A certain reaction is thermodynamically favored at temperatures below 400. K, but it is not favored at temperatures above 400. K. The value of ΔHo for the reaction is –20 kJ/mol. What is the value of ΔSo for the reaction

-0.050 kJ/mol.K

Explanation:

All in all, ΔG° = 0 at 400. K.

We can find ΔS° using the following expression.

ΔG° = ΔH° - T.ΔS°

0 = -20 kJ/mol - 400. K .ΔS°

ΔS° = -0.050 kJ/mol.K

or simply positive.

Explanation:

Hello,

In this case, one considers the Gibbs free energy for the reaction as:

Which should be negative if is spontaneous (thermodynamically favored). Thus, for the given information about the 400K, the entropy of reaction turns out:

Such value is obtained assuming a temperature below 400K and a negative value for the Gibbs free energy of such reaction, for which the entropy must be positive for all the possibilities, accomplishing the thermodynamic favorability.

Best regards,

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