Chemical Thermodynamics PYQs on the CBT Screen

Question 1 · NEET 2026 · Q48 · +4 / −1

Consider the following reaction: $2A(g) + B(g) \rightarrow 2D(g)$ . $\Delta U^{\ominus} = -10\ \text{kJ mol}^{-1}$ and $\Delta S^{\ominus} = -44\ \text{JK}^{-1}$ at 298 K. Identify the correct option with $\Delta G^{\ominus}$ for the reaction and the spontaneity of the reaction at 298 K. (Given: $R = 8.31\ \text{J mol}^{-1}\ \text{K}^{-1}$ )

Answer and explanation

Answer: (3)

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.

Question 2 · NEET 2026 · Q74 · +4 / −1

At a certain temperature, T (K), during a process, 500 J is absorbed by the system and work of 200 J is done by the system. Then change in internal energy of the system is:

Answer and explanation

Answer: (2)

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.

Question 3 · Re-NEET 2026 · Q59 · +4 / −1

Two moles of an ideal gas undergo free expansion from 10 L to 100 L at 300 K. The values of $\Delta S_{\text{system}}$ and $\Delta S_{\text{surroundings}}$ are (R is universal gas constant)

Answer and explanation

Answer: (4)

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).

Question 4 · Re-NEET 2026 · Q69 · +4 / −1

Consider the reversible processes for 1.0 mol of an ideal gas as shown in the figure.

$w_1$ , $w_2$ , $w_3$ and $w_4$ represent work done (in calories) in the processes 1, 2, 3 and 4, respectively; $\Delta U_2$ and $\Delta U_4$ are changes in the internal energy for the processes 2 and 4, respectively. [use $R = 2\ \text{cal K}^{-1}\text{mol}^{-1}$ ] The correct option is

Answer and explanation

Answer: (1)

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

Question 5 · Re-NEET 2026 · Q74 · +4 / −1

A protein undergoes reversible thermal denaturation from its initial state $\mathbf{N}$ to denatured state $\mathbf{D}$ according to $\mathrm{N} \rightleftharpoons \mathrm{D}$ . At 60 °C, the concentrations of both $\mathbf{N}$ and $\mathbf{D}$ are equal at equilibrium, and the standard enthalpy change of denaturation is 666 kJ $\text{mol}^{-1}$ . The standard entropy change ( $\Delta S^\circ$ in kJ $\text{K}^{-1}\text{mol}^{-1}$ ) of the protein upon denaturation at 60 °C is closest to

Answer and explanation

Answer: (1)

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).

Question 6 · NEET 2020 · Q97 · +4 / −1

For the reaction 2Cl(g) → Cl $_{2}$ (g), the correct option is

Answer and explanation

Answer: (3)

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.

Question 7 · NEET 2020 · Q107 · +4 / −1

The correct option for free expansion of an ideal gas under adiabatic condition is

Answer and explanation

Answer: (4)

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.

Chemical Thermodynamics, on the exam screen.