Chemical Thermodynamics: NEET previous year questions

With $\Delta n_g = 2 - 3 = -1$, $\Delta H^{\ominus} = \Delta U^{\ominus} + \Delta n_g RT = -10 - \frac{298\times8.31}{1000} = -12.48\ \text{kJ}$. Then $\Delta G^{\ominus} = \Delta H^{\ominus} - T\Delta S^{\ominus} = -12.48 - \frac{298\times(-44)}{1000} = -12.48 + 13.112 \approx +0.636\ \text{kJ mol}^{-1}$. A positive $\Delta G^{\ominus}$ means the reaction is non-spontaneous.

By the first law of thermodynamics, $\Delta U = q + w$. Heat absorbed by the system gives q = +500 J, and work done by the system gives w = -200 J. Therefore $\Delta U = 500 - 200 = 300$ J.

Free expansion is into vacuum, so no work is done and no heat is exchanged: $w = 0$, $q = 0$. The surroundings are therefore completely untouched and $\Delta S_{\text{surroundings}} = 0$. Entropy is a STATE function, so the system's change is computed along any reversible path with the same endpoints: $\Delta S = nR\ln\frac{V_2}{V_1} = 2R\ln 10 = 4.606\,R$. The net positive total confirms the irreversibility. NCERT Class 11, Thermodynamics (entropy and spontaneity).

The cycle is a Carnot-type sequence: steps 1 and 3 are isothermal (at $T_1$ and $T_2$) and steps 2 and 4 are adiabatic. For a reversible isothermal change $w = -nRT\ln\frac{V_f}{V_i}$, so with $n = 1$ and $R = 2$ the two isothermal contributions add to exactly the expression in option (1). NCERT Class 11, Thermodynamics (work in reversible isothermal and adiabatic processes).

Why the other options are wrong:

- (2) for an adiabatic step $q = 0$ so $w = \Delta U$ for EACH, giving $w_2 + w_4 = \Delta U_2 + \Delta U_4$; the minus sign is the error
- (3) gets the sign of the isothermal work backwards and wrongly treats the adiabatic $w_2$ as zero
- (4) confuses $\Delta U = 0$ over a cycle with $w = 0$; the net work equals the enclosed area and is non-zero

Equal equilibrium concentrations mean $K = 1$, and since $\Delta G^\circ = -RT\ln K$, the standard free energy change is zero. Setting $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ = 0$ gives $\Delta S^\circ = \frac{\Delta H^\circ}{T} = \frac{666}{333} = 2.0\ \text{kJ K}^{-1}\text{mol}^{-1}$, using $T = 60 + 273 = 333\ \text{K}$. NCERT Class 11, Thermodynamics (Gibbs energy and equilibrium).

The reaction $2\mathrm{Cl}(\text{g}) → \mathrm{Cl}_2(\text{g})$ involves the formation of a bond, which releases energy, making $\Delta_r H < 0$. Additionally, the reaction decreases the number of gas molecules, reducing entropy, so $\Delta_r S < 0$. Chapter: Chemical Thermodynamics.

In a free expansion of an ideal gas under adiabatic conditions, no work is done (w = 0) and no heat is exchanged (q = 0). Since the internal energy of an ideal gas depends only on temperature, $\Delta T = 0$. Chapter: Chemical Thermodynamics.