When 4.5 , text{mol} each of hydrogen and iod | vedclass

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(A) The reaction is ${H_2} + {I_2} ightleftharpoons 2HI$. Initial moles: $[H_2] = 4.5 \, 	ext{mol}$,$[I_2] = 4.5 \, 	ext{mol}$,$[HI] = 0 \, 	ext{

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(A) The reaction is ${H_2} + {I_2} ightleftharpoons 2HI$. Initial moles: $[H_2] = 4.5 \, 	ext{mol}$,$[I_2] = 4.5 \, 	ext{mol}$,$[HI] = 0 \, 	ext{mol}$. At equilibrium,$2x = 3 \, 	ext{mol}$ of $HI$ are formed,so $x = 1.5 \, 	ext{mol}$. Equilibrium moles: $[H_2] = 4.5 - 1.5 = 3 \, 	ext{mol}$,$[I_2] = 4.5 - 1.5 = 3 \, 	ext{mol}$,$[HI] = 3 \, 	ext{mol}$. Concentrations in $10 \, 	ext{L}$ volume: $[H_2] = 0.3 \, 	ext{M}$,$[I_2] = 0.3 \, 	ext{M}$,$[HI] = 0.3 \, 	ext{M}$. $K_c = \frac{[HI]^2}{[H_2][I_2]} = \frac{(0.3)^2}{(0.3)(0.3)} = \frac{0.09}{0.09} = 1$.

When $4.5 \, \text{mol}$ each of hydrogen and iodine are heated in a $10 \, \text{L}$ closed vessel,$3 \, \text{mol}$ of $HI$ are formed at equilibrium. The equilibrium constant $K_c$ for the reaction ${H_2} + {I_2} \rightleftharpoons 2HI$ is .........

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