[Noor]

So this is the question:

At 1,700 K, the equilibrium constant for the reaction below is Kc=0.506.
Br2(g) --> 2 Br(g)
What is the percent dissociation of Br2 at 1,700 K, if the initial concentration of Br2 is 1.75 mol/L?

This is what I did:

0.506 = (2Br)^2 / (Br2)
let X = moles/L of Br2 dissociating
moles of Br = 2X/L
0.506 = (2X)^2)/ (1.75 - X)
0.8855 - 0.506X = 4X^2
4X^2 + 0.506X - 0.8855 = 0
X = 0.5486
0.5486/1.75 = 31.35%

But the system keeps telling me that it's wrong!!! why? Can someone help me here???

[Dan]

4X^2 + 0.506X - 0.8855 = 0
X = 0.5486

I think your solution for X is wrong, check your maths, I get:

K = [Br.]²/{[Br₂]_i - ([Br.]/2)}

Where [Br₂]_i = 1.75 mol/L (initial Br₂ concentration)

Let X = ([Br.]/2)

Then, 0.506 = (2X)²/(1.75 - X)

so, 4X² + 0.506X - 0.8855 = 0

X = {-0.506 + (0.506² + 16(0.8855))^1/2}/8

(or X = {-0.506 - (0.506² + 16(0.8855))^1/2}/8)

=> X = 0.411  (-0.538, but negative concentration is not possible)

% dissociation is given by  X/[Br₂]_i

=> % dissociation = 23.5 %