Chemistry Form 6 Sem 2 01
Short Description
Term 2 Chem STPM...
Description
Pre – U Chemistry Semester 2
CHAPTER 1: THERMOCHEMISTRY
1.1 Introduction Energy is one of the most fundamental parts of our universe. We use energy to do work. Energy lights our cities. Energy powers our vehicles, trains, planes and rockets. Energy warms our homes, cooks our food, plays our music, gives us pictures on television. Energy powers machinery in factories and tractors on a farm. According to the conservation of energy law, energy can be neither created nor destroyed; it can only be converted from one form into another From the angle of chemistry, when a chemical reaction occur, energy changes occur generally in 2 ways, where it can be explained in terms of kinetic energy and energetic energy. In this chapter, we focus more on the study of energy changes, in the form of heat, which take place during a chemical reaction occur, which is well known as thermochemistry. In order to understand thermochemistry, we must first understand what is the difference between system and surrounding. System is the specific part of substances that involved in chemical and physical change, while surrounding is defined as the rest of the universe outside the system.
There are generally 3 types of systems. Open system
An open system can exchange mass and energy, usually in the form of heat with its surroundings
Closed system
closed system, which allows the transfer of energy (heat) but not mass.
Isolated system
isolated system, which does not allow the transfer of either mass or energy.
We shall focus more on a closed system throughout our lesson, with the assumption that energy lost by system in a chemical reaction is the same with the energy gained by surrounding. In thermochemistry energy that were gained / lost by system were measured by heat energy. In the laboratory, heat changes in physical and chemical processes are measured with a calorimeter, a closed container designed specifically for this purpose. Our discussion of calorimetry, the measurement of heat changes, will depend on an understanding of specific heat and heat capacity, The specific heat capacity (c) of a substance is the amount of heat required to raise the temperature of one gram of the substance by one degree Celsius. It has the units J g-1°C-1. The heat capacity (C) of a substance is the amount of heat required to raise the temperature of a given quantity of the substance by one degree Celsius. Its units are J °C-1. Specific heat is an intensive property whereas heat capacity is an extensive property. The relationship between the heat capacity and specific heat capacity of a substance is C = c x m (mass)
1.2 Enthalpy and Enthalpy Change Measurement of energy transferred during chemical reaction is made under control conditions. However, in a closed system, we assume that there’s no changes in the volume of a system, hence no work is done toward the heat change occur within the system. By that, we shall deduce the energy transferred in a system is corresponding to the heat transfer towards the surrounding. Heat transfer in this case is described as enthalpy, H. In a chemical reaction, where reactants products The difference of energy changes occur on a chemical reaction is known as enthalpy change, ∆H, as the difference between the enthalpies of the products and the enthalpies of the reactants ∆H = [Σ Σ ∆Hproduct – Σ ∆Hreactant]. Such enthalpy is also known as enthalpy change of reaction Since the enthalpy changes is a quantitative value use to measure the difference by the heat given off before and after a reaction, so it may be a positive value or negative value
Enthalpy – heat content of the system Enthalpy changes ; ∆H ~ heat changes occur during a chemical reaction. ∆H = [Σ Σ ∆Hproduct – Σ ∆Hreactant] Unit = kJ mol-1.
Σ ∆Hproduct > Σ ∆Hreactant
Σ ∆Hproduct < Σ ∆Hreactant
∆H = positive (+ve)
∆H = negative (–ve)
Endothermic
exothermic
Process Definition
ΔH
Energy profile
Endothermic
Process of heat absorbed by system Positive
Exothermic
Process of heat released by system Negative
Process
θ
Endothermic
Exothermic
Temperature decrease
Temperature increase
3 STEPS ON CALCULATING ENTHALPY CHANGE 1
2
3
q=mcθ
mass MV mol = or MR 1000
∆H =
q mol
Question 1 : Solution Equation : Zn (s) + H2SO4 (aq) ZnSO4 (aq) + H2 (g) Step 1 : q = m c θ @ q = (25.0) (4.18) (31.5 – 27.0) q = 470.25 J
