Chemistry - Chemical Equilibrium Question with Solution | TestHub

$$\begin{array}{c}\text{For}\text{2HI}\rightleftharpoons {\text{H}}_{\text{2}}\text{+}{\text{I}}_{\text{2}}\text{with}\alpha \text{=}\text{0.8,}\text{K}\text{=}\text{(}\alpha {\text{/2)}}^{\text{2}}\text{/(1-}\alpha {\text{)}}^{\text{2}}\text{=}{\text{(0.4)}}^{\text{2}}{\text{/(0.2)}}^{\text{2}}\text{=}\text{4}\text{For}{\text{H}}_{\text{2}}\text{+}\\ {\text{I}}_{\text{2}}\rightleftharpoons \text{2HI,}{\text{K}}_{\text{c}}\text{=}\text{1/4}\text{=}\text{0.25}\\ \text{Let}\text{x}\text{mol}\text{of}{\text{H}}_{\text{2}}\text{react:}{\text{H}}_{\text{2}}\text{=}{\text{I}}_{\text{2}}\text{=}\text{2}\text{-}\text{x,}\text{HI}\text{=}\text{2x}\\ {\text{K}}_{\text{c}}\text{=}{\text{(2x)}}^{\text{2}}{\text{/(2-x)}}^{\text{2}}\text{=}\text{0.25,}\text{so}\text{2x/(2-x)}\text{=}\text{0.5}\\ \text{2.5x}\text{=}\text{1,}\text{so}\text{x}\text{=}\text{0.4}\text{mol}\\ {\text{I}}_{\text{2}}\text{left}\text{=}\text{2}\text{-}\text{0.4}\text{=}\text{1.6}\text{mol}\\ {\text{I}}_{\text{2}}\text{+}{\text{2S}}_{\text{2}}{\text{O}}_{\text{3}}^{\text{2-}}\to {\text{2I}}^{\text{-}}\text{+}{\text{S}}_{\text{4}}{\text{O}}_{\text{6}}^{\text{2-}}\text{,}\text{so}\text{thiosulfate}\text{needed}\text{=}\text{3.2}\text{mol}\\ \text{Volume}\text{=}\text{3.2/1.6}\text{=}\text{2}\text{L}\end{array}$$