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

1: At temperatures below 542.55 K

2: At temperatures above 660 K

Step-by-step 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=(-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=(132kJ/mol)/(0.200kJ/mol-K) \\\\T=660K`

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

Best regards!