[A5HLEY]

Hi guys. I can't figure out what I'm doing wrong with this problem!

For the equilibrium below at 400K, Kc=7.0

Br2(g) + Cl2(g) <--> 2BrCl(g)

If 0.39 mol of Br2 and 0.39 mol Cl2 are introduced into a 1.0L container at 400K, what will be the equilibrium concentration of BrCl?

I know that Br2 and Cl2 have a 1:1 ratio, and they each have a 1:2 ratio when compared to BrCl.

I know that Keq=[BrCl]2/([Br2]*[Cl2]), and I can figure out the concentrations of Br2 and Cl2 since I'm given the mol. I made an ICE table, but when I tried to do the quadratic equation, I got some strange answers.

                                  Br2(g) + Cl2(g) <==> 2BrCl(g)
Initial                             .39M      .39M             0
Change                           -x           -x              +2x
Equilibrium                       .39-x      .39-x            2x

When I did the quadratic formula, I ended up getting x=-.4471185 and -3.0528814. What am I doing wrong?

[Yggdrasil]

You don't need the quadratic equation to solve the problem.  But, can you show your work so we can figure out where you went wrong in the calculation?

[A5HLEY]

Sure! For the equation, I got (4x^2)/(.78-2x)=7.0

So then I got 4x^2= 7.0(.78-2x)
4x^2=5.46-14x
4x^2 + 14x - 5.46 = 0

So I used the quadratic equation and got:
[-14 (+ or -) sqrt((14^2)-4(4)(5.46)]/8

and I ended up getting x= -.4471185 and -3.0528814

I think I went terribly, terribly wrong somewhere!

[Yggdrasil]

Where did you get the .78-2x from?  The denominator should be (.39-x)(.39-x) = .1521-.78x+x²

But an easier way to solve it is to see that:

(2x)²/(.39-x)² = 7.0

can be simplified to

(2x)/(.39-x) = sqrt(7)

[A5HLEY]

Oh, ok!

So when I solved that, here are my steps:
2x=(sqrt(7))(39-x)
2x= .39*(sqrt(7)) - (sqrt(7))x
2x + (sqrt(7))x = .39*(sqrt(7))
4.645751311x = 1.031843011
x= .2221046591