Step 2 : determine limitant mol of Zn = mass / mol mol of H2SO4 = MV /1000 = 6.00 / 65.3 = (0.100) (25.0) / 1000 = 0.0919 mol = 0.00250 mol (lim) Step 3 : ∆H = q / mol @ ∆H = 470.25 / 0.00250
∆H = 188 100 J / mol @ – 188 kJ / mol
Question 2 : Solution Equation : Na2SO4 + Ba(NO3)2 2 NaNO3 + BaSO4 Step 1 : q = m c θ @ q = (20.0 + 30.0) (4.18) (34.0 – 30.0) q = 836 J Step 2 : determine limitant mol Na2SO4 = MV /1000 mol Ba(NO3)2 = MV /1000 = (0.500) (20.0) / 1000 = (0.300) (30.0) / 1000 = 0.010 mol = 0.0090 mol (lim)
Step 3 : ∆H = q / mol @ ∆H = 836 / 0.0090 ∆H = 92889 J / mol – 92.9 kJ / mol
Question 3 : Solution Equation : 2 KI + Pb(NO3)2 2 KNO3 + PbI2 Step 1 : q = m c θ @ q = (20 + 30) (4.18) (34 – 29) q = 1045 J Step 2 : determine limitant mol KI = MV /1000 mol Pb(NO3)2 = MV /1000 = (0.18) (30) / 1000 = (0.15) (20) / 1000 = 0.0054 mol = 0.0030 mol
*Since 2 mol of KI ≡ 1 mol of Pb(NO3)2 ; KI is limitant mol of reaction = 0.0027 mol
Step 3 : ∆H = q / mol @ ∆H = 1045 / 0.0027 mol ∆H = 387037 @ = – 390 kJ / mol
Question 4 : Solution NaCl + AgNO3 NaNO3 + AgCl Step 3 : From ∆H and mol ; find q Step 2 : mol of AgNO3 = MV / 1000 mol of NaCl = MV / 1000 = (1.00)(10.0) / 1000 = (0.800)(15.0) / 1000 = 0.010 mol (lim) = 0.012 mol so q = ∆H x mol @ q = (– 63400) (0.010) q = 634 J
Step 1 : q = m c θ @
θ = q / mc θ = 634 / [(4.18) (10.0 + 15.0) θ = 6.07 oC
1.2.2 Standard condition for calculating enthalpy changes The standard conditions of temperature and pressure for thermochemical measurement are 298 K and 1 atm. Any enthalpy changes measured under these conditions is described as standard enthalpy of reaction, and the symbol is written as ∆H∅ In this Chapter, there are a total of 9 standard enthalpy change of reaction that we shall learned through. There are 3 basic rules applied when thermochemical equations were used for calculations.
The total amount of energy released or absorbed is directly proportional to the number of moles of the reactant used. For example, in the combustion of methane : CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (l) ∆H∅ = – 890 kJ mol-1 If there’s 2 mole of methane, CH4 are combusted 2 CH4 (g) + 4 O2 (g) 2 CO2 (g) + 4 H2O (l) ∆H = The enthalpy change for the reverse reaction is equal in magnitude but opposite in sign to the enthalpy change for the forward reaction. Na+ (g) + Cl– (g) NaCl (s) ∆H∅ = – 770 kJ mol-1 If the reaction is reversed NaCl (s) Na+ (g) + Cl– (g) ∆H∅ = The value of ∆H∅ for a reaction is the same whether it occurs in one step or a series of steps. This shall be further discussed on the coming sub-topic about Hess' Law
Enthalpy change of formation, ∆H∅f Energy changes occur when 1 mol of substance is formed from its individual elements under standard condition. E.g. : H2 (g) + ½ O2 (g) H2O (l) Note the following important things The physical states of the substance involved is stated accordingly under standard condition The ∆Hf∅ of water is not written as H2O (g) as it is not in gas under standard condition. 2 H2 (g) + O2 (g) 2 H2O (l) is not consider as standard as substance formed is not 1 mole. ∆H∅f of pure element is = 0 kJ / mol
Examples CO2
: C (s) + O2 (g) CO2 (g)
MgCO3 : Mg (s) + C (s) + 3/2 O2 (g) MgCO3 (s) NH3
: ½ N2 (g) + 3/2 H2 (g) NH3 (g)
NaCl
: Na (s) + ½ Cl2 (g) NaCl (s)
C6H12O6 : 6 C (s) + 6 H2 (g) + 3 O2 (g)C6H12O6 (s) SO3
: 1/8 S8 (s) + 3/2 O2 (g) SO3 (g)
CH3COOH :2C(s) + 2H2(g) + O2(g)CH3COOH (l) Al2O3
: 2 Al (s) + 3/2 O2 (g) Al2O3 (s)
∆H∅f and Stability of Compound Product formed via exothermic process are more stable than product formed via endothermic process Example : compare ∆H∅f of sodium halide NaI < NaBr < NaCl < NaF
∆H∅f more exothermic Reaction between Na and X2 more vigorous Stability of compound formed increase ∆Hrxn of a reaction can be calculated using ∆H∅f ∆Hrxn = Σ ∆H∅f (products) – Σ ∆H∅f (reactants)
Example 5
CO (g) + ½ O2 (g) CO2 (g) ∆Hrxn = Σ ∆H∅f (products) – Σ ∆H∅f (reactants) = (Hf CO2) – Hf (CO + O2) = (– 393 kJ / mol) – ( – 110 kJ / mol + 0) = – 283 kJ
Example 6
∆Hrxn
2 FeCl2 (s) + Cl2 2 FeCl3 (s) = Σ ∆H∅f (products) – Σ ∆H∅f (reactants) = (2 x Hf FeCl3) – Hf (2 FeCl2 + Cl2) =(2 x –405 kJ / mol) – ( 2 x –341 kJ / mol + 0) = – 128 kJ
*Extra Note – Cyclohexene, C6H10 contain one carbon – carbon double bond, C=C. When cyclohexene undergoes hydrogenation, the enthalpy change is –120 kJ / mol.
