For the reaction ΔH° is +137 kJ/mol and ΔS° is +120 J/K · mol. This reaction is .

Question

For the reaction  ΔH° is +137 kJ/mol and ΔS° is +120 J/K · mol. This reaction is .

Quizplus · Accepted Answer

The spontaneity of a reaction can be determined by calculating the Gibbs free energy change, ΔG°, using the equation ΔG° = ΔH° - TΔS°, where T is the temperature in Kelvin. 
If ΔG° is negative, the reaction is spontaneous; if ΔG° is positive, the reaction is nonspontaneous; if ΔG° is zero, the reaction is at equilibrium. 
In this case, ΔH° is positive (+137 kJ/mol) and ΔS° is positive (+120 J/K · mol), so the sign of ΔG° depends on temperature. 
At high temperatures, TΔS° becomes a large positive term, which can outweigh the positive ΔH° term, resulting in a negative ΔG° and spontaneous reaction. 
At low temperatures, TΔS° becomes a small positive term (or even negative), which cannot outweigh the positive ΔH° term, resulting in a positive ΔG° and nonspontaneous reaction. 
Therefore, the reaction is spontaneous only at high temperature.

For the Reaction ΔH° Is +137 KJ/mol and ΔS°