Posted by Mary on Saturday, August 16, 2008 at 7:59am.
First, change the volume concentration to molar percents. Then compare the K from the measured molar concentrations to Keq. If the measured is less, then the reaction will go forward more.
I want to add a comment about the second question: CO concentration being decreased or increased by a CATALYST...
Catalysts do not change the equilibrium concentration, they may speed it up, or reduce the activation energy, or make it more efficient, but not change the final equilibrium. However, in an exhaust system where the exaust is flowing thru at a constant rate regardless if it has reacted or not, the addition of a catalyst can change the output concentrations because more will be reacted in the short time the exhaust is in proximity with the catalyst. So in this case, the addition of a catalyst in the exhaust flow can change the concentrations because it offers a reaction path available in the short time the exhaust is available that was not otherwise available.
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?
2CO2(g) <=> 2CO(g)+ O2(g) Kp = 1 E-13 (or 1.00x10^-22)
The LEFT side is favored.
It is perfectly proper to write the reaction backwards:
2CO(g)+ O2(g) <=> 2CO2(g)
BUT now the Kp is the reciprocal of 1.00x10^-22 or
Kp = 1.00x10^22 for the reversed reaction
The above reaction, as I just wrote it, goes to completion for all practical purposes at 1200 K.
Qp for the exhaust mixture is
Qp = [(0.12)^2 / [(0.002)^2*(0.03)] = 1.20x10^5
1.20x10^5 < 1.00x10^22 or
Qp is much less tha the Kp
This means that CO will tend to oxidize to CO2 moving towards equilibrium. The speed of this trend will be enhanced by the presence of a catalyst, as BobPursley has pointed out. Since the forward reaction in the equilibrium:
2CO(g)+ O2(g) <=> 2CO2(g)
is moving toward equilibrium because of favorable conditions (concentration ratio and the catalyst), the concentration of CO will DECREASE as the exhaust gases escape.
Comment of my Qp calculation:
I assumed that those percents are volume/volume percents. Since in the gas state moles are proportional to volume, and the partial pressure of each gas is proportional to moles, the volume percents were assumed to be proportional to partial pressures and can be substituted into the Kp or Qp expressions. The concept of the volume % is confusing since all gases in the mixture have the same volume (that of the container). So, the volume % should be thought of as just a number proportional to the number of moles.
I noticed I rewrote the Kp as 1x10^-22 instead of 1x10^-13. Sorry. In the comparison of Kp with Qp for the reverse reaction we have:
Kp=10^13 > Qp=10^5, leading to the conclusion that CO is still changing to CO2 as it escapes and its concentration decreases as I suggested earlier.
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