If a benzene ring (which has 3 C=C), react with hydrogen :
supposedly the enthalpy of hydrogenation must be 3 x -120 = -360 kJ / mol. However, when experiment involving hydrogenation is carried out, the ∆H of benzene is – 208 kJ / mol, indicating that benzene molecule does not contain three double bonds in its structure. 1 mol of benzene is 152 kJ / mol more stable than 1 mole of cyclohexene. The more stable the structure, the less heat given out during a reaction. The real structure of benzene is a resonance hybrid between the structure above
1.4 Enthalpy change of combustion, ∆H∅c Energy liberated occur when 1 mol of substance is burned with excess air (oxygen) under standard condition. E.g. : CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (l) Note a few things in the thermochemical equation above : The standard combustion of substance must be 1 mole of the reactant burned. The mole of oxygen used must be balanced accordingly. Oxygen is always combust in excess For ∆H∅c is always exothermic. MAKE SURE THE ‘ – ‘ MUST BE PLACED.
Examples C : C (s) + O2 (g) CO2 (g) [= ∆H∅fof CO2] H2 : H2 (g) + ½ O2 (g) H2O (l) [= ∆H∅fof H2O] C2H5COOH : C2H5COOH (l) + 7/2 O2 (g) 3 CO2 (g) + 3 H2O (l) C2H5OH : C2H5OH (l) + 3 O2 (g) 2 CO2 (g) + 3 H2O (l) Mg : Mg (s) + ½ O2 (g) MgO (s) P : P4 (s) + 5 O2 (g) P4O10 (s) Al : Al (s) + 3/4 O2 (g) ½ Al2O3 (s) C6H12O6 : C6H12O6 (s) + 6 O2 (g) 6 CO2 (g) + 6 H2O (l)
Calorimeter ~ instrument used to measure the heat transferred during a chemical reaction. Simple calorimeter :
Advantages : Simple
to be prepared and set-up
Disadvantages : The
experimental value is always lesser than the actual ∆H∅c because of the following reason Heat is easily lost to surrounding Combustion of the sample is incomplete Combustion is not done under standard condition
Example 7
Step 1 : q = m c θ @ q = (150) (4.18) (71.0 – 27.8) q = 27.1 kJ Step 2 : calculate the mol of pentane burned mol = mass / RMM = 4.30 / 72 = 0.0597 mol Step 3 : ∆H = q / mol @ ∆H = 27.1 kJ / 0.0597 mol ∆H = – 454 kJ / mol
Bomb Calorimeter
Bomb calorimeter consist of a thick stainless steel pressure vessel called “bomb” “Bomb” is then sealed after weighted sample is placed. A volume of water is added to ensure the surface is covered Pure oxygen is pumped into the valve until 25 atm. initial temperature is recorded. Temperature of water is taken from time until it reached maximum temperature. The difference of temperature is taken as θ. Then, benzoic acid (C6H5COOH) is used to calibrate the instrument to determine the heat capacity of the instrument. Heat capacity ~ heat required to raise the temperature of the whole apparatus by 1 K enthalpy change (q )
Heat capacity , C =
temperature change (θ )
Steps of calculating ∆H∅c using bomb calorimeter Calibration Sample (using benzoic acid) (burned sample) mol = mass RMM
q = ∆H x mol
C=q/θ
∆H =
q mol
mol = mass RMM
q=Cθ
Example 10 :
ΔH = 16.2 kJ 0.0123 =-1310 kJ/ mol
