Question

1: At which temperature would a reaction withΔH = -102 kJ/mol, ΔS = -0.188 kJ/(mol×K) be spontaneous? 2: At which temperature would a reaction withΔH = 132 kJ/mol, ΔS = 0.200 kJ/(mol×K) be spontaneous?

Answer 1

Answer:

1: At temperatures below 542.55 K

2: At temperatures above 660 K

Explanation:

Hello there!

In this case, according to the thermodynamic definition of the Gibbs free energy, it is possible to write the following expression:

$\Delta G=\Delta H-T\Delta S$

Whereas ΔG=0 for the spontaneous transition. In such a way, we proceed as follows:

1:

$0=\Delta H-T\Delta S\\\\T=\frac{-102kJ/mol}{-0.188kJ/mol-K} \\\\T=542.55K$

It means that at temperatures lower than 542.55 K the reaction will be spontaneous.

2:

$0=\Delta H-T\Delta S\\\\T=\frac{132kJ/mol}{0.200kJ/mol-K} \\\\T=660K$

It means that at temperatures higher than 660 K the reaction will be spontaneous.

Best regards!