A vessel at 1000 K contains $CO_{2}$ with a pressure of 0.5 atm. Some of the $CO_{2}$ is converted into CO on the addition of graphite. If the total pressure at equilibrium is 0.8 atm, the value of $K_{p}$ is

Question

Dr. Abubacker · Accepted Answer

(a)
$CO_{2}(g) + C(s) ightleftharpoons 2CO(g)$
Initial : 0.5 atm
At equili : $(0.5 - p)$ $2p\ atm$
This is a case of heterogeneous equilibrium.
C($s$) being solid is not considered
Total pressure of $CO_{2}$ and CO gases.
$p_{CO_{2}} + p_{CO} = p_{total}$
$0.5 - p + 2p = 0.8$
$p = 0.3$ atm
$	herefore\ \ {p_{CO}}_{2} = 0.5 - 0.3 = 0.2$ atm
$p_{CO} = 2p = 0.6\ atm$
$K_{p} = \frac{p_{CO}^{2}}{{p_{CO}}_{2}} = \frac{0.6 	imes 0.6}{0.2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \  = 1.8$ atm

Q. A vessel at 1000 K contains $CO_{2}$ with a pressure of 0.5 atm. Some of the $CO_{2}$ is converted into CO on the addition of graphite. If the total pressure at equilibrium is 0.8 atm, the value of $K_{p}$ is

Explanation

Reviewed by Dr. Abubacker

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