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Dihydrogen gas used in Haber's process is produced by reacting methane from natural gas with high temperature steam. The first stage of two stage reaction involves the formation of \(\mathrm{CO}\) and \(\mathrm{H}_2\). In second stage, \(\mathrm{CO}\) formed in first stage is reacted with more steam in water gas shift reaction.
\(\mathrm{CO}(\mathrm{g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_2(\mathrm{~g})+\mathrm{H}_2(\mathrm{~g})\).
If a reaction vessel at \(400^{\circ} \mathrm{C}\) is charged with an equimolar mixture of \(\mathrm{CO}\) and steam such that \(\mathrm{p}_{\mathrm{CO}}=\mathrm{P}_{\mathrm{H}_2 \mathrm{O}}=\mathbf{4 . 0}\) bar, what will be the partial pressure of \(\mathrm{H}_2\) at equilibrium ?
\(K_p=10.1\) at \(400^{\circ} \mathrm{C}\).
\(\mathrm{CO}(\mathrm{g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_2(\mathrm{~g})+\mathrm{H}_2(\mathrm{~g})\).
If a reaction vessel at \(400^{\circ} \mathrm{C}\) is charged with an equimolar mixture of \(\mathrm{CO}\) and steam such that \(\mathrm{p}_{\mathrm{CO}}=\mathrm{P}_{\mathrm{H}_2 \mathrm{O}}=\mathbf{4 . 0}\) bar, what will be the partial pressure of \(\mathrm{H}_2\) at equilibrium ?
\(K_p=10.1\) at \(400^{\circ} \mathrm{C}\).
Solution:
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Verified Answer
Suppose the partial pressure of \(\mathrm{H}_2\) at equilibrium \(=\mathrm{p}\) bar
\(\mathrm{K}_{\mathrm{p}}=p^2 /(4-p)^2=10.1\) (Given)
\(\therefore \mathrm{p} /(4-p)=\sqrt{10.1}=3.178\)
or \(p=3.042 \mathrm{bar}\)
\(\mathrm{K}_{\mathrm{p}}=p^2 /(4-p)^2=10.1\) (Given)
\(\therefore \mathrm{p} /(4-p)=\sqrt{10.1}=3.178\)
or \(p=3.042 \mathrm{bar}\)
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