CHAPTER 2 THERMOCHEMISTRY (PRACTICE MODULE) - Flip eBook Pages 1-28 (2023)

ANSWER PRACTICE MODULE CHAPTER 2 SK 025

CHAPTER 2.0

THERMOCHEMISTRY

I Think Map 2.1 and 2.2 (✿◠‿◠)

TYPE OF ENTHALPIES CALORIMETRY

WORKSHEET 2.1 : Concept Of Enthalpy
1.

a) What are the reactants and products of the reaction?
Reactants : A+B
Products : C+D

b) What is the enthalpy change for this reaction?
20 – 40 = -20 kJ

c) Does the reaction is exothermic or endothermic reaction? Explain.
Exothermic reaction because the enthalpy change is negative.

ANSWER PRACTICE MODULE CHAPTER 2 SK 025

d) Write the thermochemical equation for this reaction
A + B → C + D ∆H = -20 kJ

2. State the enthalpy of reaction for the thermochemical equation below
a) C2H6(g) + 7/2 O2(g) → 2CO2(g) + 3H2O(l) ∆H = – 3119.6 kJ
Enthalpy of combustion of ethane
b) Ca(s) + Cl2(g) → CaCl2(s) ∆H = – 795 kJ
Enthalpy of formation of Calcium Chloride
c) KF(s) → K+(aq) + F- (aq) ∆H = +2 kJ
Enthalpy of solution of Potassium Floride
d) Ca2+ (g) → Ca2+ (aq) ∆H = – 1650kJ
Enthalpy of hydration of Calcium
e) ½ Cl2(g) → Cl(g) ∆H = +121.4 kJ
Enthalpy of atomisation of Chlorine

f) H2SO4(aq) + NaOH(aq) → Na2SO4(aq) + H2O(l) ∆H = -57.3 kJ
Enthalpy of neutralization of Sulphuric Acid and Sodium Hydroxide

3. Write the balanced thermochemical equation for the reactions below
a) Enthalpy formation for HCl, ∆Hf = -92.4 kJ mol-1
½ H2(g) + ½ Cl2(g) → HCl(g) ∆Hf = -92.4 kJ mol-1
b) Enthalpy formation for Aluminium Oxide, Al2O3, ∆Hf = -1675.5 kJ mol-1
2 Al(s) + 3/2 O2(g) → Al2O3(s) ∆Hf = -1675.5 kJ mol-1
c) Enthalpy combustion for Ethane, C2H6, ∆Hc = -1560 kJ mol-1
C2H6(g) + 7/2 O2(g) → 2CO2(g) + 3H2O(l) ∆Hc = -1560 kJ mol-1
d) Enthalpy atomisation for Nitrogen, N2, ∆Ha = 473 kJ mol-1
½ N2 (g) → N (g) ∆Ha = 473 kJ mol-1
e) Enthalpy atomisation for Magnesium ∆Ha = 150 kJ mol-1
Mg (s) → Mg (g) ∆Ha = 150 kJ mol-1
e) Enthalpy solution for Calcium Chloride, CaCl2 ∆Hsol = -120 kJ mol-1
CaCl2 (s) → Ca+ (aq) + Cl- (aq) ∆Hsol = -120 kJ mol-1

4. The complete combustion of acetic acid, HC2H3O2(l), to form H2O(l) and CO2(g) at
constant pressure release 871.7 kJ of heat per mole of HC2H3O2(l).
(a)
a) Write a balanced thermochemical equation for this reaction.
HC2H3O2(l) + 2O2(g) → 2H2O(l) + 2CO2(g) ∆H = -871.7 kJ mol-1

ANSWER PRACTICE MODULE CHAPTER 2 SK 025

b) Draw an enthalpy diagram for the reaction.

WORKSHEET 2.2 : Calorimeter

1. Define heat capacity and specific heat capacity

Tips: Heat Capacity, C is amount of heat required to raise temperature of a given

unit for heat quantity of substance by 1 K (1°C)

capacity (J°C-1)

Unit for specific Specific Heat Capacity, c is amount of heat required to raised temperature
heat capacity of 1 gram of a substance by 1 K (1°C)

( Jg-1 °C-1)

A piece of aluminum with a mass of 50g and an initial temperature of 90°C is placed

2. into 100mL of water at a temperature of 25°C. The temperature of water rises to

31.30 °C. Determine the specific heat capacity of aluminum.

