A scientist adds a small amount of catalyst to a 1 L reaction vessel containing 2.03 mol of A_{2}, 6.89 mol of B_{3}C, 1.40 mol of AB_{2}, and 2.50 mol of C_{2}. Which of the following will occur?

3 A_{2} (g) + 4 B_{3}C (g) → 6 AB_{2} (g) + 2 C_{2} (g)

Keq = 5.13 x 10^{2}

A. The reaction will more swiftly proceed to the left.

B. The reaction will more swiftly proceed to the right.

C. The addition of a catalyst does nothing to a reaction at equilibrium, so the reaction will remain at equilibrium.

D. No conclusion can be drawn.

**Explanation**

Begin by setting up the reaction quotient equation for this reaction:

Q = ([AB_{2}]^{6}[C_{2}]^{2}) / ([A_{2}]^{3}[B_{3}C]^{4})

Then plug in the values given:

Q = (1.4^{6} x 2.5^{2}) / (2.03^{3} x 6.89^{4})

The numbers are tough to deal with, so we’ll start by rounding them off:

Q = (1.5^{6} x 2.5^{2}) / (2^{3} x 7^{4})

Notice that we made the numerator bigger by rounding 1.4 up, so we need to make the denominator bigger by rounding up to 7. As much as possible, make your approximations cancel each other out.

We’ll still want to chip away at this by breaking down the numbers into small pieces:

Q = ((1.5^{2})(1.5^{2})(1.5^{2})x 2.5^{2}) / (2^{3} x (7^{2})(7^{2}))

Q = ((2.25)(2.25)(2.25)x 6.25) / (8 x (49)(49))

Round the 49 up to 50:

Q = ((2.25)(2.25)(2.25)x 6.25) / (8 x (50)(50))

Q = ((2.25)(2.25)(2.25)x 6.25) / (8 x (2,500))

Q = ((2.25)(2.25)(2.25)x 6.25) / (20,000)

Round the 2.25 down to 2 and the 6.25 up to 7 (remember, make your approximations cancel each other out):

Q = (2 x 2 x 2 x 7) / (20,000)

Q = (8 x 7) / (20,000)

Q = 56 / 20,000

Express in scientific notation:

Q = (5.6 x 10^{1}) / (2.0 x 10^{4})

Q = (5.6 / 2) x (10^{1} / 10^{4})

Q = 2.8 x 10^{-3}

Now that we’ve approximated Q, we can compare it to K_{eq} to see what will happen. Here, the Q is nearly 100,000 times smaller than the K_{eq}. That means our reaction quotient has far too many reactants (bigger denominator means a smaller number). So the reaction will move towards products, to the right.

Note that the arithmetic in this question is more difficult than you will likely see on the MCAT, simply due to the number of steps involved. The key lesson from this question, though, is that the MCAT absolutely will expect you to be able to approximate, and will expect you to solve questions that compare Q and K_{eq}.

**An alternate way to deal with this question would be simply to look at the sizes of the numbers involved. That is, you see a 7 ^{4} in the denominator. You could look at that and guess “okay this fraction will be less than 1. I don’t know how much less, but probably less than 1.” Then compare that to the K_{eq}, which is around 500, and arrive at answer (B).**

I have a question about this. Doesn’t it matter if the reactants/products are solids or not? The question does not specify if they are solids and therefore if they should/should not go into the Q and Keq equations. That’s why I thought the answer would be D. Thank you!

Hi Na, you’re right. The question should have specified the states of the reactants and products, and we’ve fixed the question to be more clear. Thanks for pointing that out!

The question says that there is 14 mol of AB2 but the question has been solved using 1.4 mol.

Hi Akriti, thank you very much for pointing this out to us! We have fixed the question so it asks about 1.4 mol, not 14 mol. If you notice any additional issues, or if we can help with anything else, please let us know!

At the end it states Keq=500, wouldn’t it be .05 since Keq=5*10^-2?

Hi Daniela, thank you for pointing this out! You are correct that 5 x 10^-2 would be equal to .05, not 500. I have changed the Keq for this reaction to 5 x 10^2. This way, you can get your answer as soon as you realize that Q will be less than 1, rather than needing to complete all of the math shown. This better approximates what you would actually need to do on the MCAT. Let us know anytime if you have additional questions!