Is it energetically favorable for ethanol combustion at 125 K, where change in enthalpy is 4000 J/mol and change in entropy is 50 J/Kmol? Why?

Answer 1

For a reaction to be energetically favorable, ∆G should be negative.

∆G = ∆H - T∆S

∆G = ?

∆H = 4000 J/mol

T = 125K

∆S = 50 J/Kmol

Solve for ∆G...

∆G = 4000 J/mol - (125K)(50 J/Kmol)

∆G = -2250 J/mol

Since ∆G is negative, the reaction would be energetically favorable.