Chemical Equilibrium Post Assessment part 3 q20

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Alana Sur 3B
Posts: 61
Joined: Fri Sep 28, 2018 12:25 am

Chemical Equilibrium Post Assessment part 3 q20

Postby Alana Sur 3B » Wed Jan 23, 2019 9:56 pm

I'm getting a complicated polynomial equation and I'm pretty sure it's incorrect. Can someone please explain how to do this question?

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 c+2x

A. [SO2] = 0.522 M, [O2] = 0.663 M, [SO3] = 5.12 x 10-11 M

B. [SO2] = 0.522 M, [O2] = 0.663 M, [SO3] = 2.02 x 10-5 M

C. [SO2] = 0.522 M, [O2] = 0.633 M, [SO3] = 5.06 x 10-6 M

D. [SO2] = 0.522 M, [O2] = 0.633 M, [SO3] = 9.88 x 10-6 M

Ashley Zhu 1A
Posts: 69
Joined: Fri Sep 28, 2018 12:16 am

Re: Chemical Equilibrium Post Assessment part 3 q20

Postby Ashley Zhu 1A » Thu Jan 24, 2019 12:34 pm

Since Kc = 5.66 x 10^-10 <<< 10^-3, you can disregard the change in molarity caused by x for SO2 and O2, so instead of having Kc = (2x)^2/(0.522-x)^2(0.633-x), you'll just have Kc = (2x)^2/(0.522)^2(0.633), which you won't even have to use the quadratic equation for.

Answer should be D.

Danielle_Gallandt3I
Posts: 70
Joined: Fri Sep 28, 2018 12:24 am

Re: Chemical Equilibrium Post Assessment part 3 q20

Postby Danielle_Gallandt3I » Thu Jan 24, 2019 12:45 pm

A good rule of thumb is that a K value that is 10^-3 or less usually means that the x value will not cause a significant difference in the initial concentration value and can be disregarded when subtracting from the initial concentration. You can also check to see if this assumption is correct if the x value is less that 5% of the initial concentration value.


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