Question 16.5: DETERMINING THE TEMPERATURE AT WHICH A REACTION BECOMES SPON...

(a) To determine whether the reaction is spontaneous at 25 °C, we need to determine the sign of ΔG = ΔH – TΔS. At 25 °C (298 K), ΔG for the reaction is

ΔG = ΔH – TΔS = (98.8 kJ) – (298 K)(0.1415 kJ/K)

= (98.8 kJ) – (42.2 kJ)

= 56.6 kJ

Because the positive ΔH term is larger than the positive TΔS term, ΔG is positive and the reaction is nonspontaneous at 298 K.

(b) At sufficiently high temperatures, TΔS becomes larger than ΔH, ΔG becomes negative, and the reaction becomes spontaneous. We can estimate the temperature at which ΔG changes from positive to negative by setting ΔG = ΔH – TΔS = 0.
Solving for T, we find that the reaction becomes spontaneous at 698 K:

T = \frac{ΔH}{ΔS}  =  \frac{98.8  kJ}{0.1415  kJ/K} = 698 K

This calculation assumes the values of ΔH and ΔS are unchanged on going from 298 K to 698 K. In general, the enthalpies and entropies of both reactants and products increase with increasing temperature, but the increases for the products tend to cancel the increases for the reactants. As a result, values of ΔH and ΔS for a reaction are relatively independent of temperature, at least over a small temperature range. In this example, the temperature range is quite large (400 K), and so the calculated value of T is only an estimate.

BALLPARK CHECK
(a) Let’s use rounded values of ΔH(100 kJ), T(300K), and ΔS(0.14 kJ/K) to estimate the relative values of ΔH and TΔS. Because TΔS = 300 K × 0.14 kJ/K = 42 kJ is smaller than Δ(100 kJ), ΔG = ΔH – TΔS is positive and the reaction is nonspontaneous at 300 K. The ballpark check agrees with the solution.
(b) The temperature at which the reaction becomes spontaneous is approximately (100 kJ)/(0.14 kJ/K) ≈ 700 K, in agreement with the solution.