If the enthalpy and entropy change for a reac | vedclass

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(D) The Gibbs free energy change is given by the equation: $\Delta G = \Delta H - T \Delta S$. Given: $\Delta H = -145 \ kJ \ mol^{-1}$,$\Delta S = -6

Vedclass · Accepted Answer

$\Delta G = +48.7 \ kJ \ mol^{-1}$,the reaction is non-spontaneous

Vedclass · Answer

(D) The Gibbs free energy change is given by the equation: $\Delta G = \Delta H - T \Delta S$. Given: $\Delta H = -145 \ kJ \ mol^{-1}$,$\Delta S = -650 \ J \ K^{-1} \ mol^{-1} = -0.650 \ kJ \ K^{-1} \ mol^{-1}$,and $T = 298 \ K$. Substituting the values: $\Delta G = -145 \ kJ \ mol^{-1} - (298 \ K 	imes -0.650 \ kJ \ K^{-1} \ mol^{-1})$. $\Delta G = -145 + 193.7 = +48.7 \ kJ \ mol^{-1}$. Since $\Delta G > 0$,the reaction is non-spontaneous.

If the enthalpy and entropy change for a reaction at $298 \ K$ are $-145 \ kJ \ mol^{-1}$ and $-650 \ J \ K^{-1} \ mol^{-1}$ respectively,which one of the following statements is correct?

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