A(s) B(g) + C(g) K P = 40 atm 2 X(s) B(g) + E(g) Above eq…

Chemistry Chemical Equilibrium General Single Correct MCQ

Published by: CGP EDU Academic Team

Published on: August 14, 2026

A(s) B(g) + C(g) K P = 40 atm 2

X(s) B(g) + E(g)

Above equilibrium is allowed to attain in a closed container and pressure of B was found to be 10 atm.

Calculate standard Gibb’s free energy change for X(s) B(g) + E(g) at 300 K (take R = 2 cal/K/mol)

A(s) B(g) + C(g) K P = 40 atm 2

X(s) B(g) + E(g)

3.5 Kcal/mol

3 Kcal/mol

2.5 Kcal/mol

2 Kcal/mol

Text Solution

Verified by Experts

The correct answer is:

A(s) B(g) + C(g) X(s) B(g) + E(g)

t = 0 – 0 0 t = 0 – 0 0

at eq. – x + y x t = eq. – (x + y) y

= P B ⋅ P C

40 = (x + y)(x)

40 = 10x

x = 4 atm

y = 6 atm

∴ = P B ⋅ P E

= (10) (6)

K P = 60 atm

Δ G 2 = –RT l nK 2

= – (300)(2.303)log60

= – (300)(2.303)(1.78)

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