(ans: 0.897 J/gºC)

Tips : q lost (Al) = q gained (Water)

q lost by mc∆t = mc∆t
something = q 50 x CAl x (90 - 31.3) = 100 x 4.18 x (31.3 - 25)
gained by (2935 CAl) = 2633.4
something CAl = 2633.4 / 2935 = 0.897 J/gºC

3. When 137 mL of water at 25 °C is mixed with 82 mL of water at 76 °C, what is the

final temperature of the water? Assume that no heat is lost to the surroundings and

that the density of water is 1.00 g/mL.

(ans: 44.1ºC)

Tips : q lost (water1) = q gained (Water2)

Cw at both side mcw∆t = mcw∆t
can be (137g)(cw)(tf - 25°C) = (82g)(cw)(76°C – tf)
cancelled 137tf - 3425 = 6232 – 82tf

219 tf = 9657

Tf = 44.1 °C

ANSWER PRACTICE MODULE CHAPTER 2 SK 025

4. An alloy of unknown composition is heated to 137 °C and placed into 100.0 g of water

at 25.0 °C. If the final temperature of the water was 36.4 °C, and the alloy weighed

2.71 g, what is the specific heat capacity of the alloy? The specific heat of water is

4.18 J/gºC.

(ans : 17.4 J/gºC)

Tips : q lost by alloy = q gained by water

Final -(mc∆t) = +mc∆t
temperature of (2.71g)(calloy)(137 – 36.4°C) = (100g)(4.18 J/gºC)(36.4 – 25°C)
alloy = final
temperature of 273.71 calloy = 4765.2
water
calloy = 17.4 J/gºC

Bomb Calorimeter – Enthalpy of combustion question

5. In an experiment, 4.5 g of xylose, C5H10O5 was completely burn in a bomb calorimeter
containing 225.0 mL of water at 29.0 0C. The maximum temperature reached after the
reaction completed was 33.0 0C.

a) Calculate the enthalpy of combustion of xylose in kJ mol-1
(ans : -125.4 kJ mol-1)

Step 1: C5H10O5(g) + 5O2(g) → 5CO2(g) + 5H2O(l)

Write the
equation
reaction

Step 2: = (12.0 x 5) + (1.0 x 10) + (16.0 x 5)
= 150.0 g mol-1
Calculate
molar mass of
C5H10O5

Step 3: = 4.5 g/150.0 gmol-1
= 0.03 mol
Find number
of moles of
C5H10O5

Step 4: = final temperature – initial temperature
= 33.0 0C – 29.0 °C
Find = 4.0 °C
temperature
change, ∆T Quantity of heat released during the combustion of C5H10O5 is absorbed
by water in bomb calorimeter,
Step 5:
q = mwater x cwater x ∆T
Calculate heat = (225.0 g)(4.18 J g-1 0C-1)(4.0 0C)
released, q = 3762.0 J

Step 6: From experiment
0.03 mol of C H O5 10 5 released 3762 J
Find the 1 mol of C H O5 10 5 released 1/0.03 x 37621 J
relationship = 125400 J
between = 125.4 kJ
experiment
and theory Enthalpy of combustion of xylose :-125.4 kJ mol-1
(from balance
equation).

ANSWER PRACTICE MODULE CHAPTER 2 SK 025

b) Write the thermochemical equation for the combustion of xylose

Tips: C5H10O5(g) + 5O2(g) → 5CO2(g) + 5H2O(l) ∆Hc = 125.4 kJ mol-1

Check

balance,
phase & ∆H

6. A 0.1375 g of Magnesium is burn in a constant volume bomb calorimeter that has a
heat capacity of 1769 J °C−1. If the calorimeter contains 300 g of water and the
temperature increases by 1.126°C, calculate the heat of combustion of magnesium in
kJ mol−1.
[The specific heat of water is 4.18 Jg−1°C−1 ; Mr Mg = 24.3 gmol-1]
(ans: −601.61 kJ mol−1)

