Thermodynamics practice · Mixed difficulty

Option D ($- 602\mathrm{kJ}/\mathrm{mol}$) is correct because it accurately calculates the molar heat of decomposition using the given temperature change and heat capacity.

Option B ($-90 \mathrm{~kJ} \mathrm{~mole}^{-1}$) is correct because at equilibrium, ∆G = -T∆S, and with the given values, ∆G = -90 kJ/mol.

Option A (-2872 J) is correct because it is the result of calculating the work done during an isothermal reversible expansion using the ideal gas law.

Option B ($-512.0 \,\mathrm{k} \,\mathrm{cal}$) is the correct enthalpy change, calculated by manipulating the given equations and applying Hess's Law.

Option A (γ) is correct because the slope of log P vs log V for a reversible adiabatic process is -γ, and for a reversible isothermal process is -1. The ratio is therefore γ.

Option B (-125 J) is correct because the change in entropy ($\Delta S^{\circ}$) is calculated using the formula $\Delta S^{\circ} = S^{\circ}_{products} - S^{\circ}_{reactants}$.

Option B (- 1420 kJ) is correct because it correctly applies the relationship between ∆H and ∆U, accounting for the change in the number of moles of gas.

Option C ($\mathrm{T} > \mathrm{425}\,\mathrm{K}$) is correct because for a reaction to be spontaneous, $\Delta G$ must be negative, and this occurs when $T > \frac{{\Delta H}}{{\Delta S}}$.

Option B (y-2x kcal) is correct because by manipulating the given equations and Hess's Law, the heat of formation of $\mathrm{SO}_2$ can be expressed as y-2x.

Option B (Density) is correct because density is an intensive property, meaning it does not depend on the amount of substance.

Option C ($\ln \mathrm{k}=\frac{\mathrm{T} \Delta \mathrm{S}^{\circ}-\Delta \mathrm{H}^{\circ}}{\mathrm{RT}}$) is correct because it represents the correct relationship between the equilibrium constant (k), Gibbs free energy change (∆G°), enthalpy change (∆H°), entropy change (∆S°), and temperature (T).

Option B (5954 Jmol⁻¹) is correct because the enthalpy change is calculated by multiplying the temperature by the change in entropy during the phase transition.

Option C (-114 kJ) is correct. The reaction involves the neutralization of a strong acid with a strong base, and since Ba(OH)₂ and H₂SO₄ are diprotic, the enthalpy change is twice that of the monoprotic reaction.

Option B (Room heated) is correct because the refrigerator expels more heat into the room than it removes, due to the energy consumed by its motor.

Option D (Zero) is correct because the work done during the expansion of a gas into a vacuum is zero, as there is no external pressure to oppose the expansion.

Option C is correct because an isolated system, by definition, does not exchange either mass or energy with its surroundings.

Option C is correct because the law of conservation of energy states that energy cannot be created or destroyed, but can be transformed from one form to another.

Option C (q+W) is correct because the sum of heat (q) and work (W) represents the change in internal energy (ΔU), which is a state function.

Option A ($2.41 \mathrm{~JK}^{-1} \mathrm{~mol}^{-1}$) is correct because it is calculated using the formula $\Delta S = \frac{q}{T}$ where T is the boiling point of water (373 K).

Option A is correct because enthalpy and internal energy are extensive properties, meaning their values depend on the amount of substance present.

Option B ($w=-11.488 \mathrm{~kJ}$) is correct because the work done in an isothermal reversible expansion is calculated using the formula $w = -nRT \ln{\frac{{V_2}}{{V_1}}}$.

Option B (-68 Kcal) is correct because the heat of combustion of CO(g) is calculated by manipulating the given reactions to isolate the combustion reaction and then summing the enthalpies.

Option B ($- 610\,\,\mathrm{kJ}\,\mathrm{mo}{\mathrm{l}^{ - 1}}$) is correct because it represents the sum of the enthalpy changes for the given steps.

Option B (+ Ve, – Ve) is correct. Boiling an egg increases disorder (+Ve ΔS), while stretching a rubber band aligns the polymer chains, decreasing disorder (–Ve ΔS).

The question is incomplete and does not provide enough information to determine the change in internal energy. It seems there is a typo and the correct answer cannot be determined with the given information.

Option C ($2 {SO}_{2({~g})}+{O}_{2({~g})} \rightarrow 2 {SO}_{3({~g})}$) is correct because the number of gaseous molecules decreases, leading to a decrease in entropy.

Option C (${Ne}$ and ${Cl}$) is correct because noble gases like Ne have a positive or near-zero electron gain enthalpy, while halogens like Cl have a highly negative electron gain enthalpy, resulting in the largest difference.

Option C (Work) is correct because work is a path function, meaning its value depends on the path taken, while enthalpy, entropy, and free energy are state functions.

Option A (-731 kJ) is correct because it is the Δ G calculated by manipulating the given reactions using Hess's Law.

Option C, ∆U, is correct because in an isochoric process (constant volume), the heat given to the system is equal to the change in internal energy.