NCERT Solutions For Class 11 Chemistry chapter 5: Thermodynamics

Question 5.1:

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

Solution:

A thermodynamic state function is a quantity whose value is independent of 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 5.2:

For the process to occur under adiabatic conditions, the correct condition is:
(i) $\Delta T = 0$
(ii) $\Delta p = 0$
(iii) $q = 0$
(iv) $w = 0$

Solution:

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 5.3:

The enthalpies of all elements in their standard states are:
(i) unity
(ii) zero
(iii) $< 0$
(iv) different for each element

Solution:

The enthalpy of all elements in their standard state is zero.
Therefore, alternative (ii) is correct.

Question 5.4:

$\Delta U^\ominus$ of combustion of methane is $-X\text{ kJ mol}^{-1}$. The value of $\Delta H^\ominus$ is
(i) $= \Delta U^\ominus$
(ii) $> \Delta U^\ominus$
(iii) $< \Delta U^\ominus$
(iv) $= 0$

Solution:

Since $\Delta H^\ominus = \Delta U^\ominus + \Delta n_g RT$ and $\Delta U^\ominus = -X\text{ kJ mol}^{-1}$
$$\Delta H^\ominus = (-X) + \Delta n_g RT$$
For combustion of methane: $\text{CH}_4(g) + 2\text{O}_2(g) \longrightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l)$, here $\Delta n_g = 1 – 3 = -2$ (negative).
$$\implies \Delta H^\ominus < \Delta U^\ominus$$
Therefore, alternative (iii) is correct.

Question 5.5:

The enthalpy of combustion of methane, graphite and dihydrogen at $298\text{ K}$ are $-890.3\text{ kJ mol}^{-1}$, $-393.5\text{ kJ mol}^{-1}$, and $-285.8\text{ kJ mol}^{-1}$ respectively. Enthalpy of formation of $\text{CH}_4$ will be
(i) $-74.8\text{ kJ mol}^{-1}$
(ii) $-52.27\text{ kJ mol}^{-1}$
(iii) $+74.8\text{ kJ mol}^{-1}$
(iv) $+52.26\text{ kJ mol}^{-1}$

Solution:

According to the question,
(i) $\text{CH}_4(g) + 2\text{O}_2(g) \longrightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l) \quad \Delta H = -890.3\text{ kJ mol}^{-1}$
(ii) $\text{C(graphite)} + \text{O}_2(g) \longrightarrow \text{CO}_2(g) \quad \Delta H = -393.5\text{ kJ mol}^{-1}$
(iii) $2\text{H}_2(g) + \text{O}_2(g) \longrightarrow 2\text{H}_2\text{O}(l) \quad \Delta H = -285.8\text{ kJ mol}^{-1}$

Thus, the desired equation representing the formation of $\text{CH}_4(g)$ is: $\text{C(graphite)} + 2\text{H}_2(g) \longrightarrow \text{CH}_4(g)$
$$\Delta_f H^\ominus = \Delta_cH(\text{C}) + 2\Delta_cH(\text{H}_2) – \Delta_cH(\text{CH}_4)$$
$$= [-393.5 + 2(-285.8) – (-890.3)]\text{ kJ mol}^{-1}$$
$$= [-393.5 – 571.6 + 890.3]\text{ kJ mol}^{-1} = -74.8\text{ kJ mol}^{-1}$$
$\therefore$ Enthalpy of formation of $\text{CH}_4(g) = -74.8\text{ kJ mol}^{-1}$.
Hence, alternative (i) is correct.

Question 5.6:

A reaction, $A + B \longrightarrow C + D + q$ 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

Solution:

For a reaction to be spontaneous, $\Delta G$ should be negative.
$$\Delta G = \Delta H – T\Delta S$$
According to the question, for the given reaction:
$\Delta S = \text{positive}$
$\Delta H = \text{negative}$ (since heat $q$ is evolved)
$\implies \Delta G = \text{negative}$ at all temperatures.
Therefore, the reaction is spontaneous at any temperature.
Hence, alternative (iv) is correct.

Question 5.7:

In a process, $701\text{ J}$ of heat is absorbed by a system and $394\text{ J}$ of work is done by the system. What is the change in internal energy for the process?

Solution:

According to the first law of thermodynamics:
$$\Delta U = q + w \quad \dots\text{(i)}$$
Where,
$\Delta U = \text{change in internal energy}$
$q = \text{heat}$
$w = \text{work}$

Given,
$q = +701\text{ J}$ (Since heat is absorbed)
$w = -394\text{ J}$ (Since work is done by the system)

Substituting the values in expression (i), we get:
$$\Delta U = 701\text{ J} + (-394\text{ J}) = 307\text{ J}$$
Hence, the change in internal energy for the given process is $307\text{ J}$.

