Hi!

Question

Thank you so much for your answer!

according to my calculations...I got:

1 bar = 750 mm Hg = 0.986 atm.

At constant T and P, volume and moles are proportional, so volume percent = mole %.

Pa = (xa)(Pt) , , ,the partial pressure of "a" is equal to the mole fraction of "a" times the total pressure. (mole fraction = %)

P CO = (0.002)(0.986) = 0.00197 atm
P O2 = (0.03)(0.986) = 0.0296 atm
P CO2 = (0.12)(0.986) = 0.118 atm

Qp = (P CO)^2 (P O2) / (P CO2)^2 = (0.00197)^2 (0.0296) / (0.118)^2 = 8.3 x 10^-6 which is much greater than Kp = 1 x 10^-13. Since Qp > Kp, the reaction will go back to the left (make more CO2). Adding a catalyst does not change the amount of CO produced, only how fast it is produced.

Is that correct?
Thanks!

Answer 1

I didn't check your math, but it looks in the ballpark. Now the last: You are not in an equilibrium here, as the exhaust is flowing thru, so the addition of a catalyst will increase the amount of CO2 (:but not to the equilibrium level). Much of the exhaust will not react because it passes through too quickly)

Answer 2

How many moles K+ are in 1.7L of the solution