Q27 / 90 · JEE Main 2019 (11 Jan Shift 2) · Chemistry

The reaction $\mathrm{MgO}(\mathrm{s})+\mathrm{C}(\mathrm{s}) \rightarrow \mathrm{Mg}(\mathrm{s})+\mathrm{CO}(\mathrm{g}),$ for which $\Delta \mathrm{H}^{\circ}=+491.1 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $\Delta \mathrm{S}^{\circ}=198.0 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$ is not feasible at $298 \mathrm{~K}$. Temperature above which reaction will be feasible is

  • A
    $2040.5 \mathrm{~K}$
  • B
    $1890.0 \mathrm{~K}$
  • Correct answer
    C
    $2480.3 \mathrm{~K}$
  • D
    $2380.5 \mathrm{~K}$

Solution

$\mathrm{MgO}(\mathrm{s})+\mathrm{C}(\mathrm{s}) \rightarrow \mathrm{Mg}(\mathrm{s})+\mathrm{CO}(\mathrm{g})$ For a reaction to be spontaneous $\Delta \mathrm{G} < 0$ $\Delta \mathrm{H}^{\circ}-\mathrm{T} \Delta \mathrm{S}^{\circ} < 0$ $\Rightarrow \mathrm{T}>\frac{\Delta \mathrm{H}^{\circ}}{\Delta \mathrm{S}^{\circ}}$ $\mathrm{T}>\frac{491.1 \times 1000}{198}$ $\mathrm{T}>2480.3 \mathrm{~K}$

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

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