For the reaction at 298K, 2A(s)+ B(g) → C(g) ∆U = -10.5 KJ and ∆S = -44.1JK-1 Calculate ∆G for the reaction and predict whether the reaction may occur spontaneously.
Answers
Answer:
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 = (–10.5 kJ) – (–1) (8.314 × 10–3 kJ K–1 mol–1 ) (298 K) = –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 = –12.98 kJ – (298 K) (–44.1 J K–1 ) = –12.98 kJ + 13.14 kJ ∆Gθ = + 0.16 kJ Since ∆Gθ for the reaction is positive, the reaction will not occur spontaneously. Read more on Sarthaks.com - https://www.sarthaks.com/8471/for-the-reaction-2a-g-b-g-2d-g-u-10-5-kj-and-s-44-1-jk-1
Explanation:
Explanation:
For the given reaction
2A
(g)
+B
(g)
⟶2D
(g)
Δn
g
=2−3=−1mole
Substituting the value of ΔU
θ
in the expansion of ΔH :
ΔH
θ
=ΔU
θ
+Δn
g
RT
=(−10.5kJ)+(+1)(8.314×10
−3
kJK
−1
)(298K)
=−10.5kJ−2.48kJ
=−12.98kJ
When we substitute the values of ΔH
θ
and ΔS
θ
in the expression of ∣deltaG
θ
then according to Gibbs' Helmholtz equation
ΔG
θ
=ΔH
θ
−TΔS
θ
=−12.98kJ−(298K)(−44.1JK
−1
)
=−12.98kJ−(13.14kJ)
ΔG
θ
=0.16kJ
Since ΔG
θ
for the reaction is positive , the reaction will not occur spontaneously .