post-module #20

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annabelchen2a
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post-module #20

Postby annabelchen2a » Mon Jan 04, 2021 1:22 pm

A vial of SO2 (0.522 mol.L-1) and O2 (0.633 mol.L-1) react and reach equilibrium. Calculate the equilibrium concentrations of the products and reactants given that KC = 5.66 x 10-10 for this reaction:
2 SO2 (g) + O2 (g) ⇌ 2 SO3 (g)
concentration SO2 O2 SO3
Initial 0.522 0.633 0
change -2x -x +2x
Equilibrium 0.522-2x 0.633-x 2x

The correct answer is [SO2] = 0.522 M, [O2] = 0.633 M, [SO3] = 9.88 x 10-6 M

I got [SO2] = 0.522 M and [O2] = 0.633 M but for some reason I'm not getting 9.88 x 10-6 M as the concentration for SO3. What am I doing wrong?

Here's my work if it helps:

5.66 x 10^-10 = [SO3]^2 / [SO2]^2 [O2] = (2x)^2 / ((0.522-2x)^2 (0.633-x))

I got x to be 6.9864 x 10^-6

[SO2] = 0.522 - 2x = 0.5219
[O2] = 0.633 - x = 0.6329
[SO3] = 2x = 2(6.9864 x 10^-5) = 1.397 x 10^-5

Any help would be appreciated, thanks!

Brittney Nguyen 2L
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Re: post-module #20

Postby Brittney Nguyen 2L » Mon Jan 04, 2021 1:33 pm

Hi!

I think where you might have went wrong is in the numerator. In the K expression, you have [SO3]^2/[O2][SO2]^2. You're right in that [SO3] must equal +2x, but when you squared 2x, you ended up with 2x^2 instead of 4x^2 (the exponent distributes to both the coefficient and the variable). Using 4x^2 should give you the correct answer of [SO3] = 9.88 x 10-6 M.

I hope this helps!

Edit: Grammar
Last edited by Brittney Nguyen 2L on Mon Jan 04, 2021 4:41 pm, edited 1 time in total.

Lea Baskin Monk 1F
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Re: post-module #20

Postby Lea Baskin Monk 1F » Mon Jan 04, 2021 1:35 pm

Your work is set up correctly but you divided by 2 instead of 4 when isolating x. (2x)^2 = 4x^2 so...

1. 5.66 x 10^-10 = [SO3]^2 / [SO2]^2 [O2] = (2x)^2 / ((0.522-2x)^2 (0.633-x))

2. 5.66 x 10^-10 = 4x^2 / ((0.522-2x)^2 (0.633-x))

I made the exact same mistake the first time I did this problem

Samantha Pedersen 2K
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Re: post-module #20

Postby Samantha Pedersen 2K » Mon Jan 04, 2021 6:29 pm

This problem works out to have a cubic equation, and K is less than 10^-4. Dr. Lavelle mentioned in one of the Audio-Visual Focus Topic videos that this means we can assume 0.522-2x is just 0.522 and that 0.633-x is just 0.633. When you make this assumption, the equation becomes
5.66 x 10^-10 = 4x^2 / ((0.522^2) x 0.633). When you solve this for X and multiply X by 2, you get 9.88 x 10^-6 M for the equilibrium concentration of SO3. I hope this helps!

annabelchen2a
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Re: post-module #20

Postby annabelchen2a » Wed Jan 06, 2021 1:48 pm

Samantha Pedersen 2K wrote:This problem works out to have a cubic equation, and K is less than 10^-4. Dr. Lavelle mentioned in one of the Audio-Visual Focus Topic videos that this means we can assume 0.522-2x is just 0.522 and that 0.633-x is just 0.633.


Oh, okay, I must've missed that! Do we assume that (that being 1.522-2x is just 0.522 and 0.633-x is just 0.633) because K is such a small number? Or am I mistaken?

Samantha Pedersen 2K
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Re: post-module #20

Postby Samantha Pedersen 2K » Thu Jan 07, 2021 11:07 pm

annabelchen2a wrote:
Samantha Pedersen 2K wrote:This problem works out to have a cubic equation, and K is less than 10^-4. Dr. Lavelle mentioned in one of the Audio-Visual Focus Topic videos that this means we can assume 0.522-2x is just 0.522 and that 0.633-x is just 0.633.


Oh, okay, I must've missed that! Do we assume that (that being 1.522-2x is just 0.522 and 0.633-x is just 0.633) because K is such a small number? Or am I mistaken?


Yes, we can make that assumption because K is less than 10^-4 (this is the cutoff that Dr. Lavelle gave in the Audio-Visual Focus Topic video). I think Dr. Lavelle will go over this again in tomorrow's lecture too!

annabelchen2a
Posts: 107
Joined: Wed Sep 30, 2020 10:01 pm

Re: post-module #20

Postby annabelchen2a » Mon Jan 11, 2021 10:19 pm

Okay, thank you! Dr. Lavelle did go over it in today's (Monday's) lecture, though the cutoff he gave was any K value less than 10^-3. Any idea as to which rule to abide by?


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