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 CO_{2}.

FeO _{(s)} + CO _{(g)} ⇌ Fe _{(s)} + CO_{2}_{(g)}; K_{p}= 0.265 at 1050 K.

What is the equilibrium partial pressures of CO and CO_{2} at 1050 K if the initial partial pressures are: p_{CO} = 1.4 atm and p_{CO2}= 0.80 atm?

Asked by Pragya Singh | 1 year ago | 73

For the given reaction,

FeO_{(g)} + CO_{(g)} ⇌ Fe_{(s)}+CO_{2(g)}

Initially, pCO = 1.4 atm and pCO_{2}= 0.80 atm

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

Since Q_{p} > K_{p , }the reaction will proceed in the backward direction.

Therefore, we can say that the pressure of CO will increase while the pressure of CO_{2 }will decrease.

Now, let the increase in pressure of CO = decrease in pressure of CO_{2} 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 CO_{2},p_{CO}=0.80-0.339=0.461 atm

And, equilibrium partial pressure of CO,p_{CO}=1.4+0.339=1.739 atm

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