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)
[tex]\tt k=\dfrac{[B]}{[A]}=2.0[/tex]
k₂ for B ⇔C :(eq 2)
[tex]\tt k=\dfrac{[C]}{[B]}=0.01[/tex]
k for 2C⇔2A :
[tex]\tt k=\dfrac{[A]^2}{[C]^2}=[\dfrac{[A]}{[C]}]^2[/tex]
k₁ x k₂ =
[tex]\tt \dfrac{[C]}{[A]}=2\times 0.01 = 0.02[/tex]
Reverse :
[tex]\tt \dfrac{[A]}{[C]}=50\rightarrow [\dfrac{[A]}{[C]}]^2=50^2=2500[/tex]