Mg(s) + 1/2O2(g) → MgO(s)

Heat released by the combustion = Heat absorbed by water + Heat
absorbed by calorimeter

-q combustion = + (q water + q calorimeter)
= mwcw∆T + Cc∆T
= (300g x 4.18 Jg−1°C−1x 1.126°C) + (1769 J°C−1x 1.126°C)

= 3403.9 J

q combustion = - 3.4039 kJ

nMg = 0.1375 g ÷ 24.3 gmol-1
= 5.658 x 10−3 mol

5.658 x10−3 mol of Mg burnt released 3.4039 kJ
1 mol Mg burnt released 601.61 kJ

∆Hc Mg = − 601.61 kJ mol−1

7. 207.5 g sample of solid aluminium is burned in a constant-volume bomb calorimeter
that has a heat capacity of 1245 J°C-1. The calorimeter contains exactly 500.0 g of

water and the temperature increases by 19.2°C. Calculate the

a) Enthalpy of combustion of aluminium (ans: -8.332 kJmol-1)

nAl = 207.5 g ÷ 27 gmol-1
= 7.685 mol

heat release by reaction = heat absorb by water and calorimeter

q reaction = q water + q calorimeter
= mwcw∆T + Cc∆T
= (500g x 4.18 Jg−1°C−1x 19.2°C) + (1245 J°C−1x 19.2°C)

= 64032 J

7.685 mol of Al burnt released 64032 J
1 mol Al burnt released 8332.08 J

ANSWER PRACTICE MODULE CHAPTER 2 SK 025

Enthalpy of combustion of aluminium = -8.332 kJmol-1

b) Enthalpy of formation of Al2O3
(ans: -16.664 kJmol-1)

Al(s) + 3/4O2(g) → 1/2Al2O3(s) ∆Hc= -8.332 kJmol-1

0.5 mol Al2O3 release 8.332 kJ
1.0 mol Al2O3 release 16.664 kJ

∆Hf Al2O3 = -16.664 kJmol-1
8.

The enthalpy of combustion of fructose, C8H12O6 is 21.2 kJ mol-1. An amount of 2.63
g of fructose was completely combusted in a bomb calorimeter at 25 °C.

a) Write the thermochemical equation for the reaction.

C6H12O6(s) + 6O2(g ) → 6CO2(g) + 6H2O(l) ΔH = -21.2 kJ/mol

b) Calculate the final temperature if the calorimeter contains 225.0 mL of water.
(ans: Tf = 25.33 °C)

Mole C6H12O6 = 2.63 ÷ 180 = 0.01461 mol

1 mol C6H12O6 releases 21.2 kJ heat
0.01461 mol C6H12O6 releases 0.01461x 21.2

= 0.3097 kJ heat

-q (heat released by combustion) = q (heat absorbed by water)
q = mcΔT

0.3097 x 103 = 225 x 4.18 x ΔT
ΔT = 0.3293 °C

0.3293 = Tf - 25
Tf = 25.33 °C

Simple Calorimeter – Enthalpy of neutralisation question

9. When a student mixes 50 ml HCl(1M) and 50 ml NaOH (0.8M) in a coffee cup
calorimeter, the temperature of the resultant solution increase from 21°C to 27.5°C.
Calculate the enthalpy of neutralization , assuming that no heat absorbed by
calorimeter.

[ Assume density of solution = density of water = 1 g/ml ; The specific heat of water
is 4.18 J g−1°C−1 ]

Step 1: HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

Write the n NaOH = (1)(0.05)
equation = 0.05 mol
reaction

Step 2:

ANSWER PRACTICE MODULE CHAPTER 2 SK 025

Calculate no

of moles of n HCl = (0.8)(0.05)
acid and base
= 0.04 mol

Step 3: n NaOH > n HCl
so, HCl is limiting reactant
Find the
limiting n H2O = n HCl = 0.04 mol
reactant
between acid Heat released by the reaction = Heat absorbed by solutions
and base (to
find mol of -qrxn = + qs
H2O) = mscs∆T
= (100 g) (4.18 J g-1 oC-1) (6.5oC)
Step 4: = 2717 J

