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