Consider the following reaction: H_2(~g)+F_2(~g) 2 HF(g) ^ =-542k… - Vidyadip

Question

Consider the following reaction: $\mathrm{H}_2(\mathrm{~g})+\mathrm{F}_2(\mathrm{~g}) \rightarrow 2 \mathrm{HF}(\mathrm{g}) \Delta\text{H}^\circ=-542\text{kJ},\ \Delta\text{S}^\circ=14\text{JK}^{-1}$
Calculate the $\Delta\text{G}^\circ$ value for the reaction and state if the reaction is spontaneous at 298K.

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Answer

$\Delta\text{G}^\circ=\Delta\text{H}^\circ-\text{T}\Delta\text{S}^\circ$$\Delta\text{G}^\circ=-542\text{kJ}-\frac{298\times14}{1000}\text{kJ}$
$=-542-4=-546\text{kJ}$
Since $\Delta\text{G}^\circ$ is negative, therefore, reaction is spontaneous at 298K.

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