Find q
reaction (heat
release by
reaction)

qrxn = - 2.717 kJ

Step 5: 0.04 mol of H2O formed release 2.717 kJ
1 mole of H2O formed release 67.93 kJ
Relate mol of
water with ∆ Hneutralization = - 67.93 kJmol-1
heat release
to calculate
enthalpy of
neutralization

10. 40.0 mL of 1.0 M NaOH were placed in a calorimeter (Ccal = 72.0 J/°C) at 22.0°C and
20.0 mL of 1.5 M H2SO4 at 22.0°C were added. The temperature of the mixture rose
to 29.0°C.

Calculate ∆H for the reaction 2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l) from

the following data.

(Ans: -113 kJ)

n NaOH = (1)(0.04)

Tips: = 0.04 mol

Cc were given

in question. n H2SO4 = (1.5)(0.02)
So, q release = 0.03 mol
= q Cc +

qsolution 2 mol NaOH ≡ 1 mol H2SO4
0.04 mol NaOH ≡ 0.02 mol H2SO4

NaOH is limiting reactant. n H2O = n NaOH = 0.04 mol

Heat released by the reaction = Heat absorbed by solutions

-qrxn = + (qCc + qs)
= Cc∆T + mscs∆T
= (72.0 J/°C x 7°C) + (60g) (4.18 J g-1 oC-1) (7oC)

ANSWER PRACTICE MODULE CHAPTER 2 SK 025

= 504 + 1755.6 J
qrxn = - 2259.6 J
2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l)

From equation,
0.04 mol of H2O formed release 2.2596 kJ

2 mole of H2O formed release 113 kJ

11. A quantity of 200 mL of 0.862M HCl is mixed with 200 mL of 0.431M Ba(OH)2 in a
constant pressure calorimeter that has a heat capacity of 453 J°C-1. The initial
temperature of the HCl and Ba(OH)2 solution is the same at 20.48°C. What is the
final temperature of the mixed solution if the enthalpy of neutralization is -56.2
kJ/mol.
[ specific heat capacity of solution= 4.18 Jg-1 °C-1, density of solution = 1 gcm-3]
(ans: 25.04°C)

Ba(OH)2(aq) + 1/2HCl(aq) → 1/2BaCl2(aq) + H2O(l)

n Ba(OH)2 = (0.431M)(0.2L)
= 0.0862 mol

n HCl = (0.826M)(0.2L)
= 0.1724 mol

1 mol HCl ≡ 1 mol H2O
0.1724 mol HCl ≡ 0.1724 mol H2O

1 mol H2O release 56.2 kJ heat
0.1724 mol H2O release 56.2 kJ x 0.1724
= 9.689 kJ heat

Heat released by the reaction = Heat absorbed by calorimeter + solution

-qrxn = +(qCc + qs)
9.689 x 10-3 J = Cc∆T + mscs∆T

= (453 J/°C x ∆T) + (400g) (4.18 J g-1 oC-1) (∆T)
∆T = 4.56°C
Tf = ∆T + Ti
= 4.56 + 20.48
= 25.04°C

12. 100 cm3 of 2.0 mol dm-3 hydrochloric acid and 100 cm3 potassium hydroxide solution,

both at initial temperature of 30.0°C are mixed in a calorimeter. The maximum

temperature of the solution is 41.0°C. Calculate the enthalpy of neutralisation for the

reaction.

(ans: ΔH = - 45.98 kJmol-1 )

Heat of reaction = mcΔT

= 200g x 4.18 Jg-1°C-1 x 11°C

= 9196 J

ANSWER PRACTICE MODULE CHAPTER 2 SK 025

No.of mole of acid = 2.0 mol dm-3 x 0.1 dm-3
= 0.2 mol

HCl + KOH → KCl + H2O
From equation,
0.2 mol produced 9196 J
1 mol produced 9196 J x ( 1 mol ÷ 0.2 mol)

