question_answer 1)

One of the reaction that takes place in producing steel from iron ore is the reduction of iron (II) oxide by carbon monoxide to give iron metal and \[C{{O}_{2}}(g)\] \[FeO(s)+CO(g)\rightleftharpoons FeO(s)+CO(g)\] \[{{K}_{p}}\]= 0.265 atm at 1050 K. What is the equilibrium partial pressure of CO and \[C{{O}_{2}}\] at1050 K if the initial partial pressure is: \[{{p}_{CO}}=1.4\]atm and \[{{p}_{C{{O}_{2}}}}=0.80\]atm?

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. \[\underset{\begin{smallmatrix} t=0 \\ {{t}_{eq}} \end{smallmatrix}}{\overset{{}}{\mathop{{}}}}\,\underset{\begin{smallmatrix} - \\ - \end{smallmatrix}}{\mathop{Fe(s)}}\,+\underset{\begin{smallmatrix} 0.80 \\ (0.80-x) \end{smallmatrix}}{\mathop{C{{O}_{2}}(g)}}\,\rightleftharpoons \underset{\begin{smallmatrix} - \\ - \end{smallmatrix}}{\mathop{FeO(s)}}\,+\underset{\begin{smallmatrix} 1.4 \\ (1.4+x) \end{smallmatrix}}{\mathop{CO(g)}}\,\] \[{{K}_{p}}=\frac{{{p}_{CO}}}{{{p}_{C{{O}_{2}}}}}\] \[\frac{1}{0.265}=\frac{1.4+x}{(0.80-x)}\] \[\therefore \] \[x=0.339\] \[{{p}_{CO}}=1.4+x=1.4+0.339=1.739atm\] \[{{p}_{C{{O}_{2}}}}=(0.80-x)=(0.80-0.339)=0.461atm\]