Question 5.8:

The reaction of cyanamide, $\text{NH}_2\text{CN}(s)$ with dioxygen was carried out in a bomb calorimeter, and $\Delta U$ was found to be $-742.7\text{ kJ mol}^{-1}$ at $298\text{ K}$. Calculate enthalpy change for the reaction at $298\text{ K}$.
$$\text{NH}_2\text{CN}(s) + \frac{3}{2}\text{O}_2(g) \longrightarrow \text{N}_2(g) + \text{CO}_2(g) + \text{H}_2\text{O}(l)$$

Solution:

Enthalpy change for a reaction ($\Delta H$) is given by the expression:
$$\Delta H = \Delta U + \Delta n_g RT$$
Where,
$\Delta U = -742.7\text{ kJ mol}^{-1}$
$\Delta n_g = \sum n_g(\text{products}) – \sum n_g(\text{reactants}) = (1 + 1) – 1.5 = 2 – 1.5 = 0.5\text{ moles}$
$T = 298\text{ K}$
$R = 8.314 \times 10^{-3}\text{ kJ mol}^{-1}\text{ K}^{-1}$

Substituting the values in the expression of $\Delta H$:
$$\Delta H = (-742.7\text{ kJ mol}^{-1}) + (0.5\text{ mol})(298\text{ K})(8.314 \times 10^{-3}\text{ kJ mol}^{-1}\text{ K}^{-1})$$
$$\Delta H = -742.7 + 1.239 = -741.5\text{ kJ mol}^{-1}$$

Question 5.9:

Calculate the number of $\text{kJ}$ of heat necessary to raise the temperature of $60.0\text{ g}$ of aluminium from $35^\circ\text{C}$ to $55^\circ\text{C}$. Molar heat capacity of Al is $24\text{ J mol}^{-1}\text{ K}^{-1}$. (Atomic mass of $\text{Al} = 27\text{ g/mol}$)

Solution:

Number of moles of Al, $n = \frac{60\text{ g}}{27\text{ g/mol}} = 2.222\text{ mol}$
From the expression of heat ($q$):
$$q = n \cdot C_m \cdot \Delta T$$
Where,
$C_m = \text{molar heat capacity} = 24\text{ J mol}^{-1}\text{ K}^{-1}$
$\Delta T = 55 – 35 = 20\text{ K}$
$$q = 2.222\text{ mol} \times 24\text{ J mol}^{-1}\text{ K}^{-1} \times 20\text{ K} = 1066.67\text{ J} = 1.07\text{ kJ}$$

Question 5.10:

Calculate the enthalpy change on freezing of $1.0\text{ mol}$ of water at $10.0^\circ\text{C}$ to ice at $-10.0^\circ\text{C}$.
$\Delta_{\text{fus}}H = 6.03\text{ kJ mol}^{-1}$ at $0^\circ\text{C}$
$C_p[\text{H}_2\text{O}(l)] = 75.3\text{ J mol}^{-1}\text{ K}^{-1}$
$C_p[\text{H}_2\text{O}(s)] = 36.8\text{ J mol}^{-1}\text{ K}^{-1}$

Solution:

Total enthalpy change involved in the transformation is the sum of the following steps:
(a) Cooling of $1\text{ mol}$ of water from $10^\circ\text{C}$ to $0^\circ\text{C}$:
$$\Delta H_1 = C_p[\text{H}_2\text{O}(l)] \Delta T = 75.3 \times (0 – 10) = -753\text{ J}$$
(b) Freezing of $1\text{ mol}$ of water at $0^\circ\text{C}$ to ice at $0^\circ\text{C}$:
$$\Delta H_2 = -\Delta_{\text{fus}}H = -6.03\text{ kJ mol}^{-1} = -6030\text{ J}$$
(c) Cooling of $1\text{ mol}$ of ice from $0^\circ\text{C}$ to $-10^\circ\text{C}$:
$$\Delta H_3 = C_p[\text{H}_2\text{O}(s)] \Delta T = 36.8 \times (-10 – 0) = -368\text{ J}$$
Total enthalpy change:
$$\Delta H = \Delta H_1 + \Delta H_2 + \Delta H_3 = -753 – 6030 – 368 = -7151\text{ J mol}^{-1} = -7.151\text{ kJ mol}^{-1}$$

Question 5.11:

Enthalpy of combustion of carbon to $\text{CO}_2$ is $-393.5\text{ kJ mol}^{-1}$. Calculate the heat released upon formation of $35.2\text{ g}$ of $\text{CO}_2$ from carbon and dioxygen gas.

