An endothermic reaction A → B proceeds to completion. Predict the sign of ∆ S .

For endothermic reaction, ∆ H is positive. So an endothermic reaction can proceed by decreasing its Gibbs energy, if ∆ H - T ∆ S is negative, which is possible only if ∆ S is positive. Hence, the sign of the entropy change is positive and the reaction is feasible.

What will be the sign of ∆ S for the reaction: N 2 ( g ) + O 2 ( g ) → 2 NO ( g ) ? Give reasons in support of your answer.

The reaction of nitrogen and oxygen is given as follows, N 2 ( g ) + O 2 ( g ) → 2 NO ( g ) . There are two moles of different reactants involved in this reaction and there are two moles of only one product. Therefore, the sign of entropy change is zero, ∆ S = 0 . There are two moles of two different gaseous reactants which have similar randomness as two moles of the gaseous product. Hence, the randomness remains same.