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In which of the following reactions, standard reaction entropy change ($\Delta S^\circ$) is positive and standard Gibb's energy change ($\Delta G^\circ$) decreases sharply with increasing temperature?
A
$\frac{1}{2}C$ (graphite) $+ \frac{1}{2}O_2(g) \rightarrow \frac{1}{2}CO_2(g)$
B
C (graphite) $+ \frac{1}{2}O_2(g) \rightarrow CO(g)$
C
$CO(g) + \frac{1}{2}O_2(g) \rightarrow CO_2(g)$
D
$Mg(s) + \frac{1}{2}O_2(g) \rightarrow MgO(s)$
Detailed Solution
$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$; if $\Delta S^\circ$ is positive, $\Delta G^\circ$ becomes more negative as T rises.
C(graphite) $+ \frac{1}{2}O_2(g) \rightarrow CO(g)$: $\Delta n_g = 1 - \frac{1}{2} = +\frac{1}{2}$, so $\Delta S^\circ$ is positive.
For $C \rightarrow CO_2$, $\Delta n_g = 0$ ($\Delta S^\circ \approx 0$); for $CO \rightarrow CO_2$ and $Mg \rightarrow MgO$, $\Delta n_g$ is negative ($\Delta S^\circ$ negative).
So the reaction forming CO from graphite is the answer.
C(graphite) $+ \frac{1}{2}O_2(g) \rightarrow CO(g)$: $\Delta n_g = 1 - \frac{1}{2} = +\frac{1}{2}$, so $\Delta S^\circ$ is positive.
For $C \rightarrow CO_2$, $\Delta n_g = 0$ ($\Delta S^\circ \approx 0$); for $CO \rightarrow CO_2$ and $Mg \rightarrow MgO$, $\Delta n_g$ is negative ($\Delta S^\circ$ negative).
So the reaction forming CO from graphite is the answer.