At 717 K,1.50 mol each of hydrogen and iodi | vedclass

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(B) The initial moles are $n(H_2) = 1.50 \ mol$ and $n(I_2) = 1.50 \ mol$ in a volume $V = 10 \ L$.At equilibrium,$n(H_2) = 1.25 \ mol$ and $n(I_2) =

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$0.16$

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(B) The initial moles are $n(H_2) = 1.50 \ mol$ and $n(I_2) = 1.50 \ mol$ in a volume $V = 10 \ L$.At equilibrium,$n(H_2) = 1.25 \ mol$ and $n(I_2) = 1.25 \ mol$.The amount of $H_2$ and $I_2$ reacted is $1.50 - 1.25 = 0.25 \ mol$ each.According to the stoichiometry of the reaction $H_2(g) + I_2(g) ightleftharpoons 2HI(g)$,the moles of $HI$ formed is $2 	imes 0.25 = 0.50 \ mol$.The equilibrium concentrations are:$[H_2] = 1.25 / 10 = 0.125 \ M$$[I_2] = 1.25 / 10 = 0.125 \ M$$[HI] = 0.50 / 10 = 0.05 \ M$The equilibrium constant $K_c$ is given by:$K_c = \frac{[HI]^2}{[H_2][I_2]} = \frac{(0.05)^2}{(0.125)(0.125)} = \frac{0.0025}{0.015625} = 0.16$.

At $717 \ K$,$1.50 \ mol$ each of hydrogen and iodine are placed in a $10 \ L$ closed vessel. At equilibrium,$1.25 \ mol$ of each hydrogen and iodine remain. The value of the equilibrium constant $K_c$ for the reaction $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$ at $717 \ K$ is .......

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