mol = 0.625 122 = 5.12 x 10-3mol
q = -3230 x 5.12 x = 16.5 kJ
10-3
C =16.5 / 1.58 = 10.5 kJ / K
mol = 0.712 58 = 0.0123 mol
q = 10.5 x 1.54 = 16.2 kJ
1.5– Hess Law ~ stated that the heat absorbed or liberated during a chemical reaction, is independent of route by which the chemical changes occur. Consider the following equation : A + B C + D required 2 steps A+BZ ZC+D A
+
B ∆HM Z ∆HN
C
+
D
Example : In the reaction of formation of SO3, it is a 2 steps reaction. Step 1 : 1/8 S8 (s) + O2 (g) SO2 (g) ∆H1 = – 297 kJ / mol Step 2 : SO2 (g) + ½ O2 (g) SO3 (g) ∆H2 = – 99 kJ Overall : 1/8 S8 (s) + 3/2 O2 (g) SO3 (g) ∆Hf∅ = – 396 kJ / mol
Energy / kJ 1/8 S8 (s) + 3/2 O2 (g)
SO2 (g) + ½ O2 (g) SO3 (g)
Using Hess’s Law, the energy required to form intermediate can also be determined. Example : In the reaction of processing ammonia, the equation is N2 (g) + 3 H2 (g) 2 NH3 (g) ∆H = – 92.2 kJ The 2 steps involve in the process of forming ammonia Step 1 : N2 (g) + 2 H2 (g) N2H4 (g) ∆H1 = x kJ/mol Step 2 : N2H4 (g) + H2 (g) 2 NH3 (g) ∆H2 = – 187 kJ / mol Since ∆Hrxn required is N2 + 2 H2 N2H4 While : N2 (g) + 3 H2 (g) 2 NH3 (g) ∆H = – 92.2 kJ Eq.2 is reversed 2 NH3 (g) N2H4 (g) + H2 (g) ∆H2 = + 187 kJ N2 (g) + 2 H2 (g) N2H4 ∆Hf = + 94.8 kJ / mol
Energy / kJ N2H4 (g) + H2 (g)
N2 (g) + 3 H2 (g) 2 NH3 (g)
Example 9 : Find H2 (g) + O2 (g) H2O2 (l) H2 (g) + ½ O2 (g) H2O (l) ∆Hf∅ = - 286 kJ/mol H2O2 (g) H2O (l) + ½ O2 (g) ∆H = - 188 kJ/mol Reverse equation (2) => equation (3) H2O (l) + ½ O2 (g) H2O2 (g) ∆H = +188 kJ (3) H2 (g) + ½ O2 (g) H2O (l) ∆Hf∅ = - 286 kJ/mol (1) So, when equation (1) + (3) H2 (g) + O2 (g) H2O2 (l) ∆H = - 98 kJ/mol
(1) (2)
Energy / kJ H2 (s) + O2 (g)
H2O2 (l)
H2O (g) + ½ O2 (g)
Example 10 : From these data, S(rhombic) + O2(g) → SO2(g) ∆Hrxn = - 296.06 kJ/mol S(monoclinic) + O2(g) → SO2(g) ∆Hrxn = - 296.36 kJ/mol Calculate the enthalpy change for the transformation S(rhombic) → S(monoclinic)
(Monoclinic and rhombic are different allotropic forms of elemental sulfur.)
Since the equation required is S(rhombic) → S(monoclinic) Make sure S(rhombic) is at the left while S(monoclinic) is at the right By reversing eq (2) and compare to eq (1) S(rhombic) + O2 (g) → SO2(g) ∆Hrxn = - 296.06 kJ/mol SO2(g) → S(monoclinic) + O2 (g) ∆Hrxn = + 296.36 kJ/mol ----------------------------------------------------------------------------------S(rhombic) → S(monoclinic) ∆Hrxn = + 0.30 kJ / mol
Energy / kJ sulphur (rhombic) + O2 (g) sulphur (monoclinic) + O2 (g)
SO2 (g)
1.5.3 Relationship between ∆Hc∅ and ∆Hf∅ using Hess Law For example, in determining the ∆Hf∅ of butane, C4H10. Given the ∆Hc∅ for C4H10, C and H2 are – 2 877 kJ / mol ; – 393 kJ / mol and -296 kJ / mol respectively. Solution : C (s) + O2 (g) CO2 (g) ∆Hc∅ = - 393 kJ / mol .. (1) H2 (g) + ½ O2 (g) H2O (l) ∆Hc∅ = - 286 kJ / mol .. (2) C4H10 (l) + 13/2 O2 (g) 4 CO2 (g) + 5 H2O (l) ∆Hc∅ = - 2877 kJ /mol .. (3) 4 C (s) + 5 H2 (g) C4H10 (l) ∆Hf∅ = ? kJ / mol \. (4) Since the equation of formation require 4 C (s) and 5 H2 (g), so the overall equation for (1) and (2) are multiply by 4 and 5 respectively, where as in equation (3) are reversed. 4 C (s) + 4 O2 (g) 4 CO2 (g) ∆Hc∅ = – 1572 kJ 5 H2 (g) + 5/2 O2 (g) 5 H2O (l) ∆Hc∅ = – 1430 kJ 4 CO2 (g) + 5 H2O (l) C4H10 (l) + 13/2 O2 (g) ∆Hc∅ = + 2877 kJ 4 C (s) + 5 H2 (g) C4H10
