Question

Consider the following reversible Hetererogenes reaction:

C(s) + CO2(g) -><- 2CO(g)

When equalibrium is reached at a certain point the total pressure of the system is found to be 4.98 atm. If the equilibrium constant Kp for this reaction is equal to 1.67 at this temperature, calculate the equilibrium partial pressures of CO2 and CO gases?

P(CO)eq= ?      P(CO2)eq=?

Answer 1

Here is what I solved before, please modify the figures as per your question. Please let me know if you have further questions. Ifthis helps then kindly rate 5-stars.

Consider the Following reversible heterogeneous reaction: C(s) + CO2 (g) --> 2CO(g) When equilibrium is reached at 700C, the total pressure of the system is found to be 4.00 atm. If the equilibrium constant Kp for this reaction is equal to 1.52, calculate the equilibrium partical pressures of CO2 and CO gases.

Answer

In equilibrium, the exp

ression for Kp would be written as:

But since the activity of any solid is equal to one, we leave the partial pressure of C, Pc out of the equation as such:
where Pco is the partial pressure of CO while Pco2 is the partial pressure of CO2.
In the problem however, partial pressures are not given. Only the total pressure is. Remember that the total pressure of the system is the sum of all the partial pressures of the components, hence:

We can substitute Pco with the variable x and Pco2 with 3.45-x and have the following solution:

Solving for x gives a value of 1.65 for Pco and subtracting this from the total pressure gives a value of 1.8 for Pco2.