(D) For a reaction to be at equilibrium, the Gibbs free energy change $\Delta G^0 = 0$.The relationship is given by $\Delta G^0 = \Delta H^0 - T\Delta

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(D) For a reaction to be at equilibrium, the Gibbs free energy change $\Delta G^0 = 0$.The relationship is given by $\Delta G^0 = \Delta H^0 - T\Delta S^0$.Setting $\Delta G^0 = 0$, we get $T = \frac{\Delta H^0}{\Delta S^0}$.Convert $\Delta H^0$ to $\text{J}$: $\Delta H^0 = -345 \times 10^3 \text{ J}$.Substitute the values: $T = \frac{-345 \times 10^3 \text{ J}}{-123 \text{ JK}^{-1}}$.$T \approx 2804.87 \text{ K} \approx 2805 \text{ K}$.

Class 11 Chemistry Thermodynamics Free energy and Work

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For a certain reaction, $\Delta H^0 = -345 \text{ kJ}$ and $\Delta S^0 = -123 \text{ JK}^{-1}$. At what temperature will the change over from spontaneous to non-spontaneous occur (in $\text{ K}$)?

MHT CET 2026 All MHT CET

A
$1052$
B
$1956$
C
$2568$
D
$2805$

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For a certain reaction, $\Delta H^0 = -345 \text{ kJ}$ and $\Delta S^0 = -123 \text{ JK}^{-1}$. At what temperature will the change over from spontaneous to non-spontaneous occur (in $\text{ K}$)?

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If for a certain reaction $\Delta_{r} H$ is $30 \; kJ \; mol^{-1}$ at $450 \; K$,the value of $\Delta_{r} S$ (in $J \; K^{-1} \; mol^{-1}$) for which the same reaction will be spontaneous at the same temperature is

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