(l)
∆Hf∅ = – 125 kJ / mol
Energy / kJ 4 C (s) + 5 H2 (g) + 13/2 O2 (g) C4H10 (l) + 13/2 O2 (g)
4 CO2 (g)
+ 5 H2 (g) + 5/2 O2 (g)
4 CO2 (g) + 5 H2O (l)
Example 11 : Given ∆Hc∅ of C2H2 and C6H6 are – 1300 kJ / mol and – 3270 kJ/mol respectively. Find 3 C2H2 (g) C6H6 (l) C2H2 (g) + 5/2 O2 (g) 2 CO2 (g) + H2O (l) ∆Hc∅ = –1300 (1) C6H6 (l) + 15/2 O2(g) 6 CO2(g) + 3 H2O(l) ∆Hc∅ = –3270 (2) Multiply equation (1) by 3 Reverse equation (2) 3C2H2 (g) + 15/2 O2 (g) 6 CO2 (g) + 3 H2O (l) ∆Hc∅ = –3900 6 CO2(g) + 3 H2O(l) C6H6 (l) + 15/2 O2 (g) ∆Hc∅ = +3270 3 C2H2 C6H6 (l)
∆H∅ = – 630 kJ
Energy / kJ 3 C2H2 (g) + 15 / 2 O2 (g) C6H6 (l) + 15/2 O2 (g)
3 H2O (g) + 6 CO2 (g)
Example 12 : C3H6 (g) + H2 (g) C3H8 (g) ∆H∅ = –124 kJ/mol C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(l) ∆Hc∅ = –2222 kJ/mol H2 (g) + ½ O2(g) H2O (l) ∆Hc∅ = – 286 kJ / mol Find for C3H6(g) + 9/2 O2(g) 3 CO2(g) + 3 H2O(l) ∆Hc∅ = ? Reverse equation (3) H2O (l) H2 (g) + ½ O2(g) ∆Hc∅ = + 286 kJ / mol C3H6 (g) + H2 (g) C3H8 (g) ∆H∅ = –124 kJ/mol C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(l) ∆Hc∅ = –2222 kJ/mol C3H6(g) + 9/2 O2(g)3 CO2(g) + 3 H2O(l) ∆Hc∅= –2060 kJ/mol
Energy / kJ C3H6 (g) + H2 (g) + 5 O2 (g) C3H8 (g) + 5 O2 (g) C3H6 (g)
+ 9/2 O2 (g) + H2O (l)
3 CO2 (g) + 4 H2O (l)
1.6 Enthalpy change of Neutralisation ∆H ∅ neut ~ amount of energy liberated when 1 mol of hydrogen ion from acid react with 1 mol of hydroxide ion from alkali to form 1 mole of water under standard condition. Equation : H+ (aq) + OH– (aq) H2O (l) ∆H∅neut for strong acid and strong base under standard condition is – 57.3 kJ / mol. The value of ∆H∅neut will be different with weak acid / base is used or if the acid used is a polyproctic acid In laboratory, ∆H∅neut can be determine using simple cup calorimeter (MPM Experiment 6) The ways of calculating ∆H∅neut is still the same as we learned previously.
Example Equation : HCl (aq) + NH3 (aq) NH4Cl (aq) Step 1 : q = m c θ @ q = (25.0 + 30.0) (4.18) (31.3 – 27.6) q = 850.63 J Step 2 : determine limitant mol of HCl = MV / 1000 = (1.00)(25.0)/1000 = 0.025 mol
mol of NH3 = MV /1000 = (0.800)(30.0) / 1000 = 0.024 mol (lim)
Step 3 : ∆H = q / mol @ ∆H = 850.63 / 0.0240 ∆H = – 35443 J / mol @ – 35.4 kJ / mol
∆H∅neut for weak acid or weak alkali reaction. If ∆H∅neut is ≠ 57.3 kJ / mol depend on :
example above, it can be tell that, the ∆H∅neut for weak acid and strong alkali is ≠ - 57.3 kJ / mol. This is due to, some heat is absorbed by CH3COO-H to break the O-H to form hydrogen ion. Therefore, it is less exothermic than the expected value.
