Answer:
The given reaction is:\[\underset{Initial\,\Pr
essure}{\mathop{{}}}\,FeO(s)+\underset{1.4atm}{\mathop{CO(g)}}\,\rightleftharpoons
Fe(s)+\underset{0.80atm}{\mathop{C{{O}_{2}}(g)}}\,\]
\[{{Q}_{p}}=\frac{{{p}_{C{{O}_{2}}}}}{{{p}_{CO}}}=\frac{0.80}{1.4}=0.571\]
\[{{Q}_{p}}>{{K}_{p}}\,\,\therefore \] the
reaction will proceed in the backward direction.
\[\underset{\begin{smallmatrix}
t=0 \\
{{t}_{eq}}
\end{smallmatrix}}{\overset{{}}{\mathop{{}}}}\,\underset{\begin{smallmatrix}
- \\
-
\end{smallmatrix}}{\mathop{Fe(s)}}\,+\underset{\begin{smallmatrix}
0.80 \\
(0.80-x)
\end{smallmatrix}}{\mathop{C{{O}_{2}}(g)}}\,\rightleftharpoons
\underset{\begin{smallmatrix}
- \\
-
\end{smallmatrix}}{\mathop{FeO(s)}}\,+\underset{\begin{smallmatrix}
1.4 \\
(1.4+x)
\end{smallmatrix}}{\mathop{CO(g)}}\,\]
\[{{K}_{p}}=\frac{{{p}_{CO}}}{{{p}_{C{{O}_{2}}}}}\]
\[\frac{1}{0.265}=\frac{1.4+x}{(0.80-x)}\]
\[\therefore \] \[x=0.339\] \[{{p}_{CO}}=1.4+x=1.4+0.339=1.739atm\]
\[{{p}_{C{{O}_{2}}}}=(0.80-x)=(0.80-0.339)=0.461atm\]
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