Question

For the reaction
\(2A(g)+B(g)→2D(g)\)
\(∆U^Θ= -10.5\ kJ\) and \(∆S^Θ= -44.1\ JK^{-1}\).
Calculate \(∆G^Θ\) for the reaction, and predict whether the reaction may occur spontaneously.

Answer 1

For the given reaction,
\(2 A(g) + B(g) → 2D(g)\)
∆n_g = 2 – (3) = –1 mole
Substituting the value of ∆U^Θ in the expression of ∆H:
∆H^Θ = ∆U^Θ + ∆n_gRT 
∆H^θ = (–10.5 kJ) – (–1) (8.314 × 10^–3 kJ K^–1mol^–1) (298 K)
∆H^θ = –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 
∆G^Θ = –12.98 kJ – (298 K) (–44.1 JK^–1)
∆G^Θ = –12.98 kJ + 13.14 kJ 
∆G^Θ = + 0.16 kJ
Since ∆G^Θ for the reaction is positive, the reaction will not occur spontaneously.