is to produce oxygen, leaving lead II oxide behind. 7. Baking soda, sodium hydrogen carbonate, is used in cooking and when heated it decomposes to form sodium carbonate, carbon dioxide and steam. 8. When carbon dioxide is bubbled through lime water (calcium hydroxide solution), milkiness is observed due to the formation of a ppt of calcium carbonate. 9. If carbon dioxide is allowed to continue to bubble through the solution from (8) for a long time, the ppt redissolves, forming calcium hydrogencarbonate. 10. In the blast furnace, carbon monoxide reduces iron III oxide to molten iron while being oxidised itself into carbon dioxide. 11. The chemical test for a sulphate is to add barium chloride solution to a solution of the test sample. A white ppt. of barium sulphate indicates a positive result. 12. Magnesium burns in an extremely exothermic reaction in which magnesium oxide is the product. 13. When conc. sulphuric acid (a liquid) is poured onto common salt, fumes of hydrogen chloride are observed, while sodium hydrogensulphate remains in the test tube. 14. A solution of hydrogen peroxide will decompose spontaneously, in the presence of manganese IV oxide catalyst, to liberate oxygen. 15. When solutions of ammonium chloride and sodium hydroxide were heated together, the smell of ammonia was observed. Sodium chloride was one of the other products.

[edit] 1.4 MASS AND GASEOUS VOLUME RELATIONSHIPS IN CHEMICAL REACTIONS (3H) [edit] Calculating theoretical yields from chemical equations. EXERCISE 9 1 What is the amount of sodium nitrate (nitrate V) (NaNO3) in 17.0g ? What mass of sodium nitrite (nitrate III) will be obtained by heating this if it decomposes according to the equation :- 2 NaNO3  2 NaNO2 + O2

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2. What mass of zinc oxide remains after heating 9.45g zinc nitrate? 2 Zn(NO3)2  2 ZnO + 4 NO2 + O2 3. What mass of sodium carbonate remains after heating 4.2g sodium hydrogencarbonate? 4. What mass of powdered zinc would react with 0.96 g of sulphur? What is the expected yield of zinc sulphide? 5. What will be the loss in mass on heating 6.2g copper carbonate until no further reaction takes place? CuCO3 (s)  CuO (s) + CO2 (g) 6. A solution contained 6.62g of lead nitrate. A solution of sodium chloride was added in order to precipitate lead chloride. What is the expected yield of lead chloride? 7. What is the amount of Mg contained in 4.8g magnesium? If this reacts with hydrochloric acid, what is the amount of hydrogen gas which will be produced? 8. An aqueous solution contains 12.6 g HNO3. What amount of HNO3 is this? Find the amount and the mass of potassium hydroxide which will just neutralise this. What mass of potassium nitrate could you expect as product? 9. 5.05 g of copper nitrate decomposes according to the following equation when it is heated 2 Cu(NO3)2  2 CuO + 4 NO2 + O2 (a) What mass of copper oxide is produced? (b) What amount of nitrogen dioxide is formed?

[edit] The limiting reactant and the reactant in excess: Theoretical, experimental and percentage yield. EXERCISE 10 1. What mass of zinc will be required to react completely with 5g of copper sulphate and what mass of copper will be formed? 2. How much iron II sulphate will be required to react with 4g sodium hydroxide and what mass of iron II hydroxide will be formed? 3. 8g of calcium are dissolved in excess hydrochloric acid. How many molar masses of acid will actually react? What amount of calcium chloride can you expect? If this crystallises out of solution as CaCl2.6H2O, what is the maximum mass of crystals that can be produced? 4. A technician used 2.4g of magnesium ribbon to produce crystals of magnesium sulphate - 7 water by reaction with dil. sulphuric acid. What should his yield (in grams) have been? If he managed to make only 16.4 g, calculate his percentage yield.

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5. A portable gas stove burns butane at the rate of 116 g per hour. What mass of water will be produced after one hour? In what physical state will this water first appear? What will eventually happen to it? If the appliance is used in a confined space, state what problems or dangers might result. 6. 8.0 g of copper oxide were dissolved in dilute sulphuric acid and after suitable treatment 20 g copper II sulphate - 5 - water were obtained. What is the percentage yield? 7. Ethanol (20 cm3) was heated with an equal volume of ethanoic acid in the presence of concentrated sulphuric acid catalyst. The yield of ethyl ethanoate was 8 cm3. (a) Calculate the mass and the amount of each reagent. (b) Which of the two reagents, ethanol or ethanoic acid is the limiting reagent i.e. which is not in excess? (c) Calculate the percentage yield of ester. [ densities in g cm-3 :- ethanol 0.79; ethanoic acid 1.05; ethyl ethanoate 0.9] 8. 8.4g of sodium hydrogen carbonate are mixed with 15g of hydrochloric acid. Explain which reagent is in excess. Find the amount of carbon dioxide produced in the reaction. Would more or less carbon dioxide have been produced by heating the hydrogencarbonate? 9. On heating 4.2g of a metal carbonate, XCO 3, 0.05 mol carbon dioxide was produced. Calculate the relative atomic mass, A r, of the metal X. XCO 3  XO + CO 3 10. Bath salts (sodium carbonate crystals) (9.53g) were gently heated in a crucible in order to drive off all their water of crystallisation. 3.53g of anhydrous salt remained. Calculate the number of molecules of water of crystallisation in the sodium carbonate and hence its formula. However when 12 g of the crystals were left in a hot room, the loss in mass was only 6.80g. What can you deduce about the formula of the sodium carbonate after it had undergone efflorescense in these conditions?

[edit] Avogadro’s law of volumes of gases. Avogadro’s law: Equal volumes of gases, at the same pressure and temperature, contain equal amounts of molecules. The law can be used to find the volumes of gases in a reaction: e.g. 1 dm3 of hydrogen will react with 1 dm3 of chlorine to give 2 dm3 of hydrogen chloride. EXERCISE 11 What volume of CO2 will be produced when 1 dm3 of methane (CH4) combines with 3 dm3 of oxygen? What volume of CO2 will be produced when 1 dm3 of ethene (C2H4) combines with 3 dm3 of oxygen?

[edit] Graphs relating to the ideal gas equation. This section meets many criteria for Topic 11.3 Graphical techniques EXERCISE 12 22

1. A sample of gas was heated while the pressure was kept constant. The changes in volume were recorded. Graph the following data: Temperature (°C) Volume (dm3) 0 1.37 100 1.84 200 2.33 300 2.79 400 3.34 500 3.81 600 4.28 Use a scale which starts at -300 °C and 0 dm3. Estimate the temperature required for the gas to have a zero volume. 2. Another gas sample was pressurised and the effect on its volume was recorded. The temperature was kept constant. Graph the following data: Pressure (bar) Volume (dm3) 1 4.64 2 2.40 3 1.60 4 1.19 5 0.94 6 0.82 7 0.67 What is the relationship between pressure and volume? 3. A gas was gradually introduced to a rigid, constant temperature, vacuum-filled container, and the pressure recorded as more gas was added. Graph the following data: Amount (mol) Pressure (bar) 0 0 1 0.500 2 0.998 3 1.510 4 1.988 5 2.517 23

6 7

3.118 3.441

What is the relationship between pressure and amount? 4. One mole of a gas was kept in a 1 m3 (1000 dm3) container as the temperature was varied. The effect on the pressure was recorded: Temperature (K) Pressure (Pa) 253 2037 273 2280 293 2501 313 2615 333 2753 353 2918 373 2925 What is the relationship between pressure and temperature? What is the gradient of the graph? Use the gradient to estimate the temperature required for the pressure to be 101325 Pa (1 atm).

[edit] Molar volume of a gas A mole of any gas at 298 K and 1 atmosphere pressure occupies 24 dm3. At 273 K the volume is 22.4 dm3. E.g. 2 moles of argon atoms occupy 48 dm3 0.5 moles of oxygen molecules occupy 12 dm3 What is the volume of carbon dioxide that is produced when 50 g of calcium carbonate decomposes? CaCO3 ® CaO + CO2

[edit] The relationship between temperature, pressure and volume for a fixed mass of an ideal gas. We can summarise the behaviour of an ideal gas with this equation: PV = nRT P is the pressure (in Pa or N m-2) V is the volume (m3) n is the number of moles of the gas 24

R is the ideal gas constant (=8.31 J K-1 mol-1) T is the absolute temperature (K) You can see from this that if pressure increase, either the volume must decrease or the temperature increases. Similarly, increasing temperature causes increased volume and/or pressure. EXERCISE 12 Use PV = nRT to calculate: a) The volume occupied by 1 mol of a gas at 25 oC (298 K) and 100000 Pa (1 bar) pressure. b) The pressure required to make 5 mol of gas occupy 50 dm3 (50 x 10-3 m3) at 35 oC. c) The number of moles of gas which occupy 25 dm3 at 85 oC and 1.05 bar. d) The temperature of a 3 mol gas sample which has a volume of 18 dm3 at 1 atm (1.013 bar) pressure. If we change the conditions for a given sample of gas, we can simplify the equation considerably. Call the initial pressure, temperature and volume P1, T1 and V1, and the values after the conditions have changed are called P2, T2 and V2. The amount of gas, n, and the constant, R, are the same in both cases. This gives us the formula: =

P1*V1 / T1 = P2*V2 / T2

The ‘1’ and ‘2’ refer to the volume (V) of the gas under different conditions of Pressure (P) and Absolute Temperature (T - measured in kelvins) Helpfully, the pressure and volume do not now have to measured in Pa and m3. Any units are suitable, so long as the same units are used for P1and P2, and for T1 and T2. EXERCISE 13 1. Use the simplified equation (and answers to the previous exercise) to calculate: a) The volume occupied by 1 mol of a gas at 0 oC and 100000 Pa (1 bar) pressure. b) The pressure required to make 5 mol of gas occupy 40 dm3 (50 x 10-3 m3) at 35 oC. c) The temperature of a 3 mol gas sample which has a volume of 25 dm3 at 1 atm (1.013 bar) pressure. d) If a sample of gas which occupies 26.25 dm3 at 120.8 oC and 1.10 bar could be the same as a sample which occupies 25 dm3 at 85 oC and 1.05 bar. 25

2. If 1 dm3 of gas is warmed from 298 to 745 K, while the pressure is kept constant at 1 atm, what is its new volume? 3. What pressure is now required to pressurise the hot gas until it reaches its original volume?

[edit] 1.5 SOLUTIONS (2H) [edit] Solute, solvent, solution and concentration Concentration (c) is measured as the quantity of solute per unit volume (usually per dm3 i.e. dm3) of solution (not solvent). Remember that 1 dm3 = 1000 cm3 The solute quantity can be measured as mass, so the units are g dm-3. cm = m/v mass concentration = mass / volume The solute quantity can be measured as amount, so the units are mol dm-3. cn = n/v molar concentration = amount / volume The molar concentration of a substance is often written in the shorthand form [substance]. E.g. Instead of writing ‘The molar concentration of NaCl’ we can write [NaCl]. EXERCISE 14 1) Write down all the possible mathematical expressions which might be needed when doing calculations based on volumetric analysis (titrations). 2) What is the mass concentration if 5 g of NaCl are dissolved in 500 cm 3 of solution? 3) What is the amount of 5 g of NaCl? 4) What is the molar concentration if 5 g of NaCl are dissolved in 500 cm 3 of solution? 5) 25 cm3 of hydrochloric acid (0.12 mol dm-3) were just neutralised by 28.4 cm3 of sodium hydroxide solution. Calculate the concentration of the base. 6) 23.2 cm3 of hydrochloric acid were required to cause the formation of the very pale grey colour - almost colourless - of screened methyl orange indicator in 25.0 cm3 of potassium hydroxide solution. Determine the concentration of the acid given that the potassium hydroxide was made by dissolving 1.372 g in 250.0 cm3 of solution. 7) A sample of lithium hydroxide was weighed as 0.62 g. It was dissolved in water & made up to 250.0 cm3. 25.00 cm3 aliquots were titrated and 21.25 cm3 of standard hydrochloric acid (0.100 mol dm-3) were required. The lithium hydroxide is thought to be contaminated with lithium chloride. Calculate the mass of lithium hydroxide in the sample and hence its percentage purity. 8) Butanedioic acid is a dibasic acid (diprotic acid) of molecular formula C4H6O4. 1.0073 g were weighed, dissolved in water and the solution made up to 250 cm 3. 25 cm3 portions required 23.95 cm3 of potassium hydroxide solution to cause the colour change of a suitable indicator. Calculate the concentration of the alkali. 26

a) What amount of butanedioic acid was used (how many moles)? b) What amount was used in each titration? c) What amount (mol) of potassium hydroxide was present in each titration? d) What is the concentration of potassium hydroxide? 9) A technician was asked to analyse a solution of commercial caustic soda for its sodium hydroxide & sodium carbonate content. She did this by titrating a 25.00 cm 3 portion with hydrochloric acid solution (0.100 mol dm-3.) using firstly phenolphthalein and then, as part of the same titration, by adding methyl orange. Results obtained were 16.90 cm3 of acid to bring about the colour change of the phenolphthalein and a total end point of 24.55 cm3. At the first colour change (phenolphthalein goes pink to colourless), the reactions taking place are: OH- + H+  H2O & also CO3-2 + H+  HCO3-2. At the second colour change (methyl orange turns green to red), the reaction taking place is HCO3-2 + H+  CO2 + H2O Thus the first part of the titration relates to the concentration of hydroxide plus carbonate. The second part refers to the concentration of the carbonate (in its hydrogencarbonate form) only. a) Calculate the number of moles of acid used in the first part and thus the total number of moles of hydroxide plus carbonate in the 25.00 cm3 portion. b) Calculate the volume and hence the number of moles of acid used in the second part of the titration and thus the number of moles of hydrogencarbonate (and thus carbonate). c) From a) & b) calculate the number of moles of hydroxide and of carbonate in the 25.00 cm3 portion. From these results calculate the concentrations of the sodium hydroxide & sodium carbonate in the solution. Give your answer in each case in mol dm -3 & also in g dm-3. 10) 1.483 g of the mineral Dolomite were added to excess hydrochloric acid (50 cm 3, 1.0 mol dm-3). After reaction was complete, the solution was filtered and made up to 250 cm 3. 25 cm3 portions of this solution containing unreacted acid were titrated against 0.1 mol dm-3 KOH when a mean titre of 26.3 cm3 was obtained. Assuming that Dolomite is essentially magnesium carbonate plus insoluble silica, calculate its percentage composition. a) What amount of potassium hydroxide was in the 26.3 cm 3? b) What amount of hydrochloric acid would have reacted with this? c) What amount of hydrocloric acid was in the 250 cm 3 solution? d) What amount of acid was taken at the start? e) What amount of acid reacted with the Dolomite? 27

f) Calculate the amount and hence the mass of magnesium carbonate which would have reacted with the hydrochloric acid. g) Calculate the percentage by mass of magnesium carbonate in Dolomite.

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IB Chemistry/Atomic Theory < IB Chemistry Jump to: navigation, search

Contents [hide] 

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1 Atomic Theory Revision Notes o 1.1 The nucleus o 1.2 Electrons 2 Wave nature of electrons 3 HL Material o 3.1 Electron shells o 3.2 The s Orbital o 3.3 The p Orbitals o 3.4 The d Orbitals 4 Material for new syllabus o 4.1 ATOMIC STRUCTURE o 4.2 SL TOPIC 2.1 THE ATOM (1H).  4.2.1 The position of protons, neutrons and electrons in the atom.  4.2.2 the relative mass and relative charge of protons, electrons and neutrons.  4.2.3 The terms mass number (A), atomic number (Z) and isotopes of an element.  4.2.4 The symbols for isotopes  4.2.5 The properties of the isotopes of an element.  4.2.6 The uses of radioisotopes 14  4.2.6.1 C in radiocarbon dating 60  4.2.6.2 Co in radiotherapy 131 125  4.2.6.3 I and I as medical tracers  4.2.7 The number of protons, electrons and neutrons in atoms and ions o 4.3 2.2 THE MASS SPECTROMETER  4.3.1 The operation of a mass spectrometer.  4.3.2 How the mass spectrometer may be used to determine relative atomic mass  4.3.3 Calculation of non-integer relative atomic masses and abundance of isotopes o 4.4 2.3 ELECTRON ARRANGEMENT (2H)  4.4.1 The electromagnetic spectrum.  4.4.2 A continuous spectrum and a line spectrum. 29

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4.4.3 How the lines in the emission spectrum of hydrogen are related to electron energy levels.  4.4.4 The electron arrangement up to Z = 20. 4.5 TOPIC 12: ATOMIC STRUCTURE (3 HOURS) 4.6 12.1 ELECTRON CONFIGURATION  4.6.1 How ionization energy data is related to the electron configuration.  4.6.2 Evidence from first ionization energies and sub-levels.  4.6.3 The relative energies of s, p, d and f orbitals.  4.6.4 The maximum number of orbitals in a given energy level.  4.6.5 The shapes of s, px, py and pz orbitals.  4.6.6 The Aufbau principle, Hund’s rule and the Pauli exclusion principle

[edit] Atomic Theory Revision Notes 2.1 The Nuclear Atom 2.1.1 : Protons and Neutrons form the nucleus of the atom, electrons orbit the nucleus in electron shells. 2.1.2 : Protons -- Mass = 1 amu , charge = +1 .. Neutrons -- Mass = 1 amu , charge = 0 .. Electron -- Mass = 1/1840 amu (usually insignificant), charge = -1

Atoms are made up of a nucleus and electrons that orbit the nucleus.

[edit] The nucleus The nucleus is made up of positively charged protons, and neutrons, which have no charge but about the same mass as a proton.

[edit] Electrons

A simple model of a lithium atom. Not to scale!

