Question

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

Answer 1

Answer:

-0.050 kJ/mol.K

Explanation:

• A certain reaction is thermodynamically favored at temperatures below 400. K, that is, ΔG° < 0 below 400. K

• The reaction is not favored at temperatures above 400. K, that is. ΔG° > 0 above 400. K

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

Answer 2

Answer:

$\Delta _RS=0.05kJ/mol$ or simply positive.

Explanation:

Hello,

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

$\Delta _RG=\Delta _RH-T\Delta _RS$

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

$-0.1kJ/mol=-20kJ/mol-399K\Delta _RS\\\Delta _RS=\frac{-0.1kJ/mol+20kJ/mol}{399K} \\\\\Delta _RS=0.05kJ/mol$

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,