$[C a {S O}_{4} \cdot 5 H_{2} O] ⇌ C a {S O}_{4} \cdot 3 H_{2} O + 2 H_{2} O (g)$ ,

The equilibrium constant for the reaction is equal to:

\frac{[C a {S O}_{4} \cdot 3 H_{2} O] {[H_{2} O]}^{2}}{[C a {S O}_{4} \cdot 5 H_{2} O]}

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\frac{[C a {S O}_{4} \cdot 3 H_{2} O]}{[C a {S O}_{4} \cdot 5 H_{2} O]}

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{[H_{2} O]}^{2}

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[H_{2} O]

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Solution

The correct option is C ${[H_{2} O]}^{2}$
The equilibrium constant for the reaction,
$[C a {S O}_{4} \cdot 5 H_{2} O] ⇌ C a {S O}_{4} \cdot 3 H_{2} O + 2 H_{2} O (g)$ , is equal to ${[H_{2} O]}^{2}$ .
Because the concentration of solids (and pure liquids) do not appear in the equilibrium constant expression.

Hence, the correct option is $C$

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Similar questions

K_{C}

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