Electrons are negatively charged and fly around the nucleus of an atom very quickly. So far, they do not appear to be made up of anything smaller: they are fundamental particles. They are extremely tiny, so small in fact that no one has managed to detect any size whatsoever. They are also very light, much much lighter than either a proton or a neutron. Hence, the weight of the electron is not included in the atomic number. An atom in its natural, uncharged state has the same number of electrons as protons. If it gains or loses electrons, the atom acquires a charge and is then referred to as an ion. The number of 30

protons in an atom defines its chemical identity (e.g. hydrogen, gold, argon, etc). Protons are not gained or lost through chemical reactions, but only through high energy nuclear processes. 2.1.3 : Mass number (A) -- Number of protons + neutrons. Atomic number (Z) -- number of proton. Isotope -- atoms with same atomic number but different mass number (i.e. different numbers of neutrons) 2.1.4 : AZX... A = mass number, Z = atomic number, X = atomic symbol. 2.1.5 : Isotopes may differ in physical properties (mass/density) and radioactivity but not generally in chemical properties. 2.1.6 : Atomic masses are the average of the atomic mass of each isotope (isotopic mass) times the isotope's relative abundance. results in non integer atomic masses 2.1.7 : Atomic number = number of protons (or number of electrons - ionic charge) , mass number - atomic number = number of neutrons. 2.2 Electron Arrangement 2.2.1 : Continuous spectrum goes continuously through red, orange, yellow, green, blue, indigo, violet. A line spectrum contains only some individual lines from this spectrum. 2.2.2 : Electrons are excited (usually by running an electric current through them). This causes electrons to 'jump' into higher electron shells ( X -> X* ) this state is only temporary, however, and the electron falls back to it's ground state. This change (When the electron falls back from the higher shell to a lower one) decreases the energy of the electron, and this energy is emitted in the form of a photon. If this photon falls into the visible spectrum of light, then it produces a visible spectrum. As electrons move further away from the nucleus, the electron shells become closer together in terms of space and energy, and so lines converge towards the end of the spectrum.

[edit] Wave nature of electrons Electrons behave as particles but also as waves. One of the results of this observation is that electrons can not orbit with any energy they like. Think of a standing wave on a guitar string. Only a whole number of half wavelengths will fit in the string to form a standing wave, likewise for an atomic shell. Since the energy is dependent on the wavelength this means that the energy of an electron in an atom (a bound electron) is quantized. This means that the energy is limited to certain distinct values, one for each shell with no middle values allowed. 2.2.3 : The main electron levels go : 2, 8, 18 etc...2n + 2 for n0, n1 and n2... 31

2.2.4 : Electrons are added from the left...after each shell is filled, move to the next...2, 8, 18...only up to Z = 20 is required.

[edit] HL Material Topic 12 is the additional HL material for Topic 2. It's not just the energy that is quantized, other properties that an electron can posess are also split into distinct units with no in betweens. The angular momentum is quantised, the spin is quantised, the component of the angular moment in any direction that you care to choose is quantised. There are in fact a whole host of rules determining the values that each of these properties can take.

[edit] Electron shells Each different shell is subdivided into one or more orbitals, each of which has a different angular momentum. Each shell in an orbital has a characteristic shape, and are named by a letter. They are: s, p, d, and f. In a one electron atom (e.g H, He+, Li++ etc.) the energy of each of the orbitals within a particular shell are all identical. However when there is more than one electron, they interact with one another and split the orbitals into slightly different energies. Within any particular shell, the energy of the orbitals depends on the angular momentum, with the s orbital having the lowest energy, then p, then d, etc.

[edit] The s Orbital Atomic 1s orbitals Stylised image of the 1s atomic orbital. The simplest orbital in the atom is the 1s orbital. The 1s orbital is simply a sphere of electron density. There is only one s orbital per shell. The s orbital can hold two electrons, as long as they have different spin quantum numbers.

[edit] The p Orbitals Stylised image of all the 2p atomic orbitals. 32

Starting from the 2nd shell, there is a set of p orbitals. The angular momentum quantum number of the electrons confined to p orbitals is 1, so each orbital has one angular node. There are 3 choices for the magnetic quantum number, which indicates 3 differently orientated p orbitals. Finally, each orbital can accommodate two electrons (with opposite spins), giving the p orbitals a total capacity of 6 electrons. The p orbitals all have two lobes of electron density pointing along each of the axes. Each one is symmetrical along its axis. The notation for the p orbitals indicate which axis it points down, i.e. px points along the x axis, p y on the y axis and pz up and down the z axis. The p orbitals are degenerate, they all have the same energy. P orbitals are very often involved in bonding.

2px 2py 2pz

[edit] The d Orbitals The first set of d orbitals is the 3d set. There are 5 choices for the magnetic quantum number, which gives rise to 5 different d orbitals. Each orbital can hold two electrons (with opposite spins), giving the d orbitals a total capacity of 10 electrons. Note that you are only required to know the shapes of s and p orbitals for the IB. In most cases, the d orbitals are degenerate, but sometimes, they can split, with the eg and t2g subsets having different energy. Crystal Field Theory predicts and accounts for this. D orbitals are sometimes involved in bonding, especially in inorganic chemistry.

[edit] Material for new syllabus [edit] ATOMIC STRUCTURE [edit] SL TOPIC 2.1 THE ATOM (1H). SEE NEUSS, P6-7 TOK: What is the significance of the model of the atom in the different areas of knowledge? Are the models and theories that scientists create accurate descriptions of the natural world, or are they primarily useful interpretations for prediction, explanation and control of the natural world?

[edit] The position of protons, neutrons and electrons in the atom. Here is a typical atom, helium: 33

TOK: None of these particles can be (or will be) directly observed. Which ways of knowing do we use to interpret indirect evidence gained through the use of technology? Do we believe or know of their existence?

[edit] the relative mass and relative charge of protons, electrons and neutrons. The accepted values are: 1 Relative Mass Charge proton 1 neutron 1 0 electron 1/1840 The mass of the atom is due to its nucleons (protons and neutrons). An atom is electrically neutral because it has equal numbers of electrons and protons. Chemistry is due to the behaviour of electrons.

[edit] The terms mass number (A), atomic number (Z) and isotopes of an element. The atomic number of an atom is the number of protons. The mass number of an atom is the number of nucleons. The atomic number defines which element we are talking about. ‘Element 16’ could be used instead of ‘sulphur’.

[edit] The symbols for isotopes 34

The following notation should be used AZX, for example 126C Give the symbols for the following isotopes: Carbon-13 Uranium-235 Strontium-90 Phosphorus-32 13 235 90 32 6C 92U 38Sr 15P

[edit] The properties of the isotopes of an element. Isotopes have the same chemical properties but different physical properties. ‘Heavy water’ (21H2O or D2O) has the following properties: Boiling point: 101.42 °C at standard pressure. Freezing point: 3.81 °C at standard pressure. Relative density: 1107 g dm-3 at STP. Hydrogen-3 (‘Tritium’, T) is radioactive with a half-life of 12.32 years. Carbon-14 is radioactive with a half-life of 5730 years. CO2 normally has a density of 1.83 g dm-3 but if made with carbon-14 it would have a density of 1.92 g dm-3 The density of chlorine-35 gas is 2.92 g dm-3 under standard conditions, but chlorine-37 gas is 3.08 g dm-3.

[edit] The uses of radioisotopes [edit] 14C in radiocarbon dating Living things constantly accumulate carbon-14 but the isotope decays with a half-life of 5730 y. After death, accumulation stops but the decay continues, so the ratio of carbon-14 to carbon-12 can be used to calculate how long ago death occurred. This can be used to date organic material from archaeological sites. [edit] 60Co in radiotherapy

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Cobalt-60 emits gamma radiation which can be directed onto tumours in an attempt to kill their cancerous cells. Whole-body irradiation can be used to destroy bone marrow before a transplant is attempted. [edit] 131I and 125I as medical tracers The thyroid is the only organ of the body which accumulates iodine, so isotopes of iodine can be used to study thyroid disorders. Iodine-131 or iodine-125 are given to patients in very low doses, and the pattern of radiation can reveal tumours or other abnormal growths. Larger doses of iodine radioisotopes can be used as therapy for thyroid cancer.

[edit] The number of protons, electrons and neutrons in atoms and ions Complete the following table: Symbol Protons Neutrons Electrons 5 6 5 5B 63 +2 29 34 27 29Cu 207 +2 82 125 80 82Pb 37 17 20 18 17Cl 197 +3 118 76 79Au 79 11

[edit] 2.2 THE MASS SPECTROMETER [edit] The operation of a mass spectrometer.

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Schematic diagram of a mass spectrometer.

[edit] How the mass spectrometer may be used to determine relative atomic mass By varying the strength of the magnetic field, ions of different masses can be brought to focus on the detector. In this way the relative abundances of ions of different masses produced from the sample can be determined. This is known as a mass spectrum. Usually the electron bombardment 37

is adjusted to produce ions with only a single charge. Any doubly charged ions will be deflected more than the singly charged ions and will in fact behave in the same way as a singly charged ion of half the mass. That is why the x-axis is labeled m/z, where m is the relative mass of the species and z its relative charge. For example, Sulfur-32 (2+) will be observed at m/z=16.

[edit] Calculation of non-integer relative atomic masses and abundance of isotopes The relative atomic masses of many elements are not whole numbers. This is because they are mixtures of isotopes. Each isotope has a molar mass which is (almost) an integer e.g. the molar mass of chlorine-35 is 35.0 g mol-1 and chlorine-37 is 37.0 g mol-1. Chlorine is a mixture of 24 % chlorine-37 and 76 % chlorine-35. The molar mass is therefore: 37 x 0.24 + 35 x 0.76 = 35.48 Secret information: Carbon-12 is the only isotope with an exact integer for its molar mass. Other isotopes have molar masses which are almost, but not quite, whole numbers. The IB don’t require you to know this – but it explains why the values you calculate don’t always match the values in your data booklet. This is the mass spectrum of mercury. There are two types of information we can find from this spectrum: Mercury has six isotopes Mercury-202 is the most common isotope The molar mass of mercury is actually the average of the molar masses of its isotopes. From the table we can read these values: Isotope Signal strength 198 33.8 % 199 56.4 % 200 77.5 % 201 44.3 % 202 100 % 204 23.0 % Note that ‘%’ does not mean ‘percent of the total mercury ions’. It means ‘percent of the most intense signal’. The total ‘percentage’ is 335 %. To find the average, we treat ‘%’ as moles: 38

33.8 moles of mercury-198 have a mass of 6692.4 g Molar mass of isotope Amount (mol) Mass (g) -1

(g mol ) 198 199 200 201 202 204 total

33.8 56.4 77.5 44.3 100 23.0 335.0 mol

6692.4 11223.6 15500 8904.3 20200 4692 67212.3 g

So we have a molar mass for the average mercury atom: = 200.6 g mol -1 Classwork 1. Copper has two stable isotopes: Isotope Abundance (%) 63Cu 69.1 65Cu 30.9 Calculate the molar mass of copper. 2. Bromine has two isotopes, 79Br and 81Br. Look up the molar mass of bromine. What is the proportion of the two isotopes? 3. Boron has two isotopes, 10B and 11B. What is the proportion? 4. Lead has four stable isotopes: Isotope Abundance (%) 204Pb 1.5 206Pb 23.6 207Pb 22.6 208Pb 52.3 Calculate the molar mass of lead. Homework – due in next week. 1 An experiment was performed to determine the density of gold. The following measurements were recorded. Mass of sample of gold = 30.923 g (to 5 sig fig), Volume of sample of gold = 1.6 cm3 (to 2 sig fig). Which of the following is the most accurate value for the density of gold (in g cm-3) which can be justified by these measurements? A. 19.327 (to 5 sig fig) B. 19.33 (to 4 sig fig) C. 19.3 (to 3 sig fig) D. 19 (to 2 sig fig) 2 The nucleus of a radon atom 22286Rn, contains A. 222 protons and 86 neutrons. B. 86 protons and 136 neutrons. C. 86 protons and 222 neutrons. D. 86 protons, 136 neutrons and 86 electrons. 3 Which of the following statements is/are true according to our current picture of the atom? I More than 90 % of the mass of a given atom is found in its nucleus. II Different atoms of an element may have different masses. 39

III The chemical properties of an element are due mainly to its electrons. A. I only B. I and II only C. II and III only D. I, II and III 4. In which pair do the species contain the same number of neutrons? A. 10846Pd and 11048Cd B. 118 120 196 198 +2 226 +2 222 50Sn and 50Sn C. 78Pt and 78Pt D. 88Ra and 86Rn 5 Which pairing of electrons and protons could represent a Sr+2 ion? Protons Electrons A. 38 36 B. 38 38 C. 38 40 D. 40 38 6 All isotopes of tin have the same I. number of protons; II. number of neutrons; III. mass number. A. I only B. II only C. III only D. I and III only

[edit] 2.3 ELECTRON ARRANGEMENT (2H) [edit] The electromagnetic spectrum. The electromagnetic spectrum unifies a vast range of ‘waves’, ‘rays’ and ‘radiation’. All electromagnetic radiation travels at 299776 ms-1 in a vacuum. Parts of the spectrum can be specified by their wavelength, frequency, or energy.

The electromagnetic spectrum. The red line indicates the room temperature thermal energy. (Opensource Handbook of Nanoscience and Nanotechnology). The energies are quoted in eV: 1 eV is 96.5 kJ mol-1. The frequency is directly proportional to the energy, and inversely proportional to the wavelength.

40

When transferring energy, electromagnetic radiation behaves as ‘packets’ of energy known as ‘photons’. Visible light has wavelengths of 400 (blue) to 700 (red) nm, which corresponds to energies of 299 (blue) to 171 (red) kJ mol-1. This is too small to interact with electrons in most bonds, but too large to set bonds resonating. Infrared light has wavelengths of 0.7 to 1000 um, and carries energies (less than 171 kJ mol -1) which can set bonds vibrating. In this way, infrared efficiently carries thermal energy. Ultraviolet light has wavelengths of 1-400 nm, which gives the waves enough energy (300+ kJ mol-1) to disrupt most chemical bonds.

[edit] A continuous spectrum and a line spectrum. A continuous emission spectrum shows emission over a wide range of wavelengths of electromagnetic radiation: A line emission spectrum only shows emissions at certain wavelengths, with no emission at intermediate wavelengths. The diagram below shows the line emission spectrum of hydrogen, and the continuous emission spectrum of a black body at 10000 K.

[edit] How the lines in the emission spectrum of hydrogen are related to electron energy levels. When hydrogen gas is stimulated, it emits a characteristic set of spectral lines. The gas is usually stimulated by passing a current through a sample of the gas at low pressure, but the same effect occurs if hydrogen gas is heated strongly. The spectral lines have the following characteristics: There are several series of lines, which become more closely packed at higher frequencies (lower wavelengths) until finally the series ends. The highest frequency series, discovered by Lyman, is in the ultraviolet. Lower frequency series are in the visible and infrared: Series Convergence limit Lyman 91 nm (UV) Balmer 365 nm (Visible) Paschen 821 nm (IR) Brackett 1.46 μm (IR) 41

Pfund

2.28 μm (IR)

42

Spectral lines of the hydrogen atom. 43

The lines are emitted by electrons in the hydrogen atoms. The electrons are ‘excited’ by the energy input to the hydrogen sample (usually electrical current). ‘Excitation’ means that the electron leaves its usual low energy orbit (its ‘ground state’) and enters a higher-energy orbit. The orbits of electrons are often called ‘shells’. Excited electrons eventually ‘relax’ to lower-energy orbits. They emit the excess energy in the form of photons. From the emission spectrum patterns we can deduce two things: a) Because we observe a line spectrum, we know that there are only a limited number of higherenergy orbits for excited electrons. b) The energy of each successive orbit/shell converges to a maximum value, because each series of lines converges at higher energy. The Lyman series is caused by electrons relaxing directly to the ground state. Ionisation energy: The first ionisation energy is the energy required to remove an electron from each atom in a mole of gaseous atoms. e.g. Ca (g) → Ca+(g) + eThe convergence limit of the Lyman series is related to the ionisation energy of the hydrogen atom: The maximum energy photon emitted in the Lyman series is the highest energy an electron can have while still remaining part of the hydrogen atom. The lower-energy series are caused by electrons relaxing, but not to the ground state. The Balmer series, for example, is due to photons emitted as electrons relax to the orbit with the secondlowest energy (to ‘the second shell’). Homework 1. The spectral line that corresponds to the electronic transition n = 3 → n = 2 in the hydrogen atom is red in colour. What type of radiation is released during the transition n = 2 → n = 1 ? A. Ultraviolet B. Red light C. Infrared 44

D. Radiowaves 2. The electron transition between which two levels releases the most energy? A. First to third B. Fourth to ninth C. Sixth to third D. Second to first 3. (a) The diagram below (not to scale) represents some of the electron energy levels in the hydrogen atom.

(i) Draw an arrow on this diagram to represent the electron transition for the ionisation of hydrogen. Label this arrow A [2] (ii) Draw an arrow on this diagram to represent the lowest energy transition in the visible emission spectrum. Label this arrow B [2]

[edit] The electron arrangement up to Z = 20. Give the ground state electron arrangements for: Sulphur atom: Potassium atom: Chloride ion: Magnesium ion: HIGHER LEVEL

[edit] TOPIC 12: ATOMIC STRUCTURE (3 HOURS) [edit] 12.1 ELECTRON CONFIGURATION [edit] How ionization energy data is related to the electron configuration. Studying the ionisation energies of an element such as calcium allows us to count the number of electrons which can occupy each shell. Ionisation Ionisation energy (kJ mol-1) log10(IE) 1st 6.00 x 102 2.78 3 2nd 1.15 x 10 3.06 3 3rd 4.91 x 10 3.69 3 4th 6.47 x 10 3.81 45

5th 6th 7th 8th 9th 10th 11th 12th 13th 14th 15th 16th 17th 18th 19th 20th

8.14 x 103 1.05 x 104 1.23 x 104 1.42 x 104 1.82 x 104 2.04 x 104 5.70 x 104 6.33 x 104 7.01 x 104 7.88 x 104 8.64 x 104 9.40 x 104 1.05 x 105 1.12 x 105 4.95 x 105 5.28 x 105

3.91 4.02 4.09 4.15 4.26 4.31 4.76 4.80 4.85 4.90 4.94 4.97 5.02 5.05 5.69 5.72

Classwork: Plot a graph of the log10(IE) against the ionisation. Why does the ionisation energy always increase? What causes the sudden jumps between the 2nd and 3rd, the 10th and 11th, and the 18th and 19th ionisations? What is the electronic configuration of calcium? Draw your prediction for the equivalent graph for sodium: Homework Which element would produce a graph like this? Explain the shape of the graph.

[edit] Evidence from first ionization energies and sub-levels. Use your data booklet to plot a graph of the first ionisation energies of the elements Li to Ne. Why do the ionisation energies generally increase from Li to Ne?