= 45980 J mol-1
Enthalpy of neutralisation = -45.98 kJmol-1

ANSWER PRACTICE MODULE CHAPTER 2 SK 025

MIND MAP 2.3 and 2.4 (✿◠‿◠)
2.3 HESS’S LAW

ANSWER PRACTICE MODULE CHAPTER 2 SK 025

WORKSHEET 2.3 : HESS’S LAW
CComplete the blanks with suitable enthalpies

Enthalpy of
hydration Na

Enthalpy of
hydration CI

2. Enthalpy of
hydration MgO
Lattice energy
of MgO

Enthalpy of
solution MgO

4 Calculate ΔHo for the conversion of methane in chloroform.

CH4(g) + 3Cl2(g) → CHCl3(g) + 3HCl(g)

By combining the following equation:

½H2(g) + ½Cl2(g) → HCl(g) ΔHo = -92.3 kJ/mol C(graphite)

+ 2H2(g) → CH4(g) ΔHo = -74.5 kJ/mol

C(graphite) + ½H2(g) + 3/2 Cl2(g) → CHCl3(g) ΔHo = -135.1 kJ/mol

ANSWER:
Step 1: Target equation

CH4(g) + 3Cl2(g) → CHCl3(g) + 3HCl(g)

Step 2: Change all thermochemical equaion & magnitud given based on
the target equation

ANSWER PRACTICE MODULE CHAPTER 2 SK 025

3/2H2(g) + 3/2Cl2(g) → 3HCl(g) ΔHo = -92.3 kJ/mol x 3

CH4(g) → C(graphite) + 2H2(g) ΔHo = +74.5 kJ/mol

C(graphite) + ½H2(g) + 3/2 Cl2(g) → CHCl3(g) ΔHo = -135.1 kJ/mol

Step 3:Terminated other than target equation

x3 3/2H2(g) + 3/2Cl2(g) → 3HCl(g) ΔHo = -92.3 kJ/mol x 3
ΔHo = +74.5 kJ/mol
reverse CH4(g) → C(graphite) + 2H2(g) ΔHo = -135.1 kJ/mol

C(graphite) + ½H2(g) + 3/2 Cl2(g) → CHCl3(g)

Step 4: Calculate the enthalpy of formation for target equation

∆H = -276.9 + 74.5 + (-135.1)
= -337.5 kJmol-1

5 Calculate and construct diagram for enthalpy solution of magnesium fluoride

based on data below:

Lattice energy of MgF2 – 3121 kJ/mol

Enthalpy of hydration Mg2+ ion –1903 kJ/mol

Enthalpy of hydration F - 461 kJ/mol

ANSWER:

+ 3121kJ

H =- H + H

soln lattice hdyr

= 3121+ (-1903) + (-461)
= +757 kJmol-1

6 Calculate and construct diagram for enthalpy hydration of KI based on data

below:

Lattice energy of KI – 632 kJ/mol

Enthalpy of solution KI + 13 kJ/mol

ANSWER PRACTICE MODULE CHAPTER 2 SK 025

Answer :

7 From the data below

(a) Construct the enthalpy (energy) cycle for sodium chloride solution

(b) Calculate the enthalpy of hydration of Cl- ion

Lattice energy of NaCl – 776 kJ/mol

Enthalpy of hydration Na+ – 390 kJ/mol

Enthalpy of solution NaCl + 6 kJ/mol

ANSWER:

+ 776kJ

ANSWER PRACTICE MODULE CHAPTER 2 SK 025

WORKSHEET 2.4 : BORN HABER CYCLE
Complete the blanks with suitable enthalpies
1

ANSWER PRACTICE MODULE CHAPTER 2 SK 025

2
3

ANSWER PRACTICE MODULE CHAPTER 2 SK 025

4
5

ANSWER PRACTICE MODULE CHAPTER 2 SK 025

Construct energy level diagram
1 By using the following data a Born-Haber cycle, calculate enthalpy formation of

CaF2(s). Draw an energy level diagram showing the energy changes.

