Thermodynamics — NCERT Solutions

Given:

• $\Delta_f H^\circ[\mathrm{CO(g)}] = -110\ \mathrm{kJ\ mol^{-1}}$
• $\Delta_f H^\circ[\mathrm{CO_2(g)}] = -393\ \mathrm{kJ\ mol^{-1}}$
• $\Delta_f H^\circ[\mathrm{N_2O(g)}] = 81\ \mathrm{kJ\ mol^{-1}}$
• $\Delta_f H^\circ[\mathrm{N_2O_4(g)}] = 9.7\ \mathrm{kJ\ mol^{-1}}$

Formula:
$$\Delta_r H = \sum \Delta_f H^\circ(\text{products}) - \sum \Delta_f H^\circ(\text{reactants})$$

Calculation:
$$\Delta_r H = [\Delta_f H^\circ(\mathrm{N_2O}) + 3\Delta_f H^\circ(\mathrm{CO_2})] - [\Delta_f H^\circ(\mathrm{N_2O_4}) + 3\Delta_f H^\circ(\mathrm{CO})]$$

$$= [81 + 3(-393)] - [9.7 + 3(-110)]$$

$$= [81 - 1179] - [9.7 - 330]$$

$$= [-1098] - [-320.3]$$

$$= -1098 + 320.3$$

$$\boxed{\Delta_r H = -777.7\ \mathrm{kJ\ mol^{-1}}}$$

Given:
$$\mathrm{N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)};\ \Delta_r H^\circ = -92.4\ \mathrm{kJ\ mol^{-1}}$$

The standard enthalpy of formation refers to the formation of 1 mole of NH₃ from its elements in their standard states:
$$\mathrm{\frac{1}{2}N_2(g) + \frac{3}{2}H_2(g) \rightarrow NH_3(g)}$$

The given reaction produces 2 moles of NH₃, so:
$$\Delta_f H^\circ[\mathrm{NH_3}] = \frac{\Delta_r H^\circ}{2} = \frac{-92.4}{2}$$

$$\boxed{\Delta_f H^\circ[\mathrm{NH_3}] = -46.2\ \mathrm{kJ\ mol^{-1}}}$$

Target reaction (formation of methanol):
$$\mathrm{C(graphite) + 2H_2(g) + \frac{1}{2}O_2(g) \rightarrow CH_3OH(l)}$$

Applying Hess's Law:

Target = (ii) + 2×(iii) − (i)

$$\Delta_f H^\circ[\mathrm{CH_3OH}] = \Delta_r H^\circ_{(ii)} + 2\Delta_r H^\circ_{(iii)} - \Delta_r H^\circ_{(i)}$$

$$= (-393) + 2(-286) - (-726)$$

$$= -393 - 572 + 726$$

$$= -965 + 726$$

$$\boxed{\Delta_f H^\circ[\mathrm{CH_3OH(l)}] = -239\ \mathrm{kJ\ mol^{-1}}}$$

Step 1: Calculate $\Delta H$ for $\mathrm{CCl_4(g) \rightarrow C(g) + 4\ Cl(g)}$

We construct this using the following thermochemical equations:

(1) $\mathrm{CCl_4(l) \rightarrow CCl_4(g)}$; $\Delta H = +30.5\ \mathrm{kJ\ mol^{-1}}$ (vaporisation)

(2) $\mathrm{C(graphite) + 2Cl_2(g) \rightarrow CCl_4(l)}$; $\Delta_f H^\circ = -135.5\ \mathrm{kJ\ mol^{-1}}$

(3) $\mathrm{C(graphite) \rightarrow C(g)}$; $\Delta_a H^\circ = +715.0\ \mathrm{kJ\ mol^{-1}}$

(4) $\mathrm{Cl_2(g) \rightarrow 2\ Cl(g)}$; $\Delta_a H^\circ = +242\ \mathrm{kJ\ mol^{-1}}$

Target: $\mathrm{CCl_4(g) \rightarrow C(g) + 4\ Cl(g)}$

Target = −(1) − (2) + (3) + 2×(4)

$$\Delta H = -30.5 - (-135.5) + 715.0 + 2(242)$$
$$= -30.5 + 135.5 + 715.0 + 484$$
$$= 1304\ \mathrm{kJ\ mol^{-1}}$$

Step 2: Calculate bond enthalpy of C–Cl

In CCl₄, there are 4 C–Cl bonds. The enthalpy change for breaking all 4 bonds:
$$\Delta H = 4 \times \text{Bond enthalpy of C–Cl}$$

$$\text{Bond enthalpy of C–Cl} = \frac{1304}{4}$$

$$\boxed{\text{Bond enthalpy of C–Cl} = 326\ \mathrm{kJ\ mol^{-1}}}$$

For an isolated system: No exchange of energy or matter with surroundings occurs.

Given: $\Delta U = 0$

For any spontaneous process in an isolated system, the entropy of the system increases. According to the second law of thermodynamics:

$$\Delta S > 0$$

Conclusion: For an isolated system with $\Delta U = 0$, the entropy change $\Delta S$ will be greater than zero ($\Delta S > 0$). The entropy of an isolated system increases for any spontaneous (irreversible) process, and remains constant only for a reversible (equilibrium) process.

