MCAT GEN CHEM COMPLEX EQUILIBRIUM QUESTION

[notahappycamper]

Page 175 Example 3.10
Consider the following complex equilibrium:
2 NO(g)+1 O2 (g) <--> 2NO2 (g) Keq1
2NO2(g) <--> 1 N2O4 (g) Keq2
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2 NO(g) +1 O2 (g) <--> 1 N2O4 (g) Keq1*keq2
When N2O4 gas is removed, how are the partial pressures of NO gas and NO2 gas affected?
correct answer: P(NO) and P(NO2) both decrease.

Solution:
The second reaction shifts to the right to compensate for the loss of N2O4, so the partial pressure of NO2 decreases. A decrease in NO2 causes the first reaction to shift to the right as well, resulting in a decrease in the partial pressure of NO gas and an increase in the partial pressure of NO2. However, the increase in NO2 from the forward shift of the first reaction is less significant than the decrease in NO2 caused by the forward shift of the second reaction. This is because the shift in the first reaction cannot completely replenish the lost NO2 without losing so much NO that the reaction is beyond equilibrium. A shift never regenerates as much as was lost. Both NO and NO2 decrease.

I get that the partial pressure of NO decreases, but can someone please explain to me what is going on with the intermediate NO2, and what this solution is saying about how its partial pressure decreases in clearer terms. I guess I just don't understand how something not involved in the equilibrium expression could be affected.

I would really appreciate any help. (this problem has been causing me a lot of stress :/)

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[notahappycamper]

sorry forgot to mention. this is from the TBR gen chem book $1

 

[lumpyduster]

So you have a two-step reaction to form N2O4. If you remove N2O4, pNO2 will decrease because that is are the intermediate that reacts to form N2O4 (Look at equation 2; not the overall reaction). What happens to the first reaction then if you're removing NO2? Well, pNO and pO2 will both decrease to form more NO2.

You can also look at the overall reaction and see that if you lose N2O4, you lose NO and O2. Then look at reaction 1. If pNO decreases, what happens to pNO2? It will decompose into NO and NO2 to re-establish equilibrium.

Even though the overall reaction does not involve NO2, it is NO2 that is reacting to form N2O4, so if you change pN2O4 you will change pNO2.

 

[notahappycamper]

thanks, I think it is starting to make more sense. But if I understood your explanation correctly, you seem to be saying that the Reaction #1 shifts to the left, when the solution seems to be saying that it is shifting to the right, except the shift to the right in reaction #1 is not as great as the shift to the right in reaction #2. This is what I don't get.

 

[lumpyduster]

I don't know if this will clear things up but this is how I saw the problem.

1. You removed some N2O4 from the reaction. What happens?
I only looked at the two separate reactions. The only one with N2O4 is reaction 2. So reaction 2 will shift to the right to form more N2O4 - this is why pNO2 decreases.

2. What happens if pNO2 decreases? Is anything else affected (or in equilibrium with) NO2? Yes. NO and O2 in reaction 1. Because pNO2 decreases, equation 1 will also shift to the right - this is why pNO decreases.
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My second explanation (where you look at the overall reaction and then gauge what happens to NO2 in reaction 1 based on what happens to NO and O2 in the last equation) isn't technically correct. I just noticed that you were looking at the overall reaction (based on your first post) and that's the roundabout, though technically incorrect, way of getting the correct answer to the question. Hopefully this kind of clears things up?

 

[notahappycamper]

I'm still kind of confused about the statement "A shift never regenerates as much as was lost," but your explanation made the problem make more sense. Thanks for your help.

 

[BerkReviewTeach]

I'm still kind of confused about the statement "A shift never regenerates as much as was lost," but your explanation made the problem make more sense. Thanks for your help.

Consider a system at equilibrium that has one mole of a product removed. The system is no longer at equilibrium, and will shift in the forward direction to attain equilibrium once again (albeit a different equilibrium, because there are now less molecules in the system). If the reaction went forward to regenerate the entire mole of lost product, then the system would have a shortage of reactant, and would not be in equilibrium. So the reaction will shift forward to reach a new equilibrium where the loss of one mole of product is shared between reactants and products both. In the end, we lose both reactants and product overall, but still reach a state where the P/R ratio has the same value as before, but with different values for the various concentrations or partial pressures.