TBR Keq Q Driving me nuts!!!

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Etorphine

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Hi guys,
Don't usually post here but this question is driving me crazy.

TBR Gen Chem I, Chp. III, passage XIV, q#95.

Consider the following endothermic equation:

1 N2O4(g) <--> 2NO2(g)

As the handle of a piston container filled with an equilibrium mixture of NO2 and N2O4 is lifted, what occurs?

I.The PNO2/PN2O4 ratio increases.
II. The piston cools down
III. The mole percent of N2O4 increases.

Answer is I+II. II is obviously true (endothermic), but here's what I don't get. The Kp (Keq) of a reaction is fixed; the ratio of products to reactants will not change except due to temperature. While it is true that there is dynamic equilibrium, and the denominator can rise if the numerator rises proportionally, the ratio cannot change.

In our piston example, as the piston handle is lifted, we now have more volume that we need to fill and maintain pressure. The equation will shift to the right.

Why does the ratio of PNO2/PN2O4 increase however if it has to obey the Keq constant?

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Hi guys,
Don't usually post here but this question is driving me crazy.

TBR Gen Chem I, Chp. III, passage XIV, q#95.

Consider the following endothermic equation:

1 N2O4(g) <--> 2NO2(g)

As the handle of a piston container filled with an equilibrium mixture of NO2 and N2O4 is lifted, what occurs?

I.The PNO2/PN2O4 ratio increases.
II. The piston cools down
III. The mole percent of N2O4 increases.

Answer is I+II. II is obviously true (endothermic), but here's what I don't get. The Kp (Keq) of a reaction is fixed; the ratio of products to reactants will not change except due to temperature. While it is true that there is dynamic equilibrium, and the denominator can rise if the numerator rises proportionally, the ratio cannot change.

In our piston example, as the piston handle is lifted, we now have more volume that we need to fill and maintain pressure. The equation will shift to the right.

Why does the ratio of PNO2/PN2O4 increase however if it has to obey the Keq constant?
I think you puting too much thoughts into the question. The concept they are basically trying to illustrate in this question is that: What will happen to the equilibrium with an increase of volume in an endothermic reaction? We know that an increase in volume will make the particles less crowded and collide less often. In that case, the equilibrium will shift to the right. Therefore, PNO2/PN2O4 ratio increases.
 
did the same passage today too lol~

got the problem wrong for a different reason though.

My reasoning for why I did not choose II (Piston cools down) is because in PV = nRT as V goes up T goes up as well.

I now realize that as V goes up, the rxn is right-shifted and since the rxn is endothermic, the piston does in fact cool down.

However, what I don't understand is why did my PV=nRT reasoning not work in this case? Maybe it's because P is not constant.. hmm
 
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First of all, II isn't obviously true if I isn't true. It's probably not hard to see that III is the opposite of I and II, so the answer you picked had to be either I and II or III, or none. No other combination would make sense. As far as why I is true though...

In this reaction, Kp is (PNO2)^2/PN2O4. From PV=nRT, pressure of any gas is nRT/V, assuming ideality. Since n/V is just molarity, PNO2 = [NO2]RT, and PN2O4 = [N2O4]RT. Substituting those expressions in gives Kp = [NO2]^2(RT)^2 / ([N2O4]RT) = [NO2]^2RT/[N2O4]. Ignoring the RT since temperature does not change, what happens if we double volume? [NO2] and [N2O4] are both halved. Notice however, that in the equation for Kp, [NO2] is squared, while [N2O4] is not. This means that the numerator will be reduced to 1/4th its value, while the denominator will only be reduced to 1/2 half its value. Qp = 1/2Kp from this, meaning that the reaction is to the LEFT of equilibrium, and needs to shift to the right (Q < K). Does this help?
 
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First of all, II isn't obviously true if I isn't true. It's probably not hard to see that III is the opposite of I and II, so the answer you picked had to be either I and II or III, or none. No other combination would make sense. As far as why I is true though...

