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Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Equilibrium
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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

Can you answer this question?
 
 

1 Answer

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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}} $

 

answered Nov 13, 2013 by mosymeow_1
edited Nov 23, 2013 by mosymeow_1
 

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