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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-

Sagot :

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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]

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