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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) $[H_2O]$

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 $[CaSO_4.5H_2O (s) ] = [CaSO_4.3H_2O (s) ] = 1$

$\therefore K = \frac{1 \times [H_2O]^2}{1} = [H_2O]^2$

edited Nov 23, 2013