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Derivation of Kp and Kc

One of the re...

Question

One of the reactions that takes place in producing steel from iron ore is the reduction of iron (II) oxide by carbon monoxide to give iron metal and $C O_{2}$ .
$F e O (s) + C O (g) \leftrightarrow F e (s) + C O_{2} (g); K_{p} = 0.265$ at 1050 K.
What are the equilibrium partial pressures of CO and $C O_{2}$ at 1050 K if the initial partial pressures are: PCO = 1.4 atm and $P C O_{2}$ = 0.80 atm?

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Solution

For the given reaction,
$\begin{matrix} F e O_{(g)} & + & C O_{(g)} & \leftrightarrow & F e_{(S)} & + & C O_{2 (g)} I n i t i a l y, & 1.4 a t m & 0.80 a t m Q_{r} = \frac{P_{c O_{2}}}{p_{C O}} = \frac{0.80}{1.4} = 0.571 \end{matrix}$
It is given that $K_{p} = 0.265$ .
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 $C O_{2}$ will decrease.
Now, let the increase in pressure of CO = decrease in pressure of $C O_{2}$ be p. Then,
we can write,
$K_{p} = \frac{P_{c o_{2}}}{P_{c o}} \Rightarrow 0.265 = \frac{0.80 - p}{1.4 + p} \Rightarrow 0.371 + 0.265 p = 0.80 - p \Rightarrow 1.265 p = 0.429 \Rightarrow p = 0.339 a t m$
Therefore, equilibrium partial of $C O_{2}, P_{c o_{2}} = 0.80 - 0.339 = 0.461 a t m$
And, equilibrium partial pressure of $C O, P_{c o} = 1.4 + 0.339 = 1.739 a t m$

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Derivation of Kp and Kc

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