# BR Gen Chem- Equilibrium Passage II # 10

Discussion in 'MCAT Study Question Q&A' started by bonitanolita, Jan 7, 2011.

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1. ### bonitanolitaNew Member

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The passage gives you the reaction

H2 (g) + Br2(l) --> 2HBr(g)

10. Which of the following starting conditions results in the GREATEST amount of H2 (g) at equlibirum?

A. .80 atm H2 and 20 g Br2
B 1.00 atm H2 and 20 g Br2
C. .80 atm H2 and 30 g Br2
D. 1.00 atm HBr and 30 g Br2

B. is the correct answer. The explanation says that the greatest amount of H2 comes form the reaction with the greatest back reaction or smallest forward reaction.A and C are eliminated because the difference is in the amount of liquid Br2, which doesn't affect the equilibrium. I understand that, but then the explanation goes on to say that if Choice D went 100% in the reverse direction, then .50 atm of H2 would be generated. This value is half of the amount of H2 in B, so D is eliminated. To generate an equivalent amount of H2 (g) at equilibrium as starting with 1.00 atm. H2 and excess Br2, requires starting with 2.00 atm HBr. Can someone please explain this to me, I don't understand how could determine this.

2. ### RaboliskNew Member

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I don't know what the passage gives you, but let's assume that you don't know the equilibrium constant. That is, you don't know the ratio of [HBr] to [H2] at equilibrium. What you know is that Br2 does not affect the equilibrium, and the stoichoimetric relationships. For every molecule of H2 consumed, 2 molecules of HBr are produced (forward direction) or for every molecule of HBr consumed, 1/2 molecule of H2 is produced (reverse direction).

In choice D, starting with HBr concentration of 1.00 atm, the maximum possible you can end up with at equilibrium is 0.50 atm of H2. Depending on what the equilibrium constant is, you will eventually end up with anywhere between 0 to 0.50 atm of H2, but never more. In choice C, because you start with 1.00 atm of H2, you can end up with anywhere between 0 to 1.00 atm of H2 at equilibrium. You might ask, why can't choice B end up with 0.10 atm of H2 and choice D end with 0.50 atm? The equilibrium constant is the same in both cases, so the ratio of H2 to HBr has to be the same. Mathematically, it is impossible for there to be more H2 at equilibrium in case D than in case B.

For example, let's say that the K (eq constant) was 1. Then in D, the equilibrium concentrations of H2 and HBr would be 0.33 atm for both. (0.33/0.33 = 1) In B, the concentrations would be 0.66 atm for both (0.66/0.66 = 1). Set up an ICE box to convince yourself that this has to be the case.
3. ### BurghStudentlurker

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I had the same problem. I need help; it's hard for me to understand partial pressures. If a problem uses partial pressures and the container is a fixed volume, I can consider them as if they're moles right? (P =nRT/V).

So, in this problem Keq = 5.44. In number 10, I got down to either choice B or D. With B, you start with 1.00atm H2, while in D, 1.00atm of HBr. How can you know without doing calculations which one will produce the highest partial pressure of H2?

I understand that in B, the H2 can only be between 0-1.00atm, and in D, 0-.5atm. But how do you know the answer is D? I feel like there's something fundamental I'm missing here.
4. ### SsinaNew Member

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the answer is actually B, in case you simply overlooked it... as Rabolisk explained, I first figured you want the least amount of Br2 for it to not go forward as much, then we lose less H2 (which I just noticed after reading his explanation Br2 is as liq so that doesn't change all that much)... but as for B vs. D... if B goes all the way forward you get 2 atm HBr (vs. D which you have 1atm) so it's more!
5. ### ljcNew Member

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To elaborate a little: Starting with choice D: D gives 1atm HBr, which by coefficients we know is only 0.5atm H2. So now you have 1atm/0.8atm H2(from the other options) vs 0.5atm H2. Additionally, the 30g Br2 in D will push the reaction to the right, which is the opposite of what we want. So we can eliminate D.

Between A and C, A is obviously the better choice for the same reason: at the same pressure, more Br2 will lower the amount of H2 (by LeChatlier's Principle), so we can eliminate C.

This leaves B and A. We start with more H2 in B, so we must end with more. So the answer is B.
Last edited: Nov 29, 2011
6. ### ErythropoietinEPO

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Yes you're right