S-Cool A Level Chemistry Practice Questions and Answers

A-Level Chemistry Question and Answers 2013

Table of Contents Atomic Structure (Answers) ................................................................................................................ 7 Atoms, Molecules and Stoichiometry (Questions) ........................................................................... 9 Atoms, Molecules and Stoichiometry (Answers) ............................................................................ 10 Chemical Equilibria (Questions) ........................................................................................................ 12 Chemical Equilibria (Answers) .......................................................................................................... 13 Electrochemistry (Questions) * ........................................................................................................ 14 Electrochemistry (Answers)............................................................................................................... 15 Group II and Group IV (Questions) ................................................................................................. 16 Group VII (Questions)........................................................................................................................ 18 Group VII (Answers) .......................................................................................................................... 19 Reaction Kinetics (Questions) ........................................................................................................... 24 States of Matter (Questions) ............................................................................................................. 26 States of Matter (Answers) ............................................................................................................... 27

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Aliphatic Compounds Aromatic and Plastics Chemical Energetics General Principles Ionic Equilibria Periodicity Transition Metals

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Atomic Structure (Questions) * 1. Chemists use a model of an atom that consists of three types of sub-atomic particle. These are protons, neutrons and electrons. a) Complete the table below, which shows the properties of these particles.

Particle Relative Charge Relative Mass

Electron

Proton

1

Neutron 0

(4 marks) b) Types of atoms and ions vary from each other by the number of sub-atomic particle found in them. What differences are there between 14

C?

(1 mark) c) Define the term 'isotope'. (1 mark) d) Write down the electronic configuration of Cr3+. (1 mark) (Marks available: 7)

2. a) The first ionisation energy for lithium is 519 kJmol-1. Write an equation to show what is meant by this statement. (1 mark)

5

12

C and

b) The second ionisation energy for lithium is 7300 kJmol-1. Explain the difference in value for the first and second ionisation energy for lithium. (1 mark) (Marks available: 2)

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Atomic Structure (Answers) Answer outline and marking scheme for question: 1 a)

Particle

Relative Charge

Relative Mass

Electron

-1

1/1820 or negligible

Proton

+1

1

Neutron

0

1

(4 marks) b) 12C has 6 protons, 6 neutrons and 6 electrons.

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C

has 6 protons, 6 electrons and 8 neutrons. The difference between this pair is that

14

C has two more neutrons than

12

C.

(1 mark) c) An isotope of an element has he same number of protons but a different number of neutrons.

(1 mark) d) 1s2 2s2 2p6 3s2 3p6 3d3 4s0

(1 mark) (Marks available: 7)

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Answer outline and marking scheme for question: 2 a) Li(g) - e- = Li+(g)

The electronic configuration of lithium is 1s2 2s1. It is relatively easy to remove the electron from the 2s orbital as it is shielded from the nuclear charge by the 1s electrons.

(1 mark) b) To remove the second electron for the second ionisation energy requires considerably more energy as this electron comes from the 1s shell which is closer to the nucleus and not subject to any shielding.

(1 mark) (Marks available: 2)

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Atoms, Molecules and Stoichiometry (Questions) 1.

Calculate the mean relative atomic mass of nickel from the mass spectrum.

(Marks available: 2)

2. In 1774 Joseph Priestly conducted one of his most famous experiments which lead to a method for the preparation of oxygen. The experiment involved heating a sample of mercury II oxide with a large lens. The equation for this reaction is shown below: 2HgO(s) = 2Hg(l) + O2(g) a) What volume of O2(g) would be obtained if 1.08g of mercury II oxide were completely decomposed? (Given that 1 mole of a gas occupies 24 dm3 under the experimental conditions) (Marks available: 3)

3. An organic compound is found to have the following composition by mass: C 72.7%; H 6.1%; N 21.2%. a) Calculate the empirical formula of the compound. (2 marks) b) The molar mass of the compound is 198 gmol-1. Calculate the molecular formula of the compound. (3 marks) (Marks available: 5)

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Atoms, Molecules and Stoichiometry (Answers) Answer outline and marking scheme for question: 1 Intensities of peaks: 58

