Thermodynamics

Correct option: (ii) whose value is independent of path.

A state function is a property whose value depends only on the current state of the system (i.e., initial and final states) and not on the path taken to reach that state. Examples include internal energy (U), enthalpy (H), entropy (S), and Gibbs energy (G).

Correct option: (iii) $q = 0$.

Adiabatic process is defined as a process in which no heat exchange takes place between the system and the surroundings. Therefore, the defining condition for an adiabatic process is $q = 0$.

Correct option: (ii) zero.

By convention, the standard enthalpy of formation ($\Delta_f H^\circ$) of every element in its most stable standard state is taken as zero. This is the reference point for all enthalpy calculations.

Correct option: (iii) < \Delta U^\circ.

Given: Combustion of methane:
$$\mathrm{CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)}$$

Calculating $\Delta n_g$:
$$\Delta n_g = \text{moles of gaseous products} - \text{moles of gaseous reactants}$$
$$\Delta n_g = 1 - (1+2) = 1 - 3 = -2$$

Using the relation:
$$\Delta H^\circ = \Delta U^\circ + \Delta n_g RT$$
$$\Delta H^\circ = -X + (-2)RT$$
$$\Delta H^\circ = -X - 2RT$$

Since 2RT > 0, we have \Delta H^\circ = -X - 2RT < -X = \Delta U^\circ.

Therefore, \Delta H^\circ < \Delta U^\circ.

Correct option: (i) $-74.8\ \mathrm{kJ\ mol^{-1}}$.

Given reactions:

(1) $\mathrm{CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)}$; $\Delta_c H^\circ = -890.3\ \mathrm{kJ\ mol^{-1}}$

(2) $\mathrm{C(graphite) + O_2(g) \rightarrow CO_2(g)}$; $\Delta_c H^\circ = -393.5\ \mathrm{kJ\ mol^{-1}}$

(3) $\mathrm{H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l)}$; $\Delta_c H^\circ = -285.8\ \mathrm{kJ\ mol^{-1}}$

Target reaction (formation of methane):
$$\mathrm{C(graphite) + 2H_2(g) \rightarrow CH_4(g)}$$

Applying Hess's Law: Target = (2) + 2×(3) − (1)

$$\Delta_f H^\circ[\mathrm{CH_4}] = \Delta_c H^\circ[\mathrm{C}] + 2\Delta_c H^\circ[\mathrm{H_2}] - \Delta_c H^\circ[\mathrm{CH_4}]$$

$$= (-393.5) + 2(-285.8) - (-890.3)$$

$$= -393.5 - 571.6 + 890.3$$

$$= -74.8\ \mathrm{kJ\ mol^{-1}}$$

Correct option: (iv) possible at any temperature.

The reaction $A + B \rightarrow C + D + q$ releases heat, so \Delta H < 0 (exothermic). It is also given that \Delta S > 0 (positive entropy change).

Using $\Delta G = \Delta H - T\Delta S$:
- $\Delta H$ is negative (−)
- $T\Delta S$ is positive (+), so $-T\Delta S$ is negative (−)

Therefore \Delta G = (\text{negative}) - (\text{positive}) < 0 at all temperatures.

Since \Delta G < 0 at every temperature, the reaction is spontaneous at any temperature.

Given:
- Heat absorbed by the system: $q = +701\ \mathrm{J}$ (positive, as heat is added to system)
- Work done by the system: $w = -394\ \mathrm{J}$ (negative, as work is done by the system)

Formula (First Law of Thermodynamics):
$$\Delta U = q + w$$

Calculation:
$$\Delta U = 701\ \mathrm{J} + (-394\ \mathrm{J})$$
$$\boxed{\Delta U = 307\ \mathrm{J}}$$

Given:
$$\mathrm{NH_2CN(s) + \frac{3}{2}O_2(g) \rightarrow N_2(g) + CO_2(g) + H_2O(l)}$$
$\Delta U = -742.7\ \mathrm{kJ\ mol^{-1}}$, $T = 298\ \mathrm{K}$

