NCERT Solution - Thermodynamics Class 11 Notes | EduRev

Class 11 : NCERT Solution - Thermodynamics Class 11 Notes | EduRev

 Page 1


  CHAPTER-6 NCERT SOLUTIONS 
 
Question 6.1: 
Choose the correct answer. A thermodynamic state function is a quantity 
(i) used to determine heat changes 
(ii) whose value is independent of path 
(iii) used to determine pressure volume work 
(iv) whose value depends on temperature only. 
 
Answer: 
A thermodynamic state function is a quantity whose value is independent of a path. 
Functions like p, V, T etc. depend only on the state of a system and not on the path. 
Hence, alternative (ii) is correct. 
 
Question 6.2: 
For the process to occur under adiabatic conditions, the correct condition is: 
(i) ?T = 0 
(ii) ?p = 0 
(iii) q = 0 
(iv) w = 0 
 
Answer: 
A system is said to be under adiabatic conditions if there is no exchange of heat between the 
system and its surroundings. Hence, under adiabatic conditions, q = 0. 
Therefore, alternative (iii) is correct. 
 
Question 6.3: 
The enthalpies of all elements in their standard states are: 
(i) unity 
(ii) zero 
(iii) < 0 
(iv) different for each element 
 
Answer: 
The enthalpy of all elements in their standard state is zero. 
Therefore, alternative (ii) is correct. 
 
Question 6.4: 
?U
?
of combustion of methane is – X kJ mol
–1
. The value of ?H
?
 is 
(i) = ?U
?
 
(ii) > ?U
?
 
(iii) < ?U
?
 
(iv) = 0 
 
Answer: 
Since ?H
?
 = ?U
?
 + ?n
g
RT and ?U
?
 = –X kJ mol
–1
, 
?H
?
 = (–X) + ?n
g
RT. 
? ?H
?
 < ?U
?
 
Therefore, alternative (iii) is correct. 
 
Question 6.5: 
The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, –890.3 kJ mol
–
1
 –393.5 kJ mol
–1
, and –285.8 kJ mol
–1
 respectively. Enthalpy of formation of CH
4(g)
 will be 
(i) –74.8 kJ mol
–1
             (ii) –52.27 kJ mol
–1
 
(iii) +74.8 kJ mol
–1
           (iv) +52.26 kJ mol
–1
. 
 
Page 2


  CHAPTER-6 NCERT SOLUTIONS 
 
Question 6.1: 
Choose the correct answer. A thermodynamic state function is a quantity 
(i) used to determine heat changes 
(ii) whose value is independent of path 
(iii) used to determine pressure volume work 
(iv) whose value depends on temperature only. 
 
Answer: 
A thermodynamic state function is a quantity whose value is independent of a path. 
Functions like p, V, T etc. depend only on the state of a system and not on the path. 
Hence, alternative (ii) is correct. 
 
Question 6.2: 
For the process to occur under adiabatic conditions, the correct condition is: 
(i) ?T = 0 
(ii) ?p = 0 
(iii) q = 0 
(iv) w = 0 
 
Answer: 
A system is said to be under adiabatic conditions if there is no exchange of heat between the 
system and its surroundings. Hence, under adiabatic conditions, q = 0. 
Therefore, alternative (iii) is correct. 
 
Question 6.3: 
The enthalpies of all elements in their standard states are: 
(i) unity 
(ii) zero 
(iii) < 0 
(iv) different for each element 
 
Answer: 
The enthalpy of all elements in their standard state is zero. 
Therefore, alternative (ii) is correct. 
 
Question 6.4: 
?U
?
of combustion of methane is – X kJ mol
–1
. The value of ?H
?
 is 
(i) = ?U
?
 
(ii) > ?U
?
 
(iii) < ?U
?
 
(iv) = 0 
 
Answer: 
Since ?H
?
 = ?U
?
 + ?n
g
RT and ?U
?
 = –X kJ mol
–1
, 
?H
?
 = (–X) + ?n
g
RT. 
? ?H
?
 < ?U
?
 
Therefore, alternative (iii) is correct. 
 
