Question

For a particular reaction, ΔH∘=67.7 kJ/molΔH∘=67.7 kJ/mol and Δ????∘=126.9 J/(mol⋅K).ΔS∘=126.9 J/(mol⋅K). Assuming these values change very little with temperature, at what temperature does the reaction change from nonspontaneous to spontaneous in the forward direction?

Answer 1

Answer:

$T=533.49\, K$

For temperatures higher than 533.49 K we will see a spontaneous reaction, and for temperatures lower than that the reaction will not be spontaneous.

Explanation:

When are chemical reactions spontaneous? To find out we need to look at the reaction's change in Gibbs Free energy:

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

When this is greater than zero, the reaction isn't spontaneous, when it is less than zero, we have a spontaneous reaction. The reaction must then change from spontaneous to non spontaneous when $\Delta G=0$. If we insert that into our equation we get:

$0=\Delta H-T\Delta S\\\\T=\frac{\Delta H}{\Delta S}$

That is the temperature at which the reaction's spontaneity will change, plugging in our values we find:

$T=533.49\, K$

At that temperature we have  $\Delta G=0$.

Now, at a temperature greater than this one, the entropy term in our equation for the Gibbs' free energy of reaction will take over, and make  $\Delta G$, thus the reaction will be spontaneous.

On the other hand, if we lower the temperature, we will have a smaller entropy term, and we will have:  $\Delta G>0$. That is, the reaction will not be spontaneous. Therefore for temperatures higher than 533.49 K we will see a spontaneous reaction, and for temperatures lower than that the reaction will not be spontaneous.