NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 exams must go through NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics . Going through the solutions provided on this page will help you to know how to approach and solve the problems.
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NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics
NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics are been solved by expert teachers of LearnCBSE.online. All the solutions given in this page are solved based on CBSE Syllabus and NCERT guidelines.
NCERT Exercises
Question 1.
Choose the correct answer. A thermodynamic state function is a quantity
(a)
used to determine heat changes
(b)
whose value is independent of path
(c)
used to determine pressure volume work
(d)
whose value depends on temperature only.
Solution.
(b):
State function is a property of the system whose value depends only upon the state of the system and is independent of the path or the manner by which the state is reached.
Question 2.
For the process to occur under adiabatic conditions, the correct condition is
(a)
ΔT = 0
(b)
Δp = 0
(c)
g = 0
(d)
w=0
Solution.
(c):
Adiabatic system does not exchange heat with the surroundings.
Question 3.
The enthalpies of all elements in their standard states are
(a)
unity
(b)
zero
(c)
<0
(d)
different for each element.
Solution.
(b) :
By convention, the standard enthalpy of formation of every element in its standard state is zero.
Question 4.
ΔU° of combustion of methane is – X kJ mol
-1
. The value of ΔH° is
(a)
= ΔU°
(b)
>ΔU°
(c)
<ΔU°
(d)
=0
Solution.
(c) :
CH
4(g)
+2O
2
→ CO
2(g)+2H
2
O
l
Δn = 1 – 3 = -2. ‘
ΔH° = ΔU° + ΔnRT = -X- 2RT
Question 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
(a)
-74.8 kJ mol
-1
(b)
-52.27 kJ mol
-1
(c)
+74.8 kJ mol
-1
(d)
+52.26 kJ mol
-1
Solution.
Question 6.
A reaction, A + B → C + D + q is found to have a positive entropy change. The reaction will be
(a)
possible at high temperature
(b)
possible only at low temperature
(c)
not possible at any temperature
(d)
possible at any temperature
Solution.
(d):
A + B → C + D + q, ΔS = +ve
Here, ΔH = -ve
ΔG = ΔH – TΔS
For reaction to be spontaneous, ΔG should be -ve. As ΔH = -ve and ΔS is +ve, ΔG will be -ve at any temperature.
Question 7.
In a process, 701 J of heat is absorbed by a system and 394 S of work is done by the system. What is the change in internal energy for the process?
Solution.
Heat absorbed by the system (q) = 701 J
Work done by the system (w) = -394 J
According to first law of thermodynamics,
ΔU = q + w = 701 + (-394) = 701 – 394 = 307 J
Question 8.
The reaction of cyanamide, NH
2
CN
(g)
, with dioxygen was carried out in a bomb calorimeter, and AU was found to be -742.7 kJ mol 1 at 298 K. Calculate enthalpy change for the reaction at 298 K.
NH
2
CN
(g)
+ \(\frac { 3 }{ 2 } { O }_{ 2(g) }\) → N
2
+CO
2(g)
+H
2
O
l
Solution.
Question 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
.
Solution.
Mass of Al = 60 g
Rise in temperature, ΔT = 55 – 35 = 20°C
Molar heat capacity of Al = 24 J mo
-1
K
-1
Specific heat capacity of Al = \(\frac { 24 }{ 27 } J{ g }^{ -1 }{ K }^{ -1 }\)
∴ Energy required = m x c x ΔT
= \(60\times \frac { 24 }{ 27 } \times 20=\frac { 28800 }{ 27 } =1066.67\quad J\)
= 1.068kJ or 1.07kJ
Question 10.
Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at -10.0°C. Δ
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
Solution.
Question 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.
Solution.
C + O
2
→ CO
2
; ΔT = -393.5 kJ
∵ When 44 g of COz is formed from carbon and dioxygen gas, heat released = 393.5 kj
∴ When 35.2 g of CO
2
is formed from carbon and dioxygen gas, heat released
= \(\frac { 393.5\times 35.2 }{ 44 } =\frac { 138551.2 }{ 44 } =314.8kJ\)
Thus, ΔH = -314.8 kJ
Question 12.
Enthalpies of formation of CO
(g)
, CO
2(g)
, N
2
O
(g)
and N
2
O
4(g)
are -110, – 393, 81 and 9.7 kJ mol
-1
respectively. Find the value of Δ
r
f for the reaction
N
2
O
4(g)
+ 3CO
(g)
→ N
2
O
(g)
+ 3CO
2(g)
Solution.
Question 13.
Given: N
2(g)
+ 3H
2(g)
→ 2NH
3(g)
; Δ
r
H° = -92.4 kJ mol
-1
What is the standard enthalpy of formation of NH
3
gas?
Solution.
N
2(g)
+ 3H
2(g)
→ 2NH
3(g)
; Δ
r
H° = -92.4 kJ mol
-1
∴ Standard enthalpy of formation of NH
3(g)
= \(\frac { -92.4 }{ 2 } =\quad -46.2\quad kJ/mol\)
Question 14.
Calculate the standard enthalpy of formation of CH
3
OH
l
from the following data :
Solution.
Question 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)
.
Δ
vap
H°(CCl
4
) = 30.5 kJ mol
-1
,
Δ
f
H°(CCl
4
) = -135.5 kJ mol
-1
,
Δ
a
H°(C) = 715.0 kJ mol
-1
, where Δ
a
H° is enthalpy of atomisation Δ
a
H°(Cl
2
) = 242 kJ mol
-1
Solution.
Question 16.
For an isolated system, ΔU = 0, what will be ΔS?
Solution.
When energy factor has no role to play, for the process to be spontaneous ΔS must be +ve i.e., ΔS > 0.
Question 17.
For the reaction at 298 K, 2A + B → C,
ΔH = 400 kJ mol
-1
and ΔS = 0.2 kJ K
-1
mol
-1
At what temperature will the reaction become spontaneous considering ΔH and ΔS to be constant over the temperature range.
Solution.
According to Gibbs Helmholtz equation, ΔG = ΔH – TΔS.
Question 18.
For the reaction, 2Cl
(g)
→ Cl
2(g)
, what are the signs of ΔH and ΔS?
Solution.
ΔH is negative because bond energy is released and ΔS is negative because there is less randomness among the molecules than among the atoms.
Question 19.
For the reaction
2A
(g)
+ B
(g)
→ 2D
(g)
, ΔU° = -10.5 kJ and ΔS° = – 44.1 JK
-1
.
Calculate ΔG° for the reaction, and predict whether the reaction may occur spontaneously.
Solution.
ΔH° = ΔU° + Δn
(g)
RT
Question 20.
The equilibrium constant for a reaction is 10. What will be the value of ΔG° ?
R = 8.314J K
-1
mol
-1
, T= 300 K.
Solution.
We know that ΔG° = – 2.303RT logK
= – 2.303 x 8.314 x 300 x log10
= – 5744.14 J/mol
Question 21.
Comment on the thermodynamic stability of NO
(g)
, given
Solution.
Since Δ
(r)
H° is +ve, i.e., enthalpy of formation of NO is positive, therefore NO is unstable. But Δ
(r)
H° is negative for the formation of NO
2
. So, NO
2
is stable.
Question 22.
Calculate the entropy change in surroundings when 1.00 mol of H
2
O
l
is formed under standard conditions.
Δ
(r)
H° = – 286 kJ mol
-1
.
Solution.
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