Chemistry - Chemical Equilibrium Question with Solution | TestHub
\begin{gathered}{}\text{Methanol,}\text{ }\text{CH}_{\text{3}}\text{OH}\text{ }\text{can}\text{ }\text{be}\text{ }\text{prepared}\text{ }\text{from}\text{ }\text{CO}\text{ }\text{and}\text{ }\text{H}_{\text{2}}\text{ }\text{as}\\{}\text{CO(g)}\text{ }\text{+}\text{ }\text{2H}_{\text{2}}\text{(g)}\text{ }\rightleftharpoons\text{ }\text{CH}_{\text{3}}\text{OH(g);}\text{ }\text{K}_{\text{p}}\text{ }\text{=}\text{ }\text{6.23}\text{ }\times\text{ }\text{10}^{\text{-3}}\text{ }\text{at}\text{ }\text{500}\text{ }\text{K.}\\{}\text{What}\text{ }\text{total}\text{ }\text{pressure}\text{ }\text{is}\text{ }\text{required}\text{ }\text{to}\text{ }\text{convert}\text{ }\text{25%}\text{ }\text{of}\text{ }\text{CO}\text{ }\text{to}\text{ }\text{CH}_{\text{3}}\text{OH}\text{ }\text{at}\text{ }\text{500}\text{ }\text{K,}\text{ }\text{if}\text{ }\text{CO}\text{ }\\{}\text{and}\text{ }\text{H}_{\text{2}}\text{ }\text{comes}\text{ }\text{from}\text{ }\text{the}\text{ }\text{reaction:}\\{}\text{CH}_{\text{4}}\text{(g)}\text{ }\text{+}\text{ }\text{H}_{\text{2}}\text{O(g)}\text{ }\to\text{ }\text{CO(g)}\text{ }\text{+}\text{ }\text{3H}_{\text{2}}\text{(g)}\end{gathered}
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\begin{gathered}{}\text{Start}\text{ }\text{with}\text{ }\text{CO}\text{ }\text{=}\text{ }\text{1,}\text{ }\text{H}_{\text{2}}\text{ }\text{=}\text{ }\text{3}\text{ }\text{mol;}\text{ }\text{25%}\text{ }\text{conversion}\text{ }\text{gives}\text{ }\text{CO}\text{ }\text{=}\text{ }\text{0.75,}\text{ }\text{H}_{\text{2}}\text{ }\text{=}\text{ }\text{2.5,}\text{ }\text{CH}_{\text{3}}\text{OH}\text{ }\\{}\text{=}\text{ }\text{0.25;}\text{ }\text{total}\text{ }\text{=}\text{ }\text{3.5}\text{ }\text{mol}\text{ }\text{Mole}\text{ }\text{fractions:}\text{ }\text{CO}\text{ }\text{=}\text{ }\text{3/14,}\text{ }\text{H}_{\text{2}}\text{ }\text{=}\text{ }\text{5/7,}\text{ }\text{CH}_{\text{3}}\text{OH}\text{ }\text{=}\text{ }\text{1/14}\\{}\text{K}_{\text{p}}\text{ }\text{=}\text{ }\text{(P/14)/[(3P/14)(5P/7)}^{\text{2}}\text{]}\text{ }\text{=}\text{ }\text{49/(75P}^{\text{2}}\text{)}\\{}\text{P}^{\text{2}}\text{ }\text{=}\text{ }\text{49/(75}\text{ }\times\text{ }\text{6.23}\text{ }\times\text{ }\text{10}^{\text{-3}}\text{)}\text{ }\text{=}\text{ }\text{104.9}\text{ }\text{gives}\text{ }\text{P}\text{ }\text{=}\text{ }\text{10.24}\text{ }\text{bar}\\{}\text{This}\text{ }\text{is}\text{ }\text{option}\text{ }\text{(c).}\end{gathered}