Solution:

Formation of $\text{CO}_2$ from carbon and dioxygen gas can be represented as:
$$\text{C}(s) + \text{O}_2(g) \longrightarrow \text{CO}_2(g) \quad \Delta_f H^\ominus = -393.5\text{ kJ mol}^{-1}$$
Molar mass of $\text{CO}_2 = 44\text{ g/mol}$
Heat released on formation of $44\text{ g }\text{CO}_2 = -393.5\text{ kJ}$
$\therefore$ Heat released on formation of $35.2\text{ g }\text{CO}_2$:
$$= \frac{-393.5\text{ kJ}}{44\text{ g}} \times 35.2\text{ g} = -314.8\text{ kJ}$$

Question 5.12:

Enthalpies of formation of $\text{N}_2\text{O}_4(g)$ and $\text{CO}(g)$, $\text{N}_2\text{O}(g)$, and $\text{CO}_2(g)$ are $9.7$, $-110$, $81$, and $-393\text{ kJ mol}^{-1}$ respectively. Find the value of $\Delta H$ for the reaction:
$$\text{N}_2\text{O}_4(g) + 3\text{CO}(g) \longrightarrow \text{N}_2\text{O}(g) + 3\text{CO}_2(g)$$

Solution:

$\Delta_r H^\ominus$ for a reaction is defined as the difference between $\Delta_f H$ value of products and $\Delta_f H$ value of reactants:
$$\Delta_r H^\ominus = \sum \Delta_f H(\text{products}) – \sum \Delta_f H(\text{reactants})$$
$$\Delta_r H^\ominus = [\Delta_f H(\text{N}_2\text{O}) + 3\Delta_f H(\text{CO}_2)] – [\Delta_f H(\text{N}_2\text{O}_4) + 3\Delta_f H(\text{CO})]$$
Given values: $\Delta_f H(\text{N}_2\text{O}_4) = 9.7\text{ kJ mol}^{-1}$, $\Delta_f H(\text{CO}) = -110\text{ kJ mol}^{-1}$, $\Delta_f H(\text{N}_2\text{O}) = 81\text{ kJ mol}^{-1}$, $\Delta_f H(\text{CO}_2) = -393\text{ kJ mol}^{-1}$:
$$\Delta_r H^\ominus = [81 + 3(-393)] – [9.7 + 3(-110)]$$
$$= [81 – 1179] – [9.7 – 330] = -1098 – (-320.3) = -777.7\text{ kJ mol}^{-1}$$
Hence, the value of $\Delta_r H^\ominus$ for the reaction is $-777.7\text{ kJ mol}^{-1}$.

Question 5.13:

Given: $\text{N}_2(g) + 3\text{H}_2(g) \longrightarrow 2\text{NH}_3(g); \Delta_r H^\ominus = -92.4\text{ kJ mol}^{-1}$. What is the standard enthalpy of formation of $\text{NH}_3$ gas?

Solution:

Standard enthalpy of formation of a compound is the enthalpy change that takes place during the formation of $1\text{ mole}$ of a substance in its standard form from its constituent elements in their standard states.
Re-writing the given equation for $1\text{ mole}$ of $\text{NH}_3(g)$:
$$\frac{1}{2}\text{N}_2(g) + \frac{3}{2}\text{H}_2(g) \longrightarrow \text{NH}_3(g)$$
$\therefore$ Standard enthalpy of formation of $\text{NH}_3(g)$ is:
$$\Delta_f H^\ominus(\text{NH}_3) = \frac{1}{2} \Delta_r H^\ominus = \frac{1}{2}(-92.4\text{ kJ mol}^{-1}) = -46.2\text{ kJ mol}^{-1}$$

Question 5.14:

Calculate the standard enthalpy of formation of $\text{CH}_3\text{OH}(l)$ from the following data:
(i) $\text{CH}_3\text{OH}(l) + \frac{3}{2}\text{O}_2(g) \longrightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l) \quad \Delta_c H^\ominus = -726\text{ kJ mol}^{-1}$
(ii) $\text{C(graphite)} + \text{O}_2(g) \longrightarrow \text{CO}_2(g) \quad \Delta_f H^\ominus = -393\text{ kJ mol}^{-1}$
(iii) $\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \longrightarrow \text{H}_2\text{O}(l) \quad \Delta_f H^\ominus = -286\text{ kJ mol}^{-1}$

Solution:

