At a certain temperature ‘T’, the endothermic reaction A →  B proceeds virtually to the end. Determine

(i) sign of ∆S for this reaction

(ii) sign of ∆G for the reaction B →A at the temperature T, and 

(iii) the possibility of reaction B → A proceeding at a low temperature.

(i) Energy factor opposes. So enthalpy factor must favour, i.e., ∆S must be positive.

(ii) For A →B, ∆G is –ve, therefore, for B →A, ∆G will be +ve,.

(iii) For B→A, ∆H = -ve and ∆S is also –ve, i.e., ∆S opposes the process but at low temp., T∆S may be so low that ∆H is greater in magnitude than T∆S and the process will be spontaneous.