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In which of the following reactions, standard reaction entropy changes $\left(\Delta S^{\circ}\right)$ is positive and standard Gibb's energy change $\left(\Delta G^{\circ}\right)$ decreases sharply with increasing temperature?
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The correct answer is:
$\mathrm{C}$ (graphite) $+\frac{1}{2} \mathrm{O}_2(g) \longrightarrow \mathrm{CO}(g)$
Among the given reactions only in the case of
$\mathrm{C} \text { (graphite) }+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{CO}(\mathrm{g})$
entropy increases because randomness (disorder) increases. Thus, standard entropy change $\left(\Delta S^{\circ}\right)$ is positive.
Moreover, it is a combustion reaction and all the combustion reactions are generally exothermic, i.e., $\Delta H^{\circ}=-$ ve
We know that
$\begin{aligned}
& \Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ} \\
& \Delta G^{\circ}=-v \mathrm{v}-T(+\mathrm{ve})
\end{aligned}$
Thus, as the temperature increases, the value of $\Delta G^{\circ}$ decreases.
$\mathrm{C} \text { (graphite) }+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{CO}(\mathrm{g})$
entropy increases because randomness (disorder) increases. Thus, standard entropy change $\left(\Delta S^{\circ}\right)$ is positive.
Moreover, it is a combustion reaction and all the combustion reactions are generally exothermic, i.e., $\Delta H^{\circ}=-$ ve
We know that
$\begin{aligned}
& \Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ} \\
& \Delta G^{\circ}=-v \mathrm{v}-T(+\mathrm{ve})
\end{aligned}$
Thus, as the temperature increases, the value of $\Delta G^{\circ}$ decreases.
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