Week 1 Sapling #10

[TiffanyBrownfield 2I]

I'm having a little trouble with the last part of number 10. Could someone explain the steps they followed or how they calculated these values?
The reaction:
N2O4↽−−⇀2NO2
is allowed to reach equilibrium in a chloroform solution at 25 ∘C . The equilibrium concentrations are 0.335 mol/L N2O4 and 1.94 mol/L NO2.
I found that my K = 11.2
Calculate the equilibrium concentrations of N2O4 and NO2 after the extra 1.00 mol NO2 is added to 1.00 L of solution.

[Juliana Rosales 1H]

I am literally having the same problem. I have the first part of the problem solved but I can't figure out where I am going wrong on the second part

[AHUNT_1A]

I am literally having the same problem. I have the first part of the problem solved but I can't figure out where I am going wrong on the second part

Same using this as a reference in case someone replies with help

[Charmaine Ng 2D]

Since 1mol was added to the 1L, (which is 1 Molarity), you add the 1.00 to the 1.94 Molarity of NO2, which means you have a starting concentration of 2.94 NO2 and 0.335 Molarity of N2O4. Since NO2 was added, it'll shift left towards the reactants. Set up an ice box with these initial concentrations, where NO2 becomes 2.94 - 2x, and N2O4 is 0.335 + x. From there, solve for x, plug it back in, and you'll have the equilibrium concentrations! Hope this helps!

[cadytran1K]

I also can't figure out the second part of this problem! I can't seem to figure out what I'm doing wrong.

[Jose Miguel Conste 3H]

so basically what you would do to solve this is using the ICE box, your initial would now be 2.94 for NO2 and 0.335 for N2O4, then you would solve it like the usual ICE box problem you do using the equilibrium constant you found for the first problem. Afterward, you would use the smallest x you solved from the ice box to solve the main problem. Which would be
[N204]= 0.335+x
[NO2]=2.94-2x

[TiffanyBrownfield 2I]

so basically what you would do to solve this is using the ICE box, your initial would now be 2.94 for NO2 and 0.335 for N2O4, then you would solve it like the usual ICE box problem you do using the equilibrium constant you found for the first problem. Afterward, you would use the smallest x you solved from the ice box to solve the main problem. Which would be
[N204]= 0.335+x
[NO2]=2.94-2x

Thank you! I was able to solve the question just in time thankfully.