Question 6.5: 
The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, –890.3 kJ mol
–
1
 –393.5 kJ mol
–1
, and –285.8 kJ mol
–1
 respectively. Enthalpy of formation of CH
4(g)
 will be 
(i) –74.8 kJ mol
–1
             (ii) –52.27 kJ mol
–1
 
(iii) +74.8 kJ mol
–1
           (iv) +52.26 kJ mol
–1
. 
 
Answer: 
According to the question, 
 
Thus, the desired equation is the one that represents the formation of CH
4
 
(g)
 i.e., 
 
 
 
Enthalpy of formation of CH
4(g) 
= –74.8 kJ mol
–1
 
Hence, alternative (i) is correct. 
 
Question 6.6: 
A reaction, A + B ? C + D + q is found to have a positive entropy change. The reaction will be 
(i) possible at high temperature 
(ii) possible only at low temperature 
(iii) not possible at any temperature 
(iv) possible at any temperature 
 
Answer: 
For a reaction to be spontaneous, ?G should be negative. 
?G = ?H – T?S 
According to the question, for the given reaction, 
?S = positive 
?H = negative (since heat is evolved) 
? ?G = negative 
Therefore, the reaction is spontaneous at any temperature. 
Hence, alternative (iv) is correct. 
 
Question 6.7: 
In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. 
What is the change in internal energy for the process? 
 
Answer: 
According to the first law of thermodynamics, 
?U = q + W (i) 
Where, 
?U = change in internal energy for a process 
q = heat 
W = work 
Given, 
q = + 701 J (Since heat is absorbed) 
W = –394 J (Since work is done by the system) 
Substituting the values in expression (i), we get 
?U = 701 J + (–394 J) 
?U = 307 J 
Page 3


  CHAPTER-6 NCERT SOLUTIONS 
 
Question 6.1: 
Choose the correct answer. A thermodynamic state function is a quantity 
(i) used to determine heat changes 
(ii) whose value is independent of path 
(iii) used to determine pressure volume work 
(iv) whose value depends on temperature only. 
 
Answer: 
A thermodynamic state function is a quantity whose value is independent of a path. 
Functions like p, V, T etc. depend only on the state of a system and not on the path. 
Hence, alternative (ii) is correct. 
 
Question 6.2: 
For the process to occur under adiabatic conditions, the correct condition is: 
(i) ?T = 0 
(ii) ?p = 0 
(iii) q = 0 
(iv) w = 0 
 
Answer: 
A system is said to be under adiabatic conditions if there is no exchange of heat between the 
system and its surroundings. Hence, under adiabatic conditions, q = 0. 
Therefore, alternative (iii) is correct. 
 
Question 6.3: 
The enthalpies of all elements in their standard states are: 
(i) unity 
(ii) zero 
(iii) < 0 
(iv) different for each element 
 
Answer: 
The enthalpy of all elements in their standard state is zero. 
Therefore, alternative (ii) is correct. 
 
Question 6.4: 
?U
?
of combustion of methane is – X kJ mol
–1
. The value of ?H
?
 is 
(i) = ?U
?
 
(ii) > ?U
?
 
(iii) < ?U
?
 
(iv) = 0 
 
Answer: 
Since ?H
?
 = ?U
?
 + ?n
g
RT and ?U
?
 = –X kJ mol
–1
, 
?H
?
 = (–X) + ?n
g
RT. 
? ?H
?
 < ?U
?
 
Therefore, alternative (iii) is correct. 
 
Question 6.5: 
The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, –890.3 kJ mol
–
1
 –393.5 kJ mol
–1
, and –285.8 kJ mol
–1
 respectively. Enthalpy of formation of CH
4(g)
 will be 
(i) –74.8 kJ mol
–1
             (ii) –52.27 kJ mol
–1
 
(iii) +74.8 kJ mol
–1
           (iv) +52.26 kJ mol
–1
. 
 
Answer: 
According to the question, 
 
Thus, the desired equation is the one that represents the formation of CH
4
 
(g)
 i.e., 
 
 
 
Enthalpy of formation of CH
4(g) 
= –74.8 kJ mol
–1
 
Hence, alternative (i) is correct. 
 
