Achieve Homework Question 10

[105561735]

Could anyone help me solve the last part of this problem The reaction N2O4↽−−⇀2NO2
is allowed to reach equilibrium in a chloroform solution at 25 ∘C. The equilibrium concentrations are 0.375 mol/L N2O4 and 2.05 mol/L NO2. Calculate the equilibrium concentrations of N2O4 and NO2 after the extra 1.00 mol NO2 is added to 1.00L of solution.

kc = 11.206

[gary_vernes]

For this part, you will need to make an ICE table with the new concentrations of reactants and products. So in your table, the initial concentration for N2O4 would remain the same, but the initial concentration of NO2 would be 3.05 mol. From here, you would solve for x as you normally do.

[Amara Martinez 3B]

Hello, I’m not entirely sure which week this achieve assignment is from, but the way I would approach teh problem is to set up an ICE table. Because you are given the equilibrium concentrations of N2O4 and NO2 and it is known that 1mol/L= 1M is added to NO2, you can set up an ice table with the concs of 0.375M on the left(Reac) and 3.05M on the right(Prod) in the initial row. Because there was product added, by rules of Le Chateliers, i would have the +x on the reactants side and -2x on the products side. This will then give you your final equib concs including x, which you can plug into the eqn you would use to find K which is [NO2]^2/[N2O4] = Kc = 11.206. Then, solve for x and plug the x back into the final concs and you should have your answers. Hope this helped!

[705903057]

Could anyone help me solve the last part of this problem The reaction N2O4↽−−⇀2NO2
is allowed to reach equilibrium in a chloroform solution at 25 ∘C. The equilibrium concentrations are 0.375 mol/L N2O4 and 2.05 mol/L NO2. Calculate the equilibrium concentrations of N2O4 and NO2 after the extra 1.00 mol NO2 is added to 1.00L of solution.

kc = 11.206

Your Kc ([NO2]^2/[N2O4]=((2.05)^2/0.375)) is 11.206. You’ll use an ice chart:

N2O4 <=> 2NO2
I(M) 0.375 2.05
C(M) +x -2x
E(M) 0.375+x 2.05 -2x
(Hopefully the format is decent)

Kc = [NO2]2 / [N2O4] = ( 2.05-2x)2 / (0.375+x) = 11.206
And then just solve for x.
To find the equilibrium concentrations,

[N2O4] = 0.375 + x = ?? mol/L

[NO2] = 2.05 - 2(x) = ?? mol/L

Hope this helps!