Thermodynamics and Thermochemistry 1 Question 11
11. The reaction, $\operatorname{MgO}(s)+\mathrm{C}(s) \rightarrow \operatorname{Mg}(s)+\mathrm{CO}(g)$,
for which $\Delta_{r} H^{\circ}=+491.1 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $\Delta_{r} S^{\circ}=198.0 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$, is not feasible at $298 \mathrm{~K}$.
Temperature above which reaction will be feasible is
(a) $2040.5 \mathrm{~K}$
(b) $1890.0 \mathrm{~K}$
(c) $2380.5 \mathrm{~K}$
(d) $2480.3 \mathrm{~K}$
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Solution:
- According to Gibbs-Helmholtz equation,
$$ \Delta_{r} G^{\circ}=\Delta_{r} H^{\circ}-T \Delta_{r} S^{\circ} $$
For a reaction to be feasible (spontaneous)
$$ \begin{aligned} & \Delta_{r} G^{\circ}<0 \ & \Delta_{r} H^{\circ}-T \Delta_{r} S^{\circ}<0 \end{aligned} $$
Given, $\Delta_{r} H^{\circ}=+491.1 \mathrm{~kJ} \mathrm{~mol}^{-1}$,
$$ \Delta_{r} S^{\circ}=198 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} $$
$\therefore 491.1 \times 10^{3}-T \times 198<0$
$$ T>\frac{491.1 \times 10^{3}}{198}=2480.3 \mathrm{~K} $$
$\therefore$ Above $2480.3 \mathrm{~K}$ reaction will become spontaneous.
