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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 $T$? $$8.9~\mathrm{kJ} = 58.1~\mathrm{kJ} -T(0.165~\mathrm{kJ}) $$
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.
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.
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
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).
