Chemical Thermodynamics MCQs for NEET — Chemistry Questions with Answers

According to Problem 5.10 (ii) and its solution, 'At 0 K, the constituent particles are static and entropy is minimum. If temperature is raised to 115 K, these begin to move and oscillate about their equilibrium positions in the lattice and system becomes more disordered, therefore entropy increases.'

The text mentions, 'theoretical arguments and practical evidences have shown that entropy of solutions and super cooled liquids is not zero at 0 K.' This implies they lack the 'perfect order' characteristic of a pure crystalline solid at absolute zero, retaining some residual disorder (multiple microstates).

The text describes 'translational, rotational and vibrational motion' as contributing to entropy. Nuclear motion is not mentioned in this context.

The text explicitly states, 'On the other hand when temperature is lowered, the entropy decreases.'

The text states, 'Therefore, increase in entropy in such systems is the natural direction of a spontaneous change. This, in fact is the second law of thermodynamics.'

$\Delta H = \Delta U + p\Delta V = \Delta U + \Delta n_g RT$.

Isothermal: constant T; Isochoric: constant V; Isobaric: constant P; Adiabatic: no Q.

Evaporation (A), decomposition producing gases (C) and atomisation (D) all increase entropy; cooling a crystal (B) decreases entropy.

Work done by gas: $W = nRT \ln(P_1/P_2) = 1\cdot 2\cdot 298\cdot \ln 2 \approx 413.14$ cal (expansion → positive).

$\Delta H$ is negative (exothermic), so the products lie below the reactants by 74.8 kJ mol⁻¹, with an activation hump in between.