question_answer 1)

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 CO and\[{{H}_{2}}\]. In second stage, CO formed in first stage is reacted with more steam in water gas shift reaction, \[CO(g)+{{H}_{2}}O(g)\rightleftharpoons C{{O}_{2}}(g)+{{H}_{2}}(g)\] If a reaction vessel at 400C is charged with an equimolar mixture of CO and steam, such that, \[{{p}_{CO}}={{p}_{{{H}_{2}}O}}=4bar\]be the partial pressure of \[{{H}_{2}}\]at equilibrium? \[{{K}_{p}}\] = 0.1 at 400C

Answer:

The given reaction is: \[\underset{{{t}_{\begin{smallmatrix} 0 \\ {{t}_{eq}} \end{smallmatrix}}}}{\overset{{}}{\mathop{{}}}}\,\underset{\begin{smallmatrix} 4 \\ (4-x) \end{smallmatrix}}{\mathop{CO(g)}}\,\,\,+\underset{\begin{smallmatrix} 4 \\ (4-x) \end{smallmatrix}}{\mathop{{{H}_{2}}O(g)}}\,\rightleftharpoons \underset{\begin{smallmatrix} 0 \\ x \end{smallmatrix}}{\mathop{C{{O}_{2}}(g)}}\,+\underset{\begin{smallmatrix} 0 \\ x \end{smallmatrix}}{\mathop{{{H}_{2}}(g)}}\,\] \[{{K}_{p}}=\frac{{{p}_{C{{O}_{2}}}}\times {{p}_{{{H}_{2}}}}}{{{p}_{CO}}\times {{p}_{{{H}_{2}}O}}}\] \[0.1=\frac{x\times x}{(4-x)(4-x)}\] \[\sqrt{0.1}=\frac{x}{(4-x)}\] \[0.316=\frac{x}{4-x}\] \[x=\frac{4\times 0.316}{1.316}=0.96\] \[{{p}_{{{H}_{2}}}}=x=0.96\,\,bar\]