State whether the sign of the entropy change expected for each of the following processes will be positive or negative, and explain your predictions. (a) PCl3(l) + Cl2(g) ⟶PCl5(s) (b) 2HgO(s) ⟶2Hg(l) + O2(g) (c) H2(g) ⟶2H(g) (d) U(s) + 3F2(g) ⟶UF6(s)

Entropy (S) is a thermodynamic parameter which measures the randomness or the disorder in a system. Greater the disorder more positive will be the value of entropy.

The extent of disorder increases as substances transition from the solid to the gaseous state. i.e.

S(solid) < S(liquid) < S(gas)

The entropy change for a given reaction is:

[tex]\Delta S = S(products)-S(reactants)----(1)[/tex]

a) [tex]PCl3(l) + Cl2(g) \rightarrow PCl5(s)[/tex]

Here the reactants are in the liquid and gas phase which have higher entropy than the product which is in the solid phase i.e. lower entropy.

Since S(product) < S(reactant), based on equation 1, ΔS will be negative.

b) [tex]2HgO(s) \rightarrow 2Hg(l) + O2(g)[/tex]

Here the products are in the liquid and gas phase which have higher entropy than the reactant which is in the solid phase i.e. lower entropy.

Since S(product) > S(reactant), based on equation 1, ΔS will be positive.

c) [tex]H2(g) \rightarrow 2H(g)[/tex]

Here the products and reactants are in the gas phase. However the number of moles of products is greater than the reactants

Since S(product) > S(reactant), based on equation 1, ΔS will be positive.

d) [tex]U(s) + 3F2(g) \rightarrow UF6(s)[/tex]

Here one of the reactants is in the gas phase which corresponds to a more positive entropy compared to the product which is in the solid phase i.e. lower entropy.

Since S(product) < S(reactant), based on equation 1, ΔS will be negative.