49.0k views
3 votes
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?

User Troy Brant
by
7.8k points

1 Answer

2 votes

Final answer:

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.

User JXITC
by
8.6k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.