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# The equilibrium constant for the reaction $A + 2B \rightleftharpoons 2C$ is 40. The Equilibrium constant for the reaction $C \rightleftharpoons B + 1/2 A$ is

(a) 1/40

(b)  $(1/40)^{1/2}$

(c) $(1/40)^2$

(d) 40

Toolbox:
• $K' = \frac {1}{K}$

Answer: $[\frac{1}{40}]^{\frac{1}{2}}$

For the reaction,
$2C \rightarrow A + 2B; K' = \frac {1}{K}$

and, for the reaction
$C \rightarrow \frac {1}{2} A + B; K'' = [\frac{1}{K}]^{\frac {1}{2}}$

Thus equilibrium constant = $[\frac{1}{40}]^{\frac{1}{2}}$

edited Nov 23, 2013