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Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Equilibrium
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The equilibrium constant for the reaction $CaSO_4.5H_2O (s) \rightleftharpoons CaSO_4.3H_2O (s) + 2H_2O (g)$ is equal to

(a) $ \frac {[CaSO_4.3H_2O] [H_2O]^2} {[CaSO_4.5H_2O]} $

(b) $ \frac {[CaSO_4.3H_2O]} {[CaSO_4.5H_2O]}$

(c) $ [H_2O]^2 $

(d) $ [H_2O] $

Can you answer this question?

1 Answer

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Answer: $[H_2O]^2$
$ K = \frac {[CaSO_4.3H_2O] [H_2O]^2} {[CaSO_4.5H_2O]} $
Concentration of solids is taken to be unity.
Thus $ [CaSO_4.5H_2O (s) ] = [CaSO_4.3H_2O (s) ] = 1$
$\therefore K = \frac{1 \times [H_2O]^2}{1} = [H_2O]^2 $


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

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