For a reaction Delta H = -30 kJ and Delta S | vedclass

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(C) The condition for spontaneity is given by the Gibbs free energy equation: $\Delta G = \Delta H - T\Delta S$.For the reaction to be at equilibrium,

Vedclass · Accepted Answer

$666.6$

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(C) The condition for spontaneity is given by the Gibbs free energy equation: $\Delta G = \Delta H - T\Delta S$.For the reaction to be at equilibrium,$\Delta G = 0$,which implies $T = \frac{\Delta H}{\Delta S}$.Given: $\Delta H = -30 \ kJ = -30000 \ J$ and $\Delta S = -45 \ J \ K^{-1}$.Substituting the values: $T = \frac{-30000 \ J}{-45 \ J \ K^{-1}} = 666.67 \ K$.Since both $\Delta H$ and $\Delta S$ are negative,the reaction is spontaneous at temperatures below $666.67 \ K$ and becomes non-spontaneous at temperatures above $666.67 \ K$.

For a reaction $\Delta H = -30 \ kJ$ and $\Delta S = -45 \ J \ K^{-1}$,at what temperature does the reaction change from spontaneous to non-spontaneous (in $K$)?

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