The values of $\Delta H$ and $\Delta S$ for the given reaction are $170 ext{ kJ}$ and $170 ext{ J K}^{-1}$

Question

The values of $\Delta H$ and $\Delta S$ for the given reaction are $170 	ext{ kJ}$ and $170 	ext{ J K}^{-1}$, respectively.
$	ext{C(graphite)} + 	ext{CO}_2	ext{(g)} ightarrow 2	ext{CO(g)}$
This reaction will be spontaneous at:

Accepted Answer

For a reaction to be spontaneous, the Gibbs free energy change ($\Delta G$) must be negative ($\Delta G < 0$). We know that, $\Delta G = \Delta H - T\Delta S$ So, $\Delta H - T\Delta S < 0 \implies T > \frac{\Delta H}{\Delta S}$ Given: $\Delta H = 170 ext{ kJ} = 170 imes 10^3 ext{ J}$ $\Delta S = 170 ext{ J K}^{-1}$ Substituting the values: $T > \frac{170 imes 10^3 ext{ J}}{170 ext{ J K}^{-1}} = 1000 ext{ K}$ Thus, the reaction will be spontaneous at any temperature strictly greater than $1000 ext{ K}$. Among the given options, $1110 ext{ K}$ is the only temperature greater than $1000 ext{ K}$.

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