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At 25 oC, Kc = 0.145 for the following reaction in the solvent CCl4. 2BrCl <----> Br2 + Cl2 A solution was prepared with the fo...Asked by Peter
At 25 oC, Kc = 0.145 for the following reaction in the solvent CCl4.
2BrCl <----> Br2 + Cl2
A solution was prepared with the following initial concentrations: [BrCl] = 0.0482 M, [Br2]= 0.0307 M, and [Cl2]= 0.0277 M.
What will their equilibrium concentrations be?
How do i know when the change in concentration is +x or -x?
I tried-2BrCl<-->Br2 Cl2
.0482 .0307 .0277
-2x +x +x
and reverse too but it's wrong. Any ideas?
2BrCl <----> Br2 + Cl2
A solution was prepared with the following initial concentrations: [BrCl] = 0.0482 M, [Br2]= 0.0307 M, and [Cl2]= 0.0277 M.
What will their equilibrium concentrations be?
How do i know when the change in concentration is +x or -x?
I tried-2BrCl<-->Br2 Cl2
.0482 .0307 .0277
-2x +x +x
and reverse too but it's wrong. Any ideas?
Answers
Answered by
DrBob222
You must have made a math error.
........2BrCl ==> Br2 + Cl2
I.......0.0482..0.0307..0.0277
First decide which direction the reaction will move? How? Qrxn.
Q = (0.0307)(0.0277)/(0.0482) = 0.0176 and compare this with Kc = 0.145. Q < K which means the fraction is too low in the numerator and too high in the denominator and that means the rxn must go to the right.
C.........-2x.......x.......x
E......0.0482-2x....x.......x
and solve for x. and 0.0482-2x.If you still have a problem post your work and I'll find the error.
........2BrCl ==> Br2 + Cl2
I.......0.0482..0.0307..0.0277
First decide which direction the reaction will move? How? Qrxn.
Q = (0.0307)(0.0277)/(0.0482) = 0.0176 and compare this with Kc = 0.145. Q < K which means the fraction is too low in the numerator and too high in the denominator and that means the rxn must go to the right.
C.........-2x.......x.......x
E......0.0482-2x....x.......x
and solve for x. and 0.0482-2x.If you still have a problem post your work and I'll find the error.
Answered by
Satan
You are wrong!!!!!!
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