ANSWER:

+ 2e
+e

ΔH = ΔH + IE1 + ΔH + EA1 + EA2 + ΔH

f a(Ca) a(F) lattice

= 148 + 318 + 738 + 1450 + (-328 x2) +( - 3121)

ΔH = - 1123 kJ/mol

ANSWER PRACTICE MODULE CHAPTER 2 SK 025

2 By using the following data a Born-Haber cycle, calculate lattice formation
energy of MgO(s). Draw an energy level diagram showing the energy changes
ANSWER:

ANSWER PRACTICE MODULE CHAPTER 2 SK 025

Construct Born Haber cycle
Worksheet 2.6
1 Using the following data, determine the enthalpy of formation of sodium chloride

with construct a diagram Born-Haber cycle.

ANSWER:

ΔH = ΔH + IE + ΔH + EA + ΔH
f a(Na) 1 a(CI) lattice

= 107 + 496 + 122 - 349 – 787

ΔH = - 411 kJ/mol

2. Using the following data, determine the lattice energy of sodium oxide, Na2O with
construct a diagram Born-Haber cycle.

Enthalpy atomisation of sodium = +108 kJ mol-1
Enthalpy of atomization of oxygen = +249.2 kJ mol-1
First ionisation energy of sodium = +496 kJ mol-1
1st electron affinity of oxygen = -141 kJ mol-1
2nd electron affinity of oxygen = +790 kJ mol-1
Enthalpy of formation of Na2O(s) = - 416 kJ mol-1

ANSWER PRACTICE MODULE CHAPTER 2 SK 025

ANSWER:

ΔH = ΔH + IEf + ΔH + EA1 + EA2 + ΔH
a(Na) 1
a(O) lattice

-416 = (108x2) + (496x2) + 249.2 - 141 +790 + ΔH

lattice

ΔH = - 2522.2 kJ/mol

lattice

3. Using the following data, determine the lattice formation energy of MgO(s) with

construct a diagram Born-Haber cycle.

Enthalpy of atomization of Mg(s) = +146 kJ mol-1
Enthalpy of atomization of O2(g) = +249.2 kJ mol-1
First ionisation energy of Mg(g) = +736 kJ mol-1
Second ionisation energy of Mg(g) = +1450 kJ mol-1
First electron affinity of O(g) = -141 kJ mol-1
Second electron affinity of O(g) = +744 kJ mol-1
Enthalpy of formation of MgO(s) = -601.7 kJ mol-1

ANSWER PRACTICE MODULE CHAPTER 2 SK 025

ANSWER:

ΔH = ΔH + IE1 + IE2 + ΔH + EA1 + EA2 + ΔH

f a(Mg) a(O) lattice

-601.7 = 146 + 736+ 1450 + 249.2 +(-141) +744 + ΔH

lattice

ΔH = - 3785.9 kJ/mol

lattice

4 The lattice energy of lithium chloride is -848 kJ mol-1 and that of sodium chloride is
-776 kJ mol-1.
a) Using sodium chloride as an example, define lattice energy.
b) Explain why the lattice energy of lithium chloride is more exothermic
than that of sodium chloride.

ANSWER:
a) Heat change when 1 mole of solid NaCI produce from Na+ and CI- in

gaseous state.
b) Because Li+ ion is smaller than Na+ ion

ANSWER PRACTICE MODULE CHAPTER 2 SK 025

WORKSHEET 2,7 (OBJECTIVE QUESTIONS)

1. The equation below represent :
Mg2+ (g) + 2CI- (g) → MgCI2(s)
A Lattice energy
B Enthalpy of formation

C Enthalpy of atomisation
D Electron affinity

2. Lattice energy becomes more exothermic when….
A Size of ions are small
B Charge of ions are small
C Attractive forces weaker
D Larger size

3. Using the equation below:

C(s) + O2(g) → CO2(g ΔH= -390 kJ
Mn(s) + O2(g) → MnO2(s) ΔH= -520kJ

What is ΔH (in kJ) for the following reaction?

MnO2(s) + C(s) → Mn(s) + CO2(g)

A 910
B 130
C -130
D -910

What is the value of enthalpy of formation of NaF(s)?

