For the reaction A longrightarrow B,Delta H^{ | vedclass

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Question

(A) At equilibrium,the change in Gibbs free energy is $\Delta G^{\circ} = 0 \, kJ \, mol^{-1}$.Using the relation $\Delta G^{\circ} = \Delta H^{\circ}

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$0 \, kJ \, mol^{-1}$ and $300 \, K$

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(A) At equilibrium,the change in Gibbs free energy is $\Delta G^{\circ} = 0 \, kJ \, mol^{-1}$.Using the relation $\Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ}$,at equilibrium,we have $0 = \Delta H^{\circ} - T \Delta S^{\circ}$.Therefore,$T = \frac{\Delta H^{\circ}}{\Delta S^{\circ}}$.Given $\Delta H^{\circ} = 7.5 \, kJ \, mol^{-1} = 7500 \, J \, mol^{-1}$ and $\Delta S^{\circ} = 25 \, J \, K^{-1} \, mol^{-1}$.$T = \frac{7500 \, J \, mol^{-1}}{25 \, J \, K^{-1} \, mol^{-1}} = 300 \, K$.Thus,the values are $0 \, kJ \, mol^{-1}$ and $300 \, K$.

For the reaction $A \longrightarrow B$,$\Delta H^{\circ} = 7.5 \, kJ \, mol^{-1}$ and $\Delta S^{\circ} = 25 \, J \, K^{-1} \, mol^{-1}$. The value of $\Delta G^{\circ}$ and the temperature at which the reaction reaches equilibrium are,respectively:

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