TMK CHAPTER 9.0 THERMOCHEMISTRY

CHEMISTRY SK025
Chapter 9.0

THERMOCHEMISTRY

TUTORIAL 9

9.1 & 9.2: Concept of Enthalpy & Calorimetry

1. (a) Sketch energy profile diagram

Energy Activated Complex

H2(g) + 1/2O2(g) Ea

Hc=-241.8kJ mol -1

H2O(l)

(b) Larger enthalpy reaction progress
Reactant

(c) Enthalpy change
ΔH = 2 (+241.8) kJ= +483.6 kJ

2. (a) Heat of combustion
ΔHc =ΣΔΗf (products) - ΣΔΗf (reactants)

= [5(-393.5) + 6(-285.8)] - [-173.2 + 0]

= - 3509.kJmol−1

(b) Thermochemical equation ∆H = -3509 kJ

C5H12 (l) + 8O2 (g) → 5CO2(g) + 6H2O(l)

(c) Heat released

3. (a) 1 mol C5H12 ≡ - 3509.1 kJ
(b) 0.2083 mol C5H12 ≡ - 731 kJ
(c)
Heat is transferred from the hot copper (hot body) to water (cold body) in which the
copper released heat and absorbed by water. Thus, the temperature of copper
decreased from 120oC to 28.5oC while water increased from 25oC to 28.5oC.

Heat absorbed by water

Heat absorbed by water, qw = mwcwΔT
= 55.5 g x 4.18 J.g-1.oC-1x(28.5-25)oC
= 811.97 J

Specific heat capacity

Heat released by copper, qCu = Heat absorbed by water, qw
mCucCuΔT = 811.97 J

811.97J
cCu = (60.0g)(91.5oC)

= 0.148J.g-1.oC-1

1

CHEMISTRY SK025
Chapter 9.0

THERMOCHEMISTRY

4. Heat of neutralisation
NaOH (aq) + HCl(aq) NaCl(aq) + H2O(l)

0.150 mol HCl ≡ 0.150 mol NaOH ≡ 0.150 mol H2O
Heat released by the reaction, qrxn -Heat absorbed by the solution,qs

0.150 mol H2O ≡ -7315 J
1 mol H2O ≡ -48766.7 J = -48.8 kJ

5. (a) Heat released when one mole of naphthalene is combusted in excess oxygen at 1 atm
and 25oC.

(b) mole of naphthalene,C10H8 = 2.05/128 = 0.0160 mol
qreaction = -(qcal)
= - (Ccal x ∆T)
= -(5.11 kJoC-1 x 7.2oC)
=-36.8 kJ
∆Hcombustion = qreaction/nnaphtalene
= -36.8kJ/0.0160 mol
= -2300 kJmol-1

9.3: Hess’s Law

1. (a) Definition
Hess’s Law states that when reactants are converted to products, the change in
enthalpy is the same whether the reaction takes place in one step or a series of

steps.

(b) Enthalpy of formation

V(s) + 2Cl2 (g) → VCl4 (l) H = -569.4 kJ

VCl3 (s) → VCl2 (g) + 1 ΔH = +436.0 kJ
2 Cl2 (g)

VCl2 (g) + VCl4 (l) → 2 VCl3(s) ΔH = -211.0 kJ

V(s) + 3 → VCl3(s) ΔH = -344.4 kJ
2 Cl2 (g)

2. Heat change

Equation 1:

C4H6 (g) + 11 → 4CO2 (g) + 3H2O(l) ΔHo =-2540.2 kJ
2 O2 (g) c

Equation 2:

2

CHEMISTRY SK025
Chapter 9.0

THERMOCHEMISTRY

C4H10 (g) + 13 → 4CO2 (g) + 5H2O(l) ΔHco =-2877.6 kJ
2 O2 (g) ΔHco =-285.8 kJ

Equation 3:

H2 (g) + 1 O2 (g) → H2O(l)
2

Calculation

Equation 1: Same

C4H6 (g) + 11 → 4CO2 (g) + 3H2O(l) ΔHo =-2540.2 kJ
2 O2 (g) c

Equation 2: reverse

4CO2 (g) + 5H2O(l) → C4H10 (g) + 13 ΔH = +2877.6 kJ
2 O2 (g)

Equation 3: ×2

2H2 (g) + O2 (g) → 2H2O(l) ΔH = (-285.8)2 kJ

C4H6 (g) + 2H2 (g) → 4C4H10 (g) ΔH = -234.2 kJ

3.

