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If the enthalpy and entropy change for a reaction at $298 \mathrm{~K}$ are $-145 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $-650 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ respectively, which one of the following statements is correct?
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The correct answer is:
$\Delta \mathrm{G}=48.7 \mathrm{~kJ} \mathrm{~mol}^{-1}$, the reaction is non-spontaneous
$\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}$
We have, $\Delta \mathrm{H}=-145 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $\Delta \mathrm{S}$
$=-650 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$
$\Rightarrow \Delta \mathrm{G}=\left(-145 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)-298\left(-650 \times 10^{-3} \mathrm{~kJ}\right.$ $\left.\mathrm{mol}^{-1} \mathrm{k}^{-1}\right)=+48.7 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and the reaction is nonSpontaneous.
We have, $\Delta \mathrm{H}=-145 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $\Delta \mathrm{S}$
$=-650 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$
$\Rightarrow \Delta \mathrm{G}=\left(-145 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)-298\left(-650 \times 10^{-3} \mathrm{~kJ}\right.$ $\left.\mathrm{mol}^{-1} \mathrm{k}^{-1}\right)=+48.7 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and the reaction is nonSpontaneous.
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