46

The charge on the nucleus increases from Li to Ne. If we subtract the charge of the inner shell of electrons we can calculate the charge exerted on the outer electron shell: The effective nuclear charge. Because each element in the period has the same number of inner-shell electrons, the effective nuclear charge increases from 1 (Li) to 8 (Ne). The increased charge holding the outer electrons in place increases the energy required to remove one of these electrons. (It also reduces the size of the atom: Li is larger than Ne). Why do the ionisation energies of boron and oxygen break the general trend? The s2 arrangement is stable (like 'noble gas configurations' are stable). Boron has a [He] 2s2 2px1 arrangement. Losing the px1 electron returns boron to this stable state, so losing this electron is suprisingly easy. Similarly, the s2 px1 py1 pz1 arrangement is stable. Oxygen has a [He] 2s2 2px2 2py1 2pz1 arrangement. Losing a p x electron returns oxygen to this stable state, so losing this electron is surprisingly easy.

[edit] The relative energies of s, p, d and f orbitals. [edit] The maximum number of orbitals in a given energy level. Each energy level (‘shell’) is made of orbitals (‘sub-shells’). Each orbital can hold two electrons. The number of orbital types is equal to the shell number e.g. shell 3 has three types of orbital, s p and d. Shell 1 1s Shell 2 2s 2p Shell 3 3s 3p 3d Shell 4 4s 4p 4d 4f Shell 5 has, in theory, five types of orbital. No known element uses its g orbitals, however. The orbitals in each shell have increasing energy. s is least energetic, then p, d, f, etc. There is only one s orbital per shell. There are three p orbitals, five d orbitals, etc.

[edit] The shapes of s, px, py and pz orbitals. 47

s orbitals are simple spheres: The three p orbitals are aligned along the x y and z axes:

The orbitals of shells 1 and 2 shown as (top) a cloud of possible electron positions and (bottom) surfaces containing most of the electron character.

[edit] The Aufbau principle, Hund’s rule and the Pauli exclusion principle The aufbau principle: To find the electron configuration of an element, we build up the electrons one by one, putting each electron into the orbital with the lowest available energy. An easy way to remember which is the lowest available orbital is to use the following diagram: 1s 2s 2p 3s 3p 3d 4s 4p 4d 4f 5s 5p 5d 5f … 6s 6p 6d … 7s 7p … Hund’s rule: If there is more than one orbital to choose from e.g. the 2p orbitals, then the orbitals are filled with one electron each, and then with pairs of electrons. 48

The electron configuration of nitrogen is: 1s 2s 2px 2py 2pz ↑↓ ↑↓ ↑ ↑ ↑ And not: 1s 2s 2px 2py 2pz ↑↓ ↑↓ ↑↓ ↑ The simplest way to write the full electronic configuration is to note the last noble gas and then to add any extra electrons like so: Vanadium: 1s2 2s2 2p6 3s2 3p6 4s2 3d3 Vanadium: [Ar] 4s2 3d3 It does not matter if you write the orbitals in the order they are filled (as in the example above) or in order of their shells: Vanadium: [Ar] 3d3 4s2 Elements 24 and 29 are special cases. The 3d5 and 3d10 configurations are so stable that an electron is taken from the 4s orbital to create 3d5 and 3d10 configurations. e.g. Chromium is not: [Ar] 4s2 3d4 Chromium is: [Ar] 4s1 3d5 Complete the following table: 1s H He 2s 2p Li [He] Be [He] B [He] C [He] N [He] O [He] F [He] Ne [He] 3s 3p Na [Ne] 49

Mg [Ne] Al [Ne] Si [Ne] P [Ne] S [Ne] Cl [Ne] Ar [Ne] 3d 4s 4p K [Ar] Ca [Ar] Sc [Ar] Ti [Ar] V [Ar] Cr [Ar] Mn [Ar] Fe [Ar] Co [Ar] Ni [Ar] Cu [Ar] Zn [Ar] Ga [Ar] Ge [Ar] As [Ar] Se [Ar] Br [Ar] 50

Kr [Ar] 4d 5s 5p Rb [Kr] Sr [Kr] Y [Kr] Zr [Kr] Nb [Kr] Mo [Kr] Tc [Kr] Ru [Kr] Rh [Kr] Pd [Kr] Ag [Kr] Cd [Kr] In [Kr] Sn [Kr] Sb [Kr] Te [Kr] I [Kr] Xe [Kr] The four blocks of the periodic table are named after the highest-energy occupied orbital:

51

IB Chemistry/Periodicity < IB Chemistry Jump to: navigation, search

Contents [hide] 

      

1 Periodicity Revision Notes o 1.1 3.1 The Periodic Table o 1.2 3.2 Physical Properties 2 Periodic Trends - Moving Across the Periodic Table 3 Periodic Trends - Moving Down the Periodic Table 4 Noble Gases 0/8A/18 5 Halogens 7A/17 6 Alkaline Earth Metals group 2A/2 7 Alkali Metals group 1A/1 8 Hydrogen, an element like no other

[edit] Periodicity Revision Notes [edit] 3.1 The Periodic Table 3.1.1 : Elements increase in atomic number across each period, and down each group. 3.1.2 : Group - the columns proceeding vertically. Period - the rows proceeding horizontally. 3.1.3 : Group = number of valence electrons in the atom. Period = number of main electron shells...s, p , d and f blocks as described above.

[edit] 3.2 Physical Properties 3.2.1 : Electron structure  

As a period is crossed, valency electrons increase As a group is descended, a shell is added (with the same number of valency electrons)

Atomic radii

52

 

Across a period, radii decreases due to an increasing nuclear charge and reduced screening (more electrons and protons in outer shell) Down a group, radii increases due to additional shells blocking nuclear charge and increased screening (shielding)

Ionic radii  

Positive ions (Na ‡ Na+ + e-) have smaller radii due to 1 less shell and reduced screening Negative ions (F + e- ‡ F-) have larger radii because the positive charge of the nucleus has less positive pulling power (PPP) due to the presence of an extra electron in this ionic state.

Ionisation energy Ionisation energy is the energy required to remove 1MOLE of electrons from 1MOLE of gaseous atoms (e.g. M ‡ M+ + e-) Successive ionisation energies are 3rd > 2nd > 1st, because of a smaller radii and greater charge. Electron affinity Electron affinity is the energy change on the addition of 1 mole of electrons to 1 mole of gaseous atoms (e.g. X(g) + e- ‡ X-(g)) Electronegativity Electronegativity is the measure of the ability of an element in a bond to attract electrons. The most electronegative elements are (in decreasing) F, O and N.  

Across a period, electronegativity increases because the increase in nuclear charge makes the nucleus more attracted to electrons Down a group, electronegativity decreases due to increased screening and less nuclear attraction

Melting Points  

Melting point decrease down group 1 due the the elements metallic structure, which is held together by attractive forces between de-localized electrons. Melting point increases down group 7 due to the strong intermolecular forces that increase in strength with more electrons.

[edit] Periodic Trends - Moving Across the Periodic Table Because of the way in which the periodic table is organized, many trends become apparent. Moving across the periodic table, proton and electrons are added. Because all of these electrons are added into the same shell, the ionization energy increases. The ionization energy is the amount of energy required to remove the outermost electrons. Because of the overall greater attraction between more protons and more electrons across the periodic table, it becomes more and more difficult to remove an electron. The atomic radius of the elements decreases. Also, metallic character decreases (electronegativity increases; electronegativity is the tendency of 53

an element to gain electrons). This traces the tendency of atoms of metals to lose electrons while nonmetals gain them.

[edit] Periodic Trends - Moving Down the Periodic Table Because elements in the same family (column) contain the same valence shell electron configuration, they tend to behave very similarly. Moving down the periodic table, electrons are added to successively higher energy levels. Because of this, atomic radius increases. Also, it is easier to remove the outermost electron of larger atoms, so ionization energy decreases as well as electro negativity.

[edit] Noble Gases 0/8A/18 helium, neon, argon, krypton, xenon, radon Noble gases except helium all share a full valence shell electron configuration ( 8 outer electrons, 2 for helium). Because this configuration is extremely stable as well as symmetrical, the noble gases are very unreactive and will only react under extremely rigorous conditions.

[edit] Halogens 7A/17 fluorine, chlorine, bromine, iodine, astatine Halogens have an electron configuration of 7, meaning they require only one more electron for a noble configuration. They have very negative electron affinities, are extremely reactive, and form ions with a -1 charge. They are so reactive that in their homogenous state UV light will catalyze a radical reaction. Fluorine is the most electronegative of the elements and is so reactive that it attacks almost any other element (noble gases, oxygen, nitrogen, and gold are the exceptions) to form fluorides. Chlorine is somewhat less reactive, bromine somewhat less reactive than chlorine, and iodine even less, but even iodine is a formidable iodizer. Strong radioactivity masks the chemical properties of astatine. All except astatine form gaseous compounds with hydrogen: hydrogen fluoride HF, hydrogen chloride HCl, hydrogen bromide HBr, and hydrogen iodide HI; these are acidic, strongly reactive substances; except for hydrogen fluoride these substances are among the strongest known acids. They free halogens and the hydrogen halides react with most metals, halides of metals being known as salts, of which sodium chloride NaCl is best known as "salt".

Electronegativity decreases in this group with increasing atomic mass, and oxygen is more electronegative than any element except fluorine; it acts much like a halogen except for its -2 54

oxidation state. Fluorine and oxygen oxidize these elements to the +6 oxidation state, resulting in such substances (for sulfur) as sulfur hexafluoride SF6 and sulfur trioxide SO3 and its derivative sulfuric acid H2SO4, one of the most heavily-used industrial chemicals. Fluorine, oxygen, chlorine, and bromine oxidize all of the elements of this group except oxygen to the +4 oxidation state. Oxides of sulfur, selenium, and tellurium are acidic. Strong radioactivity largely masks the chemical properties of polonium.

[edit] Alkaline Earth Metals group 2A/2 (beryllium, magnesium, calcium, strontium, barium, radium) These substances are all metals having two electrons in the outer shell, or the ns 2 configuration. With increasing mass these elements become softer, have lower melting and boiling points, and become more reactive. None appear uncombined in nature, and all are separated from their compounds with difficulty. All react with halogens and, except for beryllium, with water and oxygen (magnesium at temperatures higher than those of living things) All oxidize to the +2 state that represents an ion in the stable configuration of an inert gas. Beryllium is the least reactive; it is the hardest of these elements. Its oxide is amphoteric; it reacts with both strong acids and bases. Magnesium burns in hot air or steam to form alkaline oxides or hydroxides. Powders or thin slices of this metal can be ignited with a match. Magnesium hydroxide is a strong base, although its low solubility masks its alkalinity. At room temperature magnesium metal is light and strong, but it can be ignited in air. Halogens and acids attack magnesium violently with the formation of salts like magnesium chloride (MgCl2) or magnesium sulfate (MgSO4). Although magnesium would react with atmospheric oxygen, an oxide layer protects the metal from oxidation. With calcium and the heavier elements of this group, the reactions are even more violent and take place at lower temperatures; they react not only with halogens and acids but also with water, oxygen, and even nitrogen. They are such strong reducing agents that they can reduce carbon dioxide to carbon as well as many metal oxides to metals. The strong radioactivity of radium masks its chemical properties, but to the extent that its chemistry is known it fits well into the group and its pattern.

[edit] Alkali Metals group 1A/1 (lithium, sodium, potassium, rubidium, caesium (US spelling cesium), francium)

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These elements are best marked by their reactivity. Physically they are soft, shiny (when freshly prepared) solids with low melting points; they conduct electricity well. They all have one outermost electron that they lose easily to almost any electronegative substance other than nitrogen (lithium reacts with nitrogen, unlike the others). None is ever found uncombined on earth, and none is ever put to use as a structural metal. They must be kept under inert liquids such as kerosene or in inert gases (nitrogen suffices for any of these elements other than lithium. All oxidize easily to the +1 oxidation state. High reactivity masks the chemistry of francium. They react with atmospheric oxygen to form various oxides and react violently with water, halogens, and acids. A typical reaction between one of these metals and water is as such: Na(s) + H2O → Na+(aq) + OH-(aq) + 1/2 H2 (g) (WARNING: this reaction generates much heat. One should not cast any alkali metal into water or acids except with the cautions that professional chemists use. The hydrogen gas from this reaction may itself ignite with atmospheric oxygen in a dangerous flame. These metals should never allowed to touch flesh because they react with any water upon them and yield corrosive hydroxides that burn flesh. Strong solutions of alkali metal hydroxides are destructive to flesh). Hydroxides of these elements dissociate completely in water to form some of the strongest bases known. These are as strongly alkaline as any acids and react violently with acids to form halides and water in neutralization: NaOH(s) + HCl (g) → NaCl(s) + H2O (s) The results of any reactions of these metals and any acids are salts. Almost all salts of these elements are highly soluble in water and form conducting solutions, proving their ionic nature. The best known of these substances is sodium chloride, NaCl, a substance known as common salt. Salts of these elements and strong acids are neutral (for example, potassium nitrate KNO3); salts with weak acids such as acetic acid are alkaline (sodium carbonate, Na 2CO3).

[edit] Hydrogen, an element like no other Hydrogen is by far the most common element in the universe; as a gas it is too light for the Earth's gravitation to hold. It is by far the most common constituent of the Sun and all other stars and of the gas giant planets of our solar system. It exists on or just under the surface of the Earth as a component of water and in innumerable compounds of carbon, many essential to life. The heat and light from the sun (or any other star) arises largely from the nuclear fusion of hydrogen into helium. Nuclear reactions are outside of the normal scope of the discussion of chemistry, so that can be largely ignored here. Hydrogen, although having one outermost electron does not fit into the alkali metals or any other group. It deserves its own treatment. It forms compounds analogous to those of the alkali metals, 56

but such hydrogen compounds are much less alkaline (or more acidic), much less ionic, and more volatile. Sodium chloride, the stereotypical salt, is neutral and clearly ionic; hydrogen chloride is a non-ionic gas under normal conditions and is a strong acid. The hydrogen analogue of sodium hydroxide is a volatile liquid under normal situations: unlike the strongly alkaline and solid sodium hydroxide, water is feebly ionic and effectively neutral. Hydrogen is a non-metal, forming a diatomic gas which results from the sharing of the single electrons of hydrogen atoms. It can achieve a stable ionic structure (no electrons!) by losing an electron or by gaining an electron and achieving the completed shell configuration of helium. The hydrogen molecule is best described as sharing the two electrons between two hydrogen atoms. This structure is highly stable and has little inclination to form bongos between other hydrogen molecules; hydrogen is a gas down to some of the lowest temperatures known. It is also the lightest of gases, weighing less even than helium. Hydrogen readily shares its electron with a strongly-electronegative element like any halogen, oxygen, or sulfur. The combination with fluorine is particularly violent and possible down to very low temperatures: 1/2 H2 (g) + 1/2 F2(g) → HF(g) Light is enough to force combustion between hydrogen and chlorine, and a spark is enough to cause combustion between hydrogen and oxygen. In view of the Hindenburg disaster of 1936, helium has long replaced hydrogen in lighter-than-air aircraft. That reaction is as such: H2 (g) + 1/2 O2 (g) → H2O (g) (Note that at the temperatures associated with such a combustion, water is in the gaseous state!) Somewhat analogously (but not very well), hydrogen can act somewhat like a halogen, forming hydrides with some metals. Most of these react violently with water to form hydrogen gas and the metal hydroxide. Hydrogen compounds with non-metals are typically among the most volatile substances of those elements. Under pressure, in aqueous solution or non-solid acids, hydrogen is a good reducing agent. Strong acids attack most metals: Zn (s) + 2 H+ (aq) → Zn 2+ (aq) + H 2 (g) . In the atmospheres of gas giants, gaseous hydrogen under great pressure reduces nitrogen to ammonia, carbon compounds to methane and other hydrocarbons, and oxides to water. Hydrogen forms more chemical compounds than any other element including carbon (almost all carbon compounds are compounds of hydrogen as well and vice-versa, but more substances containing hydrogen but not carbon exist than do compounds of carbon but not hydrogen). Hydrogen forms bonds with most non-metals, including oxygen, nitrogen, and 57

carbon. Although a hydrogen atom can bond with only one other element, and then only in a single bond, hydrogen allows very long chains of carbon atoms to form. Most of the hydrogen compounds with carbon alone are combustible gases or volatile liquids or waxy solids that can be vaporized and burned to produce water, carbon dioxide, and much heat. Natural gas, gasoline (a mixture of liquid hydrocarbons), and waxes as found in candles make suitable fuels. With such other elements as oxygen, nitrogen, sulfur, and in some cases metals hydrogen allows the formation of substances necessary for life, including carboxylic acids, sugars, proteins, nucleic acids, hemoglobin, and chlorophyll. Such complex compounds are ordinarily discussed in organic chemistry, a study associated more obviously with carbon.

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IB Chemistry/Bonding < IB Chemistry Jump to: navigation, search

Contents [hide]  



   

1 Introduction to Bonding 2 4.1 Ionic Bond o 2.1 What are ions? o 2.2 Description of Ionic Bonds o 2.3 Formation 3 4.2 Covalent Bond o 3.1 The Valence Bond Model o 3.2 Double and Triple Bonds o 3.3 Electron Sharing and Orbitals o 3.4 The Sigma Bond o 3.5 The Pi Bond 4 4.3 Intermolecular forces 5 4.4 Metallic bonding 6 4.5 Physical Properties o 6.1 Characteristics 7 HL Material

[edit] Introduction to Bonding Put simply, chemical bonds join atoms together to form more complex structures (like molecules or crystals). Bonds can form between atoms of the same element, or between atoms of different elements. There are several types of chemical bonds which have different properties and give rise to different structures. Ionic bonds form between positive and negative ions (atoms). In an ionic solid, the ions arrange themselves into a rigid crystal lattice. NaCl (common salt) is an example of an ionic substance. Covalent bonds are formed when atoms share electrons with each other. This gives rise to two structures: molecules and covalent network solids. Methane (CH4) is a covalent molecule and glass is a covalent network solid.

59

Whether two atoms form a covalent or ionic bond can be predicted from the atoms' electronegativities: Type of bond Difference in atoms' electronegativities Example Non-polar covalent bond: 0.0-0.4 F2, CH4 Slightly polar bond: 0.5-0.9 Cl2O, NH3 Moderately polar bond: 1-1.3 CO2, SiCl4 Highly polar bond: 1.4-1.7 H2O, Al2Cl6 Slightly ionic bond: 1.8-2.2 NaCl, Al2O3 Ionic Bond: 2.3+ Na2O, CsF

Metallic bonds occur between metal atoms. In a metallically bonded substance, the atoms' outer electrons are able to freely move around - they are delocalised. Iron is a metallically bonded substance. Chemical bonding is one of the most crucial concepts in the study of Chemistry. In fact, the properties of materials are basically defined by the type and number of atoms they contain and how they are bonded together.