Question 6.6: 
A reaction, A + B ? C + D + q is found to have a positive entropy change. The reaction will be 
(i) possible at high temperature 
(ii) possible only at low temperature 
(iii) not possible at any temperature 
(iv) possible at any temperature 
 
Answer: 
For a reaction to be spontaneous, ?G should be negative. 
?G = ?H – T?S 
According to the question, for the given reaction, 
?S = positive 
?H = negative (since heat is evolved) 
? ?G = negative 
Therefore, the reaction is spontaneous at any temperature. 
Hence, alternative (iv) is correct. 
 
Question 6.7: 
In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. 
What is the change in internal energy for the process? 
 
Answer: 
According to the first law of thermodynamics, 
?U = q + W (i) 
Where, 
?U = change in internal energy for a process 
q = heat 
W = work 
Given, 
q = + 701 J (Since heat is absorbed) 
W = –394 J (Since work is done by the system) 
Substituting the values in expression (i), we get 
?U = 701 J + (–394 J) 
?U = 307 J 
Hence, the change in internal energy for the given process is 307 J. 
 
Question 6.8: 
The reaction of cyanamide, NH
2
CN
(s),
with dioxygen was carried out in a bomb calorimeter, and 
?Uwas found to be –742.7 kJ mol
–1
at 298 K. Calculate enthalpy change for the reaction at 298 
K. 
 
 
Answer: 
Enthalpy change for a reaction (?H) is given by the expression, 
?H = ?U + ?n
g
RT 
Where, 
?U = change in internal energy 
?n
g
 = change in number of moles 
For the given reaction, 
?n
g
 = ?n
g
 (products) – ?n
g
 (reactants) 
= (2 – 1.5) moles 
?n
g
 = 0.5 moles 
And, 
?U = –742.7 kJ mol
–1
 
T = 298 K 
R = 8.314 × 10
–3
 kJ mol
–1
 K
–1
 
Substituting the values in the expression of ?H: 
?H = (–742.7 kJ mol
–1
) + (0.5 mol) (298 K) (8.314 × 10
–3
 kJ mol
–1
 K
–1
) 
= –742.7 + 1.2 
?H = –741.5 kJ mol
–1
 
 
Question 6.9: 
Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminium 
from 35°C to 55°C. Molar heat capacity of Al is 24 J mol
–1 
K
–1
. 
 
Answer: 
From the expression of heat (q), 
q = m. c. ?T 
Where, 
c = molar heat capacity 
m = mass of substance 
?T = change in temperature 
Substituting the values in the expression of q: 
 
q = 1066.7 J 
q = 1.07 kJ 
 
Question 6.10: 
Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at –10.0°C. 
?
fus
H = 6.03 kJ mol
–1
 at 0°C. 
C
p
[H
2
O(l)] = 75.3 J mol
–1
 K
–1
 
C
p
[H
2
O(s)] = 36.8 J mol
–1
 K
–1
 
 
Answer: 
Total enthalpy change involved in the transformation is the sum of the following changes: 
Page 4


  CHAPTER-6 NCERT SOLUTIONS 
 
Question 6.1: 
Choose the correct answer. A thermodynamic state function is a quantity 
(i) used to determine heat changes 
(ii) whose value is independent of path 
(iii) used to determine pressure volume work 
(iv) whose value depends on temperature only. 
 
Answer: 
A thermodynamic state function is a quantity whose value is independent of a path. 
Functions like p, V, T etc. depend only on the state of a system and not on the path. 
Hence, alternative (ii) is correct. 
 
Question 6.2: 
For the process to occur under adiabatic conditions, the correct condition is: 
(i) ?T = 0 
(ii) ?p = 0 
(iii) q = 0 
(iv) w = 0 
 
Answer: 
A system is said to be under adiabatic conditions if there is no exchange of heat between the 
system and its surroundings. Hence, under adiabatic conditions, q = 0. 
Therefore, alternative (iii) is correct. 
 
Question 6.3: 
The enthalpies of all elements in their standard states are: 
(i) unity 
(ii) zero 
(iii) < 0 
(iv) different for each element 
 
Answer: 
The enthalpy of all elements in their standard state is zero. 
Therefore, alternative (ii) is correct. 
 
Question 6.4: 
?U
?
of combustion of methane is – X kJ mol
–1
. The value of ?H
?
 is 
(i) = ?U
?
 
(ii) > ?U
?
 
(iii) < ?U
?
 
