(A) For the given reaction: $CaSO_4 \cdot 5H_2O_{(s)} \rightleftharpoons CaSO_4 \cdot 3H_2O_{(s)} + 2H_2O_{(g)}$The equilibrium constant expression is

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(A) For the given reaction: $CaSO_4 \cdot 5H_2O_{(s)} \rightleftharpoons CaSO_4 \cdot 3H_2O_{(s)} + 2H_2O_{(g)}$The equilibrium constant expression is given by the ratio of the concentration of products to reactants,raised to their stoichiometric coefficients.$K_c = \frac{[CaSO_4 \cdot 3H_2O] \cdot [H_2O]^2}{[CaSO_4 \cdot 5H_2O]}$Since $CaSO_4 \cdot 5H_2O$ and $CaSO_4 \cdot 3H_2O$ are pure solids,their active masses are taken as unity $(1)$.Therefore,$K_c = 1 \cdot [H_2O]^2 / 1 = [H_2O]^2$.

Class 11 Chemistry 1 Equilibrium: Chemical Equilibrium Law of equilibrium and Equilibrium constant

Medium

The equilibrium constant $(K_c)$ for the reaction,$CaSO_4 \cdot 5H_2O_{(s)} \rightleftharpoons CaSO_4 \cdot 3H_2O_{(s)} + 2H_2O_{(g)}$ is equal to

A
$[H_2O]^2$
B
$\frac{[CaSO_4 \cdot 3H_2O]}{[CaSO_4 \cdot 5H_2O]}$
C
$[H_2O]$
D
$\frac{[CaSO_4 \cdot 3H_2O][H_2O]^2}{[CaSO_4 \cdot 5H_2O]}$

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1 Equilibrium: Chemical Equilibrium

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Law of equilibrium and Equilibrium constant

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For the reaction $CH_3COOH + H_2O \rightleftharpoons H_3O^{+} + CH_3COO^{-}$,the equilibrium constant $K$ is given by:

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The equilibrium constant $(K_c)$ for the reaction,$CaSO_4 \cdot 5H_2O_{(s)} \rightleftharpoons CaSO_4 \cdot 3H_2O_{(s)} + 2H_2O_{(g)}$ is equal to

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At $298 \ K$,for the reaction $Ag^{+} + 2NH_3 \rightleftharpoons Ag(NH_3)_2^{+}$,the concentrations of $Ag^{+}$,$Ag(NH_3)_2^{+}$,and $NH_3$ are $10^{-1} \ M$,$10^{-1} \ M$,and $10^3 \ M$ respectively. The value of $K_c$ at $298 \ K$ for this equilibrium is ...... .

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