Question

A certain reaction is spontaneous at temperatures below 400. K but is not spontaneous at temperatures above 400. K. If ΔH° for the reaction is -20. kJ mol-1 and it is assumed that ΔH° and ΔS° do not change appreciably with temperature, then the value of ΔS° for the reaction is

Answer 1

ΔS = 0.05 kJ /mol or 50 J /mol

Step-by-step explanation:

As given that the reaction is spontaneous below 400 K and non spontaneous above 400K. Thus the reaction is at equilibrium at 400 K temperature.

At equilibrium the change in free energy ΔG is zero.

The relation between free energy change, enthalpy change, entropy change and temperature is

ΔG = ΔH - TΔS

Where

ΔH = enthalpy change

T = temperature

ΔS = entropy change

at equilibrium (400 K)

ΔG = 0 = (-20) + 400XΔS

ΔS = 0.05 kJ /mol or 50 J /mol