(iv) = 0 
 
Answer: 
Since ?H
?
 = ?U
?
 + ?n
g
RT and ?U
?
 = –X kJ mol
–1
, 
?H
?
 = (–X) + ?n
g
RT. 
? ?H
?
 < ?U
?
 
Therefore, alternative (iii) is correct. 
 
Question 6.5: 
The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, –890.3 kJ mol
–
1
 –393.5 kJ mol
–1
, and –285.8 kJ mol
–1
 respectively. Enthalpy of formation of CH
4(g)
 will be 
(i) –74.8 kJ mol
–1
             (ii) –52.27 kJ mol
–1
 
(iii) +74.8 kJ mol
–1
           (iv) +52.26 kJ mol
–1
. 
 
Answer: 
According to the question, 
 
Thus, the desired equation is the one that represents the formation of CH
4
 
(g)
 i.e., 
 
 
 
Enthalpy of formation of CH
4(g) 
= –74.8 kJ mol
–1
 
Hence, alternative (i) is correct. 
 
Question 6.6: 
A reaction, A + B ? C + D + q is found to have a positive entropy change. The reaction will be 
(i) possible at high temperature 
(ii) possible only at low temperature 
(iii) not possible at any temperature 
(iv) possible at any temperature 
 
Answer: 
For a reaction to be spontaneous, ?G should be negative. 
?G = ?H – T?S 
According to the question, for the given reaction, 
?S = positive 
?H = negative (since heat is evolved) 
? ?G = negative 
Therefore, the reaction is spontaneous at any temperature. 
Hence, alternative (iv) is correct. 
 
Question 6.7: 
In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. 
What is the change in internal energy for the process? 
 
Answer: 
According to the first law of thermodynamics, 
?U = q + W (i) 
Where, 
?U = change in internal energy for a process 
q = heat 
W = work 
Given, 
q = + 701 J (Since heat is absorbed) 
W = –394 J (Since work is done by the system) 
Substituting the values in expression (i), we get 
?U = 701 J + (–394 J) 
?U = 307 J 
Hence, the change in internal energy for the given process is 307 J. 
 
Question 6.8: 
The reaction of cyanamide, NH
2
CN
(s),
with dioxygen was carried out in a bomb calorimeter, and 
?Uwas found to be –742.7 kJ mol
–1
at 298 K. Calculate enthalpy change for the reaction at 298 
K. 
 
 
Answer: 
Enthalpy change for a reaction (?H) is given by the expression, 
?H = ?U + ?n
g
RT 
Where, 
?U = change in internal energy 
?n
g
 = change in number of moles 
For the given reaction, 
?n
g
 = ?n
g
 (products) – ?n
g
 (reactants) 
= (2 – 1.5) moles 
?n
g
 = 0.5 moles 
And, 
?U = –742.7 kJ mol
–1
 
T = 298 K 
R = 8.314 × 10
–3
 kJ mol
–1
 K
–1
 
Substituting the values in the expression of ?H: 
?H = (–742.7 kJ mol
–1
) + (0.5 mol) (298 K) (8.314 × 10
–3
 kJ mol
–1
 K
–1
) 
= –742.7 + 1.2 
?H = –741.5 kJ mol
–1
 
 
Question 6.9: 
Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminium 
from 35°C to 55°C. Molar heat capacity of Al is 24 J mol
–1 
K
–1
. 
 
Answer: 
From the expression of heat (q), 
q = m. c. ?T 
Where, 
c = molar heat capacity 
m = mass of substance 
?T = change in temperature 
Substituting the values in the expression of q: 
 
q = 1066.7 J 
q = 1.07 kJ 
 
Question 6.10: 
Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at –10.0°C. 
?
fus
H = 6.03 kJ mol
–1
 at 0°C. 
C
p
[H
2
O(l)] = 75.3 J mol
–1
 K
–1
 
C
p
[H
2
O(s)] = 36.8 J mol
–1
 K
–1
 
 
Answer: 
Total enthalpy change involved in the transformation is the sum of the following changes: 
(a) Energy change involved in the transformation of 1 mol of water at 10°C to 1 mol of water at 
0°C. 
(b) Energy change involved in the transformation of 1 mol of water at 0° to 1 mol of ice at 0°C. 
(c) Energy change involved in the transformation of 1 mol of ice at 0°C to 1 mol of ice at –10°C. 
 
