Chemistry, asked by brand57awoga, 2 days ago

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

Answered by Dragonnightfury
0

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:

Answered by xxPrathuxx
0

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 .

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