TBR Gchem Chpt 3 #10

[FeinMS]

This question is regarding this reaction:
H2(g) + Br2(l) <--> 2 HBr(g)

10. Which of the following starting conditions results in
the GREATEST amount of H2(g) at equilibrium?
A. 0.80 atm. H2(g) and 20 g Br2(l)
B. 1.00 atm. H2(g) and 20 g Br2(l)
C. 0.80 atm. H2(g) and 30 g Br2(l)
D. 1.00 atm. HBr(g) and 30 g Br2(l)

The correct answer is B. TBR says that to get the greatest amount of H2(g), we need to maximize the reverse direction and minimize the forward direction. I get this part, but when explaining why D is wrong, TBR says "If choice D went one hundred percent in the reverse direction, then 0.50 atm. of H2(g) would be generated." I don't quite get how they derived the partial pressure of H2(g) to be 0.50atm.

Is this because of the stoichiometry of H2 and 2HBr? So if 1atm of HBr is totally converted to H2, you get half of the partial pressure?

Another quick question. I've been getting 60%-80% on TBR PS,BS passages, but when looking at the conversion chart for AAMC FL's, it seems like I need to get almost 90%+ to receive my a good score (my standard.) Does it mean I gotta work my butt off, or are TBR science passages naturally difficult than what I would expect on real MCAT?

Thanks for your help.

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[sunshine02]

I know this is a year old, but I'm having the exact same question right now. Does anyone know how to get the answer B without doing much calculation?

 

[Teleologist]

Calculate reaction quotient values, Q. Then think about what Q means and choose accordingly. (This only works if there is a K value provided or if a K value can be derived. I don't know since I don't have the book and passage).

 

[sunshine02]

The K value is 5.44, which was derived in the question before this one. The passage provided no further information that could help solve this question. I just don't understand why it is B. A K value of 5.44 means there are more products than reactants, so wouldn't 1 atm of H2 gas mostly shift over to products in order to establish equilibrium?

 

[Czarcasm]

The K value is 5.44, which was derived in the question before this one. The passage provided no further information that could help solve this question. I just don't understand why it is B. A K value of 5.44 means there are more products than reactants, so wouldn't 1 atm of H2 gas mostly shift over to products in order to establish equilibrium?

This is an odd question but I'll take a stab at it. First you need to realize we're at non-equilibrium conditions. Based on the conditions provided, the reaction will shift to re-establish equilbrium conditions.

The second thing you want to do is write the equilibrium expression: Keq = [HBr)^2/[H2];

Then ask yourself, what would happen in each extreme:

A. You have 0.80 atm H2. If H2 was consumed entirely such that at the end of the reaction you have no H2 remaining and entirely HBR, the most it would produce (according to the stoichiometric ratio) is 1.6 atm. Chances are though it will not react entirely, but a good bit of it will react to produce some HBr to reach equilibrium.

D. Here we have no H2 present. We're told we have 1 atm HBr. Again if we assume it reacts to completion, (because of the stoichiometric ratio 2 HBR to 1 H2), if HBR reacted completely, it would only produce 0.50 atm. But the chances of that are unlikely, even if Keq was a very small number. (Note that 0.50 atmof H2(g) is less than any of the choices available in A, B, and C). So we can eliminate D.

Because we know that some H2 will indeed react to produce HBr, why not start with the most amount possible so that at equilibrium we have the most H2 remaining. Therefore choice B is better than either A and C.

The amount of Br2 in all 4 scenarios is irrelevant since it doesn't effect equilibrium concentrations.

 

[sunshine02]

This is an odd question but I'll take a stab at it. First you need to realize we're at non-equilibrium conditions. Based on the conditions provided, the reaction will shift to re-establish equilbrium conditions.

D. Here we have no H2 present. We're told we have 1 atm HBr. Again if we assume it reacts to completion, (because of the stoichiometric ratio 2 HBR to 1 H2), if HBR reacted completely, it would only produce 0.50 atm. But the chances of that are unlikely, even if Keq was a very small number. (Note that 0.50 atmof H2(g) is less than any of the choices available in A, B, and C). So we can eliminate D.
.

Thanks! I get where you are going, but you said above that 0.5 atm of H2 is less than any of the choices available in A, B, and C. While that is true, 0.5 atm is partial pressure of H2 at equilibrium, and the partial pressure numbers in choices A, B, C are the starting concentrations, so at equilibrium, the actual partial pressure will be less. Therefore, how would you know that the final equilibrium partial pressure in choices A, B, and C doesn't go below 0.5 atm?

Again thanks so much!

 

[Czarcasm]

Thanks! I get where you are going, but you said above that 0.5 atm of H2 is less than any of the choices available in A, B, and C. While that is true, 0.5 atm is partial pressure of H2 at equilibrium, and the partial pressure numbers in choices A, B, C are the starting concentrations, so at equilibrium, the actual partial pressure will be less. Therefore, how would you know that the final equilibrium partial pressure in choices A, B, and C doesn't go below 0.5 atm?

Again thanks so much!

That's a great point. And honestly, it'd be difficult to really determine that qualitatively (based on what I know atleast). I mean by reasoning it alone, I'd look at it and see more gas molecules produced and figure it's a highly spontaneous reaction. But based on your calculation to the previous question, where you determined K was 5 (as opposed to some really huge number like 10^3), then it makes things more straightforward because now we know that equilibrium doesn't really favor either reactants or products to any significant degree (a little towards products but not to a large degree).

This is important because, if K was a huge number, for example, we might expect say:

50% of 1 atm H2 (greater shift to K) to react in choice B, leaving 0.5 atm H2 remaining.
25% of 0.8 atm H2 (smaller shift to K) to react in choice A, leaving 0.6 atm H2 remaing. (This would lead us to think choice A or C is right).

By the way, these are just random percentages (highly exaggerated) to explain my point. I highly doubt the difference would ever be as drastic though, especially since the partial pressures are relatively close to one another. If instead it was 10 atm H2 and 1 atm H2 we were comparing, then it'd be more likely that a lot more 10 atm H2 would react relative to starting at 1 atm H2. (Even still, you'd likely have more H2 remaining if 10 atm H2 were present than 1 atm H2, just because there's so much more). So in other words, because the two partial pressures are close, the change between the two wouldn't be as significant as it would if the two partial pressures were largely different.

But knowing what we now know about K, if you weren't able to reason it out, we at least know for a fact now that the change definitely won't be as drastic (only a small amount will react), so it's safe to just focus on picking the choice with the greatest amount initially (in this case, greatest partial pressure).

Hope that makes sense.

 

[Czarcasm]

Thanks! I get where you are going, but you said above that 0.5 atm of H2 is less than any of the choices available in A, B, and C. While that is true, 0.5 atm is partial pressure of H2 at equilibrium, and the partial pressure numbers in choices A, B, C are the starting concentrations, so at equilibrium, the actual partial pressure will be less. Therefore, how would you know that the final equilibrium partial pressure in choices A, B, and C doesn't go below 0.5 atm?

Again thanks so much!

Lol, disregard my response above. I totally misinterpreted your question. First, we need to realize that the chances of 0.5 atm H2 being produced in D is highly highly unlikely. The reaction as written increases entropy and probably is very exothermic (favors products at equilibrium). Therefore, if HBr were to react to produce H2, what we can definitely say is it would not react completely. (Very few reactions ever go to completion). Therefore, much less than 0.5 atm would be produced since 0.5 atm H2 represents 1 atm HBr reacting completely.

So the question then becomes, will choices A, B, and C (H2 concentration at equilibrium) be slightly greater than > 0.5 H2 atm (the extreme in choice D) and the likelyhood is yes.

Again, I'm not really a fan of the question, but I suppose it really helps to think about what's happening from a conceptual viewpoint.

 

[Chrisz]

First ignore the amount of Br2 present, it is a liquid so it does not affect the equilibrium. Second, 0.8 atm is less than 1 atm, so choice A and C are eliminated. Now we need to compare B and D. Lets assume what if 1 atm H2 reacts completely to form HBr. 1 atm would produce 2 atm HBr. Now comparing this amount with D, which is just 1 atm HBr. That means 2 atm HBr will have more H2 produced when equilibrium is reached. So answer is B

 

[Lunasly]

Calculate reaction quotient values, Q. Then think about what Q means and choose accordingly. (This only works if there is a K value provided or if a K value can be derived. I don't know since I don't have the book and passage).

I'm hoping someone can provide more insight into this problem. I'm personally not satisfied with TBR's explanation as I am not sure how we can make the assumption that H2 gas would not dip below 0.5 atm upon reaching equilibrium. My first instinct was to do what Teleologist suggested. We calculated Kp in a previous question (5.44) and the Q value using the partial pressures in this question = 6. Therefore, since Q is higher than K, the equilibrium shifts towards the reactants. The first sentence of TBR's explanation even says that the greatest amount of H2 at equilibrium will be formed with the smallest forward reaction OR the greatest backward reaction. Since Q > K, the equilibrium shifts left so we could expect less than 0.5 moles of H2 gas to form. On the other hand, I don't know how we would determine that 0.5 moles or more H2 gas will be present at equilibrium.

Thanks for the help.

 

[drechie]

any update on how to discern an answer if we are chosing between B and D

 

[DoctorInASaree]

any update on how to discern an answer if we are chosing between B and D

Two ways that I would answer this question


1) We want to maximize the amount of H2 (reactant); ignore Br2.
More H2 = yay
More product = boo

Therefore, B instead of D.

2) Le Chatalier's principle: Increase external pressure, system shifts to the left. If you want to maximize H2 which is a reactant, we need to, in turn, maximize H2's partial pressure in atm. Therefore, B!

I hope that helps.

Edit: If anyone is being confused by the answer D explanation of how TBR attained an H2 value of 0.5

Construct an ICE table and you'll see that HBR is 2x and H2 is -x, so if I have 1.00 atm of HBR I have to divide that by 2 which gives me 0.5atm H2; which is less than answer choice B's value of 1.00atm of H2

 

[TheRhymenocerous]

Am I oversimplifying this? The way I'd answer that is think, okay, that means you want the most H2 leftover unreacted. H2 and Br2 react one to one, so if you have way more H2 than Br2 it'll be in excess and not react. Bromine's atomic mass is 80, so 20 g is only .25 moles. 1 mole vs. 0.25 moles leaves a lot of H2 on the table. It wouldn't be D because even if you dissociated all of that 1 mole of HBr, that's only 0.5 moles of H2 since it's 2 moles HBr to 1 mole H2.