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7.20    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 CO2.FeO_{(s)}+CO_{(g)}\rightleftharpoons Fe_{(s)}+CO_{2}_{(g)}; Kp = 0.265 atm at 1050K What are the equilibrium partial pressures of CO and CO2 at 1050 K if the initial partial pressures are: p_{CO}= 1.4 atm and pCO_2  = 0.80 atm?

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We have,

the initial pressure of CO and CO_2 are 1.4 atm and 0.80atm resp.

The given reaction is-

                                FeO_{(s)}+CO_{(g)}\rightleftharpoons Fe_{(s)}+CO_{2}_{(g)}
initially,                                   1.4 atm                    0.80 atm

Q_p = \frac{pCO_2}{pCO}=\frac{0.80}{1.4} = 0.571
Since Q_p>K_c the reaction will proceed in the backward direction to attain equilibrium. The partial pressure of CO_2 will increase = decrease in the partial pressure of CO_2   = p                      

\therefore K_p = \frac{pCO_2}{pCO} = \frac{0.80-p}{1.4+p}=0.265

                                    By solving the above equation we get the value of p = 0.339 atm

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