Question

Zn(s) + Hg2SO4(s) + 7H2O(l) --> ZnSO4.7H2O(s) + 2Hg(l)

A) what is the sign of the entropy change and what does this indicate about the reaction?

B) What is happening to cause the sign of the entropy change? And why is the magnitude of the entropy change so large?

Answer 1

Ans. #A. The sign of entropy change for this reaction is “negative”. A negative entropy indicates that the reaction does NOT occurs under given conditions.

A reaction does NOT occur as long as its entropy change is negative.

#B. Entropy for a reaction decreases when-

I. Number of moles of reactants are greater than that of products.

In this reaction, total number of moles of reactant is 9 (= 1 mol Zn, 1 mol Hg₂SO₄ and 7 moles of H₂O). The number of moles of products formed is 3 (1 mol ZnSO₄.7H₂O and 2 mol Hg).  

Due to lowering of number of moles of product, when compared to the reactants, the system has less number of particles with randomness and less entropy.

II. Phase transition from liquid to solid.

When a liquid (aqueous) reactant is converted into solid, the entropy of reaction decreases because the resultant product has minimal extent of randomness (or freedom of random motion) due to being in solid state.

In the reactions, 7 moles of liquid water produce 2 moles of liquid Hg. There is reduction in the number of number of liquid molecules if the reaction proceeded to the right.

Therefore, due to reduction in number of moles of chemical species and prevalent formation of solid state products, the reaction has a large negative entropy change. It makes the reaction non-feasible under given conditions.