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Question

At $$973$$ K, $$K_p$$ is $$1.50$$ for the reaction:

$$C(s) + CO_2(g) \rightleftharpoons \ 2CO(g)$$

Suppose the total gas pressure at equilibrium is $$1.0$$ atm. What are the partial pressures of $$CO$$ and $$CO_2$$?

Solution
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At equilibrium,

$$C(s) + \underset{P_1}{CO_2(g)} \rightleftharpoons \underset{P_2(say)}{2CO(g)} \ \ \ \ Kp = 1.5$$

$$Kp = \cfrac{P_2^2}{P^1} = 1.5 \Rightarrow P_2^2 = 1.5 P_1$$

Also $$P_1 + P_2 = 1$$

$$P_2 = 1 - P_1$$

$$P_2^2 = 1 + P_1^2 - 2P_1 = 1.5 P_1$$

$$ 1 + P_1^2 -3.5 P_1 = 0$$

$$P_1 = 0.314 atm \Rightarrow P_{CO_2} = 0.314 atm$$

$$P_2 = 0.686 atm \Rightarrow P_{CO} = 0.686 atm$$

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