HI was heated in a closed tube at 440,^{circ} | vedclass

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(C) The dissociation reaction is: $2HI(g) ightleftharpoons H_2(g) + I_2(g)$Let the initial concentration of $HI$ be $2 \, 	ext{moles}$.At equilibri

Vedclass · Accepted Answer

$0.0199$

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(C) The dissociation reaction is: $2HI(g) ightleftharpoons H_2(g) + I_2(g)$Let the initial concentration of $HI$ be $2 \, 	ext{moles}$.At equilibrium,the amount of $HI$ dissociated is $22\%$ of $2 = 0.44 \, 	ext{moles}$.Remaining $HI = 2 - 0.44 = 1.56 \, 	ext{moles}$.According to the stoichiometry,$2 \, 	ext{moles}$ of $HI$ produce $1 \, 	ext{mole}$ of $H_2$ and $1 \, 	ext{mole}$ of $I_2$.Therefore,$0.44 \, 	ext{moles}$ of $HI$ produce $0.22 \, 	ext{moles}$ of $H_2$ and $0.22 \, 	ext{moles}$ of $I_2$.The equilibrium constant $K_c$ is given by:$K_c = \frac{[H_2][I_2]}{[HI]^2} = \frac{0.22 	imes 0.22}{(1.56)^2} = \frac{0.0484}{2.4336} \approx 0.0199$.

$HI$ was heated in a closed tube at $440\,^{\circ}C$ until equilibrium was obtained. At this temperature,$22\%$ of $HI$ was dissociated. The equilibrium constant for this dissociation will be:

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