Question

A reaction is taking place at constant pressure and 27°C temperature.∆G and ∆H for the reaction are 400KJ and 50KJ.Calculate the change in entropy.

Answer 1

Answer:

-7/6 kJ/K

Explanation:

given:

∆G=400 kJ

∆H=50 kJ

T=300 K

we know,

∆G=∆H-T∆S

=>∆S=(400-50)/-300 =-7/6 kJ/K