Question

A reaction has ΔH° = 100 kJ/mol and ΔS°=250 J/mol·K. Is the reaction spontaneous at room temperature? If not, under what temperature conditions will it become spontaneous?

Answer 1

The reaction with ΔH° = 100 kJ/mol and ΔS° = 250 J/mol·K is nonspontaneous at room temperature, as ΔG° is positive (25.5 kJ/mol). It becomes spontaneous at temperatures above 400 K, where ΔG° would be negative.

Step-by-step explanation:

To determine if a reaction with ΔH° = 100 kJ/mol and ΔS°=250 J/mol·K is spontaneous at room temperature, we must calculate the Gibbs free energy change (ΔG°) using the equation ΔG° = ΔH° - TΔS°. At room temperature (assume 298 K), ΔG° = (100 kJ/mol) - (298 K)(250 J/mol·K) = 100 kJ/mol - 74.5 kJ/mol = 25.5 kJ/mol. Because ΔG° is positive, the reaction is nonspontaneous at room temperature.

For the reaction to become spontaneous, ΔG° must be negative. We can determine the temperature at which the reaction becomes spontaneous by setting ΔG° to zero and solving for T, i.e., T = ΔH° / ΔS°. Therefore, T = (100,000 J/mol) / (250 J/mol·K) = 400 K. The reaction becomes spontaneous at temperatures above 400 K.