Answer :
Answer: The partial pressure of carbon dioxide at equilibrium is 0.0056 atm
Explanation:
The given chemical equation follows:
[tex]2CO(g)\rightleftharpoons C\text{ (graphite)}+CO_2(g)[/tex]
Initial: 4.00
At eqllm: 4.00-2x x x
The expression of [tex]K_p[/tex] for above reaction follows:
[tex]K_p=\frac{p_{CO_2}}{(p_{CO})^2}[/tex]
The partial pressure of pure solids and liquids are taken as 1 in the equilibrium constant expression.
We are given:
[tex]K_p=3.5\times 10^{-4}[/tex]
Putting values in above expression, we get:
[tex]3.5\times 10^{-4}=\frac{x}{(4-2x)^2}\\\\x=0.0056,718.28[/tex]
Neglecting the value of x = 718.28 because equilibrium pressure cannot be greater than initial pressure
Partial pressure of [tex]CO_2[/tex] = 0.0056 atm
Hence, the partial pressure of carbon dioxide at equilibrium is 0.0056 atm
