NCERT Solutions for Class 11 Chemistry Chapter 5 Thermodynamics

NCERT
SOLUTIONS
CLASS - 11th

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Class : 11th
Subject : Chemistry
Chapter : 6
Chapter Name : Thermodynamics

Q6.1 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.

Page : 189 , Block Name : Exercise

Q6.2 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,

Page : 189 , Block Name : Exercise

Q6.3 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

Page : 189 , Block Name : Exercise

Q6.4 ΔU of combustion of methane is − X kJ mol
⊖ −1 ⊖
. The value of ΔH is

(i) = ΔU ⊖

(ii) > ΔU⊖

(iii) < = ΔU ⊖

(iv) = 0

Answer. Since ΔH θ θ θ −1
= ΔU + Δng RT and ΔU = −Xk3mol
θ
ΔH = (−X) + Δng RT
θ θ
⇒ △H < ΔU

Therefore, alternative (iii) is correct.

Page : 189 , Block Name : Exercise

Q6.5 The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, –890.3
respectively. Enthalpy of formation of CH4 will be
−1 −1 −1
kJ mol − 393.5kJmol , and − 285.8kJmol

(i) −74.8kJmol −1

(ii) −52.27kJmol −1

(iii) +74.8kJmol
−1

(iv) +52.26kJmol
−1

Answer. According to the question,
CH4(g) + 2O2(g) ⟶ CO2(z) + 2H2 O(g)
(i) −1
ΔH = −890.3kJmol

C(x) + O2(y) ⟶CO2(g)
(ii) −1
ΔH = −393.5kJmol

(iii) 2H 2(g) + O2(z) ⟶ 2H2 O(g)
−1
ΔH = −285.8kJmol

Thus, the desired equation is the one that represents the formation of CH4(g) i.e..,

−1
∴ Enthalpy of formation of CH4(g) = −74.8kJmol

Hence, alternative (i) is correct.

Page : 189 , Block Name : Exercise

Q6.6 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, ΔG should be negative.

ΔG = ΔH − T ΔS

According to the question, for the given reaction,

ΔS = positive
Answer.
ΔH = negative (since heat Is evolved)

⇒ ΔG = negative

Therefore, the reaction is spontaneous at any temperature.

Hence, alternative (iv) is correct.

Page : 189 , Block Name : Exercise

Q6.7 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 rst of thermodynamics,

Where,
ΔU = change in internal energy for a process

g = 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
ΔU = 701J + (−394J)

ΔU = 307J

Hence, the charge in Internal energy for the given process is 307 J.

Page : 189 , Block Name : Exercise

Q6.8 The reaction of cyanamide,NH CN(s) with dioxygen was carried out in a bomb calorimeter, and
2

∆U was found to be –742.7 kJmol at 298 K. Calculate enthalpy change for the reaction at 298 K.
−1

3
NH2 CN(g) + O2 (g) → N2 (g) + CO2 (g) + H2 O(l)
2

Answer. Enthalpy change for a reaction (ΔH) is given by the expression,
ΔH = ΔU + Δng RT

Where,
ΔU = change in internal energy

Δηg = change in number of moles

For the given reaction,
Δng = ∑ ng ( products ) − Σng ( reactants )

= (2 − 2.5) moles

Δne = −0.5 moles

And,

−1
ΔU = −742.7kJmol

T = 298K

−3 −1 −1
R = 8.314 × 10 kJmol K

Substituting the values in the expression of ΔH:
−1 −3 −1 −1
ΔH = (−742.7kJmol ) + (−0.5mol)(298K) (8.314 × 10 kJmol K )

= −742.7 − 1.2

−1
WH = −743.9kJmol

Page : 189 , Block Name : Exercise

Q6.9 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 K . −1 −1

From the expression of heat (q)

q = m ⋅ c. ΔT

Where,

Answer. c = molar heat capacity
m = mass of substance

ΔT = change in temperature

Substituting the values in the expression of q :

60
−1 −1
q = ( mol) (24Jmol K ) (20K)
27

q = 1066.7J

q = 1.07k

Page : 190 , Block Name : Exercise

Q6.10 Calculate the enthalpy change on freezing of 1.0 mol of water at10.0°C to ice at –10.0°C.
−1 ∘
Δf us H = 6.03kJmol at 0 C
−1 −1
Cp [H2 O(I)] = 75.3Jmol K

