One of the reactions that take 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) ⇌ Fe (s) + CO2(g); Kp= 0.265 at 1050 K.

What is the equilibrium partial pressures of CO and CO2 at 1050 K if the initial partial pressures are: pCO = 1.4 atm and pCO2= 0.80 atm?

Asked by Pragya Singh | 1 year ago |  73

Solution :-

For the given reaction,

FeO(g) + CO(g) ⇌ Fe(s)+CO2(g)

Initially,  pCO = 1.4 atm and pCO2= 0.80 atm

$$Qp=\dfrac{p_{CO_{2}}}{p_{CO}}\\ \\ =\dfrac{0.80}{1.4}\\ \\$$

Since Qp > Kp , the reaction will proceed in the backward direction.

Therefore, we can say that the pressure of CO will increase while the pressure of COwill decrease.

Now, let the increase in pressure of CO = decrease in pressure of CO2 be p. Then, we can write,

$$K_{p}=\dfrac{p_{CO_{2}}}{p_{CO}}\\ \\$$

$$\Rightarrow 0.265=\dfrac{0.80-p}{1.4+p}\\ \\$$

$$\Rightarrow 0.371+0.265p=0.80-p\\ \\$$

$$\Rightarrow 1.265p=0.429\\ \\$$

$$\Rightarrow p=0.339 atm$$

Therefore, equilibrium partial of CO2,pCO=0.80-0.339=0.461 atm

And, equilibrium partial pressure of CO,pCO=1.4+0.339=1.739 atm

Answered by Abhisek | 1 year ago

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