Ni = 10.2 = 69.4% of total (abundance)

59

Ni = 3.9 = 26.5% of total (abundance)

62

Ni = 0.6 = 4.1% of total (abundance)

(Sum of intensities = 14.7) Mean relative atomic mass = (58 x 69.4) + (59 x 26.5) + (62 x 4.1) = 58.4 100 (Marks available: 2)

Answer outline and marking scheme for question: 2 Molar mass of HgO = 201 + 16 = 217 gmol-1

1.08g of HgO will contain 1.08 / 217 mols = 0.005mol From the equation, 1 mole of O2(g) is produced from 2 moles of HgO

This means that 0.005 mol of HgO will produce 0.005 / 2 mol of O2 = 0.0025 mol

0.0025 mol of O2 will occupy 0.0025 x 24dm3 = 0.06dm3 (Marks available: 3)

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Answer outline and marking scheme for question: 3 Ratio of elements in compound C : H : N

Ratio C : N : H = 72.7 / 12 : 6.1 / 1 : 21.2 / 14 = 6.1 : 6.1 : 1.5

Whole number ratio = 4 : 4 : 1

Empirical formula of compound is C4H4N (Marks available: 5)

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Chemical Equilibria (Questions) 1. Esters are a useful group of compounds due to their distinctive smells. One example of an ester is ethyl ethanoate, its formation is shown below. CH3COOH(aq) + C2H5OH(aq)λCH3COO C2H5(aq) + H2O(l) a) Systems like this are described as being a 'dynamic equilibrium'. Explain the term 'dynamic equilibrium' b) Write down the expression for the equilibrium constant, Kc, for this reaction. c) Calculate the value of Kc for this reaction given the equilibrium concentrations below. [CH3COOH] = 0.08 moldm-3 [C2H5OH] = 0.08 moldm-3 [CH3COO C2H5] = 0.25 moldm-3 [H2O] = 0.1 moldm-3 d) Concentrated sulphuric acid is added to the reaction mixture as it removes water molecules. What effect would this have on the equilibrium position of this system? (Marks available: 7)

2. The Haber-Bosch process is used for the large-scale production of ammonia from nitrogen and hydrogen gas. The reaction is shown below: N2(g) + 3H2(g)λ2NH3(g) a) Write an expression for Kp for this reaction. b) What are the units of Kp for this reaction? (Assume pressure is measured in kPa) c) When the temperature is raised for this process the proportion of NH3(g) in the mixture decreases. Explain this observation. d) What effect will an increase in the total pressure have on the equilibrium position? (Marks available: 5)

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Chemical Equilibria (Answers) Answer outline and marking scheme for question: 1 a) - concentrations of the species are constant - forward and backward reactions are continually taking place but at the same rate. b) Kc =[CH3COO C2H5] [H2O] [CH3COOH] [C2H5OH] c) Kc = [CH3COO C2H5] [H2O] = (0.25) (0.1) = 3.91 (no units)

[CH3COOH] [C2H5OH]

(0.08)(0.08)

(1 mark numerical answer, 1 mark stating no units) d) Concentrated sulphuric acid is added to the reaction mixture as it removes water molecules. What effect would this have on the equilibrium position of this system? (Marks available: 7)

Answer outline and marking scheme for question: 2 a) Kp = (p NH3)2 (1) (p N2)(pH2)2 b) Units = kPa2 = 1 / kPa2 = kPa-2 kPa x kPa3

c) Increase in temperature leading to decrease in NH3 means reaction towards right is exothermic. d) An increase in pressure will lead to an increase in the proportion of NH3(g) found in the mixture. (Marks available: 5)

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Electrochemistry (Questions) * 1. Dichromate (VI) ions are powerful oxidising agents and are reduced to chromium III ions. This colour change was once used in 'breath test' apparatus to determine if a driver had consumed excessive alcohol. Explain the term 'redox reaction' (Marks available: 1)