−1 −1
Cρ [H2 O(s)] = 36.8Jmol K

Answer. Total enthalpy change involved in the transformation is the 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.
∘ ∘

ΔH = Cp [H2 OCl] ΔT + ΔH fivering + Cρ [H2 O(s) ] ΔT

−1 −1 3 −1 −1 −1
= (75.3]mol K )(0 − 10)K + (−6.03 × 10 Jmol ) + (36.8]mol K )(−10 − 0)K

−1 −1 −1
= −7533mol − 6030Jmol − 368Jmol
−1
= −7151Jmol

−1
= −7.151kJmol

Hence, the enthalpy change involved in the transformation is −7.151kJmol .
−1

Page : 190 , Block Name : Exercise

Q6.11 Enthalpy of combustion of carbon to CO2 is –393.5 −7.151kJmol . Calculate the heat released
−1

upon formation of 35.2 g of CO from carbon and dioxygen gas.
2

Answer. Formation of CO from carbon and dioxygen gas can be represented as:
2

C(s) + O2(g) ⟶ CO2(g)
−1
Δf H = −393.5kJmol

(1 mole = 44g)

−1
Heat released on formation of 44gCO2 = −393.5kJmol

∴ Heat released on formation of 35.2gCO2
−1
= −314.8kJmol

Page : 190 , Block Name : Exercise

Q6.12 Enthalpies of formation of CO(g), CO (g), N O(g) and N O (g) are − 110, −393, 81 and 9.7
2 2 2 4

respectively. Find the value of ∆H for the reaction:
−1
kJmol

N2 O4 (g) + 3CO(g) → N2 O (g + 3CO2 (g]

Δr H for a reaction is defined as the difference between ΔH value of products and ΔH
Answer.
value of reactants.

Δ, H = ∑ Δ, H ( products ) − ∑ Δf H ( reactants )

For the given reaction,

N2 O4(g) + 3CO(g) ⟶ N2 O(g) + 3CO2(g)

Δr H = [{Δf H (N2 O) + 3ΔJ H (CO2 )} − {Δf H (N2 O4 ) + 3Δj H (CO)}]

Substituting the values of ΔH for N O, CO , N O
2 2 2 4 , and CO

From the question, we get:
−1 −1 −1 −1
Δr H = [{81kJmol + 3(−393)kJmol } − {9.7kJmol + 3(−110)kJmol }]

−1
Δr H = −7777kJmol

Hence, the value of Δ H for the reaction is −777.7kJmol
r
−1
.

Page : 190 , Block Name : Exercise

Q6.13 Given
θ −1
N2(g) + 3H2(g) ⟶ 2NH3(y) ; Δr H = −92.4kJmol

What is the standard enthalpy of formation of NH gas? 3

Answer. Standard of formation of a compound is the charge 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)

1 3
N2(g) + H2(g) ⟶ NH3(g)
2 2

∴ Standard enthalpy of formation of NH3(g)

θ
= 1/2Δr H

−1
= 1/2 (−92.4kJmol )

−1
= −46.2kJmol

Page : 190 , Block Name : Exercise

Q6.14 Calculate the standard enthalpy of formation of CH OH(l) from the following data:
3

3 ∘ −1
CH3 OH(l) + O2 (g] → CO2 (g) + 2H2 O(l) : Δ, H = −726kJmol
2
−1
C(graphite) +O2 (g) → CO2 (g] : Δe H = −393kJmol
1 −1
H2 (g) + O2 (g) → H2 O(1); Δ, H = −286kJmol
2

Answer. The reaction that takes place during the formation of CH OH 3 (l) can be written as:
1
C(s) + 2H2 O(g) + O2(G) , ⟶ CH3 OH(η) (1)
2

The reaction (I) can be obtained from the given reactions by following the algebraic calculations as:

Page : 190 , Block Name : Exercise

Calculate the enthalpy change for the process
Q6.15
CCl4(g) → C(g) + 4Cl(g)

and calculated bond enthalpy of C − Cl in CCl 4(g)

θ −1
Δvap H (C C| ) = 30.5kJmol
4

θ −1
Δf H (CCl4 ) = −135.5kJmol

is enthalpy of atomisation
θ −1 θ
Δa H (C) = 715.0kJmol , where Δa H
θ −1
Δ2 H (Cl2 ) = 242kJmol

Answer. The chemical equations imply to the given values of enthalpies are:

Page : 190 , Block Name : Exercise

Q6.16 For an isolated system, ∆U = 0, what will be ∆S ?