2S(s) + 2O2 (g) → 2SO2 (g) H = -2(296.8) kJ

2SO3(g) + 2H2O(l) → 2H2SO4 (l) ΔH = -2(227.2) kJ

2SO2 (g) + O2 (g) → 2SO3(g) ΔH = +198.2 kJ

2S(s) + 3O2 (g) + 2H2O(l) → 2H2SO4 (l) ΔH = -849.8 kJ

∆Hc = ? 2CO2(g) + H2O(l)
4. C2H2(g)
+2O2(g) +½ O2(g)
+5/2 O2
∆H1x2 ∆H3
∆H2

2C(s) + H2(g)

∆Hc = 2∆H1 + ∆H3 - ∆H2
= 2(-394) + (-286) – (227)
= -1301 kJmol-1

Na+(g) + Cl-(g) Hlattice = -776 kJ NaCl(s)

Hhyd = -390 kJ Hhyd = ?

Hsoln = -4 kJ

Na+(aq) + Cl-(aq)

ΔH Cl- = -776+4+390 = −382 kJ mol-1
hyd

3

CHEMISTRY SK025
Chapter 9.0

THERMOCHEMISTRY

9.4: Born-Haber cycle

1. (a) Lattice energy is the heat released when 1 mole of ionic compound is formed from its

gaseous ions.

(b) Cl-(g) Hlattice = -845 kJ LiCl(s)

Li+(g) +

Hhyd = -519 kJ Hhyd = -364 kJ
Hsoln = ? kJ

Li+(aq) + Cl-(aq)

∆Hsoln = (-519) + (-364) – (-845)
= -38 kJmol-1

2. (a) Name enthalpy
(b) ΔH2 = Enthalpy of atomization of sodium
ΔH4 = First ionization energy of sodium
3. (a) ΔH5 = First electron affinity of chlorine

(b) Lattice energy
ΔH1 = ΔH2 + ΔH3 + ΔH4 + ΔH5 + ΔH6
-411.3 = 107.8 + 121.3 + 495.4 -348.8 + ΔH6

ΔH6 = -787.0 kJ
∴Therefore, lattice energy is -787.0 kJmol-1

H1>H2
Reason:

• H1 is 1st IE while H2 is 2nd IE.

• when the first electron has been removed from an atom, the remaining
electrons will be pulled closer to the nucleus.

• Therefore, more energy required to remove the second electron.

Born-Haber cycle for the formation of MgCl2(s)

Energy (kJmol-1)
Mg2+(g) + 2Cl(g) + 2e

Mg2+(g) + Cl2(g) + 2e ∆HBE= +240kJ EA = -2(369) kJ
Mg2+(g) + 2Cl-(g)

Mg+(g) + Cl2(g) + e IE2 = ?

Mg(g) + Cl2(g) IE1 = +740 kJ ∆Hlattice = -3933 kJ
Mg(s) + Cl2(g) ∆Ha = +149 kJ

∆Hf = -1846 kJ
4

MgCl2 (s)

CHEMISTRY SK025
Chapter 9.0

THERMOCHEMISTRY

IE2 = -1846 –(149+740+240+2(-369)+(-3933)) = 480
= + 1696 kJmol-1

3.

Rb(s) + 1/2Br2(l) − kJ RbBr(s)

+82 kJ +112 kJ

Rb(g) Br(g) -658 kJ
+400 kJ EA = ?

Rb+(g) + Br-(g)

-389 kJ = (82 kJ) + (400 kJ) + (112 kJ) + EA + (-658 kJ)
EA = -325 kJmol-1

MEKA 9

1. Write thermochemical equation

(a) C2 H6 (g)+ 7 O2 (g) → 2CO2 (g)+3H2O(l) ΔHco = -370kJ
2

(b) Mg(s)+N2 (g)+3O2 (g) → Mg(NO3)2 (s) ΔH f = -113.0 kJ

(c) F2 (g) → 2F(g) ΔHdissociation = +244 kJ

2. (a) Definition
Standard enthalpy of combustion: the amount of heat released when 1 mol of
substance is completely burnt in oxygen at standard conditions (25oC and 1 atm).