2. Aluminium metal is extracted from molten bauxite (Al2O3.2H2O) using electrolysis. Cryolite (AlF3) added to the ore in order to lower the melting point required and thus the energy required by the process. a) Write an half equation to show how aluminium metal is produced from the ore. (1 mark) b) What mass of aluminium metal would be produced if a current of 30,000A is applied to a cell for 1 hour. (4 marks) c) In the molten mixture there is a mixture of anions which mostly consists of O2- and F-. Write an equation to show which of these anions will be oxidised in the cell? (2 marks) (Marks available: 7)

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Electrochemistry (Answers) Answer outline and marking scheme for question: 1 A redox reaction is one where electrons are transferred from one species to another. (Marks available: 1)

Answer outline and marking scheme for question: 2 a) Al3+(l) + 3e- = Al(l)

(1 mark) b) Charge passed; E = It E = 30,000 x 3600 = 1.08 x 108C (1)

1 mole of electron has a charge of 96,500C

Mols electrons = (1.08 x 108) / 96,500 = 1119 mols (1)

To reduce 1 mol of Al3+, 3 mols of electrons are required

In the cell 1119 / 3 = 373 mols of Al produced (1)

Ar Al = 27.0 Mass Al produced = 373 x 27.0 = 10071g = 10.071kg of Al (1)

(4 marks) c) 2O2-(l) - 4e- = O2(g) (1 for correct anion, 1 for correct equation)

(2 marks) (Marks available: 7)

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Group II and Group IV (Questions) Answer outline and marking scheme for question: 1 a) Ba(s) + 2H2O(aq) = Ba(OH)2(aq) + H2(g)

(1 mark) b) 2Ba(NO3)2(s) = 2BaO(s) + 4NO2(g) + O2(g)

(1 mark) c) There are two factors that affect the solubility in water; lattice energy and enthalpy of hydration. As both salts are sulphates, there is little difference in lattice enthalpy (1) The enthalpy of hydration is much higher for barium as it has a higher ionic radius (lower charge density) which makes it difficult for water to hydrate. (1) (less of an attraction between cation and polar water molecules)

(2 marks) (Marks available: 4)

Answer outline and marking scheme for question: 2 a) Covalent

(1 mark) b) SiCl4(l) + 2H2O(l) = SiO2(aq) + 4HCl(g)

(1 mark) c) The silicon has 3d electron orbitals available (1) Which can accept a lone pair of electrons from the oxygen atom of water. (1) Carbon does not have any d electron orbitals available.

(2 marks)

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d) The 6s electrons in lead are drawn in to the nucleus of the atom due to poor shielding by d and f orbital electrons. This means that they behave like an 'inert pair' of inner electrons rather than an 'active pair' of outer electrons. (1) Silicon does not have any d or f electron orbitals filled therefore the 3s electron orbitals become hybridised with the 3p electrons giving silicon a valency of 4. (1)

(2 marks) (Marks available: 6)

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Group VII (Questions) 1. Astatine (As) is a member of Group VII of the periodic table. (You may consult a data book if necessary for this question) a) What is the colour and state of Astatine at room temperature? (2 marks) b) What is the oxidation number of As in NaAsO3? (1 mark) c) Write an equation to show the reaction between astatine and cold aqueous sodium hydroxide. (1 mark) d) Predict the bond energy of H-As bond. (1 mark) e) Chlorine gas is passed through an aqueous solution of potassium astatide, KAt. Write an equation to show what reaction will take place. (1 mark) f) Under certain conditions it is possible for the following reaction to take place: 3AsO- = 2As- + AsO3What is the name given to this type of reaction? (1 mark) g) Give a method for testing for the presence of astatide ions in aqueous solution. (2 marks) (Marks available: 9 marks)

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Group VII (Answers) Answer outline and marking scheme for question: 1 a) Astatine will be a black(1) solid(1)

(2 marks) b) Oxidation number = +5

(1 mark) c) Predict the bond energy of H-As bond.