The reaction that takes place during the formation of $\text{CH}_3\text{OH}(l)$ can be written as:
$$\text{C(graphite)} + 2\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \longrightarrow \text{CH}_3\text{OH}(l)$$
This can be obtained from the given equations by: $\text{(ii)} + 2 \times \text{(iii)} – \text{(i)}$
$$\Delta_f H^\ominus[\text{CH}_3\text{OH}(l)] = \Delta_f H^\ominus(\text{C}) + 2\Delta_f H^\ominus(\text{H}_2\text{O}) – \Delta_c H^\ominus(\text{CH}_3\text{OH})$$
$$= (-393\text{ kJ mol}^{-1}) + 2(-286\text{ kJ mol}^{-1}) – (-726\text{ kJ mol}^{-1})$$
$$= -393 – 572 + 726 = -239\text{ kJ mol}^{-1}$$

Question 5.15:

Calculate the enthalpy change for the process $\text{CCl}_4(g) \longrightarrow \text{C}(g) + 4\text{Cl}(g)$ and calculate bond enthalpy of $\text{C}-\text{Cl}$ in $\text{CCl}_4(g)$ given:
$\Delta_{\text{vap}} H^\ominus(\text{CCl}_4) = 30.5\text{ kJ mol}^{-1}$
$\Delta_f H^\ominus(\text{CCl}_4) = -135.5\text{ kJ mol}^{-1}$
$\Delta_a H^\ominus(\text{C}) = 715.0\text{ kJ mol}^{-1}$ (enthalpy of atomisation of carbon)
$\Delta_a H^\ominus(\text{Cl}_2) = 242\text{ kJ mol}^{-1}$ (enthalpy of atomisation of chlorine)

Solution:

The given thermochemical equations are:
(1) $\text{CCl}_4(l) \longrightarrow \text{CCl}_4(g) \quad \Delta_{\text{vap}}H^\ominus = 30.5\text{ kJ mol}^{-1}$
(2) $\text{C}(s) \longrightarrow \text{C}(g) \quad \Delta_a H^\ominus = 715.0\text{ kJ mol}^{-1}$
(3) $\text{Cl}_2(g) \longrightarrow 2\text{Cl}(g) \quad \Delta_a H^\ominus = 242\text{ kJ mol}^{-1}$
(4) $\text{C}(s) + 2\text{Cl}_2(g) \longrightarrow \text{CCl}_4(l) \quad \Delta_f H^\ominus = -135.5\text{ kJ mol}^{-1}$

For the process $\text{CCl}_4(g) \longrightarrow \text{C}(g) + 4\text{Cl}(g)$:
$$\Delta H = \Delta_a H^\ominus(\text{C}) + 2\Delta_a H^\ominus(\text{Cl}_2) – \Delta_{\text{vap}}H^\ominus(\text{CCl}_4) – \Delta_f H^\ominus(\text{CCl}_4)$$
$$= 715.0 + 2(242) – 30.5 – (-135.5) = 715 + 484 – 30.5 + 135.5 = 1304\text{ kJ mol}^{-1}$$
Bond enthalpy of $\text{C}-\text{Cl}$ in $\text{CCl}_4(g) = \frac{1304}{4} = 326\text{ kJ mol}^{-1}$.

Question 5.16:

For an isolated system, $\Delta U = 0$, what will be $\Delta S$?

Solution:

$\Delta S$ will be positive, i.e., greater than zero ($\Delta S > 0$).
For an isolated system, spontaneous processes occur with an increase in total entropy ($\Delta S_{\text{total}} > 0$).

Question 5.17:

For the reaction at $298\text{ K}$, $2\text{A} + \text{B} \longrightarrow \text{C}$, $\Delta H = 400\text{ kJ mol}^{-1}$ and $\Delta S = 0.2\text{ kJ K}^{-1}\text{ mol}^{-1}$. At what temperature will the reaction become spontaneous considering $\Delta H$ and $\Delta S$ to be constant over the temperature range?

Solution:

From the expression $\Delta G = \Delta H – T\Delta S$
At equilibrium, $\Delta G = 0 \implies T = \frac{\Delta H}{\Delta S}$
$$T = \frac{400\text{ kJ mol}^{-1}}{0.2\text{ kJ K}^{-1}\text{ mol}^{-1}} = 2000\text{ K}$$
For the reaction to be spontaneous, $\Delta G$ must be negative. Since $\Delta H$ is positive, $T\Delta S$ must be greater than $\Delta H$. Hence, $T$ should be greater than $2000\text{ K}$.

Question 5.18:

For the reaction, $2\text{Cl}(g) \longrightarrow \text{Cl}_2(g)$, what are the signs of $\Delta H$ and $\Delta S$?

