Practice Problem 8

Predict whether the following reaction is still spontaneous at 500°C:

N₂(g) + 3 H₂(g) 2 NH₃(g)

Assume that the values of  H^o and S° used in Practice Problem 7 are still valid at this temperature.

Solution

Before we can decide whether the reaction is still spontaneous we need to calculate the temperature of the kelvin scale:

T_K = 500^o C + 273 = 773 K

We then multiply the entropy term by this temperature and subtract this quantity from the enthalpy term:

G^o₇₇₃ = H^o₂₉₈ - T S^o₂₉₈

            = 92,220 J - (773 K x -198.75 J/K)

        = 92,220 J - (-153,600 J)

        = 61,380 J

G^o = 61.4 kJ

Because the entropy term becomes larger as the temperature increases, the reaction changes from one which is favorable at low temperatures to one that is unfavorable at high temperatures.