(1 mark) d) Bond energies / kJmol-1 H-F = 562 H-Cl = 431 H-Br = 366 H-I = 299 H-As = ? From pattern between H-Cl and H-I.: Bond energy of H-As = 232 kJmol-1 (1)

(1 mark) e) Cl2(g) + 2As-(aq) = As2(s) + 2Cl-(aq)

(1 mark) f) Disproportionation

(1 mark) g) Add aqueous silver nitrate (1) Presence of astatide ions would be indicated by a dark yellow / orange precipitate of AgAs (1)

(2 marks) (Marks available: 9)

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Ionic Equilibria (Questions)

1. Benzoic acid, C6H5COOH is a weak acid. When dissolved in water the following reaction takes place: C6H5COOH(aq) + H2O(aq)λC6H5COO-(aq) + H3O+(aq) a) Explain what is meant by the term 'weak acid'. (1 mark) b) A solution is made of benzoic acid at 25°C. The acid dissociation constant for benzoic acid is 6.3 x 10-5 moldm-3. Write down an expression for Ka. (1 mark) c) Calculate the pH of a 0.005M solution of benzoic acid. (1 mark) (Marks available: 3)

2. Calculate the pH of the following solutions. a) 0.1M HCl(aq). (2 marks) b) 0.1M Be(OH)2(aq) (2 marks) (Marks available: 4)

3. Precipitation reactions are an important method of identifying the inorganic ions in a solution. For example silver nitrate is used to identify halides in solution. a) Write down the expression for the solubility product, Ksp, for silver chloride, AgCl. (1 mark)

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b) Ksp for AgCl is 1.8 x 10-10 mol2dm-6. Calculate the concentration of Ag+(aq) in a saturated solution of AgCl. (1 mark) c) What effect will the addition of HCl have on the solubility of AgCl? (2 marks)

(Marks available: 4)

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Ionic Equilibria (Answers) Answer outline and marking scheme for question: 1 a) A weak acid is one that does not completely dissociate when dissolved in aqueous solution.

(1 mark) b) Ka = [C6H5COO-][ H3O+] [C6H5COOH]

(1 mark) c) From equation we know that [C6H5COO-] = [ H3O+]. We can replace these concentrations in the expression of Ka with x2. Take [C6H5COOH] = 0.005M (as degree of dissociation is small there will be a negligible effect on this concentration). Ka = x2 / 0.005 = 6.3 x 10-5 (1) x2 = 0.005 x (6.3 x 10-5) = 3.2 x 10-7 x = 5.7 x10-4 (1) [H+] = 5.7 x 10-4 moldm-3 pH = -lg[H+] pH of solution = -lg(5.7 x 10-4) = 3.2

(1 mark) (Marks available: 3)

Answer outline and marking scheme for question: 2 a) Acid is fully dissociated - [H+] = 0.1 moldm-3 pH = -lg[H+] pH = -lg[0.1] = 1

(2 marks)

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b) Base is fully dissociated - [OH-] = 2 x 0.1 = 0.2 moldm-3+ pOH = -lg[OH-] = 0.7 pH + pOH = 14 pH = 14 -0.7 = 13.3

(2 marks) (Marks available: 4)

Answer outline and marking scheme for question: 3 a) Ksp = [Ag+(aq) ][Cl-(aq)]

(1 mark) b) From equation [Ag+(aq) ] = [Cl-(aq)] Ksp = [Ag+(aq) ]2 = 1.8 x 10-10 [Ag+(aq) ] = 1.3 x 10-5 moldm-3

(1 mark) c) Addition of HCl leads to and increase in [Cl-(aq)] This means that [Ag+(aq) ][Cl-(aq)] is greater than Ksp And AgCl will precipitate from the solution i.e. solubility has decreased.

(2 marks) (Marks available: 4)

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Reaction Kinetics (Questions) 1. Nitrogen (II) oxide reacts with oxygen in the following way: 2NO(g) + O2(g) = 2NO2(g) Experiments were carried out to determine the orders of reaction with respect to NO and O2. The results of these experiments are shown below.

Experiment [NO(g)] / moldm-3 [O2(g)] / moldm-3 Rate / moldm-3 s-1

1

1.5 x 10-5

0.5 x 10-5

2.1 x10-7

2

4.5 x 10-5

0.5 x 10-5

1.9 x10-6

3

1.5 x 10-5

2.0 x 10-5

8.4 x10-7

a) Using the data, deduce the orders of reaction with respect to NO and O 2. b) Write down the rate equation for this reaction and use it to calculate k. c) Explain the meaning of the term 'rate determining step'?

