logo

Ask Questions, Get Answers

X
 
Home  >>  JEEMAIN and NEET  >>  Chemistry  >>  Equilibrium

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] $

1 Answer

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
 

Related questions

...