In which of the following reactions, standard reaction entropy changes $\left(\Delta S^{\circ}\right)$ is…
- $\mathrm{C}$ (graphite) $+\frac{1}{2} \mathrm{O}_2(g) \longrightarrow \mathrm{CO}(g)$
- $\mathrm{CO}(g)+\frac{1}{2} \mathrm{O}_2(g) \longrightarrow \mathrm{CO}_2(g)$
- $\mathrm{Mg}(s)+\frac{1}{2} \mathrm{O}_2(g) \longrightarrow \mathrm{MgO}(s)$
- $\frac{1}{2} \mathrm{C}$ (graphite) $+\frac{1}{2} \mathrm{O}_2(g) \longrightarrow \frac{1}{2} \mathrm{CO}_2(g)$
Solution
$\mathrm{C} \text { (graphite) }+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{CO}(\mathrm{g})$
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.
Asked in: NEET 2012 (Screening)