Given:

• $\Delta H = +400\ \mathrm{kJ\ mol^{-1}}$ (endothermic)
• $\Delta S = +0.2\ \mathrm{kJ\ K^{-1}\ mol^{-1}}$

Condition for spontaneity: $\Delta G < 0$

$$\Delta G = \Delta H - T\Delta S < 0$$

$$T\Delta S > \Delta H$$

$$T > \frac{\Delta H}{\Delta S}$$

$$T > \frac{400\ \mathrm{kJ\ mol^{-1}}}{0.2\ \mathrm{kJ\ K^{-1}\ mol^{-1}}}$$

$$T > 2000\ \mathrm{K}$$

Conclusion: The reaction will become spontaneous at temperatures above 2000 K. Since both $\Delta H > 0$ and $\Delta S > 0$, the reaction is non-spontaneous at low temperatures but becomes spontaneous at high temperatures.

Analysis of the reaction: $\mathrm{2\ Cl(g) \rightarrow Cl_2(g)}$

Sign of $\Delta H$:
This reaction involves formation of a Cl–Cl bond from two chlorine atoms. Bond formation releases energy. Therefore, the reaction is exothermic.
$$\Delta H < 0\ (\text{negative})$$

Sign of $\Delta S$:
Two moles of gaseous atoms combine to form one mole of gaseous molecule. The number of moles of gas decreases from 2 to 1, which means the disorder (randomness) of the system decreases.
$$\Delta S < 0\ (\text{negative})$$

Conclusion: Both $\Delta H$ and $\Delta S$ are negative for this reaction.

Given:

• $\Delta U^\circ = -10.5\ \mathrm{kJ}$
• $\Delta S^\circ = -44.1\ \mathrm{J\ K^{-1}} = -0.0441\ \mathrm{kJ\ K^{-1}}$
• $T = 298\ \mathrm{K}$

Step 1: Calculate $\Delta H^\circ$

For the reaction $\mathrm{2A(g) + B(g) \rightarrow 2D(g)}$:
$$\Delta n_g = 2 - (2+1) = 2 - 3 = -1$$

$$\Delta H^\circ = \Delta U^\circ + \Delta n_g RT$$
$$= -10.5 + (-1)(8.314 \times 10^{-3})(298)$$
$$= -10.5 + (-2.478)$$
$$= -12.978\ \mathrm{kJ\ mol^{-1}}$$

Step 2: Calculate $\Delta G^\circ$
$$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$$
$$= -12.978 - (298)(-0.0441)$$
$$= -12.978 + 13.14$$
$$= +0.162\ \mathrm{kJ\ mol^{-1}}$$

$$\boxed{\Delta G^\circ \approx +0.16\ \mathrm{kJ\ mol^{-1}}}$$

Prediction: Since $\Delta G^\circ > 0$, the reaction is non-spontaneous under standard conditions.

Given:

• $K = 10$
• $R = 8.314\ \mathrm{J\ K^{-1}\ mol^{-1}}$
• $T = 300\ \mathrm{K}$

Formula:
$$\Delta G^\circ = -RT\ln K$$

Calculation:
$$\Delta G^\circ = -(8.314\ \mathrm{J\ K^{-1}\ mol^{-1}})(300\ \mathrm{K})(\ln 10)$$
$$= -(8.314)(300)(2.303)$$
$$= -(8.314)(300)(2.303)$$
$$= -5744.14\ \mathrm{J\ mol^{-1}}$$

$$\boxed{\Delta G^\circ = -5744\ \mathrm{J\ mol^{-1}} \approx -5.74\ \mathrm{kJ\ mol^{-1}}}$$

Reaction 1: Formation of NO(g) from elements:
$$\frac{1}{2}\mathrm{N_2(g)} + \frac{1}{2}\mathrm{O_2(g)} \rightarrow \mathrm{NO(g)};\ \Delta_r H^\circ = +90\ \mathrm{kJ\ mol^{-1}}$$

Since $\Delta_r H^\circ = +90\ \mathrm{kJ\ mol^{-1}} > 0$, the formation of NO from its elements is endothermic. This means NO has higher energy than its constituent elements. Therefore, NO is thermodynamically unstable with respect to its elements $\mathrm{N_2}$ and $\mathrm{O_2}$.

Reaction 2: Conversion of NO to NO₂:
$$\mathrm{NO(g)} + \frac{1}{2}\mathrm{O_2(g)} \rightarrow \mathrm{NO_2(g)};\ \Delta_r H^\circ = -74\ \mathrm{kJ\ mol^{-1}}$$

Since $\Delta_r H^\circ = -74\ \mathrm{kJ\ mol^{-1}} < 0$, the conversion of NO to NO₂ is exothermic. This means NO is also thermodynamically unstable with respect to NO₂.

Conclusion: NO(g) is thermodynamically unstable both with respect to its elements ($\mathrm{N_2, O_2}$) and with respect to $\mathrm{NO_2}$. However, it is kinetically stable (does not decompose rapidly at room temperature due to high activation energy).

Given:

• $\Delta_r H^\circ = -286\ \mathrm{kJ\ mol^{-1}}$ (for formation of 1 mol H₂O(l))
• $T = 298\ \mathrm{K}$ (standard conditions)

Concept: The heat released by the system is absorbed by the surroundings.

Heat absorbed by surroundings:
$$q_{\text{surr}} = -\Delta_r H^\circ_{\text{system}} = -(-286) = +286\ \mathrm{kJ\ mol^{-1}}$$

Formula for entropy change of surroundings:
$$\Delta S_{\text{surr}} = \frac{q_{\text{surr}}}{T} = \frac{+286 \times 10^3\ \mathrm{J\ mol^{-1}}}{298\ \mathrm{K}}$$

$$\Delta S_{\text{surr}} = \frac{286000}{298}$$

$$\boxed{\Delta S_{\text{surr}} = +959.73\ \mathrm{J\ K^{-1}\ mol^{-1}}}$$