Question

Predict the sign of ΔS in the following processes. Give reasons for your answer :

OR

Explain with reason sign conventions of ΔS in the following reactions.

(a) N₂O_4(g) → 2NO_2(g)

(b) Fe₂O_3(s) + 3H_2(g) → 2Fe(s) + 3H₂O_(g)

(c) N₂(g) + 3H_2(g) → 2NH_3(g)

(d) MgCO_3(s) → MgO_(s) + CO_2(g)

(e) CO_2(g) → CO_2(s)

(f) Cl_2(g) → 2Cl_(g).

Answer 1

(a) N₂O_4(g) → 2NO_2(g)

Since 1 mole N₂O₄ on dissociation gives two moles of NO₂, the number of molecules increase, disorder increases hence entropy increases, ΔS > 0.

(b) Fe₂O_3(s) + 3H_2(g) → 2Fe(s) + 3H₂O_(g)

In the reaction number of moles of gaseous reactants and products are same, hence ΔS = 0.

(c) N₂(g) + 3H_2(g) → 2NH_3(g)

In the reaction, 4 moles of gaseous reactants form 2 moles of gaseous products (Δn < 0). Therefore disorder decreases and hence entropy decreases, ΔS < 0.

(d) MgCO_3(s) → MgO_(s) + CO_2(g)

In this 1 mole of orderly solid MgCO₃ gives 1 mole of solid MgO and 1 mole of gaseous CO₂(Δn > 0) with more disorder. Hence entropy increases, ΔS > 0.

(e) CO_2(g) → CO_2(s)

In this system from higher disorder in gaseous state changes to less disorder in the solid state, hence entropy decreases, ΔS < 0.

(f) Cl_2(g) → 2Cl_(g)

Since the dissociation of Cl₂ gas gives double Cl atoms, the number of atoms increases (Δn >0) increasing the disorder of the system. Hence ΔS > 0.