Equilibrium Concentration Chemistry Problem
Need help.
Can you tell me where I go wrong?
Equation:
_ = subscript
Kp = 322
Br_2(g) + I_2(g) <-> 2IBr (g)
Initial Br_2 partial pressure = .05
Initial I_2 partial pressure = .04
ICE TABLE
I .05 .04 0
C -x -x +2x
E .05-x .04-x 2x
322 = (2x)^2 / (.05-x)(.04-x)
322(.05-x)(.04-x) = 4x^2
322(x^2 - .09x + .002) = 4x^2
(x^2 - .09x + .002) = 4x^2/322
(1 - 4/322)x^2 -.09x + .002 = 0
a= 318/322
b= -.09
c= .002
Quadratic formula gives me .0375
But solving for Kp again gives me only 180 instead of 322.
Halp
November 5, 2012
10 Comments • Newest first
Wait a minute, I used the wrong formula.
322 = [4x^2] / [(0.05-x)(0.04-x)]
x=0.0384171524
[Br_2] = 0.01158
[I_2] = 0.001583
[IBr] = 0.07683
Oh sorry. I didn't see you edited your post.
And also I see now it is rounding errors.
At these small of numbers it made a 140 difference.
Yea .0384 works.
.0375 close, is still off by 140. Crazy.
Thanks man.
Gotta make sure I don't round
[Edit] nevermind. this is wrong. brb.
My x is wrong and I don't know why. I did put the x back in.
322 = (2*.0375)^2 / (.05-.0375)(.04-.0375)
It ends up being 180=322
Can you tell me what your x was?
In the ICE table
Br_2 + I_2 <-> 2IBr
I 0.5 .04 0
C -x -x +2x
E .05-x .04-x 2x
I solved for x using the quadratic formula which was .0375.
.05 - .0375 = .0125 = concentration of Br_2 or [Br_2]
.05 - .0374 = .0025 = [I_2]
2*.0375 = .075 = [IBr]
Kp for this reaction is =
[IBr]^2 / [Br_2][I_2]
so
.075^2/ (.0125*.0025) = 180
which should be 322
Well my x turned about to be .0375.
Making the concentration of IBr = .075
Br_2 = .0125
I_2 = .0025
Resolving for Kp
.075^2 / (.0125*.0025) = 180
Original Kp = 322
hey im learning about this too!
Yea sorry.
Don't have a smart phone :[.
Its kay, I just have a quiz on Friday.
And this is the one question that I wasn't able to get right.
Thanks tho guise.
I guess I'll bring it up with my T.A. during discussion.
My professor didn't email back a friend of mine loool.
Have you considered emailing your TA, or instructor for halp?