TBR Gen Chem Equilibrium, Question 95

[leathersofa]

Here's the question and additional info for context:

95. As the handle of a piston container filled with an equilibrium mixture of NO2 and N2O4 is lifted, what occurs? (This is an endothermic reaction)

Heat + N2O4 (g) <---> 2NO2 (g)

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

A. I only
B. II only
C. I and II only
D. II and III only

I understand why statement I is right and statement III is wrong, but I have a question regarding statement II.
The answer explanation says how the piston cools down because "the reaction is shifting in the forward direction, which is endothermic, therefore heat is absorbed by the reaction. The result is that the temperature decreases."

Does the temperature get affected and consequently the Keq change whenever the equilibrium is disturbed for any reaction, or is this because the reaction is occurring in a piston?
For example, if this is a very simple, basic Le Chat's problem where NO2 is added to the reaction, would the Keq decrease because the temperature is increasing since it's shifting in the reverse direction to balance the addition of NO2? Until now, I thought that the Keq wouldn't change in this scenario but why does the temperature decrease in problem 95 then?

Thank you so much!!

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[Intraleukin-6]

Here's the question and additional info for context:

95. As the handle of a piston container filled with an equilibrium mixture of NO2 and N2O4 is lifted, what occurs? (This is an endothermic reaction)

Heat + N2O4 (g) <---> 2NO2 (g)

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

A. I only
B. II only
C. I and II only
D. II and III only

I understand why statement I is right and statement III is wrong, but I have a question regarding statement II.
The answer explanation says how the piston cools down because "the reaction is shifting in the forward direction, which is endothermic, therefore heat is absorbed by the reaction. The result is that the temperature decreases."

Does the temperature get affected and consequently the Keq change whenever the equilibrium is disturbed for any reaction, or is this because the reaction is occurring in a piston?
For example, if this is a very simple, basic Le Chat's problem where NO2 is added to the reaction, would the Keq decrease because the temperature is increasing since it's shifting in the reverse direction to balance the addition of NO2? Until now, I thought that the Keq wouldn't change in this scenario but why does the temperature decrease in problem 95 then?

Thank you so much!!

I wouldn't have even gotten this correct because I struggle with this too, but I'll give it a try.

So using dG=dH - TdS

@Keq G=0. If reaction is endothermic, entropy must be positive, so our new gibbs equation:

0= (+ive) - (T)(increases)

And because T and dS are inversely related, therefore temperature goes down as entropy is increases. I don't know if that will help..Themo wasn't really my strongest point.

 

[rmm30]

I wouldn't have even gotten this correct because I struggle with this too, but I'll give it a try.

So using dG=dH - TdS

@Keq G=0. If reaction is endothermic, entropy must be positive, so our new gibbs equation:

0= (+ive) - (T)(increases)

And because T and dS are inversely related, therefore temperature goes down as entropy is increases. I don't know if that will help..Themo wasn't really my strongest point.

I think you are going about it the wrong way. It's less of a thermodynamics problem and more of a conceptual equilibrium problem.




According to LeChatlier's principle, any time you put stress on a system at equilibrium, the system will respond to counteract that stress, ie whatever side of the equation you stress, the system will react by shifting the rxn in the OPPOSITE direction.

There are 3 basic ways to directly apply stress on a system
1) Change concentration of aqueous or gaseous products or reactants
2) Change the temperature
3) Change pressure via a volume change

But there are also more devious, less obvious ways. And that's demonstrated by this problem.

So looking at this rxn: heat + N2O4(g) <-----> 2NO2(g)

Basically what we are dealing with here is stress number 3) on the list. The volume of the container is expanding causing pressure to decrease. A decrease in pressure will cause a concomitant shift of the rxn to the side with more moles of gas (to the right in this case). (If the volume were decreased / pressure increased the rxn would shift to the side with fewer moles of gas (left). )

Here's the tricky part: It's not just the concentration / partial pressure of gaseous reactant that is changing (decreasing), it is EVERY reactant. This includes heat; heat is decreasing here and the system is cooling. If there were a bunch of solids, and liquids on the left side of the rxn, they would change too. They still wouldn't be represented in a a mass action law K but the would diminish. That's the key to this problem.

I mentioned above that heat is a reactant. While I don't know if it's technically true, but it is a very useful convention in dealing with these types of scenarios. If the rxn is said to be endothermic put 'heat' on the reactant side. If rxn is said to be exothermic put 'heat' on the product side.

In this problem it was written for you. Most cases the question would just say

N2O4 <----> 2NO2 and endothermic.