(b) Given:

Hoc of C3H8 = -220.1 kJ mol−1

i. Write thermochemical equation

C3H8 (g)+ 7 O2 (g) → 3CO2 (g)+4H2O(l) ∆H = -220.1 kJ
2

ii. Amount of heat released

(10.0 g)
( )n C3H8 =
= 0.227 mol
44 g mol−1

1 mol C3H8 release 220.1 kJ
0.227 mol C3H8 release 50.023 kJ

iii. Mass of propane
Heat released by reaction,qrxn = -heat absorbed by water,qw
= -mwcw∆T

5

CHEMISTRY SK025
Chapter 9.0

THERMOCHEMISTRY

= -(50.0 g)(4.18 Jg−1C−1)(100C - 30C)

= -14630 J

-220.1 kJ ≡ 1 mol C3H8
-14.63 kJ ≡ 0.0665 mol C3H8

Mass of C3H8 required = (0.0665 mol)(44 g mol−1) = 2.92 g

3. (a) Definition
Heat released when 1 mol of water form from the neutralization between acid and
base under stated condition.

(b) Enthalpy of neutralisation
Given:
• 50.0 mL of 1.0 M HCl, n HCl = 0.05 mol
• 50.0 mL of 0.8 M NaOH, n NaOH = 0.04 mol (Limiting reactant)
• Ti solution = 21C
• Tf solution = 27.5C
• Mass solution (Assume density solution same as water) = 100 g

Heat released by reaction,qrxn = -heat absorbed by solution,qs
= mscs∆T
= -(100 g)(4.18 Jg−1C−1)(6.5C)

= -2717 J

HCl (aq) +NaOH (aq) → NaCl (aq) + H2O (l)

1 mol of NaOH≡ 1 mol of H2O
0.04 mol of NaOHC 0.04 mol of H2O

0.04 mole of H2O ≡ -2.717 kJ
1 mole of H2O ≡ -67.93 kJ

∆ Hneutralization = - 67.93 kJmol−1

4. Heat of combustion
Given:
• Mass of sulfur = 3 g
• Burned in bomb calorimeter.
(heat released by reaction = heat absorbed by (calorimeter + water))
• Ti = 21.25C
• Tf = 26.72C
• Ccalorimeter = 923 J K−1
• Mass of water = 800 g

Heat released by reaction,qrxn =-(qc + qw)
= -(Cc∆T + mwcw∆T)
=-[ (923 J K−1)(5.47C) + (800 g)(4.18 Jg−1C−1)(5.47C)]

=- 23340.49 J

6

CHEMISTRY SK025
Chapter 9.0

THERMOCHEMISTRY

(3.0 g)
( )n of S = = 0.0117 mol
256.56 g mol−1

0.0117 mol S ≡ - 23340.49 J
1 mol S ≡ - 1996078 J

∆ Hc = - 1996 kJmol−1

5. (a) Definition
Heat change when 1 mol of substance form from its elements in their most stable
state.

(b) Enthalpy of reaction
2HCl(g) + F2 (g) → 2HF(l) + Cl2 (g)

• Eq. 1: ½
• Eq. 2: reverse and ×2
• Eq. 3: reverse

2HCl(g) + 1/ 2O2 (g) → H2O(l) + Cl2 (g) ∆H = - 74.2 kJ
H2 (g) + F2 (g) → 2HF(l) ∆H = - 1200.0 kJ
H2O(l) → H2 (g) + 1 / 2O2 (g) ∆H = + 285.8 kJ
2HCl(g) + F2 (g) → 2HF(l) + Cl2 (g) ∆H = - 988.4 kJ

6. Enthalpy formation of methane:
C(s) + 2H2 (g) → CH4 (g)

C(s) + Hf = ? CH4(g)
2H2(g)

O2(g) O2(g) 2O2(g)
-393.5 kJ -571.6 kJ -890.4 kJ

CO2(g) + 2H2O(g)

-393.5 kJ + (-571.6 kJ) = ∆Hf + (-890.4 kJ)
∆Hf = -74.7 kJ mol-1

7. (a) Definition.
Heat released when 1 mole of compound dissolve in water forming very dilute
solution under stated condition.

