JAWAPAN TUTOR, MEKA & KUMBE THERMOCHEMISTRY

ANSWER TOPIC 9
SK026

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)
H12 = 1 = .2 8
2

1 mol C5H12 ≡ - 3509.1 kJ
0.2083 mol C5H12 ≡ - 731 kJ

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.

(b) 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

(c) 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

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ANSWER TOPIC 9
SK026

THERMOCHEMISTRY

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

H = (1. )(2 1 - ) = .2
H = (1. )(1 1 - ) = .1

0.150 mol HCl ≡ 0.150 mol NaOH ≡ 0.150 mol H2O

Heat released by the reaction, qrxn -Heat absorbed by the solution,qs
=-

= -( )(4.18J -1 -1)( . )
=- 1 J

0.150 mol H2 ≡ -7315 J
1 mol H2 ≡ -48766.7 J = -48.8 kJ
H
=- -

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 )

= -(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

H ’ Lw h wh v d p d ,h h

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 Cl2 (g) ΔH = +436.0 kJ
2

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) ΔHco =-2540.2 kJ
2 O2 (g)

Equation 2:

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ANSWER TOPIC 9
SK026

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)

ΔHhydCl- = -776+4+390 = 382 kJ mol-1

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ANSWER TOPIC 9
SK026

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 - 48.8 + 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
Mg4Cl2 (s)

ANSWER TOPIC 9
SK026

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

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ANSWER TOPIC 9
SK026

THERMOCHEMISTRY

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 = 44 g mol1 = 0.227 mol

3. (a) 1 mol C3H8 release 220.1 kJ
(b) 0.227 mol C3H8 release 50.023 kJ

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

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

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

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ANSWER TOPIC 9
SK026

THERMOCHEMISTRY

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

= -2717 J

H ( q) + H ( q) → ( q) + H2O (l)

1 mol of H≡ 1 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 + wcw )
=-[ (923 J K−1)(5.47C) + (800 g)(4.18 Jg1C1)(5.47C)]

=- 23340.49 J

3.0 g = 0.0117 mol
 n of S =
256.56 g mol1

. 11 S ≡ - 23340.49 J
1 S ≡ - 1996078 J

Hc = - 1996 kJmol−

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

Enthalpy of reaction
2HCl(g)  F2 (g)  2HF(l)  Cl2 (g)

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

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2HCl(g)  1/ 2O2 (g)  H2O(l)  Cl2 (g) ANSWER TOPIC 9
H2 (g)  F2 (g)  2HF(l) SK026
H2O(l)  H2 (g)  1 / 2O2 (g)
2HCl(g)  F2 (g)  2HF(l)  Cl2 (g) THERMOCHEMISTRY

H = - 74.2 kJ

H = - 1200.0 kJ

H = + 28 .8 kJ

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 + (- 1.6 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;
 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+
U H ’ L w,

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ANSWER TOPIC 9
SK026

THERMOCHEMISTRY

The flow of reaction is start from BaCl2(s) to Ba2+(aq) and Cl−(aq).

Hsoln = L y + Hhyd.of Ba2+ + ( Hhyd. of Cl‒×2)
y +( Hhyd. of Cl‒×2))
Hhyd.of Ba2+ = Hsoln – (L
= (-40 kJ mol−1) – (3900 kJ mol−1 – (380 kJ mol−1×2))

= -3180 kJ mol−

Energy (kJmol-1)

Cu2+(g) + O2-(g)

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

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ANSWER TOPIC 9
SK026

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 = - mcwcw
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, = 1.126oC
Ccal. = 1769JoC-1,

 Water, , = 1.126oC
M=

heat released by reaction = -heat absorbed by (calorimeter + water)

=-( + )

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

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ANSWER TOPIC 9
SK026

THERMOCHEMISTRY

=- 3.403 x103 J

5.658 x 10-3 mol M ≡ -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 = + 26 6 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 = .8oC
 Mass solution = 50 g,

 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.)

= -( + )

= -[(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

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ANSWER TOPIC 9
SK026

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(- 28) + 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-.

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