Respuesta :
The given question is incomplete.. The complete question is :
Consider the following reaction:
[tex]C(s)+O_2(g)\rightarrow CO_2(g)[/tex]
At equilibrium, a 4.50 L container has 2.6 g of carbon, CO2 at a partial pressure of 0.0020 atm, and a total pressure of 0.572 atm. Calculate KP for this reaction at 725oC.
Answer: 162.45
Explanation:
Equilibrium pressure of [tex]CO_2[/tex] = 0.0020 atm
Total pressure at equilibrium = 0.572 atm
Equilibrium pressure of [tex]CO[/tex] = Total pressure at equilibrium - Equilibrium pressure of [tex]CO_2[/tex] = (0.572- 0.0020) = 0.570 atm
The given balanced equilibrium reaction is,
[tex]C(s)+CO_2(g)\rightarrow 2CO(g)[/tex]
At eqm 0 0.0020 0.570
The expression for equilibrium constant for this reaction will be,
[tex]K_p=\frac{[p_{CO}]^2}{[p_{CO_2}}[/tex]
Now put all the given values in this expression, we get :
[tex]K_p=\frac{(0.570)^2}{0.0020}[/tex]
By solving we get :
[tex]K_p=162.45[/tex]
Thus [tex]K_p[/tex] for this reaction at [tex]725^oC[/tex] is 162.45
