Hi!
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!
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)
How many moles K+ are in 1.7L of the solution
Yes, your calculations and reasoning are correct!
To determine whether the reaction will proceed towards the left or right, we compare the reaction quotient Qp to the equilibrium constant Kp. The equilibrium constant is a measure of the extent to which a reaction proceeds towards products at a particular temperature and pressure.
In this case, you correctly calculated the partial pressures of CO, O2, and CO2 using the mole fractions of each component and the total pressure. Then, you calculated the reaction quotient Qp using the formula Qp = (P_CO)^2 * (P_O2) / (P_CO2)^2.
Comparing Qp to Kp, you found that Qp is significantly larger than Kp. This means that the reaction has more products (CO2) than it should at equilibrium. According to Le Chatelier's principle, the system will shift in the direction that reduces the quotient towards the equilibrium position. In this case, the reaction will proceed towards the left, producing more CO2.
You also correctly mentioned that adding a catalyst does not change the amount of CO produced, only how fast it is produced. A catalyst provides an alternative pathway for the reaction, lowering the activation energy required for the reaction to occur.
Overall, your analysis and conclusion are accurate. Well done!