Sapling Week 1 #10

[Sean Phen]

The reaction

N2O4↽−−⇀2NO2

is allowed to reach equilibrium in a chloroform solution at 25 ∘C . The equilibrium concentrations are 0.463 mol/L N2O4 and 2.28 mol/L NO2 .

Calculate the equilibrium constant, Kc , for this reaction.

I was able to get the equation products over reactants, but is it supposed to equal something so I can solve it.

[AlbertGu_2C]

the products over reactants should equal the Kc you need to find for the question

[Sara_Lim_2C]

I'm pretty sure it's just (2.28)^2/0.463, and you just solve that.

[Giselle Granda 3F]

All you have to do is plug in the concentrations you were given into your equilibrium expression because those were explicity said to be the equilibrium concentrations. So since it's products over reactants, it will be (2.28)^2/(0.463) and solve for that which will give you your Kc. Hope this helps!

[Lucy Wang 2J]

To solve for the equilibrium constant, Kc, you do [products]/[reactants]. So for this it would be (2.28)^2/(0.463)

[Sondia Luong 1C]

Since you are given the concentrations for both N2O4 and NO2, you would plug them into the equation for the equilibrium constant, which should be [NO2]^2/[N2O4].

[Jose Miguel Conste 3H]

so basically you're trying to find what the new equilibrium concentration would be if you added 1.0 M of NO2. So with the Kc you found from the original, you would then use icebox to solve what the new concentrations would be with 1.0 M added to NO2. Aka the initial would be 3.28 for NO2 and .463 for N2O4. Use icebox to solve this and set it equal to the Kc to find the NEW concentration.

[Arnav Saud 2C]

For that part of the problem, you just have to put the products/reactants and then solve for kc.
In this case it would be [NO2]^2/[N2O4]. Just plug in the given values and you'll get the kc.

[Ariana Flores 3F]

Because you know the concentrations for N2O4 and NO2, you just plug them into the equation for the equilibrium constant --> (NO2)^2/(N2O4)