[edit] 4.1 Ionic Bond 4.1.1 : Ionic bond - +ve (cations) and -ve (anions) ions are attracted to each other and form a continuous ionic lattice.

[edit] What are ions? Ions are atoms or molecules which are electrically charged. Cations are positively charged and anions carry a negative charge. Ions form when atoms gain or lose electrons. Since electrons are negatively charged, an atom that loses an electron will become positively charged (similarly an atom that gains one or more electrons becomes negatively charged).

[edit] Description of Ionic Bonds Ionic bonds form between positive and negatively charged ions. These oppositely charged ions attract each other and remain close together - they become ionically bonded. The law of electrostatic explains why this happens: opposite charges attract and like charges repel. When many ions attract each other, they form into large, orderly crystal lattices in which each ion is surrounded by ions of the opposite charge. When a metal forms an ionic bond with a non-metal, electrons are transfered from one element to another. When Ions are formed it is called Ionization A diagram of an ionic solid should go here 60

[edit] Formation

Lewis structure of the ionic bond between sodium and chlorine. Ionic bonds form when metals and non-metals chemically react. By definition, a metal is relatively stable if it loses electrons to form a complete valence shell and becomes positively charged. Likewise, a non-metal is relatively happy to gain electrons to complete its valence shell and become negatively charged. When a metal and a non-metal come into contact, the metal loses electrons by transferring them to the non-metal, which gains them. Consequently, ions are formed, which instantly attract each other and form an ionic bond.

4.1.2 : Group 1 metals form +1 ions, group 2 metals form +2 ions, metals in group 3 form +3 ions. Examples : Li+, Mg2+, Al3+...Greater ease of ionisation Li->Cs is due to the increased electron shielding of the nuclear attraction caused by additional inner shells of electrons. The easier atoms are to ionise, the more reactive they will be because less energy is required to ionise them, and so they react faster. 4.1.3 : Group 6 ions will form 2- ions, Group 7 ions will form 1- ions. Examples : O2-, Cl-... 4.1.4 : The transitions metals (elements from Ti to Cu, ignore Sc and Zn) can form multiple ions (ie Fe2+, Fe3+) (due to proximity of 4s and 3d shells) 4.1.5 : The ionic or covalent nature of the bonding in a binary compound is a result in the difference between their electronegativity...NaCl (s) is ionic, HCl(g) is (polar) covalent (also, covalent molecules tend to be gases/liquids, ionic tends to be solid...except network covalent which will be solid). In general, if the difference between electronegativities is greater than 1.7, the bond will be more than 50% ionic. 4.1.6 : Take the name of the group 1,2, or 3 metal and add...fluoride, chloride, bromide, iodide etc , oxide, sulfide etc...Nitride and phosphide...how exciting :)

[edit] 4.2 Covalent Bond One useful model of covalent bonding is called the Valence Bond model. It states that covalent bonds form when atoms share electrons with each other in order to complete their valence (outer) electron shells. They are mainly formed between non-metals (i.e. chlorine, sulfur, carbon etc.).

[edit] The Valence Bond Model 61

An example of a covalently bonded substance is hydrogen gas (H2). A hydrogen atom on its own has one electron – there is room for two to complete its valence shell. When two hydrogen atoms bond, each one shares its electron with the other, i.e. the electrons are attracted by two nuclei instead of just one and so releasing energy. Both atoms now have access to two electrons: they become a stable H2 molecule joined by a single covalent bond.

[edit] Double and Triple Bonds Covalent bonds can also form between other non-metals, for example chlorine. A chlorine atom has 7 electrons in its valence shell — it needs 8 to complete it. Two chlorine atoms can share 1 electron each to form a single covalent bond. They become a Cl2 molecule. Oxygen can also form covalent bonds; however, it needs a further 2 electrons to complete its valence shell (it has 6). Two oxygen atoms must share 2 electrons each to complete each other's shells, making a total of 4 shared electrons. Because twice as many electrons are shared, this is called a double covalent bond. Furthermore, nitrogen has 5 valence electrons (it needs a further 3). Two nitrogen atoms can share 3 electrons each (6 in total) to make a N2 molecule joined by a triple covalent bond.

[edit] Electron Sharing and Orbitals Carbon, contrary to the trend, does not share four electrons to make a quadruple bond. The reason for this is that the fourth pair of electrons in carbon cannot physically move close enough to be shared. The valence bond model explains this by considering the orbitals involved. Also more energy can be released by making 4 single bonds to 4 other carbon atoms to form a diamond structure. Recall that electrons exist as clouds of electron density (orbitals) in atoms. The valence bond model works on the principle that orbitals on different atoms must overlap to form a bond. There are several different ways that the orbitals can overlap, forming several distinct kinds of covalent bonds.

[edit] The Sigma Bond The first, and simplest kind of overlap is when two s orbitals come together. It is called a sigma bond (sigma, or 'σ', is the greek equivalent of 's'). Sigma bonds can also form between two p orbitals that lie pointing towards each other. 62

picture of sigma bonds- please press the no.1O sigma bond.jpg

[edit] The Pi Bond The second, and equally important kind of overlap is between two parallel p orbitals. Instead of overlapping, head-to-head (as in the sigma bond), they join side-to-side, forming two areas of electron density above and below the molecule(delocalization). This type of overlap is referred to as a pi ('π', from the greek equivalent of p) bond. 4.2.1 : Covalent bonds are where two atoms each donate 1 electron to form a pair held between the two atoms...Such bonds are generally formed by atoms with little difference in electronegativity...ie C, H and O in organic chemistry. 4.2.2 : All electrons must be paired...Lewis diagrams are the element symbol with the outer (valence) shell of electrons left over and spare electrons pair up...in general C forms 4 bonds, N forms 3, O forms 2, halogens form 1, H forms 1...(Li would form 1, Be 2, and B 3 but they don't usually...metallic or ionic bonding) 4.2.3 : Electronegativity values range from 0.7 to 4...from bottom left to top right respectively (hydrogen falls B and C with a electronegativity of 2.1... 4.2.4 : When covalent molecules have a difference in electronegativity (between the two bonding atoms) then the pair will be held closer to the more electronegative atom...resulting in a small -ve charge on the more electronegative atom, and a small +ve charge on the other...results in polar bonds 4.2.5 : Shape of molecule with 4 electron pairs depends on number of lone pairs. 3 lone pairs -> linear, 2 lone pairs -> bent, 1 lone pair -> trigonal pyramid, No lone pairs -> tetrahedral 4.2.6 : The polarity of a molecule depends on both the shape and the polarity of the bonds...1) if there are no polar bonds, it's not polar. 2) if there are polar bonds, but the shape is symmetrical, it's not polar (think about it like 3D vector addition...if they add to zero, then it's not polar). 3) if there are polar bonds, and it's not symmetric, then the molecule is polar

[edit] 4.3 Intermolecular forces 4.3.1 : van der Waal's forces -- Electrons will not be evenly spread around an atom/molecule at any given time, meaning the molecule will have a slight postitive charge on one end, and a negative at the other. This temporary state may cause attraction between two molecules, pulling them together (also known as London Dispersion Forces). Polar molecules, when properly oriented, will attract each other as a result of Dipole-Dipole forces. Dipole-Dipole forces are stronger than van der Waal's forces. Hydrogen bonding is when hydrogen is bonded to nitrogen, oxygen or fluorine, and a very strong dipole is formed, making the hydrogen very strongly 63

positive. This hydrogen is then attracted to the lone pairs on other similar molecules--nitrogen, oxygen and fluorine all have lone pairs--forming a hydrogen bond, which is stronger than van der Waal's or dipole-dipole forces, but weaker than covalent bonding. 4.3.2 : Structural features -- Nonpolar molecules have van der Waal's forces only. This is also present in all other molecules, though its strength is often insignificant compared to the others. Polar molecules have dipole-dipole forces, which arise from polar bonds and asymmetry in molecules. Hydrogen bonds result from strongly delta positive hydrogen. This results in molecules with hydrogen bonding exhibiting stronger intermolecular forces, i.e. higher boiling/melting points etc. For example, H2O has a higher bp then H2S due to hydrogen bonding. Neutral molecules don't conduct electricity but some polar molecules exchange protons to form ions e.g. 2H2O makes H3O+ and OH-

[edit] 4.4 Metallic bonding 4.4.1 : The metal atoms lose their outer electrons which then become delocalized, and free to move throughout the entire metal. These -ve delocalized electrons hold the metal cations together strongly. Since these electrons can flow, atoms with metallic bonding exhibit high electrical conductivity. The number of valence electrons involved in the bonding and the strength of the nucleus charge determines the strength. Unlike ionic bonding, distorting the atoms does not cause repulsion so metallic substances are ductile (can be stretched into wires) and malleable (can be made into flat sheets). The free moving electrons also allow for high thermal conductivity, and the electrons can carry the heat energy rather than it being transferred slowly through atoms vibrating. Metallic bonds occur among metal atoms. A sea of valence electrons surrounds positive metal ions. The electrons are free to move throughout the resulting crystal. The delocalized nature of the electrons explains a number of unique characteristics of metals: they are good conductors of electricity, they are ductile, meaning they can be made into wires, and they are malleable, meaning they can easily be hammered into thin sheets.

[edit] 4.5 Physical Properties Melting and Boiling point: High with Ionic, Metallic Bonding and Network Covalent. Low with Covalent Molecular Bonding. Volatility: Covalent Molecular Substances are volatile, others are not. Conductivity: Metallic substances conduct. Polar molecular substances conduct, non-polar ones don't. Ionic substances do conduct when molten or dissolved in water but never when solid. Solubility: Ionic substances --> generally dissolve in polar solvents (like water). Metallic substances --> soluble in liquid metal. Non-polar molecules are generally soluble in non-polar solvents, and polar in polar. Organic molecules with a polar head --> Short chain molecules are 64

solubility in polar solvents but long chains can eventually outweigh the polar 'head' and will dissolve in non-polar solvents.

[edit] Characteristics . Ionically bonded substances typically have the following characteristics.      

High melting point (solid at room temp) Hard Brittle (can shatter) Some dissolve in water Conduct electricity when dissolved or melted Typically stronger than covalent bonds.

65

IB Chemistry/States of Matter < IB Chemistry This page may need to be reviewed for quality. Jump to: navigation, search

Contents [hide]  

1 States of Matter Revision Notes 2 5.1 States of matter o 2.1 5.1.1 Solids o 2.2 5.1.2 Liquids o 2.3 5.1.3 Gas o 2.4 5.1.4 Transitions o 2.5 5.1.5 Increasing the temperature o 2.6 5.1.6 Diffusion

[edit] States of Matter Revision Notes [edit] 5.1 States of matter [edit] 5.1.1 Solids In a solid-state molecules/atoms are tightly packed. The force of attraction between molecules/atoms overcomes any translational motion of molecules (they do, however, vibrate in position).

[edit] 5.1.2 Liquids Liquids are particles that are held close together, but not as strongly as solids in that they are free to move, but not to escape the liquid (except for fast traveling particles -> evaporation).

[edit] 5.1.3 Gas Gas consists of particles that can move independently and randomly, with no significant forces between particles, and a large (comparatively) amount of space between them. An ideal gas is composed of randomly moving point masses occupying no space and with no forces between 66

masses. The average (rms) speed of the movement of particles is proportional to temperature (in K). As a result, the kinetic energy of the particles is also proportional to temperature.

[edit] 5.1.4 Transitions Solid into liquid , the rigid structure of the solid is overcome due to increased vibration of particles (due to energy being added in the form of heat). The particles cannot escape from the liquid, but can move within it (thus dissolving in it). Liquid to gas, as energy (in the form of heat) is given to the liquid, the particles gain enough energy to escape the liquid, and become a gas, with particles widely spaced gas can take tremendous volumes. Gas into liquid, as heat (energy) is removed, the particles slow down to the point where the forces between particles are strong enough to hold them together, a liquid is formed! Liquid into solid, even more energy is removed and particles move slower. Stronger forces build a more rigid structure for the material.

[edit] 5.1.5 Increasing the temperature Causes the material to become larger in volume or higher pressure. An increase in volume (which holds the material) will translate itself in a decrease in temperature or a decrease in pressure. An increase in pressure(on the other hand) will cause a decrease in volume or increase in temperature.

[edit] 5.1.6 Diffusion Since the particles are moving at random (here: in a gas), two separated samples of gas will eventually mix causing diffusion. This will occur at a higher rate with higher temperature since the particles are moving faster.

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IB Chemistry/Energetics < IB Chemistry Jump to: navigation, search

Contents [hide]       

1 6.1 Exothermic and Endothermic Reactions 2 6.2 Calculation of enthalpy changes 3 6.3 Hess's Law 4 6.4 Bond enthalpies 5 6.5 6 6.6 7 HL Material

[edit] 6.1 Exothermic and Endothermic Reactions 6.1.1 : If the reaction produces heat (increases the temperature of the surroundings) then it's exothermic. If it decreases the temp (that is, absorbs heat) then it's endothermic. Also, the yield of an equilibrium reaction which is exothermic will be increased if it occurs at low temps, and so for endothermic reactions at high temps. 6.1.2 : Exothermic -> a reaction which produces heat. Endothermic -> a reaction which absorbs heat. Enthalpy of reaction -> the change in internal energy (H) through a reaction is ∆H. 6.1.3 : ∆H will be negative for exothermic reactions (because internal heat is being lost) and positive for endothermic reactions (because internal energy is being gained). 6.1.4 : The most stable state is where all energy has been released...therefore when going to a more stable state, energy will be released, and when going to a less stable state, energy will be gained. On an enthalpy level diagrams, higher positions will be less stable (with more internal energy) therefore, if the product is lower, heat is released (more stable, ∆H is -ve) but if it is higher, heat is gained (less stable, ∆H is +ve). 6.1.5 : Formation of bonds -> release of energy. Breaking of bonds -> gain / absorption of energy.

[edit] 6.2 Calculation of enthalpy changes 68

6.2.1 : change in energy = mass x specific heat capacity x change in temperature --> E = mc∆T 6.2.2 : Enthalpy changes (∆H) are related to the number of moles in the reaction...if all the coefficients are doubled, then the value of ∆H will be doubled (attention must be paid to limiting reagents though). 6.2.3 : When a reaction is carried out in water, the water will gain or lose heat from (or to) the reaction, with hopefully little escaping the water. Therefore, the change in energy, and so the ∆H value, can be calculated with E = mc∆T where E is equal to ∆H, m is the mass of water present, and c = 4.18 kJ Kg-1 K-1. This ∆H value can then be calculated back to find the enthalpy change for each mol of reactants. 6.2.4 : The solution should be placed in a container as insulated as possible, to keep as much heat as possible from escaping. The temperature should be measured continuously , and the value used in the equation is the maximum change in temp from the initial position. 6.2.5 : The results will be a change in temperature. this can be converted into a change in heat (or energy) by using the above equation and a known mass of water. this can be used to calculate the ∆H for the amount of reactants present, and then this can be used to calculate for a given number of mols.

[edit] 6.3 Hess's Law 6.3.1 : Hess' Law states that the total enthalpy change between given reactants and products is that same regardless of any intermediate steps (or the reaction pathway). To calculate: 1) Reverse any reactions which are going the wrong way and invert the sign of their ∆H values (multiply by -1) 2) Divide or multiply the reactions until the intermediate products will cancel out when the reactions are vertically added (always multiply/divide the ∆H value by the same number) 3) Vertically add the reactions. 4) Divide or multiply the resulting reaction to the correct coefficients.

[edit] 6.4 Bond enthalpies 6.4.1 : Bond enthalpy (aka dissociation enthalpy) -- the enthalpy change when one mole of bonds are broken homolitically in the gas phase. ie X-Y(g) -> X(g) + Y(g) : ∆H(dissociation). Molecules such as CH4 have multiple C-H bonds to be broken, and so the bond enthalpy for C-H is actually an average value. These values can be used to calculate unknown enthalpy changes in reactions where only a few bonds are being formed/broken. 69

6.4.2 : If the reaction can be expressed in terms of the breaking and formation of bonds in a gaseous state, then by adding (or subtracting when bonds are formed) the ∆H values the total enthalpy of reaction can be found.

[edit] 6.5 6.5.1 An increased disorder (entropy) can be caused by mixing two different types of particles, increased movement of particles (including state changes), or increased number of particles. Increasing the number of particles in a gaseous state gives the largest change in entropy. 6.5.2 Since ∆S = -∆H/T, if ∆H is -, then ∆s is positive, and visa versa.

[edit] 6.6 6.6.1 Standard free energy of reaction is the free energy that a reaction takes or gives at standard values of temperature and pressure,. 6.6.2 If ∆G is negative, the reaction is spontaneous. If ∆G is positive, the reaction is not spontaneous. 6.6.3 If ∆sT > ∆H, the reaction is spontaneous. If temperature drops so that ∆sT < ∆H, then the reaction is non-spontaneous.

70

IB Chemistry/Kinetics < IB Chemistry Jump to: navigation, search Rates of reaction 7.1.1 : Rate of reaction is concerned with how quickly a reaction reaches a certain point (not to be confused with how far a reaction goes...i.e. equilibrium). 7.1.2 : Interpretation of rate graphs. Reaction rate graphs will generally be graphed with time on the x-axis and some measure of how far the reaction has gone (ie concentration, volume, mass loss etc) on the y-axis. This will generally produce a curve with, for example, the concentration of the reactants approaching zero. Collision theory 7.2.1 : Collision theory -- reactions take place as a result of particles (atoms or molecules) colliding and then undergoing a reaction. Not all collisions cause reaction, however, even in a system where the reaction is spontaneous. The particles must have sufficient kinetic energy, and the correct orientation with respect to each other for the two to react. Even then, the transition state may revert to the reactant molecules instead of forming the product molecules. 7.2.2 : Higher temperature causes a greater average kinetic energy of the particles in a material. This leads to a faster reaction because there are more collisions, and each collision is more likely to succeed. Higher concentrations cause more collisions and therefore a faster reaction. Catalysts may provide an alternative pathway with lower activation energy and increase the probability of proper orientation. Each collision is more likely to succeed and this results in a faster reaction. In heterogeneous reactions (where the reactants are in different states) the size of the particles of a solid may change reaction rate, since the surface is where the reaction takes place, and the surface area is increased when the particles are more finely divided; therefore smaller solid particles in a heterogeneous reaction cause a faster reaction. 7.2.3 : Most reactions involve several steps, which can be individually slow or fast, and which, all together, make up the complete reaction. The slowest of these steps is called the rate determining step, as is determines how fast the reaction will go. It is also not necessary that all the reactants are involved in ever step, and so the rate determining step may not involve all the 71

reactants. as a result, increasing their concentration (for example) of a reactant which is not involved in the rate determining step will not change the overall reaction rate.

