achieve week 1, #10

[Vicky Tong 3E]

On achieve, the last question asks this:
The reaction

N2O4↽−−⇀2NO2

is allowed to reach equilibrium in a chloroform solution at 25 ∘C. The equilibrium concentrations are 0.435 mol/L N2O4 and 2.21 mol/L NO2.

Calculate the equilibrium concentrations of N2O4 and NO2 after the extra 1.00 mol NO2 is added to 1.00 L of solution.

N2O4​
.435
-x
.435-x

2NO2
3.21
+2x
3.21+2x

I got 11.22 for Kc, and when I did the new ICE chart, I ended up with x=3.47. Obviously, this is incorrect, but I can't figure out what I'm doing wrong. I would greatly appreciate the help!

[Anne Mariano 3F]

Did you solve this equation in the process of finding x?
Kc=[NO2]^2 / [N2O4]
11.2=(3.21−2x)^2 / (0.435+x)

(11.2) * (0.435+x)=(3.21−2x)^2
11.2x + 4.872 = 4x^2 − 12.8x + 10.3
4x^2 − 24x + 5.43 = 0

Make sure to double-check your numbers and to input parentheses correctly.
Using the quadratic formula, you will get two answers for x, the lower one will be the only possible answer.

[Yajing Feng 2J]

Hi,

I had different given values but my ice table was set up like this:
initial: N2O4=0.343M, 2NO2=2.96M
change: N2O4= +x, 2NO2= -2x
equilibrium: N2O4=0.435 +x, 2NO2= 3.21 -2x

**The difference between our ICE tables is the change row. My change row has the reactant side as positive (+x) and the product side as negative (-2x). I did this because the problem said that "extra 1.00 mol NO2 is added" which means that the reaction will want to get back to equilibrium but shifting to the left and use that extra 1.00 mol of NO2 to create more reactant, hence why my ICE table's reactant side as positive (+x) and the product side as negative (-2x). Then continue the work you were doing.

Hope this helps.

[Allison Lee 2C]

Your Kc value should be correct; however, your Change and Equilibrium rows for the ICE table are where there are errors. Because we increased the concentration for NO2 (the product side), we know that the reaction will shift left to the reactant side so equilibrium can be reached again. Thus, the change row should have -2x (not +2x) for the 2NO2 column, and +x for the N204 column, since more reactants are forming. Thus, the equilibrium row should be 0.435 + X for N204, and 3.21-2x for 2NO2. From there, you should be able to solve for X and find the new concentrations. Hope that helps!