Chemical Thermodynamics — NEET UG practice

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Sample questions with solutions

Q1 · 2026

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

A.
11.1
B.
2.0
C.
2000.0
D.
333.0

Answer: B

Since the concentrations of N and D are equal at $60^{\circ}C$ , the system is at equilibrium at this temperature, which means the standard Gibbs free energy change is zero.

\Delta G^{\circ} = 0

Using the relation between free energy, enthalpy, and entropy,

\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}

Since $\Delta G^{\circ} = 0$ , this simplifies to

\Delta H^{\circ} = T\Delta S^{\circ}

Converting the given temperature to Kelvin, $T = 60 + 273 = 333\ K$ , and rearranging for $\Delta S^{\circ}$ ,

\Delta S^{\circ} = \frac{\Delta H^{\circ}}{T} = \frac{666}{333}

Calculating this ratio,

\Delta S^{\circ} = 2.0\ \text{kJ K}^{-1}mol^{-1}

Hence, the correct answer is B.

Q2 · 2026

Consider the following reaction :

2A(g) + B(g) \rightarrow 2D(g)

\Delta U^{\ominus} = -10\ \text{kJ mol}^{-1} \text{ and } \Delta S^{\ominus} = -44\ JK^{-1} \text{ at } 298\ K.

Identify the correct option with $\Delta G^{\ominus}$ for the reaction and spontaneity of the reaction at 298 K.

(Given : $R = 8.31\ J\ mol^{-1}K^{-1}$ )

A.
$-1.635\ \text{kJ mol}^{-1}$ , spontaneous
B.
$-0.63568\ \text{kJ mol}^{-1}$ , spontaneous
C.
$+0.63568\ \text{kJ mol}^{-1}$ , non-spontaneous
D.
$+1.635\ \text{kJ mol}^{-1}$ , non-spontaneous

Answer: C

To connect $\Delta U^{\ominus}$ with $\Delta H^{\ominus}$ , use the relation between enthalpy change and internal energy change for a reaction involving gases,

\Delta H^{\ominus} = \Delta U^{\ominus} + \Delta n_g RT

Here $\Delta n_g$ is the change in the number of moles of gas, calculated as moles of gaseous products minus moles of gaseous reactants. For $2A(g)+B(g)\rightarrow 2D(g)$ ,

\Delta n_g = 2 - (2+1) = -1

Substituting $\Delta U^{\ominus} = -10000\ J/mol$ , $\Delta n_g = -1$ , $R = 8.31\ J\ mol^{-1}K^{-1}$ , and $T = 298\ K$ ,

\Delta H^{\ominus} = -10000 + (-1)(8.31)(298) = -12476\ J/mol \approx -12.48\ kJ/mol

To find whether the reaction is spontaneous, use the Gibbs-Helmholtz equation,

\Delta G^{\ominus} = \Delta H^{\ominus} - T\Delta S^{\ominus}

Substituting $\Delta H^{\ominus} = -12.48\ kJ/mol$ , $T = 298\ K$ , and $\Delta S^{\ominus} = -0.044\ kJ/K$ ,

\Delta G^{\ominus} = -12.48 - (298)(-0.044) = -12.48 + 13.112 = +0.632\ kJ/mol

Since $\Delta G^{\ominus}$ comes out positive, the reaction is non-spontaneous at 298 K.

Hence, the correct option is C.

Q3 · 2026

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 :

A.
400 J
B.
300 J
C.
700 J
D.
500 J

Answer: B

The first law of thermodynamics relates the change in internal energy to heat and work as

\Delta U = q + w

Since heat is absorbed by the system, $q$ is taken as positive,

q = +500\ J

Since work is done by the system (not on it), $w$ is taken as negative,

w = -200\ J

Substituting these values into the first law equation,

\Delta U = 500 + (-200) = 300\ J

Hence, the change in internal energy is 300 J, so the correct option is B.

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