72

IB Chemistry/Equilibrium < IB Chemistry This page may need to be reviewed for quality. Jump to: navigation, search 8.1 Dynamic equilibrium 8.1.1 : In all reactions, there are in fact two reactions occurring, one where the reactants produce the products, and the other where the products react to form the reactants. In some reactions, this second reaction is insignificant, but in others there comes a point where the two reactions exactly cancel each other out...thus the reactants and products remain in equal proportions, though both are continually being used up and produced at the same time. 8.2 The position of equilibrium 8.2.1 : The equilibrium constant Kc is a constant which represents how far the reaction will proceed at a given temperature. 8.2.2 : When Kc is greater than 1, products exceed reactants (at equilibrium). When much greater than 1, the reaction goes almost to completion. When Kc is less than 1, reactants exceed products. When much less than 1 (Kc can never be negative...so when it is close to zero) the reaction hardly occurs at all. 8.2.3 : The only thing which can change the value of Kc for a given reaction is a change in temperature. The position of equilibrium, however, can change without a change in the value of Kc. Effect of Temperature : The effect of a change of temperature on a reaction will depend on whether the reaction is exothermic or endothermic. When the temperature increases, Le Castellani principle says the reaction will proceed in such a way as to counteract this change, ie lower the temperature. Therefore, endothermic reactions will move forward, and exothermic reactions will move backwards (thus becoming endothermic). The reverse is true for a lowering of temperature. Effect of Concentration : When the concentration of a product is increased, the reaction proceeds in reverse to decrease the concentration of the products. When the concentration of a reactant is increased, the reaction proceeds forward to decrease the concentration of reactants. Effect of Pressure : In reactions where gases are produced (or there are more mols of gas on the right), and increase in pressure will force the reaction to move to the left (in reverse). If pressure 73

is decreased, the reaction will proceed forward to increase pressure. If there are more mols of gas on the left of the equation, this is all reversed. 8.2.4 : Based on the previous section, you should be able to predict what's going to happen given a reaction if the temperature, pressure, or concentration is changed. 8.2.5 : A catalyst does not effect either Kc or the position of equilibrium, it only effects the rate of reaction. 8.2.6 : N2(g) + 3H2(g) 2NH3(g) : delta-H = -92.4 kJ mol-1 as can be seen, there are more mols of gas on the left than the right, so a greater yield will be produced at high pressure. The reaction is exothermic, therefore it will give a greater yield at low temperatures, however this is not possible as the rate of reaction becomes too low, and the temperature must actually be increased. A catalyst of finely divided iron is also used to help speed the reaction (finely divided to maximize the surface area).

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IB Chemistry/Acids and Bases < IB Chemistry Jump to: navigation, search This is chapter 9, and this space will hopefully house an introduction to the chapter. To do: Add titles to subheadings and fill in #HL Material

Contents [hide] 



 





1 Properties of acids and bases o 1.1 Properties of acids and bases in aqueous solution o 1.2 Experimental properties of acids and bases 2 Brønsted-Lowry acids and bases o 2.1 o 2.2 o 2.3 o 2.4 3 Lewis theory o 3.1 4 Strong and weak acids and bases o 4.1 o 4.2 Examples of strong and weak acids and bases o 4.3 5 The pH scale o 5.1 o 5.2 6 HL Material o 6.1 Calculations involving acids and bases o 6.2 Salt hydrolysis o 6.3 Acid-base titrations o 6.4 Indicators

[edit] Properties of acids and bases 75

[edit] Properties of acids and bases in aqueous solution A reaction between a strong acid and a strong base produces water and a salt. A reaction between an acid that is not strong and a base that is not strong (can be only 1) produces only a salt. For example:

Acids reacting with metals will produce hydrogen gas, e.g.

.

Acids reacting with carbonates will produce water and carbon dioxide, e.g. .

[edit] Experimental properties of acids and bases To do: Make this paragraph coherent When acids and bases neutralize, the reaction is noticeably exothermic (i.e., heat can be felt coming from the reaction). Obviously, they will have an effect on the color of indicators as described below. The hydrogen produced in the reaction of acids with metal will produce a 'pop' sound if a match is held to it, and the CO2 from the carbonate reaction will turn limewater a milky white when bubbled trough it. Examples of some acids: Examples of bases:

,

,

,

.

, NH3, CH3COO-.

[edit] Brønsted-Lowry acids and bases [edit] According to the Brønsted-Lowry theory, acids are defined as proton (H+ ion) donors and bases are defined as proton acceptors.

[edit] For a compound to act as a B-L acid, it must have a hydrogen atom in it, which it is capable or losing while remaining fairly stable. A BL base must be capable of accepting a hydrogen ion while remaining relatively stable (or reacting to form a stable compound, e.g. water and a salt). 76

Some compounds (such as water) may act as both, e.g. (H2O-> OH- or H3O+). These substances are called "amphoteric".

[edit] Acid base reactions always involve an acid-base conjugate pair. If the reactant is an acid, the matching product is its conjugate base, e.g. HCl/Cl -, CH3COOH/CH3COO-, NH4+/NH3.

[edit] The conjugate base will always have one less H atom than the acid (or the acid one more than the base). In compounds where there are many hydrogen atoms, the one which is held the weakest is generally the one which is lost, and this must be reflected in the writing of the compound, as in the CH3COOH example above.

[edit] Lewis theory [edit] Lewis defined acids as electron pair acceptors and bases as electron pair donors. A simple proton-hydroxide reaction shows that this is equivalent to Brønsted-Lowry theory: H+ + :OH- --> H-OH The proton is accepted because the hydroxide has a pair of electrons with which it forms a covalent bond to the proton. Lewis theory goes beyond Brønsted-Lowry theory to describe non-proton acid-base reactions: F3B + :NH3 --> F3B-NH3 Ligands and nucleophiles can be classed as Lewis bases. Lewis acids include metal ions and electrophiles. Ag+ + 2 :NH3 --> [Ag(NH3)2]+

[edit] Strong and weak acids and bases [edit] 77

Strong and weak acids are defined by their ease of losing (or donating) a proton. A strong acid, when placed in water, will almost fully ionise/dissociate straight away, producing H 3O+ ions from water. a weak acid will, however, only partially do this, leaving some unreacted acid remaining. This is set up as an equilibrium, and so when some of the H3O+ ions produced by a weak acid are reacted, LCP means that more of the acid will react to form H3O+ ions. This means that, given an equal number of moles of acid, they will be neutralized by the same amount of strong base, but their solutions will have different pH values. A weak base is the same as this, only it accepts protons and so produces OH- ions from water rather than H3O+. Any solution's ability to conduct electricity is defined by its ionic charges. As a result, a strong acid will produce more charged ions than a weak one, and so its solution will be a better electrical conductor than a weak acid. The same goes for strong/weak bases.

[edit] Examples of strong and weak acids and bases Strong acids : HCl, HNO3, H2SO4, HClO3 . Weak acids : CH3COOH, H2CO3. Strong bases : group 1 hydroxides (e.g. NaOH, etc.), BaOH. Weak bases : NH3, CH3CH2NH2.

[edit] The strength of an acid or base can obviously be measured with a universal indicator or a pH meter, and also the rate of reaction. Hydrogen production with metals or CO 2 with CaCO3 will reveal the strength of an acid. The relative acidities can also be found by neutralizing two acids with a strong base in the presence of an indicator.

[edit] The pH scale [edit] pH vales range up and down from 7 (being the neutral value of pure water at 25°C and 1 atmosphere). Lower pH value are acidic; higher values are basic. pH can be measured with a pH meter, or with pH paper (paper containing a mixture of indicators to cause a continuous color change). pH is a measure of the concentration of H3O+ ions.

[edit] If we have two solutions with their pH values, the lower one will be more acidic and the higher one will be more basic (though they could both still be basic/acidic with respect to water – pH 7). 9.4.3 : a change of 1 in the pH scale represents a 10 times change in the acidity or basicity of the solution (because it is a log scale). Concentration is equal to 10 -pH or pH=-log10[H3O+] where [H3O+] is the concentration of hydrogen ions in solution. 78

[edit] HL Material Topic 18 is the additional HL material for Topic 9. Brønsted-Lowry acids and bases Lewis theory

[edit] Calculations involving acids and bases This HL Sub-topic is Sub-topic 5 of SL Option A, with one difference: The SL Option does not require knowedge of Ka x Kb = Kw.

[edit] Salt hydrolysis A reader requests that this page be expanded to include more material. You can help by adding new material (learn how) or ask for assistance in the reading room.

[edit] Acid-base titrations [edit] Indicators Indicators change colour depending in the presence of acid or base. Here is a table of common indicators: Indicator methyl orange bromophenol blue bromophenol green methyl red bromothymol blue phenol red phenolphthalein

pKa 3.7 4.0 4.7 5.1 7.0 7.9 9.3

pH range 3.1-4.4 3.0-4.6 3.8-5.4 4.2-6.3 6.0-7.6 6.8-8.4 8.3-10.0

Acid colour red yellow yellow red yellow yellow colourless

Alkali colour yellow blue blue yellow blue red red

Table adapted from: Stark JG & Wallace, HG (1982) Chemistry Data Book p103 2nd ed, John Murray, London.

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IB Chemistry/Oxidation and Reduction < IB Chemistry Jump to: navigation, search Oxidation refers to the loss of electrons, while reduction refers to the gain of electrons. A full reaction will have a net change of 0 electrons, meaning that the amount of electrons lost in one portion of the equation will equal the amount of electrons gained in another. In other words, there will be a reduced portion and an oxidized portion. Collectively, this is referred to as a redox reaction.

Contents [hide]     



1 Oxidation and Reduction Half Equations 2 Oxidation Numbers 3 Reactivity o 3.1 Reactivity Series 4 Voltaic Cells 5 Electrolysis o 5.1 Electrorefining o 5.2 Factors affecting discharge of ions o 5.3 Standard electrode potentials  5.3.1 Hydrogen Electrode 6 HL Material

[edit] Oxidation and Reduction Half Equations Each full reaction can be broken apart to its component half reactions, the oxidation and reduction parts. In a half reaction, electrons are involved as part of the reaction. We will consider the reaction of magnesium and oxygen to form magnesium oxide. Since this is an ionic compound, it will have a negative ion (O2-) and a positive ion (Mg2+). The first step will be the formation of these ions. Mg → Mg2+ + 2eO2 + 4e- → 2O2As stated earlier, the loss and gain of electrons must be balanced. In this situation, we will double the entire oxidation reaction (that involving magnesium): 80

2Mg → 2Mg2+ +4eIf we were to combine these half reactions, we would obtain the full equation: 2Mg + O2 → 2MgO There are several mnemonics for oxidation and reduction, such as OILRIG (Oxidation Is Loss, Reduction Is Gain), and LEO GER (Lose Electrons: Oxidation, Gain Electrons: Reduction).

[edit] Oxidation Numbers It is easy to see which element is getting oxidized and which is getting reduced by using Oxidation Numbers, which are assigned by some rules which you must learn. 

The oxidation number of an atom, or a single-atom ion, is the same as the charge: Species Oxidation number



Au

0

Au+

+1

Au+3

+3

Au-

-1

Consider every compound to be made of ions. Consider covalent bonds to break with the two electrons assigned to the higher electronegativity atom. If two identical atoms are covalently bonded, consider the bond's electrons to be assigned one to each atom. Species Oxidation number H2

0

HCl

+1 (H) and -1 (Cl)

H2O

+1 (H) and -2 (O)

O2

0

Consequently:    

The sum of all the oxidation numbers in an ion is the charge of the ion. The sum of all the oxidation numbers in a compound is 0. Elements not combined with anything have an oxidation number of 0. In compounds, the following rules are observed: o Group 1 always form +1 ions, and Group 2 always form +2 ions. 81

o o o

Fluorine is always -1 Oxygen is always -2, except when in a peroxide (oxidation number is -1), or oxygen fluoride (oxidation number is +1) Hydrogen is always +1, except when in a hydride, for example sodium hydride (NaH), (oxidation number is -1)

We can quickly assign oxidation states using the observation above: NaHSO4 Na will be +1 (Group 1 elements only ever form +1 ions). H will be +1 (It is not a hydride). O will be -2. The sum of the oxidation numbers is 0, so S must be +6. Reduction occurs when the oxidation number decreases, and oxidation is when it increases. The change will be the number of electrons involved in the half equation. Keep this in mind that when something is being neutralized or precipitated it is not a redox reaction. An oxidizing agent is that which oxidizes something (it takes its electrons, thereby getting reduced). Oppositely, a reducing agent is one that reduces something (it loses its electrons, thereby getting oxidized). It is possible for an atom to have an oxidation number of 0, for example in Ni(CO) 4, due to the ligand nature of the bonds.

[edit] Reactivity The reactivity of a metal is related to how readily it loses its outer electrons. This means that metals higher in the series will displace metals in an ionic reaction lower down. Metals high in the reactivity series are also better reducing agents, due to the fact that they lose their outer electrons more easily. As learned with periodicity, the electronegativity increases down the series. The reactivity series is below, which the better reducing agents are

[edit] Reactivity Series          

K Na Li Sr Ca Mg Al C Zn Cr 82

           

Fe Cd Co Ni Sn Pb H Cu Ag Hg Au Pt

[edit] Voltaic Cells A half cell is a metal in contact with a solution containing its own ions (for example, Zn in a solution of Zn2+). A voltaic cell is when there are two half cells connected together with a wire and a salt bridge. Electrons are liberated during an oxidation in one half cell, and then used in the reduction in the other half cell. This flow of electrons is the current from the voltaic cell. The salt bridge is used to allow for the ion transfer to eliminate the build up of charge in a half cell. In many simple cases, this can be something such as a porous cup.

[edit] Electrolysis Electricity can be used to force a non-spontaneous redox reaction to occur. This is done in the electrolytic cell. Electricity is passed through the electrolyte. If the electrodes are inert, such as graphite, and the electrolye is a molten compound then the products will be the oxidized and reduced forms of each. The anode (positive electrode) is where oxidation occurs as negative ions are attracted and lose electrons. The electron flow is therefore from the positive to the negative electrode (which to me doesn't seem logical -- negative electrodes should be attracted?) If the electrolyte is aqueous then there is a different, more complicated scenario as there is oxygen and hydrogen that may also be formed at the electrodes. The lower in the reactivity series a metal is, the greater its ability to gain electrons (high electronegativity). Thus, if hydrogen is lower then it will be reduced at the cathode in preference to another positive ion. This means that copper and silver will be the only metal ions that wil be reduced in preference to hydrogen at the cathode. However, if one ion is more concentrated than another it will be discharged. For example, in dilute solutions of sodium chloride, oxygen gas is given off in preference as there is simply more of it. If the electrode is made of one of the metal of one of the ions in the electrolyte, ie there are copper electrodes and the electrolyte is CuSO4, then the positive electrode is oxidized, and the negative electrode is reduced. This means that the concentration of the metal ion is constant as it is being removed and deposited at the same rate. 83

[edit] Electrorefining Anode made from impure metal, cathode = sheets of pure metal, the electrolyte is a compound of the metal. Eg. copper: Oxidation at anode produces copper (II) ions and 2eat cathode the impure copper(II) joins with 2e- produced at anode to create copper metal. the impurities drop to the bottom of the cell where they can be removed.

[edit] Factors affecting discharge of ions   

Duration - More time means more electrons can be passed from one electrode to the other. Current - More current, more electrons available, more products Charge of the ion - Ions with 2+ charge require 2mol of electrons/ mol of products, therefore smaller charge = more products / electron flow.

To find how much is produced = Current*time/96500 = moles of electrons. Then you can work out how many moles of electrons you need per mol of product, and therefore find how much product will be produced, and the weight.

[edit] Standard electrode potentials The standard electrode potential E0 (a table is found in the IB data booklet), is worked out by using the hydrogen electrode. Which is defined as the emf generated when it is connected to the standard hydrogen electrode by an external circuit and a salt bridge, measured under standard conditions. (All solution must have a concentration of 1.0mol/dm3. All gases must be at pressure 100kPa. All substance must be pure. Temperature at 298K.) [edit] Hydrogen Electrode The hydrogen electrode is made up of the H half-cell, which is H2 gas at 1 atm pressure, 298K, 1.0mol/dm3 hydrogen ion content with a Pt electrode. This is assigned 0V. If the metal in the other half cell is above hydrogen in the reactivity series, then these lose electrons to hydrogen and the electrode potential is negative. If the half-cell has a metal below hydrogen, then electrons will flow from H to the metal and produce a positive value. --E0cell is the difference between electrode potentials in a cell. If the cell is written with reduction on the right in the standard cell notation (Metal/Ions||Ions/Metal), then the E0 will be the left hand side subtracted from the right hand side. In reality, the easiest way to remember it is that it 84

is always More Positive - More Negative. You should normally get a positive value since a positive E0 means a negative Gibbs free energy, which means the reaction is spontaneous. ELECTRONS ALWAYS FLOW FROM THE MORE NEGATIVE HALF-CELL TO THE MORE POSITIVE HALF-CELL. Thus you can always tell which is undergoing oxidation and which has reduction.

