Calculate the equilibrium partial pressure of CO if initially PCO2 = 1.32 atm and PCO = 4.85 atm. Pure graphite is present initially and when equilibrium is achieved.

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 = 1.32  atm and P_CO = 4.85 atm. Pure graphite is present initially and when equilibrium is achieved.

B. Calculate the equilibrium partial pressure of CO₂.

Answer 1

CO₂​(g) + C(s) <==> 2CO(g)

1.32......................... 4.85.............Initial

Q = (CO)² / CO₂) = (4.85)² / 1.32 = 17.8

Compared to Kp (1.50), reaction is far to the right, so reaction will proceed to the left.

CO₂​(g) + C(s) <==> 2CO(g)

1.32......................... 4.85.............Initial

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

1.32+x....................4.85-2x..........Equilibrium

Kp = (CO)² / CO₂)

1.50 = (4.85-2x)² / (1.32+x)

1.50x + 1.98 = 4x² - 19.4x + 23.5

4x² - 20.9x + 21.52 = 0

x = 1.41

(A) Partial pressure CO @ equilibrium = 4.85 - 2.82 = 2.03 atm

(B) Partial pressure CO₂ @ equilibrium = 1.32 + 1.41 = 2.73 atm