How to calculate the temperature at which a reaction becomes spontaneous? 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.