This becomes useful in questions like this but also when examining stress 2): Changing the temperature of the system. If you were to increase the temperature of a rxn that was endothermic, the rxn would shift right (as in the OP question). If you were to decrease the temperature of an endothermic rxn, the rxn would shift left. In an exothermic rxn, the temperature would be a product. So increasing the temp in exothermic rxn would shift the rxn lft?

But the temperature isn't being manipulated in this question. It's changing as a result of the entire rxn shifting to the right (all the reactants being diminished) including heat. That's what makes this problem so tricky and that's why the explanation in TBR says 'don't think in terms of PV=nRT' bc Temp isn't directly being manipulated.

This explanation makes sense to me. I could be wrong. I hope not.

 

[Intraleukin-6]

I think you are going about it the wrong way. It's less of a thermodynamics problem and more of a conceptual equilibrium problem.




According to LeChatlier's principle, any time you put stress on a system at equilibrium, the system will respond to counteract that stress, ie whatever side of the equation you stress, the system will react by shifting the rxn in the OPPOSITE direction.

There are 3 basic ways to directly apply stress on a system
1) Change concentration of aqueous or gaseous products or reactants
2) Change the temperature
3) Change pressure via a volume change

Equilibrium isn't thermodynamics? At equilibrium delta G is going to be zero. Gibbs free energy relates the entropy, enthalpy, and temperature. I don't see what's wrong with looking at it from that perspective, and applying LeChatlier's principle that way.

 

[leathersofa]

But the temperature isn't being manipulated in this question. It's changing as a result of the entire rxn shifting to the right (all the reactants being diminished) including heat. That's what makes this problem so tricky and that's why the explanation in TBR says 'don't think in terms of PV=nRT' bc Temp isn't directly being manipulated.

This explanation makes sense to me. I could be wrong. I hope not.

When the entire reaction is shifting to the right including the heat, the temperature is decreasing. I get that part. But, would the Keq also increase in this scenario because the temperature is decreasing?

Also, another question, you mentioned how the third way to apply stress on a system is to "change pressure via a volume change." However, this is a piston so the Pressure is constant though the volume changes. With that in mind, would that 3rd scenario still apply here?

 

[sazerac]

This piston is having an outside force lift it - it is definitely a pressure change / pressure drop.

 

[rmm30]

When the entire reaction is shifting to the right including the heat, the temperature is decreasing. I get that part. But, would the Keq also increase in this scenario because the temperature is decreasing?

Also, another question, you mentioned how the third way to apply stress on a system is to "change pressure via a volume change." However, this is a piston so the Pressure is constant though the volume changes. With that in mind, would that 3rd scenario still apply here?

For the second part of the question: The pressure isn't constant. The mols of gas is. The piston has a movable handle top. So imagine a fixed number of gas molecules in a cylinder with a movable lid that reaches up to the ceiling. As you move that lid down toward the floor the pressure within the cylinder will increase. This is exactly the type of scenario that applies here. You will hardly ever see these questions posed another way.

Here is something tricky to look out for. If instead of increasing the pressure by decreasing the volume, you increased the pressure inside the cylinder by adding a bunch of Inert gas (He). You might think that the rxn would shift in the direction with less mols of gas but this is not the case. In fact, the rxn doesn't shift at all. I think the reasoning is because while you increase the overall pressure of the system, the ratios of partial pressure of gaseous reactants to gaseous products remains the same. Just FYI

For the first part of the question: Yes. Changing temperature changes the equilibrium constant. It is the ONLY variable that changes the equilibrium constant.
For exothermic rxn: increasing temperature decreases Keq
endothermic rxn: increasing temperature increases Keq

So in this problem, there would be a new, lower Keq.

But the problem doesn't ask that question. I don't know the reasoning and I actually think it's beyond the scope of the mcat and prep book lessons.

 

[leathersofa]

Alright, thank you so much everyone. I think i have a better understanding now.

for inert gases, as long as the volume stays the same, inert gases do not disturb equilibrium.

and yes, the piston is shifting to the right and having a pressure change so that it can come back to equilibrium. Lifting the piston's handle caused the volume to increase and pressure to decrease, so it shifts right to reestablish equilibrium. Once equilibrium is reestablished and the piston's lid is stationary again, the internal pressure equals the external pressure.

i think i was coming at this question the wrong way. the ONLY way equilibrium can be changed is through a temperature increase or decrease. While adding more product (if the reaction is endothermic) may evolve heat, this doesn't necessarily mean that the Keq will change. The only way Keq can change is through a direct temperature shift. Is this sort of what you guys were trying to say?

Please let me know, thank you so much