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.

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.

DrBob222 · Answer

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

josh · Answer

do you have to use limiting reagent for this problem