Question

For a chemical reaction the values of ∆H and ∆S at 300 K are – 10 kJ mol^-1 and – 20 JK^-1 mol^-1 respectively. What is the value of ∆G of the reaction? Calculate the ∆G of a reaction at 600K assuming ∆H and ∆S values are constant. Predict the nature of the reaction?

Answer 1

Given: 

∆H = -10 kJ mol-1 = -10000 J mol-1 

∆S = – 20 JK^-1 mol-1 

T = 300 K

∆G? 

∆G = ∆H – T∆S 

∆G = -10 kJ mol^-1 – 300 K × (-20 × 10^-3 ) kJ K^-1 mol^-1 

∆G = (-10 + 6) kJ mol^-1 

∆G = – 4 kJ mol^-1

At 600 K, 

∆G = – 10 kJ mol^-1 – 600 K × (-20 × 10^-3) k^-1 mol^-1 

∆G = (-10 + 12) kJ mol^-1 

∆G + 2 kJ mol^-1

The value of ∆G is negative at 300K and the reaction is spontaneous, but at 600K the value ∆G becomes positive and the reaction is nonspontaneous.