In this reaction, Kp is (PNO2)^2/PN2O4. From PV=nRT, pressure of any gas is nRT/V, assuming ideality. Since n/V is just molarity, PNO2 = [NO2]RT, and PN2O4 = [N2O4]RT. Substituting those expressions in gives Kp = [NO2]^2(RT)^2 / ([N2O4]RT) = [NO2]^2RT/[N2O4]. Ignoring the RT since temperature does not change, what happens if we double volume? [NO2] and [N2O4] are both halved. Notice however, that in the equation for Kp, [NO2] is squared, while [N2O4] is not. This means that the numerator will be reduced to 1/4th its value, while the denominator will only be reduced to 1/2 half its value. Qp = 1/2Kp from this, meaning that the reaction is to the LEFT of equilibrium, and needs to shift to the right (Q < K). Does this help?

Well done on the explanation. The one thing I'd add is that the reaction has to be endothermic, because you are breaking a bond (without forming a new bond) as the reaction goes in the forward direction. Bond breaking is an endothermic process, so the reaction must be endothermic. They also tell you it's endothermic (in a roundabout way) at the end of the last paragraph of the passage.
 
First of all, II isn't obviously true if I isn't true. It's probably not hard to see that III is the opposite of I and II, so the answer you picked had to be either I and II or III, or none. No other combination would make sense. As far as why I is true though...

In this reaction, Kp is (PNO2)^2/PN2O4. From PV=nRT, pressure of any gas is nRT/V, assuming ideality. Since n/V is just molarity, PNO2 = [NO2]RT, and PN2O4 = [N2O4]RT. Substituting those expressions in gives Kp = [NO2]^2(RT)^2 / ([N2O4]RT) = [NO2]^2RT/[N2O4]. Ignoring the RT since temperature does not change, what happens if we double volume? [NO2] and [N2O4] are both halved. Notice however, that in the equation for Kp, [NO2] is squared, while [N2O4] is not. This means that the numerator will be reduced to 1/4th its value, while the denominator will only be reduced to 1/2 half its value. Qp = 1/2Kp from this, meaning that the reaction is to the LEFT of equilibrium, and needs to shift to the right (Q < K). Does this help?
More complex explanation; however, makes perfect sense.
 
First of all, II isn't obviously true if I isn't true. It's probably not hard to see that III is the opposite of I and II, so the answer you picked had to be either I and II or III, or none. No other combination would make sense. As far as why I is true though...

In this reaction, Kp is (PNO2)^2/PN2O4. From PV=nRT, pressure of any gas is nRT/V, assuming ideality. Since n/V is just molarity, PNO2 = [NO2]RT, and PN2O4 = [N2O4]RT. Substituting those expressions in gives Kp = [NO2]^2(RT)^2 / ([N2O4]RT) = [NO2]^2RT/[N2O4]. Ignoring the RT since temperature does not change, what happens if we double volume? [NO2] and [N2O4] are both halved. Notice however, that in the equation for Kp, [NO2] is squared, while [N2O4] is not. This means that the numerator will be reduced to 1/4th its value, while the denominator will only be reduced to 1/2 half its value. Qp = 1/2Kp from this, meaning that the reaction is to the LEFT of equilibrium, and needs to shift to the right (Q < K). Does this help?

Ahhhh that's it! Very eloquently put, that helps a lot. Thanks!!!
 
Hi, I'm sorry to resurface an old thread, but I'm really struggling with one thing on this question: How is it that volume increases when pressure is increased? I understand that the piston moves up to re-establish equilibrium, which in turn increases volume, but I can't help but think of Boyle's law (the inverse relationship between the two)...and I get confused. Does it not apply to pistons?

Thanks in advance!
 
The change to the system is pulling the lid of the piston up by the handle, meaning you are expanding the volume of the cylinder. This increase in volume decreases the total pressure (V goes up and P goes down inside). It is that drop in internal pressure that is the driving force for the reaction to move from the side with one gas molecule (N2O4) to the side with two gas molecules (2 NO2).
 
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