Consider the following reaction:A_{(g)} + 3 B | vedclass

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(B) The relationship between Gibbs free energy,enthalpy,and entropy is given by the equation: $\Delta G^{\ominus} = \Delta H^{\ominus} - T \Delta S^{\

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$-50.33$

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(B) The relationship between Gibbs free energy,enthalpy,and entropy is given by the equation: $\Delta G^{\ominus} = \Delta H^{\ominus} - T \Delta S^{\ominus}$.Given values are: $\Delta G^{\ominus} = -9 \ kJ = -9000 \ J$,$\Delta H^{\ominus} = -24 \ kJ = -24000 \ J$,and $T = 25 + 273 = 298 \ K$.Substituting these values into the equation: $-9000 = -24000 - (298 	imes \Delta S^{\ominus})$.Rearranging the terms: $298 	imes \Delta S^{\ominus} = -24000 + 9000 = -15000$.Solving for $\Delta S^{\ominus}$: $\Delta S^{\ominus} = -15000 / 298 \approx -50.33 \ J \ K^{-1}$.

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:

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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: