TBR passage - La Chatelier's principle and external pressure increase

[Monkeymaniac]

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So in one of the TBR passages, the following reaction is given.

A(g)+B(g)<->C(g)

Since the passage talks about equilibrium, I'm assuming that the reaction takes place in a closed system.

Once the reaction reaches equilibrium, stress is applied to the system, that is external pressure is doubled.

Before this change, partial pressure of each gas was 1 atm, after the stress is applied, it doubled to 2 atm each. But by La Chatelier's principle, the reaction shifts to right (because it's the side with less number of gas molecules) to reduce the pressure of the system.

My questions are

1) What is external pressure? Say if the reaction takes place in a closed jar, can external pressure be the atmostpheric pressure surrounding the jar?
2) Why does increasing the external pressure also increase the system partial pressures by exactly same degree (x2) at least for initially?
3) Shouldn't (External pressure) = (Total system pressure: sum of all partial pressures)? Initially total system pressure is 3 atm, so external pressure must be 3 atm. Later, the external pressure is jumped to 6 atm, but the system pressure would be less than 6 (due to shifted equilibrium, the particular example in the book shows 5.56 atm as the last total pressure). How is this discrepancy accounted for?

I'd apprecaite for any comments.

 

[Aljazera]

I'm just taking a stab at this since i finished this whole section (including all 100 questions) yesterday:

The external pressure applies work on the system and if the closed system did not change pressure to equivilate with the outside pressure, the closed system would implode.

So the internal pressure must equivilate with the outside pressure by raising its partial pressures to equal the new double pressure amount.

Also what page/question is this I can look it up for you.

 

[Isoprop]

So in one of the TBR passages, the following reaction is given.

A(g)+B(g)<->C(g)

Since the passage talks about equilibrium, I'm assuming that the reaction takes place in a closed system.

Once the reaction reaches equilibrium, stress is applied to the system, that is external pressure is doubled.

Before this change, partial pressure of each gas was 1 atm, after the stress is applied, it doubled to 2 atm each. But by La Chatelier's principle, the reaction shifts to right (because it's the side with less number of gas molecules) to reduce the pressure of the system.

My questions are

1) What is external pressure? Say if the reaction takes place in a closed jar, can external pressure be the atmostpheric pressure surrounding the jar?
2) Why does increasing the external pressure also increase the system partial pressures by exactly same degree (x2) at least for initially?
3) Shouldn't (External pressure) = (Total system pressure: sum of all partial pressures)? Initially total system pressure is 3 atm, so external pressure must be 3 atm. Later, the external pressure is jumped to 6 atm, but the system pressure would be less than 6 (due to shifted equilibrium, the particular example in the book shows 5.56 atm as the last total pressure). How is this discrepancy accounted for?

I'd apprecaite for any comments.

I think by "external pressure" we are talking about increasing the pressure in the reaction chamber "from the outside." I think the book is trying to differentiate this by increasing the pressure of by putting in an inert gas inside the reaction chamber. If you increase the pressure this way, you won't get any shift.

So when we are talking about increasing the external pressure, I think we just translate this into increasing the pressure of the system.

2) Why does increasing the external pressure also increase the system partial pressures by exactly same degree (x2) at least for initially?

This can be easily explained by remember the formula for partial pressures.

PT = XAPA + XAPB

PT = total pressure
PA and PB = partial pressures of gas A and B, respectively
XA and XB = mole fractions of gas A or B, respectively

If PT is doubled, we must double the right side of the equation too.

2PT = 2XAPA + 2XAPB

Since mole fractions are constant, what must double is the partial pressures (each by a factor of 2).

3) Shouldn't (External pressure) = (Total system pressure: sum of all partial pressures)? Initially total system pressure is 3 atm, so external pressure must be 3 atm. Later, the external pressure is jumped to 6 atm, but the system pressure would be less than 6 (due to shifted equilibrium, the particular example in the book shows 5.56 atm as the last total pressure). How is this discrepancy accounted for?

There is no discrepancy. We apply an external pressure on the system, and the system shifts to minimize as much as possible the total pressure of the system. So if we double the external pressure, according to Le Chat, the system shifts so that the final total pressure is not doubled also (if it can).

 

[Monkeymaniac]

This can be easily explained by remember the formula for partial pressures.

PT = XAPA + XAPB

PT = total pressure
PA and PB = partial pressures of gas A and B, respectively
XA and XB = mole fractions of gas A or B, respectively

If PT is doubled, we must double the right side of the equation too.

2PT = 2XAPA + 2XAPB

Since mole fractions are constant, what must double is the partial pressures (each by a factor of 2).

Although I agree that in the isochoric reaction above, total pressure would depend on the number of gas molecules, I don't think the equation applies to the situation above. The example talks about a jar with 3 types of gasses, that is no liquid is involved. The euqation you mentioned is used to predict the total vapor pressure above a body of liquids.

 

[Monkeymaniac]

There is no discrepancy. We apply an external pressure on the system, and the system shifts to minimize as much as possible the total pressure of the system. So if we double the external pressure, according to Le Chat, the system shifts so that the final total pressure is not doubled also (if it can).

As Aljazera pointed out, if the total system pressure is less than outside pressure, the system would implode. Think about a glass jar for instance. The reason the jar retains its glass exterior is because air pressure inside equals outside pressure. If the force pressuring the jar from outside is greater than the force inside that pushes out the glass exterior, the glass wall would crack. I was asking how interior pressure could be 5.64 atm when external pressure is 6.0 in the example.

Aljazera, the example is listed in page 178 of gen chem part I. Thanks!

 

[Isoprop]

Although I agree that in the isochoric reaction above, total pressure would depend on the number of gas molecules, I don't think the equation applies to the situation above. The example talks about a jar with 3 types of gasses, that is no liquid is involved. The euqation you mentioned is used to predict the total vapor pressure above a body of liquids.

http://en.wikipedia.org/wiki/Dalton's_law
http://en.wikipedia.org/wiki/Partial_pressure

The equation can be applied to predict vapor pressure, but Dalton's Law is used for partial pressures in general.

As Aljazera pointed out, if the total system pressure is less than outside pressure, the system would implode. Think about a glass jar for instance. The reason the jar retains its glass exterior is because air pressure inside equals outside pressure. If the force pressuring the jar from outside is greater than the force inside that pushes out the glass exterior, the glass wall would crack. I was asking how interior pressure could be 5.64 atm when external pressure is 6.0 in the example.

Aljazera, the example is listed in page 178 of gen chem part I. Thanks!

(Caveat: I don't have the example, and I'm not confident in my answer.)

First of all, I believe you are mistaken that the jar will crack or implode if there is a pressure difference. Since we are talking about a rigid body, it should resist to some degree a difference in pressure. Think of SCUBA gear. The pressure inside is close to 1 atm but the pressure outside underwater is many times that.

Also, we are not doubling the external pressure by doubling the pressure outside the reaction container (I don't think). I think external is a misnomer here. I believe, but not certain, that external pressure is pressure change NOT coming from adding an inert gas. Adding a gas to the reaction chamber would increase the total pressure but not change equilibrium.

In an isochoric process, we can change pressure by changing the number of moles of gas molecules or by changing temperature. We aren't changing the number of moles of gas; so we are changing pressure by changing temperature. That's how we increase pressure, not by increasing the pressure outside the reaction chamber or jar.

 

[Monkeymaniac]

The equation can be applied to predict vapor pressure, but Dalton's Law is used for partial pressures in general.

I am aware of what Dalton's law is used for. The equation you listed above is used to predict vapor pressure. The equation isn't Dalton's law.

 

[Monkeymaniac]

I think by "external pressure" we are talking about increasing the pressure in the reaction chamber "from the outside." I think the book is trying to differentiate this by increasing the pressure of by putting in an inert gas inside the reaction chamber. If you increase the pressure this way, you won't get any shift.

Also, we are not doubling the external pressure by doubling the pressure outside the reaction container (I don't think). I think external is a misnomer here. I believe, but not certain, that external pressure is pressure change NOT coming from adding an inert gas.

Wrong, wrong, wrong, wrong and wrong! Seriously, please restrain from posting something unless you are absolutely certain that they are right. This would send people to all kinds of wrong directions, and they would have to spend meaningless hours in attempting to disprove some fraking BS theories. I'm sorry if I am being confrontational, but I've been a victim of this for so many times.

So you think pressure increase is not coming from doubling the "external" pressure or by adding inert gases. Then how would you expect the pressure to increase? By temperatuer increase? That would increase pressure inside an isochoric system, but why would the passage call it external then?

Also, who said anything about external pressure increased by the addition of inert gas? The term external pressure is NOT, I reapeat, NOT a misnomer here. Think of a piston. We can increase the external pressure by putting mroe weights on the lid of the piston. Although external pressure is constant after the weigts being added, volume is free to vary, so the system pressure decrease through equilibrium could be achieved without that implosion Aljazera mentioned. Yes, the pressure insdie SCUBA dive suit may be at SP but its volume is subject to change. That's why it doesn't implode, period.

 

[Isoprop]

I am aware of what Dalton's law is used for. The equation you listed above is used to predict vapor pressure. The equation isn't Dalton's law.

You're right. I switched Partial pressures with mole fractions. My mistake.

But the answer is the same.

PA = (XA)(PT)

Doubling PT
2PA = (XA)(2PT)

Since mole fraction stays the same, partial pressure must double also.

So you think pressure increase is not coming from doubling the "external" pressure or by adding inert gases. Then how would you expect the pressure to increase? By temperatuer increase? That would increase pressure inside an isochoric system, but why would the passage call it external then?

Also, who said anything about external pressure increased by the addition of inert gas? The term external pressure is NOT, I reapeat, NOT a misnomer here. Think of a piston. We can increase the external pressure by putting mroe weights on the lid of the piston. Although external pressure is constant after the weigts being added, volume is free to vary, so the system pressure decrease through equilibrium could be achieved without that implosion Aljazera mentioned. Yes, the pressure insdie SCUBA dive suit may be at SP but its volume is subject to change. That's why it doesn't implode, period.

Your piston analogy doesn't work because this is an isochoric process.

Like I said, this is my best guess. You ask a lot of good questions. But I think that your assumption that a rigid container must have equal pressures on the outside and inside is flawed.

The deepest diving submarine is ~6500 m. Some quick math:

Pressure = (density of water)*g*d (d = depth).
= (1000 kg/m^3)(10 m/s^2)(6500 m)
= 65000 kPa

That's 65000 kPa MORE THAN 1 atm (~100 Pa).

We can assume that the inside of the submarine is at about 1 atm. And I'm pretty sure the change in volume is negligible.

 

[Isoprop]

So you think pressure increase is not coming from doubling the "external" pressure or by adding inert gases. Then how would you expect the pressure to increase? By temperatuer increase?

The only way I know how to change the pressure without changing the volume is by changing the number of moles of gas, by changing the volume of the container, or by changing the temperature.

I know for certain the adding an inert gas does not change equilibrium. We are also not messing with the number of moles of gases. And you mention that this is a rigid chamber. So I'm guessing that this has to be a change in temp. But we even have to be careful with this because changing the temperature also changes the equilibrium constant, K.

I'm trying to make sense of the "external" too. Because I don't know.

I've never heard of "external pressure" with respect to Le Chat's principle. In all my texts (I'm looking at my undergrad text right now), it's been referred to as simply pressure.

 

[BerkReviewTeach]

In an attempt to play UN here:

I think the problem here is that there are several correct statements being made, but in the process, the facts about the question have been lost.

In the question (a sample question in the text on page 178 in the 2008 version), the system experiences a doubling of the external pressure. It does not say anything about being isochoric, so the doubling of the pressure must in fact be compressing the container, thereby increasing the internal pressure. Thus, the stress applied to the system is either a doubling of the internal pressure or the halving of the volume, depending on your prefered perspective.

The reaction shifts accordingly to try to alleviate the pressure increase, by moving to the side of the reaction with fewer gas molecules.

• In response to the difference between this question and a later piston question, I assume that's because this question assumed ideal gas behavior and the other did not, but I didn't look at the exact question to be sure.

 

[chemtopper]

let me try to help you
let us say we have a closed container of volume V1 and pressure P1 in which the reaction mentioned is carried out
A(g)+B(g)<->C(g)
now if you reduce its volume to V2 than we can say pressure inside the container has increased because now same amount of gases are kept in less volume . So more number of molecules per unit area and more collision per unit area . This in turn increase the number of effective collisions among the molecules of A and B to give D . The net effect will appear as shifting of equilibrium to right till inside pressure becomes equal to external atmospheric pressure .
Now the confusion is of external pressure mentioned by you (which is actually internal pressure in the container ) and atmospheric pressure (actually external pressure on the container )
Internal pressure is the pressure due to collisions of the gaseous molecules with the walls of the container which can be increased by decreasing the volume or adding an inert gas .Atmospheric pressure is the pressure on the container from outside .Shifting of Equilibrium is trying to nullify the effect of any increase in pressure inside the jar so that it will not explode . Le Chattlier principle on pressures is mainly based on equalizing the internal pressure and external pressure(atmospheric or reaction pressure ).
When you say that stress is applied on system it means you have tried to increase or decrease the pressure inside the jar in which reaction is carried out(system) and not of the surroundings.Surroundings pressure is still atmospheric pressure and now system has more pressure than outside .So equilibrium shifted so that pressure inside the system again can be balanced with outer atmospheric pressure.
I hope this will be helpful .