= (75.3 J mol
–1
 K
–1
) (0 – 10)K + (–6.03 × 10
3
 J mol
–1
) + (36.8 J mol
–1
 K
–1
) (–10 – 0)K 
= –753 J mol
–1
 – 6030 J mol
–1
 – 368 J mol
–1
 
= –7151 J mol
–1
 
= –7.151 kJ mol
–1
 
Hence, the enthalpy change involved in the transformation is –7.151 kJ mol
–1
. 
 
Question 6.11: 
Enthalpy of combustion of carbon to CO
2
 is –393.5 kJ mol
–1
. Calculate the heat released upon 
formation of 35.2 g of CO
2
 from carbon and dioxygen gas. 
 
Answer: 
Formation of CO
2
 from carbon and dioxygen gas can be represented as: 
 
(1 mole = 44 g) 
Heat released on formation of 44 g CO
2
 = –393.5 kJ mol
–1
 
Heat released on formation of 35.2 g CO
2
 
 
= –314.8 kJ mol
–1
 
 
Question 6.12: 
Enthalpies of formation of CO
(g), 
CO
2(g), 
N
2
O
(g) 
and N
2
O
4(g) 
are –110 kJ mol
–1
, – 393 kJ mol
–1
, 81 
kJ mol
–1
 and 9.7 kJ mol
–1
 respectively. Find the value of ?
r
H for the reaction: 
N
2
O
4(g)
 + 3CO
(g)
 N
2
O
(g)
 + 3CO
2(g)
 
 
Answer: 
?
r
H for a reaction is defined as the difference between ?
f
H value of products and ?
f
H value of 
reactants. 
 
For the given reaction, 
N
2
O
4(g)
 + 3CO
(g)
  N
2
O
(g)
 + 3CO
2(g)
 
 
Substituting the values of ?
f
H for N
2
O, CO
2
, N
2
O
4,
 and CO from the question, we get: 
 
Hence, the value of ?
r
H for the reaction is . 
 
Question 6.13: 
Given 
; ?
r
H
?
 = –92.4 kJ mol
–1
 
Page 5


  CHAPTER-6 NCERT SOLUTIONS 
 
Question 6.1: 
Choose the correct answer. A thermodynamic state function is a quantity 
(i) used to determine heat changes 
(ii) whose value is independent of path 
(iii) used to determine pressure volume work 
(iv) whose value depends on temperature only. 
 
Answer: 
A thermodynamic state function is a quantity whose value is independent of a path. 
Functions like p, V, T etc. depend only on the state of a system and not on the path. 
Hence, alternative (ii) is correct. 
 
Question 6.2: 
For the process to occur under adiabatic conditions, the correct condition is: 
(i) ?T = 0 
(ii) ?p = 0 
(iii) q = 0 
(iv) w = 0 
 
Answer: 
A system is said to be under adiabatic conditions if there is no exchange of heat between the 
system and its surroundings. Hence, under adiabatic conditions, q = 0. 
Therefore, alternative (iii) is correct. 
 
Question 6.3: 
The enthalpies of all elements in their standard states are: 
(i) unity 
(ii) zero 
(iii) < 0 
(iv) different for each element 
 
Answer: 
The enthalpy of all elements in their standard state is zero. 
Therefore, alternative (ii) is correct. 
 
Question 6.4: 
?U
?
of combustion of methane is – X kJ mol
–1
. The value of ?H
?
 is 
(i) = ?U
?
 
(ii) > ?U
?
 
(iii) < ?U
?
 
(iv) = 0 
 
Answer: 
Since ?H
?
 = ?U
?
 + ?n
g
RT and ?U
?
 = –X kJ mol
–1
, 
?H
?
 = (–X) + ?n
g
RT. 
? ?H
?
 < ?U
?
 
Therefore, alternative (iii) is correct. 
 
Question 6.5: 
The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, –890.3 kJ mol
–
1
 –393.5 kJ mol
–1
, and –285.8 kJ mol
–1
 respectively. Enthalpy of formation of CH
4(g)
 will be 
(i) –74.8 kJ mol
–1
             (ii) –52.27 kJ mol
–1
 
(iii) +74.8 kJ mol
–1
           (iv) +52.26 kJ mol
–1
. 
 
