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,
$2 A(g) + B(g) → 2D(g)$
∆ng = 2 – (3) = –1 mole
Substituting the value of ∆UΘ in the expression of ∆H:
∆HΘ = ∆UΘ + ∆ngRT
∆Hθ = (–10.5 kJ) – (–1) (8.314 × 10–3 kJ K–1mol–1) (298 K)
∆Hθ = –10.5 kJ – 2.48 kJ
∆Hθ = –12.98 kJ
Substituting the values of ∆HΘ and ∆SΘ in the expression of ∆GΘ:
∆GΘ = ∆HΘ – T∆SΘ
∆GΘ = –12.98 kJ – (298 K) (–44.1 JK–1)
∆GΘ = –12.98 kJ + 13.14 kJ
∆GΘ = + 0.16 kJ
Since ∆GΘ for the reaction is positive, the reaction will not occur spontaneously.