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.
To calculate the concentration of BrCl once equilibrium has been reestablished, you can make use of the equilibrium constant expression and the given equilibrium concentrations. Here's how you can approach the problem:
1. Write the balanced chemical equation and the expression for the equilibrium constant (Kc):
Br2 + Cl2 --> 2BrCl
Kc = [BrCl]^2 / ([Br2] * [Cl2])
2. Substitute the given equilibrium concentrations into the Kc expression:
Kc = (0.0114^2) / (0.00430 * 0.00430)
Kc = 7.03
3. Set up an ICE table (initial, change, equilibrium) to establish the changes in concentrations after the addition of Br2:
Br2 + Cl2 --> 2BrCl
0.00430 + 0.00430 --> 0.0114
-0.100 0 +0.200 (twice the amount of Br2 added)
4. Write the equilibrium expression with the new equilibrium concentrations:
Kc = (0.0114 + 2x)^2 / (0.00430 - x)(0.00430 - x)
5. Solve the quadratic equation (0.003022356 - 0.808912959x + 3.03x^2 = 0) to find the value of x. Select the positive root that ensures none of the concentrations become negative.
6. Substituting the value of x into the expression for BrCl concentration, calculate the new concentration of BrCl:
[BrCl] = 0.0114 + 2(0.0037900876)
[BrCl] = 0.0114 + 0.0075801752
[BrCl] = 0.0190 M
Therefore, the concentration of BrCl once equilibrium has been reestablished is 0.0190 M.