Chemistry - Chemical Equilibrium Question with Solution | TestHub
\begin{gathered}\text{At }444^\circ\mathrm C,\ \mathrm{HI}\text{ dissociates to the extent of 22.22% at equilibrium.}\\[8pt]\text{What is the equilibrium constant of the reaction: 2HI(g) }\rightleftharpoons\text{ }\mathrm{H_2}\text{(g) + }\mathrm{I_2}\text{(g), at}\\\text{444°C?}\end{gathered}
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Solution:
\begin{gathered}{}\text{Take}\text{ }\text{1}\text{ }\text{mol}\text{ }\text{of}\text{ }\text{HI,}\text{ }\alpha\text{ }\text{=}\text{ }\text{22.22%}\text{ }\text{=}\text{ }\text{2/9}\text{ }\text{At}\text{ }\text{equilibrium:}\text{ }\text{HI}\text{ }\text{=}\text{ }\text{1}\text{ }\text{-}\text{ }\text{2/9}\text{ }\text{=}\text{ }\text{7/9}\text{ }\text{mol,}\text{ }\text{H}_{\text{2}}\text{ }\text{=}\text{ }\\{}\text{I}_{\text{2}}\text{ }\text{=}\text{ }\text{1/9}\text{ }\text{mol}\\{}\text{Since}\text{ }\Delta\text{n\_g}\text{ }\text{=}\text{ }\text{0,}\text{ }\text{volume}\text{ }\text{cancels}\\{}\text{K}_{\text{c}}\text{ }\text{=}\text{ }\text{(1/9)(1/9)/(7/9)}^{\text{2}}\text{ }\text{=}\text{ }\text{1/49}\text{ }\text{=}\text{ }\text{0.02}\\{}\text{This}\text{ }\text{is}\text{ }\text{option}\text{ }\text{(b).}\end{gathered}
