Temperature Affect on Equilibrium Constant

Halogens can react with each other, as follows:
Br2(g) + Cl2(g) --> 2 BrCl(g)

At 350K, this reaction was carried out in a reaction vessel and came to equilibrium, with the following concentrations: [Br2] = 0.0550M, [Cl2] = 0.0850M; [BrCl] = 0.0150M.
No chemicals were added to or removed from the reaction vessel, but the temperature was then increased to 380K, at which temperature the new equilibrium constant was
Kc = 0.0175. Reaction occurred to reach the new position of equilibrium. Determine the equilibrium concentrations of Br2 and BrCl at 380K.

Thanks so much for any help!

wow... i should know how to do this. ugh let me think it through...

k well i have an idea, but i dont know if its right...

the Kc of the 350K is (.015*.015)/(.0850*.0550)= 0.048128342

since the new Kc is lower than the original, we know that the equilibrium must have shifted to the left (since the reactions are on the bottom of the Keq equation, by shifting to the left, the Keq gets smaller)
Br2 Cl2 2BrCl
I .0550M .0850M .0150M
C + x +x -2x
E .055+x .085+x .015-2x

Kc=.0175= [(.015-2x)^2]/[(.085+x)(.055+x)]
.0175= [4x^2-.06x+.000225]/[x^2+.14x+.004675]
cross multiply and simplify:
3.9825x^2-.05755x+.0001431875
quadratic equation time... x=[-b+/-(b^2-4ac)^.5]/2a x=.011256696 or 0.003194025487
it cant be the first option because .0150-2(.011256696) < 0. You cant have a negative concentration. So it must be 0.003194025487.

[Br2]=.055+x= .058194025
[BrCl]= .0150-2x= .008611949026
[Cl2]= .085 + x = .088194025

Check:
Kc= (.008611949026^2)/(.088194025*.058194025… uhh... that is not right!!!

alright so there is a really good chance that i messed up with the numbers. but i believe that my method was correct. try it yourself and hopefully your numbers work out.