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# 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.

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

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

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

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.

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

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

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

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
.

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

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?

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.

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

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

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

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
.

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

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?

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.

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
.

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

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.

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

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

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

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
.

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

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?

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.

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
.

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

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.

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)

?
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.

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

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

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

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
.

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

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?

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.

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
.

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

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.

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)

?
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?

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
.

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

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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