A -928

ANSWER PRACTICE MODULE CHAPTER 2 SK 025

B -574
C +107
D +496

5 Which equation represent the process that occurs when the standard enthalpy of
atomisation of iodine is measured?

A I2 (s) → I(g)

B ½ I2 (s) → 2I(g)
C ½ I2 (g) → I(g)
D I2 (s) → 2I(g)

6 Living plants produce glucose in the process of photosynthesis according to this
equation:
6CO2(g) + 6H2O(l) + energy → ¼ C6H12O6(s) + 6O2(g)

Is this reaction endothermic or exothermic, and is the value of ∆Hº positive or
negative?
A endothermic, positive
B endothermic, negative
C exothermic, positive
D exothermic, negative

7. In order to produce 972 kJ of heat, how many grams of H2 must burn?

H2(g) + ½O2(g) ¼ H2O(g) H= + 243 kJ

A 0.250 g

B 4.04 g

C 8.08 g

D 16.0 g

8. What is 4.18 J?

A The heat required to raise the temperature of one gram of water by one
Celsius degree.

B The heat required to raise the temperature of one mole of water by one Celsius
degree.

C The heat required to raise the temperature of one gram of substance by one
Celsius degree.

D The heat required to raise the temperature of one mole of substance by one
Celsius degree.

9. What happens to the water in a calorimeter when an exothermic reaction occurs in it?
A It absorbs heat, and a drop in temperature is observed.
B It absorbs heat, and a rise in temperature is observed.
C It releases heat, and a drop in temperature is observed.
D It releases heat, and a rise in temperature is observed.

10 Which is true for an exothermic reaction?
A The ∆H is positive.
B The products have less potential energy than the reactants.
C The reactants have more kinetic energy than the products.
D The reactants are below the products in the potential energy diagram.

ANSWER PRACTICE MODULE CHAPTER 2 SK 025

WORKSHEET 2.8 ( MODEL PSPM)
Q1: clone PSPM 2007/2008

The enthalpy of combustion of fructose, C6H12O6 is 21.2 kJ mol -1. An amount of 2.63
g of C6H12O6 was completely combusted in a bomb calorimeter at 25 °C.

[ 5 marks]
i) Write the thermochemical equation for the reaction.
ii) Calculate the final temperature if the calorimeter contains 225.0 mL of water.

(ans: 25.33°C)
Answer:
i) C6H12O6(s) + 6O2(g ) ⎯→ 6CO2(g) + 6H2O(l) ΔH = -21.2 kJ

ii) Mole C H O6 12 6 = mol

1 mol C H O6 12 6 released 21.2 kJ heat

0.01461 mol C H O6 12 6 released 0.01461× 21.2

= 0.3097 kJ heat

-q (heat released by combustion) = q (heat absorbed by water)

q = mcΔT
0.3097 × 103 = 225 × 4.18 × ΔT

ΔT = 0.3293 ° C

0.3293 = T2 - 25
T2 = 25.33 ° C

Q2: clone PSPM 2011/2012

The enthalpy of combustion of benzoic acid is -3226.7 kJmol-1. When 3.2 g benzoic

acid, C6H5COOH is completely combusted in a bomb calorimeter containing 2.0 kg

of water, the temperature of the water increased by 3.8°C. Calculate the heat

capacity of the calorimeter.

(ans: C = 1.39 x 104 J˚C-1)

[5 marks]

Answer:

mol of C6H5COOH = 3.2g = 0.0262 mol

122g/mol

1 mol C6H5COOH released 3226.7 kJ

0.0262 mol C6H5COOH released (3226.7 kJ x 0.0262) /1

= -84.63 kJ

q = mcΔT + CΔT

84.63 = (2.0 x 10-3)(4.18)(3.8) + C(3.8)
C = 1.39 x 104 J˚C-1

ANSWER PRACTICE MODULE CHAPTER 2 SK 025

Q3: clone PSPM 2019/2020

a) Ethanol, CH3CH2OH is produced from fermentation of sugarcane and widely used as
an alternative fuel to replace petrol. A 1.00 g sample of ethanol was burned in a
calorimeter, which has a total heat capacity of 11.0 kJ oC-1. The temperature of the
calorimeter and its contents increased from 25.0 oC to 27.7 oC.