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IB Chemistry/Organic Chemistry < IB Chemistry Jump to: navigation, search Organic chemistry is the study of compounds which contain carbon. The name 'organic' refers to the historical link between these compounds and living organisms. 'Inorganic' compounds were those which could be derived from non-living sources. Confusingly, some inorganic compounds contain carbon. This material in this chapter is restricted to that needed by SL Chemistry students For further material see Organic Chemistry

[edit] 10.1 10.1.1 Describe the features of a homologous series. This is how organic compounds are classified into 'families'. Successive members of a homologous series differ by a -CH2. Molecular mass increases by a fixed amount as you go up the series. For example, alkanes. Methane (CH4), ethane (C2H6, propane (C3H8), butane (C4H10). General formula: CnH2n+2 Other series have an additional 'functional group' which is also shown in the general formula. The functional group is usually a small group of atoms attached to a carbon atom in the molecule. For example, alcohols have the functional group -OH. General formula: CnH2n+1OH. 10.1.2 Predict and explain the trends in boiling points of members of a homologous series. Since successive members differ by -CH2, their carbon chains will continue to increase in length. Boiling point increases with increasing carbon number. Note that if you know the chemical characteristics of a functional group, you can predict the chemical properties of the series. 10.1.3 Distinguish between empirical, molecular and structural formulas. The empirical formula is the lowest whole number ratio of the atoms. The molecular formula is the actual number of atoms. The structural formula is a representation of how the molecules are bonded together. Full structural shows all bonds, whereas condensed structural omits bonds where they can be assumed. Ethane... Empirical: CH3, Molecular: C2H6, Condensed structural: CH3CH3 Full structural: H |

H |

H—C—C—H |

|

86

H

H

10.1.4 Describe structural isomers. Compounds with the same molecular formula but different arrangements of atoms. 10.1.5 Deduce structural formulas for the isomers of the non-cyclic alkanes up to C6. C4H10: Butane and 2-methyl propane. H |

H |

H |

H |

H |

H |

H |

H—C—C—C—C—HH—C—C—C—H | H

| H

| H

| H

| | | H | H H — C — H | H

C5H12: Pentane, 2-methylbutane, 2,2-dimethylpropane. H |

H |

H |

H |

H |

H |

H |

H |

H |

H |

H—C—C—C—C—C—HH—C—C—C—C—HH—C—H | H

| H

| H

| H

| H

| | | H | H H — C — H

| H

H | H | | | H — C — C — C —

H | H

| | | H | H H — C — H | H

[edit] HL Material Naming Organic Compounds Organic chemistry is concerned with the compounds of carbon. Since there are more compounds of carbon known than all the other elements together, it is helpful to have a systemic way of naming them. Identify longest carbon chain 1 carbon = meth- 2 carbons = eth- 3 carbons = prop- 4 carbons = but- 5 carbons = pent- 6 carbons = hex- 7 carbons = hept- 8 carbons = oct-

87

Identify type of bonding in the chain All single bonds in the carbon chain = -an- One double bond in the carbon chain = -en- One triple bond in the carbon chain = -ynIdentify functioning group Formula Name Example R-H Alkane Methane R-OH Alcohol ethanol R-NH3 Amine Ethylamine R-X Halogenoalkane Bromoethane (Where X = F, Cl, Br, I) R-C=O Aldehyde Ethanal | H

R-C-R Ketone Propanone || O

R-C-OH Carboxylix Acid Ethanoic Acid || O

R-C-OR Ester Ethyl Ethanoate || O

Numbers are used to give the positions of groups or bonds along the chain. Homologous Series The alkanes form a series of compounds all with the same general formula CnH2n+2 e.g. Methane CH4 Ethane C2H6 Propane C3H8 If one of the hydrogen atoms is removed what is left is known as an alkyl radical R- (e.g. methyl CH3- ) When other atoms or groups are attached to an alkyl radical they can form different series of compounds. These atoms or groups are known as functioning goups and the series formed are all homologous eries. Homologous series have the same general formula with the neighbouring members of the series differing by by -CH2: for example the general formula of alcohols is CnH2n+1OH. The chemical properties of the individual members of an homologous series are similar and they show a gradual change in physical properties.

Boiling Points As the carbon chain gets longer the mass of the moleculaes increases and the van der Waals' forces of attraction increase. A plot of boiling points against the number of carbon atoms show a sharp increse at first, as the percentage increase in mass is high, but as successive CH2- groups are added the rate of increase in boiling point decreases. When branching occurs the molecules become more spherical in shape, which reduces the contact surface area between them and lowers the boiling point. Other homologous series show similar trends but the actual temeratures at which the compounds boilwill depend on the type of attractive forces between the 88

molecules. The volatility of the compounds also follows the same pattern. The lower members of the alkanes are all gases as the attractive forces are weak and the next few members are volatile liquids. Methanol, the first member of the alcohols is a liquid at room temperature, due to the presence of hydrogen bonding. Methanol is calssed as volatile as its boiling point is 337.5K but when there are four or more carbon atoms in the chain the boiling points exceed 373K and the hugher alcohols have low volatility. Functional Group Strongest type of intermolecular attraction Alkane van der Waals' Alkene van der Waals' Alkyne van der Waals' Ester Dipole:Dipole Aldehyde Dipole:Dipole Ketone Dipole:Dipole Amine Hydrogen bonding Alcohol Hydrogen bonding Carboxylix Acid Hydrogen bonding

Low Reactivity of Alkanes Because of the strong C-H and C-C in alkanes they have low reactivity and only usually undergo combustion in the presence of Oxygen and substitution reactions with halogens in ultra-violet light. MECHANISM OF SUBSTITUTION REACTION WITH HALOGENS Chemical bonds can break homolytically or heterolytically. Heterolytically is when both of the shared electrons go to one of the atoms resulting in a positive and one negative ion. When halogen are exposed to ultraviolet light the halogen molecule bond can break homololytically meaning each of the atoms retains one of the pair of previously shared electrons. The bond between two halogen atoms is weaker than C-H or C-C bond in alkanes and so is more likely to become a free radical due to homolytic fission of the covalent bond. When a Chlorine molecule is exposed to ultraviolet rays and becomes two free radical chlorine atoms it is called the intiation stage. Free radicals have an unpaired electron and are therefore very reactive.

Topic 20 is the additional HL material for Topic 11. Determination of structure This HL Sub-topic is required for Sub-topic 1 of SL Option A Hydrocarbons The material on benzene in this HL Sub-topic is required for Sub-topic 1 of SL Option A Nucleophilic substitution reactions This HL Sub-topic is Sub-topic 4 of SL Option A Alcohols 89

[edit] Option G Syllabus G.1.1: Describe and explain the electrophilic addition Mechanisms of the reactions of alkenes with halogens and hydrogen halides. G.1.2: Predict and explain the formation of the major products in terms of the relative stabilities of carbocations. Markovnikov’s Rule(See Mechanism above as example): In the addition of HX to an unsymmetrical alkene, the H attaches to the C with fewer alkyl substituents, producing the more stable carbocation (Tertiary carbocations are more stable than secondary which are more stable than primary). The X then attaches to the C with more alkyl groups.

G.2.1: Describe, using equations, the addition of hydrogen cyanide to aldehydes and ketones. G.2.2: Describe and explain the mechanism for the addition of hydrogen cyanide to aldehydes and ketones. G.2.3: Describe, using equations, the hydrolysis of cyanohydrins to form carboxylic acids. G.3.1: Describe, using equations, the dehydration reactions of alcohols with phosphoric acid to form alkenes. G.3.2: Describe and explain the mechanism for the elimination of water from alcohols. G.4.1: Describe, using equations the reactions of 2,4-dinitrophenylhydrazine with aldehydes and ketones. G.5.1: Describe and explain the structure of benzene using physical and chemical evidence. Physical Evidence: All bonds of benzene are of equal length—shorter than single bonds, but longer than double. 90

Chemical Evidence: Benzene does not undergo addition reactions Heat of hydrogenation of benzene is endothermic whereas the heat of hydrogenation of a triene would be extremely exothermic

G.5.2: Describe and explain the relative rates of hydrolysis of benzene compounds halogenated in the ring and in the side chain. The rate of hydrolysis in the side chain is much faster than benzene compounds halogenated in the ring because benzene does not undergo addition reactions. Chlorobenzene is much slower than alkanes in hydrolysis and does not undergo nucleophilic substitution for two main reasons: 1. The nonbonding pair of election of the chlorine may react with the pi bond, and become delocalized, thus strengthening (and possibly depolarizing) the C-Cl bond. 2. The high electron density of the delocalized pi bond repels the nucleophile and blocks it from the slight positive carbon atom attached to the chlorine.

G.6.1: Outline the formation of Grignard Reagents. G.6.2: Describe, using equations, the reactions of Grignard reagents with water, carbon dioxide, aldehydes, and ketones G.7.1: Deduce the reaction pathways given the starting materials and the product. G.8.1: Describe and explain the acidic properties of phenol and substituted phenols in terms of bonding. Alcohols are not very acidic because they produce an anion which contains only one resonance structure, and localized electrons. Phenol is acidic due to its electron-withdrawing groups. As an anion, it becomes more stable due to the delocalized electrons which produce various resonance structures. 2,4,6-trinitrophenol contains many electron-withdrawing groups which cause it to be very acidic due to further delocalization of the electrons.

G.8.2: Describe and explain the acidic properties of substituted carboxylic acid in terms of bonding. The adjacent C=O bond weakens the normally strong O-H bond. 91

The ion formed by the removal of the H from a carboxylic acid is more stable because of the delocalized electrons forming several resonance structures.

G.8.3: Compare and explain the relative basicities of ammonia and amines. Ammonia is a weak base Amines are more basic than ammonia because the inductive effect of the alkyl group pushes the electrons towards the nitrogen atom, increasing the electron density of the nonbonding electron pairs of the nitrogen.

G.9.1: Describe, using equations, the reactions of acid anhydrides with nucleophiles to form carboxylic acids, esters, amides, and substituted amides. G.9.2: Describe, using equations, the reactions of acyl chlorides with nucleophiles to form carboxylic acids, esters, amides, and substitutes amides. G.9.3: Explain the reactions of acyl chlorides with nucleophiles in terms of an addition-elimination mechanism. G.10.1: Describe, using equations, the nitration, chlorination, alkylation, and acylation of benzene. G.10.2: Describe and explain the mechanism for the nitration, chlorination, alkylation, and acylation of benzene. G.10.3: Describe, using equations, the nitration, chlorination, alkylation, and acylation of methylbenzene. The methyl group is an electron supplier and an activator, so it is an ortho/para-director.

G.10.4: Describe and explain the directing effects and relative rates of reaction of different substituents on a benzene ring. G.11.1: Deduce reaction pathways given the starting materials and the product.

92

IB Chemistry/Human Biochemistry < IB Chemistry This page may need to be reviewed for quality. Jump to: navigation, search Option C: Human Biochemistry

Contents [hide]         

1 C.1 Diet (2h) 2 C.2 Proteins (3h) 3 C.3 Carbohydrates (2.5h) 4 C.4 Lipids (2.5h) 5 B.5 Micronutrients and Macronutrients 6 C.6 Hormones (2.5h) 7 B.7 Enzymes 8 B.8 Nucleic Acids 9 B.9 Respiration

[edit] C.1 Diet (2h) In a calorimeter experiment, the food is completely combusted. The energy released is used to heat a sample of water. The heat energy can be calculated using this equation: H = m cp T H is the heat released (J) m is the mass of water heated (g) cp is the specific heat capacity of water (4.18 J g-1 K-1) T is the change in water temperature (oC or K) The value calculated by this equation will typically need to be divided by the number of moles, or the number of grams of the food which has been burnt. Example: 0.547 g of sucrose, C12H22O11, was completely combusted in a food calorimeter. The heat evolved was equivalent to increasing the temperature of 250 g of water from 22.22 °C to 30.81 °C. Calculate the calorific value of sucrose (in kJ g-1 and kJ mol−1) given the specific heat capacity of water in Table 2 of the Data Booklet. Using the formula H = m cp T H is the heat released (J) m is the mass of water heated (250 g) cp is the specific heat capacity of water (4.18 J g-1 K-1) T is the change in wa ter temperature (8.59 oC or K) H = 8.98 kJ 93

This is due to combustion of 0.547 g of sucrose. Dividing by this mass gives us: 16.4 kJ g-1. 1 mole of sucrose has a mass of 342 g, so the molar enthalpy change is 5612 kJ mol-1. (342 g x 16.4 kJ g-1 or 8.98 kJ ÷ 1.60 x 10-3 mol.)

Questions 03NS3 C1. (a) State how genetically modified food differs from unmodified food. [1] (b) List two benefits and two concerns of using genetically modified crops. [4] C2 (d) 1.00 g of sucrose, C12H22O11, was completely combusted in a food calorimeter. The heat evolved was equivalent to increasing the temperature of 631 g of water from 18.36 °C to 24.58 °C. Calculate the calorific value of sucrose (in kJ mol−1) given the specific heat capacity of water in Table 2 of the Data Booklet. [3] 05MS3 C1. (c) The energy content of a vegetable oil was determined using a calorimeter. A 5.00 g sample of the oil was completely combusted in a calorimeter containing 1 000 g of water at an initial temperature of 18.0 oC. On complete combustion of the oil, the temperature of the water rose to 65.3 oC. Calculate the calorific value of the oil in kJ g−1 [4]

[edit] C.2 Proteins (3h) 2-aminopropanoic acid shows the essential structure shared by all 2-amino acids: H2NCH(CH3)-COOH The –CH3 group can be changed for one of 19 other groups to make the 20 amino acids encoded in our genes. The –NH2 group of one amino acid can join to the –COOH group of another. This condensation reaction forms a peptide bond and a water molecule. To use either of these techniques the peptide bonds in the proteins must first be hydrolysed to release individual amino acids. Include the use of Rf values in paper chromatography. Given isoelectric points, students should be able to determine a suitable pH to achieve good separation in electrophoresis. The amino acids of a protein can be identified if the protein is hydrolysed. Hydrolysis usually involves refluxing the protein with acid or alkali to give the separate amino acids. Chromatography separates the amino acids according to whether they prefer to dissolve in a solvent, or remain attached to a solid. Careful choice of solid and solvent allow amino acids to be separated as the solvent moves over the solid. The ratio of the movement of the amino acid, to the movement of the solvent, is called the Rf value. 94

Example: Using paper as the solid phase and a 8:1:1 mix of ethanol:water:ammonia as the solvent, the Rf values of Aspartic acid, lysine and leucine were found to be 0.13, 0.35 and 0.84 respectively. Draw a chromatogram showing the expected pattern of spots for the three pure amino acids and a mixture of all three. Electrophoresis uses an electric field to separate the amino acids. Positively-charged amino acids will tend towards the cathode and negatively charged amino acids will move towards the anode. Each amino acid has an isoelectric point, given in table 20 of your data book. This is the pH which gives the amino acid an overall neutral charge. Above this pH, the amino acid is negative (deprotonated) and below this pH the amino acid is positive (protonated). For example, at pH 5.4: Charge Structure Movement Asparagine

Cysteine

Glutamine



Note how neutral amino acids form zwitterions.

The sequence of amino acids in a protein is called its primary structure. The sequence is given from the free amine end (N-terminal) to the free acid end (C-terminal). The backbone of the protein can form two regular patterns due to hydrogen bonding. These are called secondary structures: A single chain can form an -helix. Hydrogen bonds form between the C=O of one amino acid and the H-N of the amino acid four residues further down the chain. When two chains run alongside one another, the C=O groups of one chain and N-H groups of the other chain can be aligned and hydrogen bonds form. This forms a -sheet. Tertiary structure is the folding of the protein into its final shape. The main driving force is for oily amino acids to be buried within the structure and polar amino acids to be exposed on the 95

surface. Other forces involved are hydrogen bonds, salt bridges (ionic bonds between acidic and basic amino acids) and disulphide bridges (covalent bonds between cysteine residues). A fictional peptide showing two disulphide bridges, a salt bridge, and inter-residue hydrogen bonding. Some proteins may be formed from multiple polypeptide chains. The assembly of two or more chains to form one protein is called quaternary structure. Like tertiary structure, the bonding may involve hydrophobic, ionic, covalent and hydrogen bonding. These are structure, biological catalysts (enzymes) and energy sources. Proteins are involved in holding biological structures together (e.g. collagen, keratin) Enzymes are biological catalysts which control metabolism. They are all proteins. Protein can be used as a source of energy but is generally only used if fat and carbohydrate is not available.

Questions 06MS3 C1. (a) (i) Deduce the structure of one of the dipeptides that can be formed when the two aminoacids below react together.[2] (ii) State the name given to this type of reaction and identify the other product of the reaction. [2] (b) Describe how a mixture of aminoacids can be analysed using electrophoresis. [4] (c) (i) Explain what is meant by the primary structure of proteins. [1] (ii) Explain, with reference to hydrogen bonding, why the α-helix and β-sheet secondary structures of proteins are different. [2] (iii) Identify three types of interactions responsible for the tertiary structure of proteins. [2]

[edit] C.3 Carbohydrates (2.5h) Monosaccharides contain a carbonyl group (C=O) and at least two -OH groups, and have the empirical formula CH2O. Monosaccharides are organic molecules with three or more carbon atoms in a straight chain. Each carbon has a single –OH group, except one which has a C=O group. Aldose sugars have an aldehyde group (i.e. C=O is on carbon number 1). Ketose sugars have their C=O in the middle of the chain. Fructose: Glucose: Note that glucose and galactose are identical except that one chiral centre is different: Galactose: Having four chiral centres, glucose has seven isomers which differ in the orientation at one or two of the chiral centres. Galactose is just one these. Glucose has an enantiomer (L-glucose) which differs at all four chiral centres. You need to know the straight-chain formula of glucose. Monosaccharides in solution form a ring structure involving the C=O group: 96

α-glucose straight-chain version of glucose β-glucose You need to know these formulae of glucose. Galactose undergoes a similar reaction: The –OH group on carbon 1 of -glucose can link to the –OH group on carbon 6 of fructose. H2O is lost and a bridging –O- links the two molecules. This disaccharide is called sucrose. If the –OH group on carbon 1 of -galactose links to the –OH group on carbon 4 of glucose, a similar reaction occurs resulting in the disaccharide lactose. A chain of -glucose molecules linked carbon 1 to carbon 4 is called amylose. It is part of starch and glycogen. Amylose chains can be linked together by connecting the carbon-1 at one end to the carbon-6 of another chain. This branched polymer is called amylopectin and it is the second component of starch and glycogen. These are energy sources, energy reserves (eg glycogen) and precursors for other biologically important molecules. Glucose is broken down into water and carbon dioxide to yeld metabolic energy. Other carbohydrates are converted into glucose and metabolized in a similar manner. Starch and glycogen are used as stores of glucose in plants and animals respectively. Carbohydrates are used in other roles but are usually modified beforehand e.g. as glycoproteins, glycolipids, hormones.

