Question

At 750°C, the Kp for the reaction 2 CO + O2➞ 2 CO2 is 2.72 x 10^11. ( 2.72 times ten to the eleventh) At that temperature, CO and O2 are added to a constant volume vessel. The initial pressure of the CO is 2.00 atm, and that of the O2 is 1.50 atm. Calculate the pressures of all three components in the equilibrium mixture.

Answer 1

Kp = p(CO2)^2/p(CO)^2*p(O2)

Make an ICE chart.
initial:
CO2 = 2
P2 = 1.5
CO2 = 0

change:
CO2 = 2x
O2 = -x
CO = -2x

equilibrium:
CO2 = +x
CO = 2-2x
O2 = 1.5-x

Substitute into the Kp exression and solve for x.
2x will be CO2
2-2x will be CO
1.5-x will be O2

Answer 2

do you have to use limiting reagent for this problem