In the reaction,H_2 + I_2 rightleftharpoons 2 | vedclass

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(D) The reaction is $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$.Initial moles: $H_2 = 0.4 \ mol$,$I_2 = 0.4 \ mol$,$HI = 0 \ mol$.At equilibrium,$0.5

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$11.1$

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(D) The reaction is $H_2(g) + I_2(g) ightleftharpoons 2HI(g)$.Initial moles: $H_2 = 0.4 \ mol$,$I_2 = 0.4 \ mol$,$HI = 0 \ mol$.At equilibrium,$0.5 \ mol$ of $HI$ is formed. Since $2 \ mol$ of $HI$ are produced from $1 \ mol$ of $H_2$ and $1 \ mol$ of $I_2$,the moles of $H_2$ and $I_2$ consumed are $0.5/2 = 0.25 \ mol$ each.Equilibrium moles: $H_2 = 0.4 - 0.25 = 0.15 \ mol$,$I_2 = 0.4 - 0.25 = 0.15 \ mol$,$HI = 0.5 \ mol$.Equilibrium concentrations in $2 \ L$ volume: $[H_2] = 0.15/2 \ M$,$[I_2] = 0.15/2 \ M$,$[HI] = 0.5/2 \ M$.$K_c = \frac{[HI]^2}{[H_2][I_2]} = \frac{(0.5/2)^2}{(0.15/2) 	imes (0.15/2)} = \frac{0.5^2}{0.15 	imes 0.15} = \frac{0.25}{0.0225} \approx 11.11$.

In the reaction,$H_2 + I_2 \rightleftharpoons 2HI$. In a $2 \ L$ flask,$0.4 \ mol$ of each $H_2$ and $I_2$ are taken. At equilibrium,$0.5 \ mol$ of $HI$ are formed. What will be the value of equilibrium constant,$K_c$?

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