Question

Given that Kp = 3.5 x 10-4 for the reaction 2 CO(g) <=> C(graphite) + CO2(g), what is the partial pressure of CO2(g) at equilibrium if initially 4.00 atm of CO(g) is in contact with graphite?

Answer 1

Answer: The partial pressure of carbon dioxide at equilibrium is 0.0056 atm

Explanation:

The given chemical equation follows:

                     $2CO(g)\rightleftharpoons C\text{ (graphite)}+CO_2(g)$

Initial:             4.00

At eqllm:       4.00-2x        x                 x

The expression of $K_p$ for above reaction follows:

$K_p=\frac{p_{CO_2}}{(p_{CO})^2}$

The partial pressure of pure solids and liquids are taken as 1 in the equilibrium constant expression.

We are given:

$K_p=3.5\times 10^{-4}$

Putting values in above expression, we get:

$3.5\times 10^{-4}=\frac{x}{(4-2x)^2}\\\\x=0.0056,718.28$

Neglecting the value of x = 718.28 because equilibrium pressure cannot be greater than initial pressure

Partial pressure of $CO_2$ = 0.0056 atm

Hence, the partial pressure of carbon dioxide at equilibrium is 0.0056 atm