## Chem. Equilibrium Part 3 Post-Assess. #19

Grace Lee 3G
Posts: 19
Joined: Fri Sep 29, 2017 7:07 am

### Chem. Equilibrium Part 3 Post-Assess. #19

19. 0.482 mol N2 and 0.933 mol O2 are placed in a 10.0 L reaction vessel and form N2O (dinitrogen oxide): 2N2(g) + O2(g) ⇌ 2N2O(g) KC = 2.0 x 10^-37
What is the composition of the equilibrium mixture?

A. [N2] = 0.0482 mol.L-1; [O2] = 0.0933 mol.L-1; [N2O] = 6.6 x 10-21 mol.L-1

B. [N2] = 0.0933 mol.L-1; [O2] = 0.0482 mol.L-1; [N2O] = 6.6 x 10-21 mol.L-1

C. [N2] = 0.482 mol.L-1; [O2] = 0.933 mol.L-1; [N2O] = 6.6 x 10-21 mol.L-1

D. [N2] = 6.6 x 10-21 mol.L-1; [O2] = 0.0933 mol.L-1; [N2O] = 0.0482 mol.L-1

I tried using the 5% rule because Kc is such a small number but I'm getting weird and different numbers each time.
How do I do this?

Suhail Zaveri
Posts: 24
Joined: Sat Jul 22, 2017 3:01 am

### Re: Chem. Equilibrium Part 3 Post-Assess. #19

Hello,

Ok this might be confusing but please bear with me. First you have to convert moles to molarity by dividing by the 10 liters. Secondly, since nitrogen has a 2 as a coefficient you have to square the initial molarity of nitrogen you solved above. Also you have to square the x for the change for n20 so the equation will look like ((x^2)/(.0482^2-x)(.0933-x) equal to 2.0 x 10^-37 and then you can assume x is small to get (x^2)/(.0933)(.0482) equal to the K value. Now you can get the x value and the answer, which is A.

Hope this helps :)

Michelle Dong 1F
Posts: 110
Joined: Fri Sep 29, 2017 7:04 am

### Re: Chem. Equilibrium Part 3 Post-Assess. #19

We only assume X is small because the K equilibrium constant is small enough to consider the change in concentration for the products negligible.

Grace Lee 3G
Posts: 19
Joined: Fri Sep 29, 2017 7:07 am