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Reaction Quotient

At 817oC, K...

Question

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

A

P_{T o t a l} = 0.34 a t m

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B

P_{T o t a l} = 0.68 a t m

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C

P_{T o t a l} = 1.3 a t m

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D

None of these

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Solution

The correct option is B $P_{T o t a l} = 0.68 a t m$
$C (s) + C O_{2} (g) ⇌ 2 C O (g)$
Let P be the total pressure.
$P_{C O_{2}} = 0.06 P$ atm and $P_{C O} = 0.94 P$ atm

$K_{P} = \frac{P_{C O}^{2}}{P_{C O_{2}}} = 10$ atm

$K_{P} = \frac{(0.94 P)^{2}}{0.06 P} = 10$ atm

$P = 0.68$ atm.

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