Module 3 #19

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Haley Dveirin 1E
Posts: 101
Joined: Sat Jul 20, 2019 12:17 am

Module 3 #19

Postby Haley Dveirin 1E » Tue Jan 14, 2020 8:45 pm

Can someone explain how to solve this and what the answer is:

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?

Brooke Yasuda 2J
Posts: 102
Joined: Sat Jul 20, 2019 12:17 am

Re: Module 3 #19

Postby Brooke Yasuda 2J » Tue Jan 14, 2020 8:50 pm

So the moles of the reactants are given and the volume is also given. With this, you can calculate the concentration of N2 and O2. Then, you set up an equilibrium ICE table. When you do the "change" part of ICE, remember to look at the molar ratios of the reactants and products. Then, calculate the equilibrium constant and equate it to 2.0 x 10^-37. From here you can solve for the change x and find the equilibrium concentrations of the reactants and products.

faithkim1L
Posts: 105
Joined: Fri Aug 09, 2019 12:17 am

Re: Module 3 #19

Postby faithkim1L » Wed Jan 15, 2020 1:53 pm

First, convert moles given to molarity. It's a 10.0 L container, so the molarities of N2 and O2 are .0482 and .0933 respectively.

With the equation given, Kc = [NO2]^2/[O2][N2]^2

Create the ICE table.
N2 O2 N2O
I .0482 .0933 0
C -2x -x 2x
E .0482 - 2x .0933 - x 2x

Kc = 2.0 x 10^-37
Kc = (2x)^2/(.933 - x)(.482 - 2x)^2

*Since Kc is so small, we can assume that at equilibrium, there is a large amount of reactant and a very small amount of product. We can assume that the value of x makes an inconsequential different in the initial molarities of the reactants, so we can just treat "x" as 0.

Kc = 4x^2/(.933)(.482)^2
x = 3.3x10^-21

Use this x value and plug it into the equilibrium molarities.


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