Making CO from CO2 is a potential energy source. The value of Kp for the reaction is is 1.50 at 700.0°C.

Question

Making CO from CO₂ is a potential energy source. The value of K_p for the reaction is is 1.50 at 700.0°C.

CO2​(g)+C(s)2CO(g)

a. Calculate the equilibrium partial pressure of CO if initially P_CO2 = 1.35  atm and P_CO = 5.31 atm. Pure graphite is present initially and when equilibrium is achieved.

b. Calculate the equilibrium partial pressure of CO₂.

Answer 1

CO_2​(g) + C(s) <==> 2CO(g)

First, let us determine in which direction the reaction will proceed. We will compare Q to Kp.

Q = (CO)² / (CO₂) = (5.31)² / 1.35

Q = 20.9 which is much greater than Kp of 1.50. This tells us the reaction will proceed to the LEFT, or will go back to make reactant in order to attain equilibrium.

Now, we can set up the ICE table as follows:

CO_2​(g) + C(s) <==> 2CO(g)

1.35.........................5.31..........Initial

+x...........................-2x.............Change

1.35+x....................5.31-2x......Equilibrium

Kp = (5.31-2x)² / 1.35-x

1.50 = (5.31-2x)² / 1.35-x

2.03 + 1.50x = 4x² - 21.2x + 28.2

4x² - 22.7x + 26.2 = 0 and solve using the quadratic equation

x = 1.61 atm

(a) Equilibrium partial pressure of CO = 5.31 - 2x = 5.31 - 3.22 = 2.09 atm

(b) Equilibrium partial pressure of CO₂ = 1.35 + x = 2.96 atm

Check if this is feasible by plugging these values into the Kp expression and solving for Kp:

Kp = (2.09)² / (2.96)

Kp = 1.48 ≈ 1.50 which is the value given as the Kp.