Answer. ∆S will be positive i.e., greater than zero
Since ∆U = 0, ∆S will be positive and the reaction will be spontaneous.

Page : 190 , Block Name : Exercise

Q6.17 For the reaction at 298 K,
2A + B → C
∆H = 400 kJmol and ∆S = 0.2 kJmol
−1 −1

At what temperature will the reaction become spontaneous considering ∆H and ∆S to be constant over
the temperature range.

Answer. From the expression
ΔG = ΔH − T ΔS

Assuming the reaction at equilibrium, δ T for the reaction would be:
1
T = (ΔH − ΔG)
ΔS

ΔH
= (ΔG = 0 at equilibrium)
ΔS

−1
400kJmol
=
−1 −1
0.2kJK mol

T = 2000K

For the reaction to be spontaneous, ΔG must be negative. Hence, for the given reaction to be

spontaneous, T should be greater than 2000 K.

Page : 190 , Block Name : Exercise

Q6.18 For the reaction, 2Cl (g)
→ Cl2(g) , what are the signs of ∆H and ∆S ?

Answer. ∆H and ∆S are negative
The given reaction represents the formation of chlorine molecule from chlorine atoms. Here, bond
formation is taking place. Therefore, energy Is being released. Hence ∆H is negative.
Also, two moles of atoms have more randomness than one mole of a molecule. Since spontaneity is
decreased, ∆S is negative for the given reaction.

Page : 190 , Block Name : Exercise

Q6.19 For the reaction
2A(g) + B(g) → 2D(g)

e ∘ −1
ΔU = −10.5kJ and ΔS = −44.1JK

Calculate ΔG for the reaction, and predict whether the reaction may occur spontaneously
⊖

Answer. For the given reaction,
2A(g) + B(g) → 2D(g)

Δηg = 2 − (3)

= −1 mole

Substituting the value of ΔU in the expression of ΔH:
θ

θ θ
ΔH = ΔU + Δng RT

−3 −1 −1
= (−10.5kJ ) − (−1) (8.314 × 10 kJK mol ) (298K)

= −10.5kJ − 2.48kJ
⊖
ΔH = −12.98kJ

Substituting the values of ΔH ⊖
and ΔS
⊖
in the expression of ΔG
⊖

θ θ θ
ΔG = △H − T ΔS

−1
= −12.98kJ − (298K) (−44.1JK )

= −12.98kJ + 13.14kJ
⊖
ΔG = +0.16kJ

Since ΔG for the reaction is positive, the reaction will not occur spontaneously.
θ

Page : 190 , Block Name : Exercise

Q6.20 The equilibrium constant for a reaction is 10. What will be the value of ∆G ? R = 8.314 ⊖

T = 300 K.
−1 −1
JK mol

From the expression,
Answer. θ
ΔG = −2.303 RT logk
eq

θ
ΔG for the reaction,

−1 −1
= (2.303) (8.314JK mol ) (300K) log 10

−1
= −5744.14Jmol

−1
= −5.744kkmol

Page : 191 , Block Name : Exercise

Q6.21 Comment on the thermodynamic stability of NO g), given (

1 1 ⊖ −1
N2 (g) + O2 (g) → NO(g) ; Δr H = 90kJmol
2 2

1 e −1
NO(g) + O2 (g) → NO2 (g) : Δr H = −74kJmol
2

Answer. The positive value of Δ H indicates that heat is absorbed during the formation of NO(g), j. This
r

means that NO(g) has higher than the reactants (N and O ). 2 2

Hence, NO(g) is unstable.
The negative value of Δ H H indicates that heat is evolved during the formation of NO
r from 2(g)

NO(g) and O2(g) . The product, NO 2(g) is stabilized with minimum energy.
Hence, unstable NO (g) changes to unstable NO 2(g) .

Page : 191 , Block Name : Exercise

Q6.22 Calculate the entropy change in surroundings when 1.00 mol of H O 2 (l)
is formed under standard
conditions. ΔH θ −1
= −286kJmol

Answer. It is given that 286 kJ mol of heat is evolved the formation of 1 mol of H O . Thus, an equal
−1
2 (l)

amount of heat will be absorbed by the surroundings.
−1
qsurr = +286kJmol

Entropy change (ΔS for the surroundings =
qsurr

surr )
7
−1
286kJmol
=
298k

−1 −1
∴ ΔSsurt = 959.73Jmol K

Page : 191 , Block Name : Exercise