The equilibrium constant for the reaction H_{ | vedclass

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(D) The equilibrium constant expression for the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$ is given by: $K_c = \frac{[HI]^2}{[H_2][I_

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$4.0 	imes 10^{-3} \ M$

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(D) The equilibrium constant expression for the reaction $H_{2(g)} + I_{2(g)} ightleftharpoons 2HI_{(g)}$ is given by: $K_c = \frac{[HI]^2}{[H_2][I_2]}$ Given: $K_c = 32$ $[HI] = 8 	imes 10^{-3} \ M$ $[I_2] = 0.5 	imes 10^{-3} \ M$ Substituting the values into the expression: $32 = \frac{(8 	imes 10^{-3})^2}{[H_2] 	imes (0.5 	imes 10^{-3})}$ $32 = \frac{64 	imes 10^{-6}}{[H_2] 	imes 0.5 	imes 10^{-3}}$ $[H_2] = \frac{64 	imes 10^{-6}}{32 	imes 0.5 	imes 10^{-3}}$ $[H_2] = \frac{64 	imes 10^{-6}}{16 	imes 10^{-3}}$ $[H_2] = 4 	imes 10^{-3} \ M$

The equilibrium constant for the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$ is $32$ at a given temperature. The equilibrium concentrations of $I_2$ and $HI$ are $0.5 \times 10^{-3} \ M$ and $8 \times 10^{-3} \ M$ respectively. The equilibrium concentration of $H_2$ is:

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