the
Basicity of an acid : HCl H+ + Cl– [monoproctic acid] H2SO4 2 H+ + SO42- [diproctic acid] H3PO4 3 H+ + PO43- [triproctic acid] For example, when NaOH (aq) react with H2SO4 (aq) Stage 1 :H2SO4 (aq) + NaOH (aq) NaHSO4 (aq) + H2O (l) ∆H∅neut = –61.95 kJ / mol Stage 2 : NaHSO4 (aq) + NaOH (aq) Na2SO4 (aq) + H2O (l) ∆H∅neut = –70.90 kJ / mol Overall : 2 NaOH (aq) + H2SO4 (aq) Na2SO4 (aq) + 2 H2O (l) ∆H∅neut = [-61.95 + (-70.90)]= - 132.85 kJ
Reaction involving HF : HF (aq) + NaOH (aq) NaF (aq) + H2O (l) ∆H∅neut= –102.4 kJ / mol The reaction become more exothermic than expected despite that HF is consider as a weak acid. When HF is dissolve in water, H-F dissociate in water to form H+ and F-. The enthalpy of hydration, ∆H∅hyd of the fluoride ion is very exothermic, making the overall process to be much exothermic F- (g) + water F- (aq) ∆H∅hyd = – 63.4 kJ/mol
1.6 Standard Enthalpy Change of Atomisation, ∆H∅atom ~ energy absorbed when 1 mole of gaseous atoms are formed from its element under standard condition. Eq : A (s) A (g) ∆H∅atom = + ve kJ/mol Example : Mg (s) Mg (g) ¼ P4 (s) P (g) ½ Cl2 (g) Cl (g)
1/8 S (s) S (g)
CH4 (g) C (g) + 4 H (g)
PBr3 (s) P (g) + 3 Br (g)
Since the reaction required the substance involve to become gaseous atom, so the process involved an endothermic process. For a solid, the ∆H∅atom involves 2 processes. For example, in sodium, Na, to become a gaseous sodium, the solid metal undergoes melting process before vapourising to gas. Energy required to change 1 mol of solid to liquid is named as enthalpy change of fusion, while the energy required to change 1 mol of liquid to gas to called as enthalpy change of vapourisation, according to the following equation
Na (s) → Na (l) Na (l) → Na (g)
Na (s) Na (g)
∆Hfusion ∆Hvapourisation
∆H∅atom
Since noble gas exist naturally as monoatom gas the Enthalpy Change of Atomisation for noble gas 0 As for the Bond enthalpy, it is the energy required to break the bond between 2 covalently bond atoms. For example, the bonding enthalpy of chlorine gas Cl – Cl (g) 2 Cl (g) ∆H∅BE = + 242 kJ / mol Compare to the , ∆H∅atom of chlorine atom ; ½ Cl2 (g) Cl (g) ∆H∅atom = + 121 kJ / mol
1.7 Ionisation energy, ∆H∅IE ~ energy absorbed when 1 mole of electron is removed from a gaseous atom under standard condition. Eq : A (g) A+ (g) + e∆H∅IE = + ve kJ/mol The process is always endothermic as heat is absorbed to free one mole of electron from an atom (to overcome the electrostatic forces of attraction between the nucleus and outermost electron) Generally, when goes down to Group, ionisation energy decrease, while across the Period, ionisation energy increase. These trend shall be further discussed in Chapter 3 It is believed that, when enormous amount of energies is supplied, electrons in an atom can be removed completely from an atom. The energies required to consecutively remove the electrons from an atom is called as successive ionisation energies
The total ionisation energy is the sum of all the successive ionisation of the element involve. Example 1st IE of Al : Al (g) Al+ (g) + e– ∆H∅IE = + 577 kJ / mol 2nd IE of Al : Al+ (g) Al2+(g) + e– ∆H∅IE = + 1820 kJ / mol 3rd IE of Al : Al2+ (g) Al3+(g) + e– ∆H∅IE = + 2740 kJ / mol Overall : Al (g) Al3+ (g) + 3e– ∆HIE = + 5137 kJ The information of the 1st until the 4th ionisation energy of elements can be obtained through Data Booklet supplied during examination
1.8 Electron Affinity ∆H∅EA ~ energy liberated when 1 mole of electron is received from gaseous atom under standard condition. Eq : O (g) + e- O– (g) ∆H∅EA = – X kJ / mol For 1st Electron Affinity, the process is always exothermic, since upon receive an electron, the energy carries by the electron is released upon combining with the gaseous atom. The trend of of 1st electron affinity is the same as in Ionisation energy, where 1st electron affinity decrease when going down to group, whereas the 1st electron affinity increase when going across Period. However, unlike 2nd ionisation energy, after an atom received an electron an form negative charged ion, upon receiving the second electron, a repulsion forces is felt between the anion and electron receive, due to the mutual charge between both substance. Hence, for second electron affinity, heat is absorbed (endothermic) by the anion to overcome the repulsion forces between the anion and electron.