2. Carbon dating using

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C is a highly important tool in archaeology. Using this technique it is possible

to determine the length of time since an organism has died. The half-life of

14

C is 5600 years.

a) Define the term 'half-life'. b) A sample of wood taken from an archaeological dig is analysed and found to have a radioactive count of 4 counts per minute per gram of carbon. A new sample of the same type of wood has a count of 16 c.p.m.g-1. What is the approximate age of the archaeological sample? c) Which order of kinetics does radioactive decay follow? (Marks available: 3)

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Reaction Kinetics (Answers) Answer outline and marking scheme for question: 1 a) For NO - examine experiments 1 and 2. A 3-fold increase in the concentration of NO leads to a 9-fold increase in rate. (1) This means that the reaction must be second order with respect to NO. (1) For O2 - examine experiments 1 and 3. A 4-fold increase in O2 concentration leads to a 4-fold increase in rate. (1) This means that the reaction must be first order with respect to O2. (1) b) Rate = k[NO]2[O2] (1) 2.1 x10-7 = k (1.5 x10-5)2 (0.5 x10-5) k = (2.1 x10-7) / (1.5 x10-5)2 (0.5 x10-5) k = 1.9 x 108 mol-2dm6s-2 c) The rate-determining step is the slowest section of the mechanism of the reaction when compared with the other steps.

Answer outline and marking scheme for question: 2 a) The half-life of a substance is the time taken for its concentration to decrease by 50%. b) After 5600 years the count form the piece of new wood would drop to 8 c.p.m, after a further 5600 years it will drop to 4 c.p.m. The old wood is therefore approximately two half-lives old = 11200 years. c) First order (Marks available: 3)

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States of Matter (Questions) 1. Hydrogen generated from the reaction between magnesium and nitric acid was collected. At a temperature of 30°C and pressure of 1.2 x 105 Nm-2, 340cm3 of gas was collected. What would the volume of this gas be at s.t.p.? (Marks available: 2)

2. The ideal gas equation can be written as 'PV = nRT' a) Use this equation to calculate the volume of 0.5 moles of an ideal gas at 350K and 1x 105Nm-2 pressure. (2 marks) b) A compound was vaporised and it was found that 0.15g of the gas occupied 120cm3 at 323K and 100 kPa. Calculate the molecular mass of the compound. (2 marks) (Marks available: 4)

3. Explain in terms of structure and bonding: a) Why graphite conducts electricity but diamond does not. (3 marks)

b) Why water is a liquid at room temperature but hydrogen sulphide is a gas. (2 marks) (Marks available: 5)

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States of Matter (Answers) Answer outline and marking scheme for question: 1 Reaction conditions: P1 = 1.2 x 105 Nm-2 ; T1 = 303K ; V1 = 340cm3

S.t.p.

P2 = 1.01 x105 Nm-2 ; T2 = 273K (1) P1V1 = P2V2 (1.2 x 105)(340) = (1.01 x105)V2 T1 T2 303 273 V2 = 364.0 cm3 (Marks available: 2)

Answer outline and marking scheme for question: 2 a) Rearranging the equation: V = nRT/P V = 0.5 (8.314)(350) / (1 x 105)

V = 0.015 m3 (numerical = 1, units = 1)

(2 marks) b) Using PV = nRT and n = m / Mr

n = PV / RT = (1 x 105)( 0.00012) / (8.314)(323)

n = 0.004

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Mr = 0.15 / 0.004 = 37.5

(2 marks) (Marks available: 4)

Answer outline and marking scheme for question: 3 a) The carbon atoms in graphite are covalently bonded to three other carbon atoms and have delocalised electrons in between the sheets of atoms.

(1 mark) These delocalised electrons are able to move when a p.d. is applied

(1 mark) The carbons in diamond are covalently bonded to four other carbon atoms so there are no electrons able to move freely.

(1 mark) (max 3 marks) b) Water (H2O) has a higher boiling point due to there being a lot of hydrogen bonding between the molecules. Hydrogen sulphide (H2S) has a much lower boiling point because there is a much lower degree of hydrogen bonding.

(2 marks) (Marks available: 5)

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