For the given reaction: H_{2(g)} + S_{(s)} to | vedclass

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(B) For a reaction to occur reversibly,the change in Gibbs free energy must be zero,i.e.,$\Delta G = 0$. Using the relation $\Delta G = \Delta H - T\D

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$250$

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(B) For a reaction to occur reversibly,the change in Gibbs free energy must be zero,i.e.,$\Delta G = 0$. Using the relation $\Delta G = \Delta H - T\Delta S$,we set $\Delta H - T\Delta S = 0$. Given $\Delta H_r = 100\, kJ/mol = 100,000\, J/mol$ and $\Delta S_r = 400\, J/mol\cdot K$. Substituting the values: $100,000 - T(400) = 0$. $T = \frac{100,000}{400} = 250\, K$.

For the given reaction: $H_{2(g)} + S_{(s)} \to H_2S_{(g)}$; $\Delta H_r = 100\, kJ/mol$ and $\Delta S_r = 400\, J/mol\cdot K$. The temperature at which the above reaction occurs reversibly is (Assuming $\Delta H_r$ and $\Delta S_r$ are independent of temperature) ..... $K$.

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