Question

For the Equilibrium find the equilibrium concentrations of reactants and product

Answer 1

Equilibrium equation
\[Br2 + Cl2 -> <- 2BrCl \] all gases
at 400 K Kc= 7

If .25 mol Br2 and . 55 mol Cl2 are introduced into a 3.0 L container at 400 K, what will be the equilibrium concentration of Br2? Cl2? and BrCl2?

So i made an ICE Table
\[Br2 + Cl2 -> <- 2BrCl \]
I .083 M .183 M 0M
C -x -x +2x
E .083-x .183-x 2x

But i can't figure the math out for some reason...i know you have to use Kc= [Brcl] ^2/ [Br2] [Cl2]

Answer 2

YOu could look up some videos on youtube. Youtube has videos for like everything and can explain it good.

Answer 3

I understand that...I would prefer if someone explained this problem with these numbers though, it would make way more sense to me...Im not too good at math

Answer 4

I can't because I am not sure

Answer 5

Ok...Someone else might though

Answer 6

@myininaya @Nnesha @bibby

Answer 7

Follow from the E row values from your ice table. Use the equation you wrote exactly in the end of your post and equate it to 7. Get an equation and find x.
After that, solve the expression of concentration of each product and reactant after reaction, using the x value you found.

Answer 8

if they're gases then you need to use \(K_p\) and not \(K_c\)

the relationship relating them is \(K_p=K_c*RT^{\Delta n}\)

http://bilbo.chm.uri.edu/CHM112/lectures/KpKc.htm

Since \(\Delta n=0\), then \(K_p=K_c\), so it's fine to use the same value.

\(K_p=7=\dfrac{[BrCl]^2}{[Br_2][Cl_2]}\)

\(K_p=7=\dfrac{(2x)^2}{(x-0.083)(x-.0183)}=\dfrac{4x^2}{(x-0.083)(x-.0183)}\)

\(7=\dfrac{4x^2}{(x-0.083)(x-.0183)}\)


x=0.0160892577080661, x=0.2202774089586005

only 0.016 makes sense because 0.22 exceeds the 0.083 value.

Plug in x into the E expressions (from ICE table) to find the concentration at eq.

Answer 9

thank you but how did you get .0183? it is .183

Answer 10

oh damn, i miscopied that from your post. you should rework the value for x

Answer 11

that's the part i'm not good at...i don't understand math very well

Answer 12

you'll end up getting a quadratic equation, and therefore 2 x's. You have to then pick one after logical reasoning. If you're having trouble with the math... you can use wolfram and it'll give you the resulting x's... but basically you expand the denominator, and then bring it the left side, and then bring everything to the left side to have the right side =0. you then factor using the quadratic equation.

Answer 13

ok thank you