(b) Energy cycle
Using BaCl2,
Dissolution process of BaCl2 involve;

7

CHEMISTRY SK025
Chapter 9.0

THERMOCHEMISTRY

• Lattice Energy
BaCl2 (s) → Ba2+ (g) + 2Cl− (g)

• ∆ Hhyd. for Ba2+ and Cl−
Ba2+ (g) → Ba2+ (aq)
Cl− (g) → Cl− (aq)

lattice energy Ba2+(g) + 2Cl-(g)

BaCl2(s)

Hs ol n 2+ -

Hhyd of Ba Hhyd of Cl x 2

Ba2+(aq) + 2Cl-(aq)

(c) Enthalpy of hydration of Ba2+
Using Hess’ Law,
The flow of reaction is start from BaCl2(s) to Ba2+(aq) and Cl−(aq).

∆ Hsoln = Lattice energy + ∆ Hhyd.of Ba2+ + (∆ Hhyd. of Cl‒×2)
∆ Hhyd.of Ba2+ = ∆ Hsoln – (Lattice energy +(∆ Hhyd. of Cl‒×2))

= (-40 kJ mol−1) – (3900 kJ mol−1 – (380 kJ mol−1×2))
= -3180 kJ mol−1

Energy (kJmol-1)

Cu2+(g) + O2-(g)

Cu2+(g) + O(g) + 2e EA2 = +791 kJ
Cu2+(g) + O-(g) + e
∆Ha = +248 kJ EA1 = -141 kJ
Cu2+(g) + ½ O2(g) + 2e

IE2 = +1960 kJ ∆Hlattice = ? kJ

Cu+(g) + ½ O2(g) + e

IE1 = +745 kJ
Cu(g) + ½ O2(g)

∆Ha = +339 kJ
Cu(s) + ½ O2(g)

∆Hf = -155 kJ

CuO (s)

∆Hlattice = (-155) – ( 339 + 745 +1960 +248 – 141 + 791)
= -4097 kJmol-

8

CHEMISTRY SK025
Chapter 9.0

THERMOCHEMISTRY

KUMBE 9

1. (a) Final temperature
Given:

• Mass cold water = 400 g
• Mass hot water = 600 g
• Ti cold water = 25oC
• Ti hot water = 60oC
• Heat will be transfer from hot water to cold water until Tf equal

heat released by hot water,qhw = -heat absorbed by cold water,qcw
mhwcw∆T= - mcwcw∆T

600 g(Tf – 60 oC) = -400 g(Tf – 25oC)

Tf= 46 oC

(b) Given:
• Combustion of x molAl produced 0.25 mol Al2O3 = - 419 kJ

i. Standard enthalpy of combustion of aluminium

Al(s) + 3 O2 ( g ) → 1 Al2O3 (s)
4 2

From equation,
0.5 mol Al2O3 ≡ 1 mol Al
0.25 mol Al2O3 ≡ 0.5 mol Al

0.5 mol Al release 419 kJ
1 mol Al release 838 kJ

ΔHoc of Al = -838 kJ mol-1

ii. enthalpy of formation

1 mol Al ≡ 838 kJ

2 mol Al ≡ 1676 kJ

ΔHof of Al2O3 = -1676 kJ mol-1

Thermochemical equation Al2O3 (s) ΔHof = -1676 kJ
2Al(s) + 3/2 O2 (g)

2. Heat of combustion of Mg
Given: bomb calorimeter:
(heat released by reaction = heat absorbed by (calorimeter + water))

• Mass of Mg = 0.1375 g
n of Mg = 5.658 x 10-3 mol

• Calorimeter,
Ccal. = 1769JoC-1, ∆T = 1.126oC

• Water,
Mass = 300 g, ∆T = 1.126oC

heat released by reaction = -heat absorbed by (calorimeter + water)
=-( C∆T + mc∆T)

