Question

38. For a chemical reaction of the type

AB
K=2.0 and BC K=0.01
Equilibrium constant for the reaction 2C 2 A is:
A. 25
B. 50 C. 2500 D.4 x 10-

Answer 1

Equilibrium constant for the reaction 2C ⇔2 A is : 2500

Further explanation

Given

A⇔B, k = 2.0

B⇔C, k = 0.01

Required

k for 2C⇔2A

Solution

k₁ for A⇔B :(eq 1)

`\tt k=([B])/([A])=2.0`

k₂ for B ⇔C :(eq 2)

`\tt k=([C])/([B])=0.01`

k for 2C⇔2A :

`\tt k=([A]^2)/([C]^2)=[([A])/([C])]^2`

k₁ x k₂ =

`\tt ([C])/([A])=2* 0.01 = 0.02`

Reverse :

`\tt ([A])/([C])=50\rightarrow [([A])/([C])]^2=50^2=2500`