[ 6 marks]
i)Write a balanced chemical equationthat takes place in the bomb calorimeter.
ii) Calculate the enthalpy of combustion of ethanol per mol.

(ans: -1368.66 kJ)
b) Born haber cycle for the formation of calcium sulphide is given in FIGURE 2.

( The cycle is not drawn to scale)
[6marks]

Ca2+(g) + S2-(g)

Ca2+(g) + S(g)

ΔH5 -200 kJ mol-1 ΔH6
ΔH4 +1145 kJ mol-1 Ca2+(g) + X

Ca+(g) + S(g)

ΔH3 +590 kJ mol-1 -3013 kJ mol-1 ΔH7
Ca(g) + S(g)

ΔH2 +279 kJ mol-1
Ca(g) + S(s)

ΔH1 +178 kJ mol-1
Ca(s) + S(s)

ΔH8 -482 kJ mol-1 CaS(s)

i) Name the enthalpy changes for ∆H7 and ∆H8.
ii) Identify the species X formed from ∆H5.
iii)Use the data in FIGURE 2 to calculate the value of ∆H6.

iv) Will the lattice energy for CaO be larger or smaller than CaS? Explain briefly.

ANSWER PRACTICE MODULE CHAPTER 2 SK 025

ANSWER:
a) i) CH3CH2OH + 7/2O2 → 2CO2 + 3H2O

ii) Mol CH CH OH = 1.00 g
32 46.0 g mol−1

= 0.0217 mol

q = (mcT + CT)
rxn
= CT
= 11.0 kJ C x (27.7 C − 25.0 C)
= 29.7 kJ

0.0217 mol ethanol released 29.7 kJ
1mol ethanol released 1368.66 kJ

H = −1368.66 kJ mol−1
C

b) i) ∆H7 = Enthalpy of lattice energy of CaS
∆H8 = Enthalpy formation of CaS.
ii) S-(g)

iii) ∆H8 = Σ (∆H1 +∆H2 +∆H3 + ∆H4 + ∆H5 + ∆H6 + ∆H7)
-482 = 178 + 279+ 590 + 1145 +(-200) + ∆H6 + (-3013)
∆H6 = +539 kJmol-1

iv) - CaO has larger lattice energy
- Because O2- is smaller size than S2-. Attractive forces between Ca2+ and O2-
are stronger than between Ca2+ and S2-.

Q4: clone PSPM 2020/2021

a) TABLE 1 shows the lists of standard enthalpy change associated with the solubility

of silver chloride, AgCl.

TABLE 1

Enthalpy ∆ ° (kJ mol-1)

Lattice energy of AgCl +915

Hydration energy of Ag+ -465

Hydration of Cl- -364

i) Construct the thermochemical cycle to show the relationship between lattice energy,

hydration energy and enthalpy of solution for AgCl.

ii) Calculate the enthalpy of solution of AgCl.

ANSWER PRACTICE MODULE CHAPTER 2 SK 025

b) Arrange the following substances in ascending order of lattice energy. Explain your
answer.
AlCl3, MgCl2, NaCl

ANSWER: ∆ ° = -915 kJ mol-1
ai)
+ Cl- (g) AgCl(s)
Ag+(g)

∆ ° + ∆ ° − ∆ °

= -465 kJ mol-1 = -364 kJ mol-1

Ag+(aq) + Cl- (aq)

ii) ∆ ° = ∆ ° + + ∆ ° − −∆ ° @

-915 = −465 + (-364) −∆ °

∆ ° = +86 kJ/mol

b) NaCl < MgCl2 < AlCl3

Charges of cations increase from Na+ to Mg2+ to Al3+ @
Ionic radius of the cation decreases from Na+ to Mg2+ to Al3+

The higher the charge of the cations, the stronger the attraction towards the Cl- ion
@ the smaller the ionic radius of the cations, the stronger the attraction towards the
Cl-.
Thus, lattice energy increases.

Prepared by : Mdm Suhaibah Binti Mustafa & Mdm Siti Fatimah Binti Md Sollhi
Checked by : Mdm Noor Shuhada bt Rahim & Mdm Wan Rosilah bt Wan Llah