= -[(1769JoC-1)(1.126oC) + (300g)(4.18 Jg-1oC-1)(1.126oC)]

9

CHEMISTRY SK025
Chapter 9.0

THERMOCHEMISTRY

=- 3.403 x103 J

5.658 x 10-3 mol Mg ≡ -3.403 x 103 J
1 molMg ≡ -602 kJ

ΔHc of Mg = - 602 kJ mol-1

3. Heat of formation
4P(s) + 5O2 (g) → P4O10 (s)
• Eq. 1: ×5
• Eq. 2: Reverse
• Eq. 3: ×5
• Eq. 4: reverse and ×3

10PCl3(l) + 5O2 (g) → 10POCl3(l) ∆H = - 2935 kJ
10POCl3 (l) → P4O10 (s) + 6PCl5 (s) ∆H = + 419 kJ
10 4P(s) + 15Cl2 (g) → 10PCl3(l)
6PCl5 (s) → 6P(s) + 15Cl2 (g) ∆H = - 3430 kJ
4P(s) + 5O2 (g) → P4O10 (s) ∆H = + 2676 kJ
∆H = - 3270 kJ

4. (a) Given:
• 20.0 mL, 2.0 M NaOH

nNaOH = 0.04 mol
• 30.0 mL, 1.0 M CH3COOH

nCH3COOH = 0.03 mol
• Mass solution = 50 g, ∆T = 5.8oC
• Plastic cup calorimeter,

(heat released by rxn = -heat absorbed by (calorimeter + soln.))
Ccal = 37.30 J oC-1

i. State type of enthalpy
Enthalpy of neutralization

ii. Standard enthalpy of reaction
nNaOH = 0.04 mol (available)
n CH3COOH = 0.03 mol (available)

CH3COOH (aq) + NaOH (aq) → CH3COONa(aq) + H2O(l)

Limiting reactant is CH3COOH
Heat released = -heat absorbed by (calorimeter + soln.)

= -(C∆T + mc∆T)
= -[(37.30J oC-1×5.8oC) + (50.0 g×4.18Jg-1oC-1×5.8oC)]
= -1428.54 J
= -1.4285 kJ

0.03 mol H2O releases 1.4285 kJ
1 mol H2O releases 47.62kJ

ΔHoNeutralization =-47.62 kJ/mol

10

CHEMISTRY SK025
Chapter 9.0

THERMOCHEMISTRY

(b) i. Born-Haber cycle
5. (a)
Mg(s) + F2(l) − kJ
(b) MgF2(s)

+148 kJ +159 kJ x 2

Mg(g) 2F(g) Lattice Energy
+738 kJ -328 kJ x 2

+
Mg (g)

+1450 kJ

Mg2+(g) + 2F-(g)

ii. Lattice energy of MgF2
From cycle,
-1123 = 148 + (2 x 159) + 738 + 1450 + 2(-328) + ΔHo lattice
ΔHo lattice = (-1123) - (148 + (2 x 159) + 738 + 1450 + 2(-328)
= -3121 kJmol-1

amount of heat released
C8H18 (l) + 25 / 2O2 (g) → 8CO2 (g) + 9H2O(l)
• Eq. 1: ×9
• Eq. 2: ×8
• Eq. 3: reverse

9H2 (g) + 9 / 2O2 (g) → 9H2O(l) ∆H = - 2574 kJ
8C(s) + 8O2 (g) → 8CO2 (g) ∆H = - 3136 kJ
C8H18 (l) → 8C(s) + 9H2 (l) ∆H = + 250 kJ
C8H18 (l) + 25 / 2O2 (g) → 8CO2 (g) + 9H2O(l) ∆H = - 5460 kJ

n C8H18= 0.8772 mol

1 molC8H18 releases 5460 kJ
0.8772molC8H18 releases 4789.5 kJ

i. Name enthalpy
ΔH1 - enthalpy of atomisation of Mg
ΔH3 - second ionisation energy of Mg
ΔH7 - lattice energy of MgO

ii. Lattice energy of MgO is higher than BaO.
Reason:
• Size of Mg2+ is smaller than Ba2+.
• Ionic bond between Mg2+ and O2- are stronger than between Ba2+
and O2-.

11