Question

At which temperature would a reaction with h = -92 kj/mol, s = -0.199 kj/(molk) be spontaneous?

Answer 1

Answer: 400 K is the answer for this question

Answer 2

The relationship between change in Gibb's free energy, change in enthalpy and change in entropy is:

Δ
`G^(0)` = Δ
`H^(0) - T`Δ
`S^(0)`

For a reaction at equilibrium, Δ
`G^(0)` = 0

So Δ
`H^(0) - T`Δ
`S^(0)` = 0

Δ
`H^(0) = T`Δ
`S^(0)`

-92 kJ/mol = T(-0.199kJ/(mol.K))

T = 462 K

At temperatures greater than 462 K, the value T [/tex]Δ
`S^(0)` would become more than Δ
`H^(0)`, giving a negative value for Δ
`G^(0)` . Therefore, the reaction would become spontaneous at temperatures > 462 K