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At $25~^\circ\mathrm{C}$, $298~\mathrm{K}$, the reduction of copper(I) oxide, $ \Delta H = 58.1~\mathrm{kJ}$, $\Delta S = 165~\mathrm{J/K}$, is nonspontaneous, $ \Delta G = 8.9~\mathrm{kJ}$.
Calculate the temperature at which the reaction becomes spontaneous.

So for this question I'm thinking I use the equation

$$ \Delta G = \Delta H - T\Delta S$$

Then plug in the values and solve for $T$? $$8.9~\mathrm{kJ} = 58.1~\mathrm{kJ} -T(0.165~\mathrm{kJ}) $$

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  • $\begingroup$ the formula you want is : T = delta H/delta S the one that said to set G to zero was correct but they didn't show how that formula looks when you do that, so here it is. $\endgroup$ Mar 31, 2020 at 1:49

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You have the right equation. But the problem already gives you the temperature at which $\Delta G$ is 8.9 kJ.

Assume that $\Delta H$ and $\Delta S$ are constant and do not vary with temperature. $\Delta G$ still will though because of the $T$ in the $\Delta G = \Delta H - T\Delta S$ equation.

So plug in the values for $\Delta H$ and $\Delta S$, and also the $\Delta G$ value that you think represents the transition from spontaneous to non-spontaneous. You should be able to take it from there.

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Actually he had the wrong equation. The correct equation to use was $$\frac{d\left(\frac{\Delta G}{T}\right)}{dT}=-\frac{\Delta H}{T^2}$$

Sign error corrected on 10/16/15.

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What you really want to do here is use $ΔG = ΔH - TΔS$, use the given data for enthalpy and entropy. Set $ΔG$ to $0$, and solve for temperature, as that is the criteria for equilibrium and the point for $ΔG$ below which the reaction will be spontaneous

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Any value above 352K is the answer. Sum of temperature times delta S must be over 58.1(total of delta H) so that gibbs free energy is 0 or less . Thus divide 58.2 by 0.165 to find energy needed for delta G =0 any temp above this will result in a negative (spontaneous reaction).

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