TBR- gen chem equilibrium help

[Chelsea FC]

If a flask were filled with PCI3 and CI2 to a total
pressure of 1.50 atm. such that the mole fraction of
PCI3 is twice that of CI2, then what is the total
pressure of the system once at equilibrium?
A. Less than 1.00 atm.
B. Between 1.00 atm. and 1.25 atm.
C. Between 1.25 atm. and 1.50 atm.
D. Greater than 1.50 atm.

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[NextStepTutor_2]

If a flask were filled with PCI3 and CI2 to a total
pressure of 1.50 atm. such that the mole fraction of
PCI3 is twice that of CI2, then what is the total
pressure of the system once at equilibrium?
A. Less than 1.00 atm.
B. Between 1.00 atm. and 1.25 atm.
C. Between 1.25 atm. and 1.50 atm.
D. Greater than 1.50 atm.

Hi @Chelsea FC !

this question has been pored over previously on SDN so I will summarize what has been discussed.

Answer: B, To solve this question, use the formula X Pj = Ptotal- ln this case the following holds true:
Ptotal = PPCI3 + PC12 + pPCl5 = (1 " x) + (0-5 - x) + x = 1.5 - x.
If everything reacted to the limiting reagent capacity, then x is 0.5. But the reaction does not go to completion, so the value of x falls between 0 and 0.5. To solve this question, you must estimate the magnitude of x. The value of x is greater than 0.25, because over half shifts to products to reach an equilibrium constant of 1.95. This makes choice B the best answer.


You know that you start with 1.00 atm of PCl3 and 0.5 atm Cl2. Eliminate choice D immediately since you know total pressure will decrease as PCl5 is formed.

Total pressure at equilibrium = pressure of PCl3 + pressure of Cl2 + pressure PCl5 = (1-x) + (.5-x) + (0+x) = 1.5 - x

Note that x can't be greater than 0.5, since that means that Cl2 would have a negative partial pressure at equilibrium

So now you know that total pressure at equilibrium MUST be greater than 1.0 (since 1.5 - 0.5 = 1.0). Eliminate A.

How do you decide between choices B and C? If you are given that Keq = 1.95

Look at the case where x = .25. In that case, Q for the reaction will be (.25)/(.25)(.75) which is only 1.33. Since Q < K the reaction will keep shifting right. The answer must be B. Since Q < K, that means that the reaction needs to shift to the right. Since the reaction has to shift to the right, more product has to be made. For more product to be made, the x value has to increase and be bigger than 0.25. An increase in the x-value will decrease the expression "1.5 - x". This is why the answer is B between 1 to 1.25 atm.

hope this helps, good luck!

 

[Chelsea FC]

Hi @Chelsea FC !

this question has been pored over previously on SDN so I will summarize what has been discussed.

Answer: B, To solve this question, use the formula X Pj = Ptotal- ln this case the following holds true:
Ptotal = PPCI3 + PC12 + pPCl5 = (1 " x) + (0-5 - x) + x = 1.5 - x.
If everything reacted to the limiting reagent capacity, then x is 0.5. But the reaction does not go to completion, so the value of x falls between 0 and 0.5. To solve this question, you must estimate the magnitude of x. The value of x is greater than 0.25, because over half shifts to products to reach an equilibrium constant of 1.95. This makes choice B the best answer.


You know that you start with 1.00 atm of PCl3 and 0.5 atm Cl2. Eliminate choice D immediately since you know total pressure will decrease as PCl5 is formed.

Total pressure at equilibrium = pressure of PCl3 + pressure of Cl2 + pressure PCl5 = (1-x) + (.5-x) + (0+x) = 1.5 - x

Note that x can't be greater than 0.5, since that means that Cl2 would have a negative partial pressure at equilibrium

So now you know that total pressure at equilibrium MUST be greater than 1.0 (since 1.5 - 0.5 = 1.0). Eliminate A.

How do you decide between choices B and C? If you are given that Keq = 1.95

Look at the case where x = .25. In that case, Q for the reaction will be (.25)/(.25)(.75) which is only 1.33. Since Q < K the reaction will keep shifting right. The answer must be B. Since Q < K, that means that the reaction needs to shift to the right. Since the reaction has to shift to the right, more product has to be made. For more product to be made, the x value has to increase and be bigger than 0.25. An increase in the x-value will decrease the expression "1.5 - x". This is why the answer is B between 1 to 1.25 atm.

hope this helps, good luck!

Thank you ! but this method doesnt always work . For example this question

At 773K, the Kp for the following reaction is 3.0 x10"5. If the partial pressure of
N2(g) is initially 3.75 atm., and the partial pressure of H2(g) is initially 2.0 atm,;
what is the partial pressure of NH3(g) once equilibrium is established, assuming
there is no ammonia in the system initially?
N2(g) + 3H2(g) <-------> 2NH3(g)
A. 0.0900 atm.
B. 0.0300 atm.
C. 0.0100 atm.
D. 0.0010 atm.

When I do that method I come up with Ptotal = N2+ 3H2+2NH3 = 3.75 -x + 2-3x +2x = 5.75-2x .. This still doesnt work in this instances so I have to set up the quotient equation (2x^2)/(3.75)x(2^3) to find x.

 

[BerkReviewTeach]

In question #17, the answer choices were in terms of a range, so you had to think in a general fashion. The question can be thought of as:

If a flask were filled with PCI3 and CI2 to a total pressure of 1.50 atm. such that the mole fraction of
PCI3 is twice that of CI2, then what is the total pressure of the system once at equilibrium?
A. A shift larger than the limiting reagent would allow to the side with PCl5.
B. A shift to the side with PCl5 that would result if Keq > 1.
C. A shift to the side with PCl5 that would result if Keq < 1.
D. A shift to the side with PCl3 and Cl2, which is not possible unless you have PCl5.

There is no exact calculation necessary here. So intuition is your best approach, given that calculations take so much time. This is what you need to develop for the MCAT.

In Example 3.8, the answer choices are very similar numbers, so you cannot apply intuition to narrow down to a single answer. You have to do a calculation. So you are correct that you cannot always uses logic; sometimes you'll need to do math.

However, the answer explanation says at the end, it is more math than expected on the MCAT and is there to refamiliarize you with what you did in school. Once you do a few questions in detail, the shortcuts and strategies become more clear. That is why the first one you posted (#17) was in our practice passages, where you emphasize logic and speed. The second one you posted was in our text section, where it is meant to be review. Once you've done Example 3.8 to get reacquainted with what you did in general chemistry, you should not revisit that question. But as far as the passage questions go, we spent a great deal of time picking questions exactly like #17 to emphasize the power of reasoning. You should focus more on the passages. Text is for review and some very helpful test strategies.