When forming O2– from O– (2nd EA), electron is received by negative ion. erepulsion forces formed between anion and electron received. Heat is absorbed to overcome the forces of repulsion. 1st EA : O (g) + e– O– (g) ∆H∅EA= – 142 kJ / mol 2nd EA : O– (g) + e– O2– (g) ∆H∅EA = + 844 kJ / mol Overall : O (g) + 2e– O2– (g) ∆HEA = + 702 kJ
1.9 Lattice Energy, ∆H∅LE ~ energy liberated when 1 mole of solid crystal lattice is formed from oppositely charged gaseous ions under standard condition. Eq : M+ (g) + X– (g) MX (s) ∆H∅LE = –X kJ/mol LE – always negative (exothermic) : heat is released when ionic bond is formed. Examples of writing thermochemical equation :
NaF : Na+ (g) + F- (g) NaF (s) MgO : Mg2+ (g) + O2- (g) MgO (s) CaCl2 : Ca2+ (g) + 2 Cl- (g) CaCl2 (s) K2O : 2 K+ (g) + O2- (g) K2O (s) Al2O3 : 2 Al3+ (g) + 3 O2- (g) Al2O3 (s) AlN : Al3+ (g) + N3- (g) AlN (s)
Factors influencing Lattice Energy – i) charge of ion ii) inter-ionic distance
Charge of ions (Zn+ . Zn–)
Greater the charge ; greater the forces of attraction ; greater the value of Lattice Energy (more exothermic)
Lattice energy
Inter-ionic distance (r+ + r–)
∝
Smaller the distance, greater the attraction forces between ions, greater the lattice energy
Z n+ • Z n− r+ + r−
Compound
Total charge
∑ Ionic radius
Compound
Total charge
Ionic radius
NaF
1
0.231
NaCl
1
0.276
KBr
1
0.328
KCl
1
0.314
CaO
4
0.239
MgO
4
0.205
Al2O3
6
0.190
K2O
2
0.273
The trend of lattice energy of these 8 compounds are
KBr < KCl < NaCl < NaF < K2O < CaO < MgO < Al2O3 Lattice energy increase
1.10 Born Haber Cycle Lattice energy cannot be determined experimentally. They can only be obtained by applying Hess’s Law in an energy cycle called Born-Haber Cycle, which is a cycle of reactions used for calculating the lattice energies of ionic crystalline solids. There are basically 5 types of Born Haber Cycle which is mostly tested all times. i) A+B- ii) A2+B2- iii) A2+B2- iv) A2+B2- v) A23+B32 To build the Born Haber cycle, students must be able to write ∆Hf∅ of the compound and ∆H∅LE. Here, we are going to build the Born Haber cycle using the 5 examples aboveSodium chloride, NaCl Calcium chloride, CaCl2 Potassium oxide, K2O Magnesium oxide, MgO Chromium (III) oxide, Cr2O3
Na+ (g) + Cl (g) + e∆HEA of Cl
∆Hatom of Cl
Na+ (g) + ½ Cl2 (g) + eNa+ (g) + Cl- (g)
∆HIE of Na
Na (g) + ½ Cl2 (g) ∆Hatom of Na
∆HLEof NaCl
Na (s) + ½ Cl2 (g) ∆Hf of NaCl
NaCl (s) ∆H∅f = ∆H∅LE + [∆H∅atom Na + ∆H∅atom Cl + ∆H∅1st IE Na + ∆H∅1st EA Cl] ∆H∅LE = (-411) – [(+108) + (+121) + (+494) + (-364)] = – 770 kJ/mol
Ca2+ (g) + 2 Cl (g) + 2 e2 x ∆Hatom of Cl
2 x ∆HEA of Cl
Ca2+ (g) + Cl2 (g) + 2 e∆H1st IE of Ca + ∆H2nd IE of Ca
Ca2+ (g) + 2 Cl- (g)
Ca (g) + Cl2 (g) ∆Hatom of Ca
∆HLEof CaCl2
Ca (s) + Cl2 (g) ∆Hf of CaCl2
CaCl2 (s) ∆H∅f = ∆H∅LE + [∆H∅atom Ca + 2∆H∅atom Cl + ∆H∅1st IE Ca + ∆H∅2nd IE Ca + 2∆H∅1st EA Cl] ∆H∅LE = (-795) – [(+132) + 2(+121) +(+590) +(1150) + 2(-364)]
= – 2181 kJ/mol
2 K+ (g) + O2- (g) ∆H1st EA + ∆H2nd EA
2 K+ (g) + O (g) + 2 e∆Hatom of O
2 K+ (g) + ½ O2 (g) + 2 e2 X ∆H1st IE of K
∆HLEof K2O
2 K (g) + ½ O2 (g) 2 X ∆Hatom of K
2 K (s) + ½ O2 (g) ∆Hf of K2O
K2O (s) ∆H∅f = ∆H∅LE + [2∆H∅atom K + ½∆H∅BE O + 2∆H∅1st IE K + ∆H∅1st EA O + ∆H∅2nd EA O]
∆H∅LE = (-362) – [2(+129) + ½(+498) + 2(418) + (-141)+(+844)] = – 2408 kJ/mol
Mg2+ (g) + O2- (g) ∆H1st EA + ∆H2nd EA of O
Mg2+ (g) + O (g) + 2 e∆Hatom of O
Mg2+ (g) + ½ O2 (g) + 2 e∆H1st IE of Mg + ∆H2nd IE of Mg
∆HLEof MgO
Mg (g) + ½ O2 (g) ∆Hatom of Mg
Mg (s) + ½ O2 (g) ∆Hf of MgO
MgO (s) ∆H∅f = ∆H∅LE + [∆H∅atom Mg + ½∆H∅BE O +∆H∅1st IE Mg + ∆H∅2nd IEMg + ∆H∅1st EA O + ∆H∅2nd EA O] ∆H∅LE = (-612) – [(+146) + ½(+498)+(736) + (1450) + (-141)+(+844)] = – 3896 kJ/mol
