Given ∆H=+30.0kJmol−1 ∆S=0.06kJK−1mol−1 p=1atm ∆G=0 We know that,
∆G=∆H−T∆S⇒0=∆H−T∆S
if, we put temperature 227∘C or 227+273=500K In equation then the value of free energy is obtained will be zero. Hence, ∆G=30−500×0.06 ∆G=30−30=0 and nature of reaction will be non-spontaneous.