For the given reaction,

2 A_{(g)} + B_{(g)} → 2D_{(g) }

∆ng = 2 – (3)

= –1 mole

Substituting the value of ∆U^{θ} in the expression of ∆H:

∆H^{θ} = ∆U^{θ} + ∆ngRT

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

= –10.5 kJ – 2.48 kJ ∆H^{θ} = –12.98 kJ

Substituting the values of ∆H^{θ} and ∆S^{θ} in the expression of

∆G^{θ} : ∆G^{θ} = ∆H^{θ} – T∆S^{θ}

= –12.98 kJ – (298 K) (–44.1 J K^{–1} )

= –12.98 kJ + 13.14 kJ

∆G^{θ} = + 0.16 kJ

Since ∆Gθ for the reaction is positive, the reaction will not occur spontaneously.