The reaction below is spontaneous at room temperature but becomes nonspontaneous at a much higher temperature.

Question

From this fact alone, obtain the the signs of H° and S°, assuming that H° and S° do not change much with temperature.

N2(g) + 3 H2(g) 2 NH3(g)

H°
positive
negative

S°
positive
negative
Explain your reasoning.

Answer 1

If you are having a hard time conceptualizing this, just pick numbers and see how it works.

deltaG=deltaH-TdeltaS (the reaction is spontaneous if G<0)

So, if the reaction is spontaneous at room temp but not at higher temps, it means that delta S is unfavorable (negative). You know this because the entropy term TdeltaS becomes greater in absolute magnitude as the temperature increases. Also, just by looking at the equation, 2 molecules are produced from 3, so the entropy of the system decreased.

This also tells you that deltaH must be negative (favorable). If S and H were both unfavorable, the reaction would never be spontaneous.

Answer 2

The important relation here is:

Delta G = DH - TDS

At room temp the reaction is spontaneous, so DG is negative. But, as the temp increases, it becomes non-spontaneous. So, the term TDS must be negative, so DS must be negative. DH must also be negative so that at low temperatures, DG can be negative.

If this doesn't make sense, send me a message, and I'll try again...