Question

For the reaction

\[\ce{2A_{(g)} + B_{(g)} → 2D_{(g)}}\]

ΔU^θ = –10.5 kJ and ΔS^θ= –44.1 JK^–1.

Calculate ΔG^θ for the reaction, and predict whether the reaction may occur spontaneously.

Answer 1

For the given reaction,

\[\ce{2A_{(g)} + B_{(g)} → 2D_{(g)}}\]

Δn_g = 2 – (3)

= –1 mole

Substituting the value of ΔU^θ in the expression of ΔH:

ΔH^θ = ΔU^θ + Δn_gRT

= (–10.5 kJ) – (–1) (8.314 × 10^–3 kJ K^–1 mol^–1) (298 K)

= –10.5 kJ – 2.48 kJ

ΔH^θ = –12.98 kJ

Substituting the values of ΔH^θ and ΔS^θ in the expression of ΔG^θ:

ΔG^θ = ΔH^θ – TΔS^θ

= –12.98 kJ – (298 K) (–44.1 J K^–1)

= –12.98 kJ + 13.14 kJ

ΔG^θ = + 0.16 kJ

Since ΔG^θ for the reaction is positive, the reaction will not occur spontaneously.