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

• 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. Mar 31, 2020 at 1:49

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).