Questions 03NS3 C2. (a) Draw the straight chain structure of glucose. [1] (b) The structure of α-glucose is shown below. Outline the structural difference between α-glucose and β-glucose. [1] (c) Glucose molecules can condense to form starch which can exist in two forms, amylose and amylopectin. Describe the structural differences between the two forms. [2] 05NS3 C1. (a) The equilibria which exist in an aqueous solution of glucose is shown in the structures below. (i) Identify the α and β forms of glucose [2] (ii) State, with a reason, whether or not the two ring forms of glucose are enantiomers. [1] (iii) In structure B identify, by stating the numbers, the carbon atoms which are not chiral. [1] (b) The structure of lactose, a disaccharide formed from glucose and galactose, is shown in the Data Booklet. Draw the ring structure of galactose and state whether it is an α or β isomer. [2] (c) State one major function of polysaccharides such as starch and glycogen. [1]

[edit] C.4 Lipids (2.5h) Lipids generally do not dissolve in water. There are 3 different kinds of lipids in the human body: 97

- Triglycerides - Phospholipids - Steroids

Tricglycerides(Oils) Fats and oils (collectively known as lipids) have a variety of chemical structures, but they are all insoluble in water. The most common lipid structure is the triacylglyceride (TAG) which is an ester. Three longchained acids form ester bonds to a single molecule of glycerol (propane-1,2,3-triol). Saturated fatty acids have no C=C double bonds. Unsaturated fatty acids have one or more C=C double bonds. Polyunsaturates have more than one C=C. Saturated fatty acids tend to have higher melting points. The intermolecular bonds must be greater. This is due to the more efficient van der Waals' forces that are possible when the flexible saturated chains lie alongside one another – the contact surface is maximised. In unsaturated fatty acids the double bonds reduce the flexibility of the molecules, and this prevents the molecules aligning so efficiently. Margarine manufacturers use unsaturated vegetable oils and hydrogenate them until they become saturated enough to have the right semi-solid consistency at room temperature. The number of C=C bonds can be determined from the number of moles of I2 which add to one mole of fat. Like Br2 in the test for alkenes, I2 will add to the C=C double bonds in unsaturated fatty acids. Unlike Br2 it only reacts with C=C bonds. The iodine number is the mass of iodine which reacts with 100 g of a lipid. For saturated fatty acids such as palmitic acid (C15H31COOH) the iodine number is zero. For unsaturated fatty acids like hexadecadienoic acid C15H27COOH the iodine number is calculated as follows: The number of double bonds is calculated by comparing the fatty acid to the appropriate saturated fatty acid. For every two hydrogens missing, there must be one double bond. In this case, compare C15H27COOH with C15H31COOH: There are four missing hydrogens and therefore two C=C bonds.

Example 98

Calculate the molar mass of the fatty acid and the iodine: C15H27COOH 252 g mol-1 I2 254 g mol-1 Every C=C will react with one iodine, so: one mole of C15H27COOH will react with two moles of I2 252 g of C15H27COOH will react with 508 g of I2 100 g of C15H27COOH will react with 202 g of I2

Hydrolysing TAGs into glycerol and fatty acids is achieved by refluxing with sodium hydroxide. The fatty acid salts which result acts as soaps. The ionic salt end is water-soluble but the long alkane chain is oily. The molecule can therefore bridge between water and oily impurities. C.4.5 List the major functions of fats in the body. These are energy sources, insulation and cell membranes. Fats are used as a source of energy. They are more slowly broken down than carbohydrates so they are not as useful in an emergency as glycogen. On the other hand, fats do not accumulate water so they are more efficient energy stores, pound for pound, than glycogen. Fat is a good thermal insulator (blubber).

Phospholipids Made of three structures: -Glycerol(backbone) -2 fatty acid chains -phosphate chain. It is amphipathic meaning the head, glycerol is polar but the chains are not. Phosopholipids form bilayers. The bilayers form cell membranes

Steroids 3 Cyclohexane rings and 1 cyclopentane ring bonded together Examples of steroids are: Vitamins, hormones and cholesterol.

Questions 03NS3 C3. Linoleic acid (C17H31COOH, Mr = 280) and stearic acid (C17H35COOH, Mr = 284) both contain eighteen carbon atoms and have similar molar masses. (a) Explain why the melting point of linoleic acid is considerably lower than the melting point of stearic acid. [3] (b) Determine the maximum mass of iodine (I2 Mr = 254) that can add to (i) 100 g of stearic acid: 99

[1] (ii) 100 g of linoleic acid: [2] (c) (i) Draw the simplified structural formula of a fat containing one stearic acid and two linoleic acid residues. [1] (ii) Give the formulas of the products formed when this fat is hydrolyzed by sodium hydroxide. [1] 05MS3 C1. (a) A brand of vegetable fat consists of 88 % unsaturated fats and 12 % saturated fats. State the major structural difference between unsaturated and saturated fats. [1] (b) Linoleic acid, CH3(CH2)4CH=CHCH2CH=CH(CH2)7COOH, and palmitic acid, CH3(CH2)14COOH, are components of vegetable fat. Explain why palmitic acid has the higher melting point. [3] (d) List two functions of fats in the human body. [2] 06MS3 C2. (a) The formula of oleic acid is CH3(CH2)7CH=CH(CH3)7COOH. Using R to represent CH3(CH2)7CH=CH(CH3)7, show the structure of the triglyceride formed from this acid. [1] (b) Explain why some triglycerides that are liquid at room temperature become solids when they are completely hydrogenated. [3] 05NS3 C2. (a) The general formula for saturated fatty acids is CnH2nO2. The molecular formula of linoleic acid is C18H32O2. (i) Determine the number of carbon to carbon double bonds in linoleic acid. [1] (ii) Iodine number is defined as the number of grams of iodine that adds to 100 g of a fat or an oil in an addition reaction. Determine the iodine number of linoleic acid. [2] (b) (i) State one structural similarity between fats and oils. [1] (ii) Explain, by referring to their structures, why fats are solid at room temperature, but oils are liquid. [3] (c) When a fat is reacted with aqueous sodium hydroxide, soap and one other product are formed. (i) State the condition required for this reaction. [1] (ii) Draw the structural formula of the other product. [1]

[edit] B.5

Micronutrients and Macronutrients

B.5.1 - Outline the difference between micronutrients and macronutrients 

Micronutrients are substances required in very small amounts (mg or ug) and that mainly function as a co-factor of enzymes (.005% body weight). Examples include proteins, fats, carbohydrates, and minerals (Na, Mg, K, Ca, P, S and Cl)

B.5.2 - Compare the structures of retinol (vitamin A), calciferol (vitamin D) and ascorbic acid (vitamin C) 

Vitamin C - Has hydroxyl groups (4), Ester within in the ring, polar because hydroxl groups make it easy to dissolve in water.



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Vitamin A - Only 1 hydroxyl group, long polyene chain. Has small polar parts, but overall non-polar (therefore doesn't dissolve in water, fat soluable) Vitamin D - Only one hydroxyl group. Some alkene groups, small parts are polar areas but overall no big difference. Overall non-polar due to the carbon chains and cyclocarbons.

B.5.3 - Deduce whether a vitamin is water or fat soluable from its structure Water soluble molecules have a high proportion of electronegative atoms in their structures. Fat soluble vitamins are relatively non-polar and consist mainly of hydrocarbon. 

Water-soluable are vitamins B and C Fat-soluable are vitamins A, D, E, K

B.5.4 - Discuss the causes and effects of nutrient deficiencies in different countires and suggest solutions. Micronutrient defiencies include: 

Iron - anemia Iodine - goitre Retinol (vitamin A) - Xerophythalmia, night blindness Niacin (Vitamin B3) - Pellagra Thiamin (Vitamin B1) - Beriberi Ascorbic Acid (Vitamin C) - Scurvy Calciferol (Vitamin D) - Rickets

Macronutrient definciencies 

Protein - Marasmus and kwashiorkor

Solutions include 

Providing food rations that are composed of fresh and vitamin -and mineral- rich foods Adding nutrients missing in commonly consumed foods Genetic modification of foods Providing nutrition supplements Providing selenium supplements to people eating foods grown in selenium-poor soil

Questions 04MS3 C1. (a) The structures of three important vitamins are shown in Table 22 of the Data Booklet. State the name of each one and deduce whether each is water-soluble or fat-soluble, explaining your choices by reference to their structures. [5]

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(b) Identify the metal ion needed for the maintenance of healthy bones and state the name of the vitamin needed for its uptake. [2] (c) State the name of the vitamin responsible for maintaining healthy eyesight and the name of the functional group which is most common in this vitamin. [2] (d) Identify one major function of vitamin C in the human body and state the name of the most common disease caused by deficiency of this vitamin. [2] (e) Fresh fruits and vegetables are good sources of vitamin C. Explain why some meals made from these foods may contain little vitamin C. [2]

05NS3 C3. (a) State the name of a disease which results from the deficiency of each of the following vitamins. [2] vitamin A vitamin C vitamin D (b) A person consumes an excess of both vitamin A and C. State, with a reason, which one is more likely to be stored in the body and which is more likely to be excreted. [2] 06MS3 C3. The structures of vitamins A and C are shown in Table 22 of the Data Booklet. State, with a reason, whether each is fat soluble or water soluble. [3]

[edit] C.6 Hormones (2.5h) Questions 04MS3 C2. The structures of two sex hormones, progesterone and testosterone, are shown in Table 22 of the Data Booklet. (a) State the names of two functional groups that are present in both hormones. [2] (b) Identify which of the two hormones is the female sex hormone and where in the human body it is produced. [2]

(c) Outline the mode of action of oral contraceptives. [3]

05MS3 C2. (a) By referring to Table 22 of the Data Booklet, identify one vitamin that is water soluble and one vitamin that is fat soluble. Explain the differences in solubility in terms of their structures and intermolecular forces. [4] (b) Vitamins C and D are vital in a balanced diet. State one major function of each of these vitamins and state a disease that results from the deficiency of each one. [4]

[edit] B.7 Enzymes Made of amino acid chains

[edit] B.8 Nucleic Acids 102

DNA RNA

[edit] B.9 Respiration

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IB Chemistry/Environmental Chemistry < IB Chemistry This page may need to be reviewed for quality. Jump to: navigation, search

Contents [hide] 

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1 E1 Air Pollution o 1.1 E.1.1 Describe the main sources of carbon monoxide (CO), oxides of nitrogen and sulfur, particulates and volatile organic compounds in the atmosphere. o 1.2 E.1.2 Evaluate the current methods for the reduction of air pollution. 2 E2 Acid Deposition o 2.1 E.2.1 State what is meant by the term acid deposition and outline its origins. o 2.2 E.2.2 Discuss the environmental effects of acid deposition and possible methods to counteract them. 3 E3 Greenhouse Effect o 3.1 E.3.1 Describe the greenhouse effect. o 3.2 E.3.2 List the main greenhouse gases and their sources, and discuss their relative effects. o 3.3 E.3.3 Discuss the influence of increasing amounts of greenhouse gases on the atmosphere. 4 E4 Ozone Depletion o 4.1 E.4.1 Describe the formation and depletion of ozone in the stratosphere by natural processes. o 4.2 A.4.2 List the ozone-depleting pollutants and their sources. o 4.3 E.4.3 Discuss the alternatives to CFCs in terms of properties. 5 E5 Dissolved oxygen in water o 5.1 E.5.1 Outline biochemical oxygen demand (BOD) as a measure of oxygendemanding wastes in water. o 5.2 E.5.2 Distinguish between aerobic and anaerobic decomposition of organic material in water. o 5.3 E.5.3 Describe the process of eutrophication and its effects. o 5.4 E.5.4 Describe the source and effects of thermal pollution in water. 6 E6 Water Treatment o 6.1 E.6.1 List the primary pollutants found in waste water and identify their sources. o 6.2 E.6.2 Outline primary, secondary, and tertiary stages of waste water treatment, and state the substance that is removed during each stage.  6.2.1 Primary Treatment: the removal of large solids  6.2.2 Secondary Treatment: the removal of organic materials using microbes 104

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6.2.3 Tertiary Treatment: the removal of remaining organics, nutrients and toxic heavy metal ions o 6.3 E.6.3 Evaluate the process to obtain fresh water from sea water using multistage distillation and reverse osmosis. 7 E7 Soil o 7.1 E.7.1 Discuss salinization, nutrient depletion and soil pollution as causes of soil degradation. o 7.2 E.7.2 Describe the relevance of the soil organic matter (SOM) in preventing soil degradation, and outline its physical and biological functions. o 7.3 E.7.3 List common organic soil pollutants and their sources 8 E8 Waste o 8.1 E.8.1 Outline and compare various methods for waste disposal. o 8.2 E.8.2 Describe the recycling of metal, glass, plastic and paper products, and outline its benefits. o 8.3 A.8.3 Describe the characteristics and sources of different types of radioactive waste. o 8.4 A.8.4 Compare the storage and disposal methods for different types of radioactive waste. 9 E9 Ozone Depletion (HL) o 9.1 E.9.1 Explain the dependence of O2 and O3 dissociation on the wavelength of light. o 9.2 E.9.2 Describe the mechanism in the catalysis of O3 depletions by CFCs and NOx o 9.3 E.9.3 Outline the reasons for greater ozone depletion in the polar regions. 10 E10 Smog (HL) o 10.1 E.10.1 State the source of primary pollutants and the conditions necessary for the formation of photochemical smog. o 10.2 E.10.2 Outline the formation of secondary pollutants in photochemical smog. 11 E11 Acid Deposition (HL) o 11.1 E.11.1 Describe the mechanism of acid deposition caused by the oxides of nitrogen and oxides of sulfur. o 11.2 E.11.2 Explain the role of ammonia in acid deposition. 12 E12 Water and Soil (HL) o 12.1 E.12.1 Solve problems relating to the removal of heavy-metal ions, the phosphates and nitrates from water by chemical precipitation. o 12.2 E.12.2 State what is meant by the term cation-exchange capacity (CEC) and outline its importance. o 12.3 E.12.3 Discuss the effects of soil pH on cation-exchange capacity and availability of nutrients. o 12.4 E.12.4 Describe the chemical functions of soil organic matter (SOM).

[edit] E1 Air Pollution 105

[edit] E.1.1 Describe the main sources of carbon monoxide (CO), oxides of nitrogen and sulfur, particulates and volatile organic compounds in the atmosphere. see chart below

[edit] E.1.2 Evaluate the current methods for the reduction of air pollution.

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[edit] E2 Acid Deposition [edit] E.2.1 State what is meant by the term acid deposition and outline its origins. Acid deposition is the process by which acidic particles, gases, and precipitation leave the atmosphere. Rain is naturally acidic with a pH of about 5.6 due to dissolved CO2, but acid rain has a pH below 5.6 and is caused by oxides of sulfur and nitrogen. These oxides react with rain water to form acids: CO2 + H2O → H2CO3 NO + H2O → HNO3 NO2 + H2O → HNO3 SO2 + H2O → H2SO3 SO3 + H2O → H2SO4

[edit] E.2.2 Discuss the environmental effects of acid deposition and possible methods to counteract them. Some effects of acid deposition include: 

Leeches important nutrients from soil such as Ca 2+, Mg2+, and K+ which can lead to reduction in chlorophyll and therefore the ability to photosynthesize.

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Can kill aquatic life in lakes and rivers, and nitrates can lead to eutrophication.

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Erosion of stone which contains calcium carbonate (such as marble)

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Irritation of the mucus membranes increases the risk of respiratory illness such as asthma, bronchitis and emphysema

Acid deposition can be counteracted by lower the amount of sulfur and nitrogen oxides with: 

Improved engine design

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Catalytic converters

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Removing sulfur before, during, and after use of sulfur-containing fuels

It can also include the reduction of the amount of fuel burned, alternative energy methods and the use of mass transportation. Furthermore, alkaline scrubbers, such as CaCO 3 and CaO, to remove the oxides.

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Adding CaO or CaOH to lakes may also neutralize acidity, increases amount of calcium ions, and precipitate Al from the solution.

[edit] E3 Greenhouse Effect [edit] E.3.1 Describe the greenhouse effect. Greenhouse gases allow the passage of incoming solar short-wave radiation but absorb the longer-wavelength radiation from the earth. Some of the absorbed re-radiation is re-radiated back to earth.

[edit] E.3.2 List the main greenhouse gases and their sources, and discuss their relative effects. Gas

CH4

H2O

CO2

N2O O3 CFCs

Heat trapping compared to CO2

Source    

Anaerobic decay Termites Rice fields Petroleum and natural gas production

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Evaporation Combustion of hydrocarbons

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Combustion of fossil fuels, biomass Decay of plants and animals Oxidation of soil Forest fires Internal combustion engines

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Bacterial action Fertilizers

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Secondary pollutant smog

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Refrigerants Propellants

Contribution to global warming

30x

18%

0.1x

>1%

1x

50%

150x

6%

2000x

12%

2500-10000x

14%

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[edit] E.3.3 Discuss the influence of increasing amounts of greenhouse gases on the atmosphere. Increasing greenhouse gases could increase the earth’s natural greenhouse effect and lead to global warming. The oceans may expand with an increase in temperature, and polar ice caps may melt. Also, changes in temperature and precipitation, thus leading to changes in crop production may result from a possible global warming.

[edit] E4 Ozone Depletion [edit] E.4.1 Describe the formation and depletion of ozone in the stratosphere by natural processes. The ozone layer occurs in the stratosphere between 12km and 50 km from the surface of the Earth. Stratospheric ozone is in dynamic equilibrium with oxygen and is continually being formed and decomposed.

Formation: O2 + UV → 2O◦ O2 + O◦ → O3

Depletion: O3 + UV → O2 + O◦ O3 + O◦ → 2O2

[edit] A.4.2 List the ozone-depleting pollutants and their sources. Chlorofluorocarbons were previously used as refrigerants, propellants, and cleaning solvents. Unfortunately, these molecules can destroy the ozone layer.

