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Question

(C) The equilibrium reaction is $2HX_{(g)} \rightleftharpoons H_{2(g)} + X_{2(g)}$.The expression for the equilibrium constant $K$ is $K = \frac{[H_2]

Vedclass · Accepted Answer

$12 	imes 10^{-2} \ M$

Vedclass · Answer

(C) The equilibrium reaction is $2HX_{(g)} ightleftharpoons H_{2(g)} + X_{2(g)}$.The expression for the equilibrium constant $K$ is $K = \frac{[H_2][X_2]}{[HX]^2}$.Given $K = 1 	imes 10^{-5}$,$[H_2] = 1.2 	imes 10^{-3} \ M$,and $[X_2] = 1.2 	imes 10^{-4} \ M$.Substituting the values: $10^{-5} = \frac{(1.2 	imes 10^{-3}) 	imes (1.2 	imes 10^{-4})}{[HX]^2}$.$[HX]^2 = \frac{1.44 	imes 10^{-7}}{10^{-5}} = 1.44 	imes 10^{-2}$.$[HX] = \sqrt{1.44 	imes 10^{-2}} = 1.2 	imes 10^{-1} \ M$.Converting to the format of the options: $1.2 	imes 10^{-1} \ M = 12 	imes 10^{-2} \ M$.

The equilibrium constant for the equilibrium $2HX_{(g)} \rightleftharpoons H_{2(g)} + X_{2(g)}$ is $1 \times 10^{-5}$. What will be the equilibrium concentration of $HX$ if the equilibrium concentrations for $H_2$ and $X_2$ are $1.2 \times 10^{-3} \ M$ and $1.2 \times 10^{-4} \ M$ respectively?

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