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)

$\tt k=\dfrac{[B]}{[A]}=2.0$

k₂ for B ⇔C :(eq 2)

$\tt k=\dfrac{[C]}{[B]}=0.01$

k for 2C⇔2A :

$\tt k=\dfrac{[A]^2}{[C]^2}=[\dfrac{[A]}{[C]}]^2$

k₁ x k₂ =

$\tt \dfrac{[C]}{[A]}=2\times 0.01 = 0.02$

Reverse :

$\tt \dfrac{[A]}{[C]}=50\rightarrow [\dfrac{[A]}{[C]}]^2=50^2=2500$