Question

Hey guys stuck again!
Bromine and chlorine dissolve in carbon tetrachloride and react to form BrCl:
Br2 + Cl2--> 2BrCl (in CCl4)
Under equilibrium conditions [Br2] = [Cl2] = 0.00430 M and [BrCl] = 0.0114 M.
The addition of 0.100 moles of Br2 to exactly one liter of this mixture disturbs the equilibrium. Assuming a negligible volume change, calculate the concentration (in mol/liter) of BrCl once equilibrium has been reestablished.

My maybe answer:
Kc=7.03 from Kc=[.0114]^2/[.00430]^2
ICE Table
Br2 Cl2 BrCl
.10430 .00430 .0114
-x -x 2x
.10430-x .00430-x ,0114+2x

7.03=(,0114+2x)^2/(.10430-x)(.00430-x)
After a lot of algebra:
.003022356-.808912959x+3.03x^2=0
Quadratic formula yields x=.263 and x=.0037900876
The second x works with nothing being negative, so
BrCl=.0114+2(.00379)=.0190M
I didn't go through the math but I plugged the new values for products and reactants into the Kc expression and obtained 7.03 so I would have confidence that these were right.

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