Answer: 
According to the question, 
 
Thus, the desired equation is the one that represents the formation of CH
4
 
(g)
 i.e., 
 
 
 
Enthalpy of formation of CH
4(g) 
= –74.8 kJ mol
–1
 
Hence, alternative (i) is correct. 
 
Question 6.6: 
A reaction, A + B ? C + D + q is found to have a positive entropy change. The reaction will be 
(i) possible at high temperature 
(ii) possible only at low temperature 
(iii) not possible at any temperature 
(iv) possible at any temperature 
 
Answer: 
For a reaction to be spontaneous, ?G should be negative. 
?G = ?H – T?S 
According to the question, for the given reaction, 
?S = positive 
?H = negative (since heat is evolved) 
? ?G = negative 
Therefore, the reaction is spontaneous at any temperature. 
Hence, alternative (iv) is correct. 
 
Question 6.7: 
In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. 
What is the change in internal energy for the process? 
 
Answer: 
According to the first law of thermodynamics, 
?U = q + W (i) 
Where, 
?U = change in internal energy for a process 
q = heat 
W = work 
Given, 
q = + 701 J (Since heat is absorbed) 
W = –394 J (Since work is done by the system) 
Substituting the values in expression (i), we get 
?U = 701 J + (–394 J) 
?U = 307 J 
Hence, the change in internal energy for the given process is 307 J. 
 
Question 6.8: 
The reaction of cyanamide, NH
2
CN
(s),
with dioxygen was carried out in a bomb calorimeter, and 
?Uwas found to be –742.7 kJ mol
–1
at 298 K. Calculate enthalpy change for the reaction at 298 
K. 
 
 
Answer: 
Enthalpy change for a reaction (?H) is given by the expression, 
?H = ?U + ?n
g
RT 
Where, 
?U = change in internal energy 
?n
g
 = change in number of moles 
For the given reaction, 
?n
g
 = ?n
g
 (products) – ?n
g
 (reactants) 
= (2 – 1.5) moles 
?n
g
 = 0.5 moles 
And, 
?U = –742.7 kJ mol
–1
 
T = 298 K 
R = 8.314 × 10
–3
 kJ mol
–1
 K
–1
 
Substituting the values in the expression of ?H: 
?H = (–742.7 kJ mol
–1
) + (0.5 mol) (298 K) (8.314 × 10
–3
 kJ mol
–1
 K
–1
) 
= –742.7 + 1.2 
?H = –741.5 kJ mol
–1
 
 
Question 6.9: 
Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminium 
from 35°C to 55°C. Molar heat capacity of Al is 24 J mol
–1 
K
–1
. 
 
Answer: 
From the expression of heat (q), 
q = m. c. ?T 
Where, 
c = molar heat capacity 
m = mass of substance 
?T = change in temperature 
Substituting the values in the expression of q: 
 
q = 1066.7 J 
q = 1.07 kJ 
 
Question 6.10: 
Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at –10.0°C. 
?
fus
H = 6.03 kJ mol
–1
 at 0°C. 
C
p
[H
2
O(l)] = 75.3 J mol
–1
 K
–1
 
C
p
[H
2
O(s)] = 36.8 J mol
–1
 K
–1
 
 
Answer: 
Total enthalpy change involved in the transformation is the sum of the following changes: 
(a) Energy change involved in the transformation of 1 mol of water at 10°C to 1 mol of water at 
0°C. 
(b) Energy change involved in the transformation of 1 mol of water at 0° to 1 mol of ice at 0°C. 
(c) Energy change involved in the transformation of 1 mol of ice at 0°C to 1 mol of ice at –10°C. 
 
= (75.3 J mol
–1
 K
–1
) (0 – 10)K + (–6.03 × 10
3
 J mol
–1
) + (36.8 J mol
–1
 K
–1
) (–10 – 0)K 
= –753 J mol
–1
 – 6030 J mol
–1
 – 368 J mol
–1
 
= –7151 J mol
–1
 
= –7.151 kJ mol
–1
 
Hence, the enthalpy change involved in the transformation is –7.151 kJ mol
–1
. 
 
Question 6.11: 
Enthalpy of combustion of carbon to CO
2
 is –393.5 kJ mol
–1
. Calculate the heat released upon 
formation of 35.2 g of CO
2
 from carbon and dioxygen gas. 
 
