1 mole of HI is heated in a closed container | vedclass

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(B) The reaction is: $2HI \rightleftharpoons H_2 + I_2$Initial moles: $1 \ mol$ of $HI$,$0 \ mol$ of $H_2$,$0 \ mol$ of $I_2$.At equilibrium,$0.5 \ mo

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$0.25$

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(B) The reaction is: $2HI ightleftharpoons H_2 + I_2$Initial moles: $1 \ mol$ of $HI$,$0 \ mol$ of $H_2$,$0 \ mol$ of $I_2$.At equilibrium,$0.5 \ mol$ of $HI$ is dissociated. Therefore,remaining $HI = 1 - 0.5 = 0.5 \ mol$.According to stoichiometry,$2 \ mol$ of $HI$ produces $1 \ mol$ of $H_2$ and $1 \ mol$ of $I_2$.So,$0.5 \ mol$ of $HI$ produces $0.25 \ mol$ of $H_2$ and $0.25 \ mol$ of $I_2$.Equilibrium concentrations in $2 \ L$ container:$[HI] = 0.5 / 2 = 0.25 \ M$$[H_2] = 0.25 / 2 = 0.125 \ M$$[I_2] = 0.25 / 2 = 0.125 \ M$$K_C = \frac{[H_2][I_2]}{[HI]^2} = \frac{0.125 	imes 0.125}{(0.25)^2} = \frac{0.015625}{0.0625} = 0.25$

$1 \ mole$ of $HI$ is heated in a closed container of capacity $2 \ L$. At equilibrium,half a mole of $HI$ is dissociated. The equilibrium constant of the reaction is

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