A process has Delta H = 200 J mol^{-1} and | vedclass

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(C) For a process to be spontaneous,the Gibbs free energy change $\Delta G$ must be negative,i.e.,$\Delta G < 0$.We know that $\Delta G = \Delta H - T

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$5$

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(C) For a process to be spontaneous,the Gibbs free energy change $\Delta G$ must be negative,i.e.,$\Delta G We know that $\Delta G = \Delta H - T\Delta S$.Setting $\Delta G = 0$ gives the equilibrium temperature: $0 = \Delta H - T\Delta S$.Therefore,$T = \frac{\Delta H}{\Delta S}$.Substituting the given values: $T = \frac{200 \ J \ mol^{-1}}{40 \ J \ K^{-1} \ mol^{-1}} = 5 \ K$.For the process to be spontaneous,$T$ must be greater than $5 \ K$ (since $\Delta H$ and $\Delta S$ are both positive,the process is spontaneous at higher temperatures).

$A$ process has $\Delta H = 200 \ J \ mol^{-1}$ and $\Delta S = 40 \ J \ K^{-1} \ mol^{-1}$. Choose the minimum temperature above which the process will be spontaneous.

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