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.

how do u solve this should i use limiting reagent to find the excess or change in O2

Answer 1

Assume the partial pressure of O2 is reduced by x atm due to the reaction. You will be left with

Po2 = 1.50 - x atm
Pco = 2.00 - 2x
Pco2 = 2x
Solve this equation for x:

Pco2^2/[Po2*Pco^2]
= 4x^2/[(1.5-x)(2.0-2x)^2] = Kp
= 2.72 x 10^11

You end up with a cubic equation for x. You may have to solve it graphically. Take the root that is between 0 and 1.
As x -> 1, the term on the left approaches
4/[(0.5)(2-2x)^2 ] = 2.27*10^11
2/(1-x)^2 = 2.27*10^11
(1-x) = 3*10^-6
x = 0.999997
You are left with 0.500003 moles O2 and
10^-5 moles CO (the limiting reactant). 1.99999 moles of CO2 are formed

This reaction will, however, be very slow at room temperature. The reaction does not proceed kinetically as written, and requires some O or OH as a chain carrier.