Calculating $\Delta n_g$:
$$\Delta n_g = \text{moles of gaseous products} - \text{moles of gaseous reactants}$$

Gaseous products: $\mathrm{N_2(g)}$ = 1 mol, $\mathrm{CO_2(g)}$ = 1 mol → total = 2 mol

Gaseous reactants: $\mathrm{O_2(g)}$ = $\frac{3}{2}$ mol (NH₂CN is solid)

$$\Delta n_g = 2 - \frac{3}{2} = +\frac{1}{2}$$

Using the relation:
$$\Delta H = \Delta U + \Delta n_g RT$$
$$\Delta H = -742.7 + \left(\frac{1}{2}\right)(8.314 \times 10^{-3}\ \mathrm{kJ\ mol^{-1}K^{-1}})(298\ \mathrm{K})$$
$$\Delta H = -742.7 + (0.5)(2.477)$$
$$\Delta H = -742.7 + 1.239$$
$$\boxed{\Delta H = -741.5\ \mathrm{kJ\ mol^{-1}}}$$

Given:
- Mass of Al = 60.0 g
- Molar mass of Al = 27 g mol⁻¹
- Molar heat capacity, $C_p = 24\ \mathrm{J\ mol^{-1}\ K^{-1}}$
- $\Delta T = 55 - 35 = 20\ \mathrm{K}$

Number of moles of Al:
$$n = \frac{60.0}{27} = 2.222\ \mathrm{mol}$$

Formula:
$$q = n \times C_p \times \Delta T$$

Calculation:
$$q = 2.222\ \mathrm{mol} \times 24\ \mathrm{J\ mol^{-1}\ K^{-1}} \times 20\ \mathrm{K}$$
$$q = 2.222 \times 480$$
$$q = 1066.56\ \mathrm{J}$$
$$\boxed{q \approx 1.067\ \mathrm{kJ}}$$

The overall process is carried out in three steps:

Step 1: Cooling liquid water from 10°C to 0°C
$$\Delta H_1 = n \times C_p[\mathrm{H_2O(l)}] \times \Delta T = 1.0 \times 75.3 \times (0 - 10)$$
$$\Delta H_1 = 75.3 \times (-10) = -753\ \mathrm{J} = -0.753\ \mathrm{kJ}$$

Step 2: Freezing water at 0°C (phase change)
$$\Delta H_2 = -\Delta_{fus}H = -6.03\ \mathrm{kJ\ mol^{-1}}$$
(Negative because freezing is the reverse of fusion)

Step 3: Cooling ice from 0°C to −10°C
$$\Delta H_3 = n \times C_p[\mathrm{H_2O(s)}] \times \Delta T = 1.0 \times 36.8 \times (-10 - 0)$$
$$\Delta H_3 = 36.8 \times (-10) = -368\ \mathrm{J} = -0.368\ \mathrm{kJ}$$

Total enthalpy change:
$$\Delta H_{total} = \Delta H_1 + \Delta H_2 + \Delta H_3$$
$$\Delta H_{total} = -0.753 + (-6.03) + (-0.368)$$
$$\boxed{\Delta H_{total} = -7.151\ \mathrm{kJ\ mol^{-1}}}$$

Given:
$$\mathrm{C(graphite) + O_2(g) \rightarrow CO_2(g)};\ \Delta_c H^\circ = -393.5\ \mathrm{kJ\ mol^{-1}}$$

Molar mass of CO₂ = 12 + 32 = 44 g mol⁻¹

Moles of CO₂ formed:
$$n = \frac{35.2\ \mathrm{g}}{44\ \mathrm{g\ mol^{-1}}} = 0.8\ \mathrm{mol}$$

Heat released:
$$q = n \times |\Delta_c H^\circ| = 0.8\ \mathrm{mol} \times 393.5\ \mathrm{kJ\ mol^{-1}}$$
$$\boxed{q = 314.8\ \mathrm{kJ}}$$

The heat released upon formation of 35.2 g of CO₂ is 314.8 kJ.

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

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

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