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(A) For a reaction to be spontaneous,the Gibbs free energy change $\Delta G^{\circ}$ must be less than $0$.Given the equation: $\Delta G^{\circ} = \De

Vedclass · Accepted Answer

$1118$

Vedclass · Answer

(A) For a reaction to be spontaneous,the Gibbs free energy change $\Delta G^{\circ}$ must be less than $0$.Given the equation: $\Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ}$.At equilibrium,$\Delta G^{\circ} = 0$,so $T = \frac{\Delta H^{\circ}}{\Delta S^{\circ}}$.Given values: $\Delta H^{\circ} = 179.1 \ kJ \ mol^{-1} = 179100 \ J \ mol^{-1}$ and $\Delta S^{\circ} = 160.2 \ J \ K^{-1} \ mol^{-1}$.Substituting these values into the equation:$T = \frac{179100 \ J \ mol^{-1}}{160.2 \ J \ K^{-1} \ mol^{-1}} \approx 1117.97 \ K$.Since the reaction is endothermic and entropy increases,the reaction becomes spontaneous at temperatures above $1118 \ K$.

Similar Questions

Consider the following reaction:
$A_{(g)} + 3 B_{(g)} \longrightarrow 2 C_{(g)} ; \Delta H^{\ominus} = -24 \ kJ$.
At $25^{\circ} C$,if $\Delta G^{\ominus}$ of the reaction is $-9 \ kJ$,the standard entropy change (in $J \ K^{-1}$) of the same reaction at the same temperature is:

At a temperature of $298 \ K$,if $\Delta H < T \Delta S$,what is the value of the equilibrium constant $K$?

For the oxidation of ammonia at $298 \ K$,the standard enthalpy and standard entropy changes are $-382.64 \ kJ \ mol^{-1}$ and $-145.6 \ J \ K^{-1} \ mol^{-1}$ respectively. The standard Gibbs energy change for the same reaction at $298 \ K$ is $..... \ kJ \ mol^{-1}$.

For the reaction occurring in the gaseous phase: $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$,which of the following conditions is correct for the reaction to be spontaneous at high temperatures?

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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