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In which of the following reactions, standard entropy change $\left(\Delta \mathrm{S}^{\circ}\right)$ is positive and standard Gibb's energy change $\left(\Delta \mathrm{G}^{\circ}\right)$ decreases sharply with increasing temperature?
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The correct answer is:
$\mathrm{C}$ (graphite) $+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}(\mathrm{g})$
Since, in the first reaction gaseous products are forming from solid carbon hence entropy will increase i.e. $\Delta \mathrm{S}=+\mathrm{ve}$.
$\mathrm{C}(\mathrm{gr} .)+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}(\mathrm{g}) ; \Delta \mathrm{S}^{\circ}=+\mathrm{ve}$
Since, $\Delta \overrightarrow{\mathrm{G}^{\circ}}=\Delta \mathrm{H}^{\circ}-\mathrm{T} \Delta \mathrm{S}$ hence the value of
$\Delta$ G decrease on increasing temperature.
$\mathrm{C}(\mathrm{gr} .)+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}(\mathrm{g}) ; \Delta \mathrm{S}^{\circ}=+\mathrm{ve}$
Since, $\Delta \overrightarrow{\mathrm{G}^{\circ}}=\Delta \mathrm{H}^{\circ}-\mathrm{T} \Delta \mathrm{S}$ hence the value of
$\Delta$ G decrease on increasing temperature.
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