Q65 / 180 · NEET 2026 (Cancelled) · Chemistry

Consider the following reaction :
$2\text{A (g)} + \text{B (g)} \rightarrow 2\text{D (g)}$
$\Delta U^{\ominus} = -10$ kJ mol$^{-1}$ and $\Delta S^{\ominus} = -44$ J K$^{-1}$ 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
    $-1.635$ kJ mol$^{-1}$, spontaneous
  • B
    $-0.63568$ kJ mol$^{-1}$, spontaneous
  • Correct answer
    C
    $+0.63568$ kJ mol$^{-1}$, non-spontaneous
  • D
    $+1.635$ kJ mol$^{-1}$, non-spontaneous

Solution

The change in the number of moles of gaseous species is:

$\Delta n_g = n_p - n_r = 2 - (2 + 1) = -1$

The standard enthalpy change $\Delta H^{\ominus}$ is given by:

$\Delta H^{\ominus} = \Delta U^{\ominus} + \Delta n_g R T$

Substituting the given values:

$\Delta H^{\ominus} = -10 + (-1) \times (8.31 \times 10^{-3}) \times 298$

$\Delta H^{\ominus} = -10 - 2.47638 = -12.47638 \text{ kJ mol}^{-1}$

The standard Gibbs free energy change $\Delta G^{\ominus}$ is:

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

$\Delta G^{\ominus} = -12.47638 - 298 \times (-44 \times 10^{-3})$

$\Delta G^{\ominus} = -12.47638 + 13.112 = +0.63562 \text{ kJ mol}^{-1}$

Since $\Delta G^{\ominus} > 0$, the reaction is non-spontaneous.

Answer: $+0.63568$ kJ mol$^{-1}$, non-spontaneous

Hence, the correct answer is option: 'C'.

Relevant Questions