Solution:

$\Delta H$ and $\Delta S$ are both negative.
The reaction represents the formation of a chlorine molecule from chlorine atoms. Bond formation releases energy, so $\Delta H$ is negative.
Two moles of separate atoms have more randomness (higher entropy) than one mole of a molecule. Thus, randomness decreases, so $\Delta S$ is negative.

Question 5.19:

For the reaction $2\text{A}(g) + \text{B}(g) \longrightarrow 2\text{D}(g)$, $\Delta U^\ominus = -10.5\text{ kJ}$ and $\Delta S^\ominus = -44.1\text{ J K}^{-1}$. Calculate $\Delta G^\ominus$ for the reaction, and predict whether the reaction may occur spontaneously at $298\text{ K}$.

Solution:

$$\Delta n_g = 2 – (2 + 1) = -1\text{ mole}$$
First calculate $\Delta H^\ominus$:
$$\Delta H^\ominus = \Delta U^\ominus + \Delta n_g RT = -10.5\text{ kJ} + (-1)(8.314 \times 10^{-3}\text{ kJ K}^{-1}\text{ mol}^{-1})(298\text{ K})$$
$$\Delta H^\ominus = -10.5 – 2.478 = -12.978\text{ kJ}$$
Now calculate $\Delta G^\ominus$:
$$\Delta G^\ominus = \Delta H^\ominus – T\Delta S^\ominus = -12.978\text{ kJ} – (298\text{ K})(-44.1 \times 10^{-3}\text{ kJ K}^{-1})$$
$$\Delta G^\ominus = -12.978 + 13.142 = +0.164\text{ kJ}$$
Since $\Delta G^\ominus$ for the reaction is positive, the reaction will not occur spontaneously.

Question 5.20:

The equilibrium constant for a reaction is $10$. What will be the value of $\Delta G^\ominus$; given $R = 8.314\text{ J K}^{-1}\text{ mol}^{-1}$, $T = 300\text{ K}$.

Solution:

From the expression:
$$\Delta G^\ominus = -2.303 RT \log K_{\text{eq}}$$
$$\Delta G^\ominus = -2.303 \times (8.314\text{ J K}^{-1}\text{ mol}^{-1}) \times (300\text{ K}) \times \log(10)$$
$$\Delta G^\ominus = -2.303 \times 8.314 \times 300 \times 1 = -5744.14\text{ J mol}^{-1} = -5.74\text{ kJ mol}^{-1}$$

Question 5.21:

Comment on the thermodynamic stability of $\text{NO}(g)$, given:
(1) $\frac{1}{2}\text{N}_2(g) + \frac{1}{2}\text{O}_2(g) \longrightarrow \text{NO}(g) \quad \Delta_r H^\ominus = 90\text{ kJ mol}^{-1}$
(2) $\text{NO}(g) + \frac{1}{2}\text{O}_2(g) \longrightarrow \text{NO}_2(g) \quad \Delta_r H^\ominus = -74\text{ kJ mol}^{-1}$

Solution:

The positive value of $\Delta_r H^\ominus$ ($90\text{ kJ mol}^{-1}$) for equation (1) indicates that heat is absorbed during the formation of $\text{NO}(g)$, meaning $\text{NO}(g)$ has higher energy than its reactants ($\text{N}_2$ and $\text{O}_2$). Hence, $\text{NO}(g)$ is thermodynamically unstable.

The negative value of $\Delta_r H^\ominus$ ($-74\text{ kJ mol}^{-1}$) for equation (2) indicates that heat is evolved during the formation of $\text{NO}_2(g)$ from $\text{NO}(g)$ and $\text{O}_2(g)$, yielding a more stable product. Thus, unstable $\text{NO}(g)$ tends to react further to form stable $\text{NO}_2(g)$.

Question 5.22:

Calculate the entropy change in surroundings when $1.00\text{ mol}$ of $\text{H}_2\text{O}(l)$ is formed under standard conditions. $\Delta_f H^\ominus = -286\text{ kJ mol}^{-1}$.

Solution:

$286\text{ kJ mol}^{-1}$ of heat is evolved during the formation of $1\text{ mol}$ of $\text{H}_2\text{O}(l)$.
Thus, an equal amount of heat will be absorbed by the surroundings: $q_{\text{surr}} = +286\text{ kJ mol}^{-1} = 286000\text{ J mol}^{-1}$
Standard temperature $T = 298\text{ K}$
$$\Delta S_{\text{surr}} = \frac{q_{\text{surr}}}{T} = \frac{286000\text{ J mol}^{-1}}{298\text{ K}} = 959.73\text{ J mol}^{-1}\text{ K}^{-1}$$