Answer: 
Formation of CO
2
 from carbon and dioxygen gas can be represented as: 
 
(1 mole = 44 g) 
Heat released on formation of 44 g CO
2
 = –393.5 kJ mol
–1
 
Heat released on formation of 35.2 g CO
2
 
 
= –314.8 kJ mol
–1
 
 
Question 6.12: 
Enthalpies of formation of CO
(g), 
CO
2(g), 
N
2
O
(g) 
and N
2
O
4(g) 
are –110 kJ mol
–1
, – 393 kJ mol
–1
, 81 
kJ mol
–1
 and 9.7 kJ mol
–1
 respectively. Find the value of ?
r
H for the reaction: 
N
2
O
4(g)
 + 3CO
(g)
 N
2
O
(g)
 + 3CO
2(g)
 
 
Answer: 
?
r
H for a reaction is defined as the difference between ?
f
H value of products and ?
f
H value of 
reactants. 
 
For the given reaction, 
N
2
O
4(g)
 + 3CO
(g)
  N
2
O
(g)
 + 3CO
2(g)
 
 
Substituting the values of ?
f
H for N
2
O, CO
2
, N
2
O
4,
 and CO from the question, we get: 
 
Hence, the value of ?
r
H for the reaction is . 
 
Question 6.13: 
Given 
; ?
r
H
?
 = –92.4 kJ mol
–1
 
What is the standard enthalpy of formation of NH
3
 gas? 
 
Answer: 
Standard enthalpy of formation of a compound is the change in enthalpy that takes place during 
the formation of 1 mole of a substance in its standard form from its constituent elements in their 
standard state. 
Re-writing the given equation for 1 mole of NH
3(g),
 
 
Standard enthalpy of formation of NH
3(g)
 
= ½ ?
r
H
?
 
= ½ (–92.4 kJ mol
–1
) 
= –46.2 kJ mol
–1
 
 
Question 6.14: 
Calculate the standard enthalpy of formation of CH
3
OH
(l)
 from the following data: 
CH
3
OH
(l)
 + O
2(g)
  CO
2(g)
 + 2H
2
O
(l)
 ; ?
r
H
?
 = –726 kJ mol
–1
 
C
(g)
 + O
2(g)
  CO
2(g)
 ; ?
c
H
?
 = –393 kJ mol
–1
 
H
2(g)
 + O
2(g)
  H
2
O
(l)
 ; ?
f
H
?
 = –286 kJ mol
–1
. 
 
Answer: 
The reaction that takes place during the formation of CH
3
OH
(l)
 can be written as: 
C
(s)
 + 2H
2
O
(g)
 + O
2(g)
  CH
3
OH
(l)
 (1) 
The reaction (1) can be obtained from the given reactions by following the algebraic calculations 
as: 
Equation (ii) + 2 × equation (iii) – equation (i) 
?
f
H
?
 [CH
3
OH
(l)
] = ?
c
H
? 
+ 2?
f
H
?
 [H
2
O
(l)
] – ?
r
H
?
 
= (–393 kJ mol
–1
) + 2(–286 kJ mol
–1
) – (–726 kJ mol
–1
) 
= (–393 – 572 + 726) kJ mol
–1
 
?
f
H
?
 [CH
3
OH
(l)
] = –239 kJ mol
–1
 
 
Question 6.15: 
Calculate the enthalpy change for the process 
CCl
4(g)
 ? C
(g)
 + 4Cl
(g)
 
and calculate bond enthalpy of C–Cl in CCl
4(g).
 
?
vap
H
?
 (CCl
4
) = 30.5 kJ mol
–1
. 
?
f
H
?
 (CCl
4
) = –135.5 kJ mol
–1
. 
?
a
H
?
 (C) = 715.0 kJ mol
–1
, where ?
a
H
?
 is enthalpy of atomisation 
?
a
H
?
 (Cl
2
) = 242 kJ mol
–1
 
 
Answer: 
The chemical equations implying to the given values of enthalpies are: 
?
vap
H
?
 = 30.5 kJ mol
–1
 
?
a
H
?
 = 715.0 kJ mol
–1
 
 ?
a
H
?
 = 242 kJ mol
–1
 
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