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  • 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 CO_2. FeO (s) + CO (g) Fe (s) + CO_2 (g); K_p = 0.265 atm at 1050K

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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best_answer

We have,

K_p=0.265
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
                                   =0.371+.265p=0.80-p

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

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manish

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