Initiation: CF2Cl2 + UV → Cl◦ + CF2Cl◦

Propagation: Cl◦ + O3 → Cl◦ + O2 ClO◦ + O◦ → O2 + Cl◦

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Termination: ClO◦ + ClO◦ → 2Cl◦ + O2 In this way, the CFC is acting as a catalyst—destroying the existing O3 and preventing the formation of O3 without being consumed. NOx can also react catalytically with O3. NO + O3 → NO2 + O2 NO2 + O◦ → NO + O2 Net effect: O3 + O◦ → 2O2 NO2 + UV → NO + O◦ O3 + O◦ → 2O2

[edit] E.4.3 Discuss the alternatives to CFCs in terms of properties. Some options include HCFCs (Hydrochlorofluorocarbons), HFCs (Hydrofluorocarbons), and other non-chlorine containing hydrocarbons. Examples include: Chlorotrifluoromethane, 1,1,1,2tetrafluoroethane, and 2-methylpropane. By replacing some of the chlorine atoms with fluorine, which requires more energy in breaking the bond, there is less radicalization taking place.

[edit] E5 Dissolved oxygen in water [edit] E.5.1 Outline biochemical oxygen demand (BOD) as a measure of oxygendemanding wastes in water. BOD is the measure of dissolved oxygen (in parts per million) required to decompose all organic waste and ammonia in water biologically over a 5 day period at 20⁰C. The wastes demand oxygen to be decomposed.

[edit] E.5.2 Distinguish between aerobic and anaerobic decomposition of organic material in water. If there’s sufficient oxygen present in the water, organic matter is broken down by microbes aerobically. This oxidizes the C, N, P, S, and H to produce CO2, NO−3, PO3−4, SO2−4, and H2O. If there’s an insufficient amount of oxygen present in the water, organic matter is decomposed by microbes that don’t require oxygen. They break down C, N, S, and P to form CH4, NH3, H2S, and PH3. Element Aerobic product Anaerobic product C CO2 CH4 N NO−3 NH3 P PO3−4 PH3 110

S

SO2−4

H2S

[edit] E.5.3 Describe the process of eutrophication and its effects. Nitrates from fertilizers and phosphates from detergents can accumulate in lakes and streams. These nutrients can increase the growth of plants and algae. This impacts the BOD because if plant growth increases too fast and the DO is not sufficient to decompose all organic material and waste by aerobic decomposition, anaerobic decomposition will occur. More species will die as a result of the anaerobic decay. The lake will become stagnant and devoid of life. Eutrophication:

[edit] E.5.4 Describe the source and effects of thermal pollution in water. If water is heated, the solubility of oxygen in the water decreases. At the same time, fish are cold-blooded, so as the temperature of the water increases, their metabolism increases. This forms a dilemma since the DO decreases as the BOD increases. This process helps to contribute to red tide.

[edit] E6 Water Treatment [edit] E.6.1 List the primary pollutants found in waste water and identify their sources. Waste water contains floating, suspended, and colloidal organic matter, dissolved ions with a wide range of microorganisms and bacteria as well as miscellaneous grit, trash, grease and other chemicals. Pesticides: DDt, herbicides, paraquat, fungicides Dioxins: formed when organochlorine compounds are not incinerated at high enough temperatures. Very toxic and can accumulate in the liver Polychlorobiphenyls (PCBs): used in transformers and capacitors. Persists in the environment and can accumulate in the liver, also carcinogenic 111

Nitrates: from fertilisers or acid rain. they are toxic at high levels, especially to babies because they have less stomach acid than adults, can cause blue baby syndrome Heavy metals: Cadmium (Cd) (from recharageable batteries), Mercury (Hg) (from batteries), Copper (Cu) (from household plumbing), Lead (Pb)

[edit] E.6.2 Outline primary, secondary, and tertiary stages of waste water treatment, and state the substance that is removed during each stage. [edit] Primary Treatment: the removal of large solids Primary treatment removes 60% of the solid material and a third of the BOD waste in the water. However, afterwards the water will still not be safe to drink. Primary treatment involves running water through the below mechanisms in order: 1. Bar screens: these remove large objects and debris from the surface of the water and remove floating solids. 2. Settling tanks: these are used to settle out sand, grit, and small objects from the water (as they sink to the bottom); these particles are then sent to landfills. 3. Sedimentation tanks: Alum (Ca(OH)2 and Al2(SO4)3) precipitates out and carry with them solid suspended particles (this process is called flocculation) [edit] Secondary Treatment: the removal of organic materials using microbes 

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Activated sludge process: o Air is bubbled into sewage which has been mixed with bacteria-laden sludge. o Aerobic bacteria oxidize organic material in the sewage. o Water-containing decomposed suspended particles are passed through the sedimentation tanks where the activated sludge is collected. o Some of the sludge is recycled, and some is sent to landfills. o This removes 90% of organic oxygen-demanding waste, 50% of nitrogen, and 30% of phosphates Effluent is then treated with chlorine or ozone to kill pathogenic bacteria before releasing the water to lakes or rivers Other methods include a carbon bed to remove the remaining organics, ion exchange which removes many soluble ions, reverse osmosis and electrodialysis.

[edit] Tertiary Treatment: the removal of remaining organics, nutrients and toxic heavy metal ions 

Heavy metal ions and phosphates are removed by precipitation, for example, nickel: 112

Ni2+(aq) + OH−(aq) → Ni(OH)2 (s) 

Aluminum sulfate and phosphates are removed by precipitation: Al3+(aq) + PO3−4 (aq) → AlPO4 (s) Al3+(aq) + SO2−4 (aq) → Al2(SO4)3 (s)

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Aluminum sulfate and calcium oxide can be used to remove phosphates: 3CaO(aq) + 2PO3−4 (aq) + 3H2O → Ca3(PO4)2 (s) + 6OH−(aq)

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Heavy metals will precipitate in the presence of hydroxide: Cr3+(aq) + 3OH−(aq) → Cr(OH)3 (s)

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Nitrates are more difficult to remove by precipitation because they’re quite soluble, however, there are some ways to remove them: o Anaerobic denitrifying bacteria can reduce nitrates into nitrogen 2NO2−3 (aq) → N2 (g) + 3O2 (g)

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Another method is to pass them into algae ponds where algae uses nitrate as a nutrient

[edit] E.6.3 Evaluate the process to obtain fresh water from sea water using multi-stage distillation and reverse osmosis. There are also a few other treatments, such as distillation. In distillation, sea water is pumped into a reservoir, at which point it is heated. The pure water which evaporates condenses on the cool water being pumped in, leaving a salty brine, which is then pumped out. Another method used is the reverse osmosis system. In this type of system, there is a semipermeable membrane which the water is pumped through, thereby being the opposite of a normal osmosis system (in which water would flow from low concentration to high concentration).

[edit] E7 Soil [edit] E.7.1 Discuss salinization, nutrient depletion and soil pollution as causes of soil degradation. Soil is a complex mixture of inorganic and organic materials, including living organisms. Soil degradation lowers crop production and is caused by a variety of human factors including; acidification, salinization, contamination, desertification, erosion.

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We are interested in the following factors: 

Salinization: the result of continual irrigation of soil; In poorly drained soil, after the water evaporates, salt is left behind, and plants die because they are unable to take water away from the salty soil.

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Nutrient Depletion: plants remove nutrients and minerals from soil as they grow. If not properly managed by crop rotation or fertilizing the soil, nutrients will become depleted.

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Soil Depletion: caused by improper use of pesticides and over-fertilizing; chemicals can disrupt the food web, reducing soil’s biodiversity, and ultimately ruining the soil.

[edit] E.7.2 Describe the relevance of the soil organic matter (SOM) in preventing soil degradation, and outline its physical and biological functions. SOM refers to the organic constituents in the soil. This includes plant and animal tissue, partial decomposition products and soil biomass. Chemicals found in SOM from decomposition of plants are high molecular mass organics such as Polysaccharides, proteins, sugars, and amino acids. The end product of decomposition is humus. Humus is the organic decomposition layer which plants live on. It has a mixture of simple and more complex organic chemicals from plants, animals, or microbial origin.

How SOM prevents soil degradation:    

helps soil to retain moisture, and dark color helps to retain heat and warm the soil during the spring. contains mineral nutrients that it exchanges with plants (at the roots). it improves the soil structure it reduces soil erosion.

Biological functions of SOM: 

humus provides a source of nutrients (such as N, P, and S) to the soil. Nitrogen provides proteins, Phosphoros provides enzymes, Sulfur provides amino acids.

Physical functions of SOM: 

SOM can retain several times its mass of water (like a sponge). Therefore more SOM means more water, making the soil more stable.

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Chemically, SOM acts like clay with cation exchange capacity (CEC): it contains active sites that enable it to bind to nutrient cations. Humus also has the ability to maintain a constant pH by acting as a buffer.

[edit] E.7.3 List common organic soil pollutants and their sources Here is a list of common soil pollutants and their major sources:     

Agrichemicals: from pesticides, herbicides and fungicides. Polyaromatic hydrocarbons: from incomplete combustion of coal, oil, gas, wood and garbage. Polychlorinated biphenyls (PCBs): from transformers and generators (they are used as a coolant). Organotin compounds: from bactericides and fungicides (used in paper, wood, textile and anti-fouling paint). Hydrocarbons and other VOCS: from transport, solvents and industrial processes.

[edit] E8 Waste [edit] E.8.1 Outline and compare various methods for waste disposal. Method of disposal | Advantages (+) | Disadvantages (-) Landfill | (+)Cheap, leaves large amount of land reused after fill | (-)Leaches into soil and ground water; needs time to settle, maintenance for methane Open Dumping | (+)Extremely cheap, convenient | (-)Unsightly; causes disease, odor, ground water pollution Ocean Dumping | (+)Cheap, convenient | (-)Toxic in oceans, dangerous to fish, pollutes the sea Incineration | (+)Provides source of energy, takes up little space, has stable residue | (-)Causes air pollution Recycling | (+)Produces new raw materials, creates a sustainable environment | (-)Expensive, still causes some air pollution

[edit] E.8.2 Describe the recycling of metal, glass, plastic and paper products, and outline its benefits. There are 3 main benefits to recycling that apply to metal, glass, plastic and paper. These are: 

Saving raw materials 115

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Saving energy (as energy is required to produce new materials) Saving space (in landfills)

In addition, glass and metals can be constantly recycled (over and over) without much degradation in the material.

The processes of recycling for each of the materials are as follows:    

Metals: sorted (by magnets or flotation) --> melted --> re-moulded --> re-used. Glass: sorted (colour) --> washed --> crushed --> re-moulded --> re-used. Plastics: sorted --> degraded to monomers (through pyrolysis, hdrogenation, gasification and thermal cracking) --> repolymerised --> re-used. Paper: mixed into water and chemicals (to form pulp) --> pulp is spun (removes staples/paper clips) --> washed to remove ink --> dried and bleached white --> re-used.

[edit] A.8.3 Describe the characteristics and sources of different types of radioactive waste. Low-level waste includes any gloves, paper towels or protective clothing that has been used in areas where radioactive materials have been handled. The level of activity is low and the half lives are short. This waste generally comes from hospitals due to cancer treatment, and includes any items that have come in contact with the radioactive material. High-level waste is generated by nuclear power plants and the military. It demonstrates a high level of activity and generally isotopes have long half-lives. High-level waste also comes from fuel rods or the reprocessing of spent fuel (power companies, military)

[edit] A.8.4 Compare the storage and disposal methods for different types of radioactive waste. The nuclear decay process produces heat and energy. Low-level waste is stored in cooling ponds until the activity has fallen to safe levels (generally a few years). The water is then passed through ion exchange resins which remove isotopes responsible for activity. The water is then diluted and released into the sea. High-level waste takes thousands of years to lose activity. Much of spent radioactive fuel is recovered for reuse. If not, the waste, generally a liquid mixture of radioactive waste, is converted into a solid glass component through a vitrification process: The waste is dried in a furnace and fed into a melting pot together with glass-making material (sand). The molten material is then poured into a stainless steel container where it cools and solidifies. These containers will remain radioactive for thousands of years. The containers are currently stored in concrete vaults, but it is hoped that they will later be transferred to salt chambers one day to be stored for thousands of years until the activity falls to safe levels. 116

[edit] E9 Ozone Depletion (HL) [edit] E.9.1 Explain the dependence of O2 and O3 dissociation on the wavelength of light. O2 has a bond order of 2 (a double bond), therefore it is more difficult to break. This means that it will require more energy to do so, so a shorter wavelength (242 nm). O2 + UV (242 nm) → 2O◦ O3, due to its resonant structure, has a bond order of 1.5, meaning it is less difficult to break than the double bond in O2. This means that it will require less energy to do so, meaning a longer wavelength (330 nm). O3 + UV (330 nm) → O2 + O◦

[edit] E.9.2 Describe the mechanism in the catalysis of O3 depletions by CFCs and NOx See A.4.2

[edit] E.9.3 Outline the reasons for greater ozone depletion in the polar regions. A hole in the ozone layer is found above Antarctica. Depletion is seasonal with the largest holes occurring during the early spring (October/November). This decrease is due chemicals produced by man. In the winter (Jun-Sept), NO2 and CH4 are trapped with ClO and Cl2 on the surface of ice. A catalytic reaction on the surface of the ice converts the ClO into Cl2 and HClO forming a “chlorine reservoir.” This temporarily lessens the amount of chlorine released into the atmosphere. When the ice melts in the spring, the chlorine contained in the ice crystals surges into the atmosphere, thus causing the hole in the ozone layer to temporarily expand.

[edit] E10 Smog (HL) [edit] E.10.1 State the source of primary pollutants and the conditions necessary for the formation of photochemical smog. Smog is a poisonous combination of smoke, fog, air and other chemicals. Photochemical smog occurs in cities where exhaust from internal combustion engines concentrates in the atmosphere. Oxides of nitrogen and hydrocarbon give the air a characteristic yellow/brown color. In the 117

sunlight, these chemicals are converted into secondary pollutants. Smog tends to form in large cities and is favored by a lack of wind. It also occurs more often in bowl-shaped cities because the higher ground surrounding these cities prevents the movement of air. Smogs typically occur where there is a temperature inversion. Normally, the temperature decreases with altitude. Warm air typically rises, takes the pollutants with it, and is then replaced by cleaner cooler air. However, typically in areas which are notorious for smog, the atmospheric conditions cause a layer of still warm air to layer a blanket of cooler air. The trapped pollutants cannot rise, and if the condition persists, the amount of pollutants in the warm air near the ground can rise to dangerous levels.

[edit] E.10.2 Outline the formation of secondary pollutants in photochemical smog. Photochemical smog is a chemical soup containing hundreds of different chemicals formed in the atmosphere as a result of free radical reactions caused by UV light. In the early morning hours, a buildup of hydrocarbons (VOCs) and NOx from car exhaust occurs. As the sun comes out, the NO2 absorbs sunlight and forms free radicals. N2 + O2 → NO2 (Primary Pollutant) NO2 + UV → NO + O◦ These radicals react with O2 to form ozone and water or to form hydroxyl radicals. O◦ + O2 → O3 (Secondary Pollutant) O◦ + H2O → 2OH◦ Secondary photochemical oxidants react with a variety of molecules and hydrocarbons to form peroxides, aldehydes and ketones. OH◦ + NO2 → HNO3 OH◦ + RH → R◦ + H2O R◦ + O2 → ROO◦ (Peroxide Radical) Chain termination can occur when peroxide radicals react with NO2 to form Peroxyacylnitrates (PANs) which are irritants to the eyes and skin. ROO◦ + NO2 → ROONO2 (PAN)

[edit] E11 Acid Deposition (HL) 118

[edit] E.11.1 Describe the mechanism of acid deposition caused by the oxides of nitrogen and oxides of sulfur. NOx and SOx are converted into acid via free radical reactions: H2O + O3 → 2HO◦ + O2 The hydroxyl radicals then react directly with SOx and NOx in the presence of water to form acids. HO◦ + NO2 → HNO3 HO◦ + NO → HNO2 or, HO◦ + SO2 → HOSO2◦ HOSO2◦ + O2 → HO2◦ + SO3 SO3 + H2O → H2SO4

[edit] E.11.2 Explain the role of ammonia in acid deposition. Ammonia gas can to a small extent neutralize the effect of acid rain in the atmosphere by the formation of (NH4)2SO4 and NH4NO3: 2NH3 + H2SO4 → (NH4)2SO4 NH3 + HNO3 → NH4NO3 NH+4 is a strong conjugate acid, so when ammonium salts sink into the ground, the NH+4 enters the soil where acidification and nitration can occur.

[edit] E12 Water and Soil (HL) [edit] E.12.1 Solve problems relating to the removal of heavy-metal ions, the phosphates and nitrates from water by chemical precipitation. Some of the salt in the soil dissolves, however most remains in solid form, thus creating equilibrium. Remember these 5 steps: 1. 2. 3. 4.

Write a balanced equation Find the equilibrium equation (Ksp = [ion conc.][ion conc.]) Identify the direction of the change in equilibrium ICE it! Initial, Change, Equilibrium

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5. Plug the E line into the equilibrium equation and solve for x, then follow through to find what the question is asking for (i.e. plug back into 2x if concentration, or find Ksp if appropriate.)

[edit] E.12.2 State what is meant by the term cation-exchange capacity (CEC) and outline its importance. Both SOM and clay have negatively charged particles which will bond to cations such as Ca+2, Mg+2, Na+, K+, Al+3. The amount of positively charged cations that soil can hold is called the cation-exchange capacity(CEC). A larger CEC indicates a larger capacity to hold cations. These cations are exchanged with cations such as hydrogen on the root hairs of a plant to provide it with nutrients.

[edit] E.12.3 Discuss the effects of soil pH on cation-exchange capacity and availability of nutrients. If soil is more acidic, there is a higher percentage of acidic cations found in the soil. Soil pH is important because acid cations such as aluminum ions are harmful to the plants. Although soil has some buffering capacity, it is sometimes necessary to add lime to the soil to raise the pH and increase the concentration of basic cations held by the clay and the SOM. When soil is analyzed, the total concentration of basic cations is compared to the total concentration of acidic cations. Cations such as Al+3 are harmful to plants. Soil pH is important because above pH=5, Al+3 will precipitate out of solution. If there is acid in the rain which lowers the pH of the soil, the Al+3 will no longer precipitate out of the solution. This cation is toxic to plants, so in essence, acid rain would be killing the plants.

[edit] E.12.4 Describe the chemical functions of soil organic matter (SOM). In addition to the nutrient cations required by the plants and organic matter, SOM can also bind to organic and inorganic compounds in the soil which helps to reduce the negative environmental effects of contaminants such as pesticides, heavy metal ions and other pollutants. SOM contributes to CEC, enhances the ability of soil to buffer changes in pH, and forms stable complexes with cations. SOM also reduces the effect of pesticide, heavy metals, and other pollutants.

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