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Characteristics of Equilibrium Constant

At 8170C, K...

Question

At $817^{0} C, K_{p}$ for the reaction between pure $C O_{2}$ and excess hot graphite to form $2 C O (g)$ is 10 atm.
At what total pressure will be gas mixture analyze 6%, $C O_{2}$ by volume?

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Solution

$C O_{2} + C \to 2 C O$
$P_{O}$ O
$P_{0} - P$ 2P
$k_{p} = \frac{(2 p)^{2}}{P_{O} - P} \Rightarrow k_{p} - \frac{4 P^{2}}{P_{0} - P}$
$\frac{P_{O} - P}{P_{O}} \times 100 = 6$ %
$P_{O} - P = 0.06 P_{O}$
$0.94 P_{O} = P$
$10 = \frac{4 P^{2}}{P_{O} - 0.94 P_{O}}$
$4 P^{2} = 0.06 P_{O}$
$P = 0.16 a t m$
$P_{O} = 0.17 a t m$
total pressure = $P_{O} + P$
$= 0.17 + 0.16 = 0.33 a t m$

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