2 Cr3+ (g) + 3 O2- (g) 3 x (∆H1st EA O + ∆H2nd EA of O)
2 Cr3+ (g) + 3 O (g) + 6 e3 x ∆Hatom of O
2 Cr3+ (g) + 3/2 O2 (g) + 6 e2 x (∆H1st IE of Cr + ∆H2nd IE of Cr + ∆H3rd IE of Cr)
∆HLEof Cr2O3
2 Cr (g) + 3/2 O2 (g) 2 x ∆Hatom of Cr
2 Cr (s) + 3/2 O2 (g) ∆Hf of Cr2O3
Cr2O3 (s) ∆H∅LE = – 16408 kJ/mol
1.12 Enthalpy Change of Hydration, ∆hhyd In terms of Thermochemistry, the solubility of ionic compound in water depend on 2 factors The enthalpy change of hydration Lattice energy of the salt involved
Standard enthalpy change of hydration, ∆H∅hyd is ~ energy liberated when one mole of gaseous ion is hydrated by \\\\\\\\\\\\\\\\\..\\\\\\\\\\\\
water. \\\\\\\\\\\\\\\\\...under standard condition. n+ (g) + water Mn+ (aq) M ∆Hhyd = – x kJ/mol Equation :
Qn- (g) + water Qn- (aq) ∆Hhyd = – x kJ/mol Intermolecular forces occur during hydration of ions are ion-dipole forces, which were stronger than hydrogen bonding. Diagram below shows the ion-dipole forces between a positively and negatively charged ion with water respectively.
Since the intermolecular forces between ion and water is strong, the ∆H∅hyd is always exothermic. Similar to lattice energy, the magnitude of ∆H∅hyd depends on 2 factors : Charge of ion - Greater the charge of ion, stronger the attraction between the water and ion, the more exothermic it is enthalpy change of hydration of ions Size of ion - Smaller the size of ion, stronger the attraction between the ions and water, the more exothermic it is the enthalpy change of hydration of ions
δ+
δ-
Mn+
∆Hhyd for cation increase Na+ < Mg2+ < Al3+
Qn-
∆Hhyd for anion increase l– < Br– < Cl–
1.12 Enthalpy change of solution, ∆Hsoln Energy change when 1 mole of solute is dissolved in a large excess water to form an infinite dilute solution. For ionic substance : MX (s) + water M+ (aq) + X- (aq) Some covalent subs : C6H12O6 (s) + water C6H12O6 (aq) ∆Hsoln is determined by ∆Hhyd and ∆HLE ∆HLE : M+ (g) + X- (g) MX (s) [reverse] ∆Hhyd : M+ (g) + X- (g) + water M+ (aq) + X- (aq) – ∆HLE : MX (s) M+ (g) + X- (g) MX (s) + water M+ (aq) + X- (aq) As a conclusion, ∆Hsoln = ∆Hhyd + (– ∆HLE) If ∆Hsoln = - ve, then the salt is soluble in water If ∆Hsoln = + ve, then the salt is insoluble in water
In the solubility of Group 2 sulphate both lattice energy and enthalpy change of hydration are proportional to of the ions. Hence, when going down to Group 2 decrease As the size of sulphate, both of these energies \\\\\\. increase metal ion \\\\\\\\.. Less significant However, the rate decrease in lattice energy is \\\\\\\\ than the rate of decrease in ∆H∅hyd This is because the size of sulphate ion is much larger than the size of metal ions, so even though the size of cation increases, the increase of (r+ + r-) is very small. This makes the lattice energy changes become less significant when goes down to Group 2. While in ∆H∅hyd it depend on both cation and anion. Since the ∆H∅hyd for anion is constant, so the ∆H∅hyd is mainly depend on the size of cation. When goes down to Group 2, the metal ion size increase less \\\\\\\\\. , making ∆H∅hyd become \\\\\ decrease of the heat become more exothermic. So, the \\\\\. significant thus causing the rate of ∆H∅hyd is greater than lattice energy.
Group 2 sulphate
Be SO4
Mg SO4
Ca SO4
Sr SO4
Ba SO4
∆Hsolution (kJ / mol)
-95.3
-91.2
+ 17.8
+ 18.70
+19.4
Solubility (g / 100mL)
41.0
36.4
0.21
0.010
0.00025
∆Hhydration
∆Hlattice energy
BeSO4
Mg SO4
CaSO4
Sr SO4
BaSO4
Example : Solubility of Group 2 sulphate :
Sr2+ Ba2+
Be2+ Ca2+
Mg2+
SO42-
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