Bromine gas is allowed to reach equilibrium according to the equation.

Question

Br2=2Br
Kc=0.0011 at 1280°C
Initial concentration of br2 is 0.063M and Br is 0.012M. How to calculate the concentration of these species at equilibrium. Please help me .

Answer 1

...............Br2 ==> 2Br
I............0.063....0.012

First you must determine which way the reaction will proceed; i.e., to the left or to the right.
Qc = (Br2)/(Br)^2 = 0.063/(0.012)^2 = about 440 and kc = 0.0011 so Qc is larger which means Br2 is too high and Br is too low. The reaction will proceed to the right.
............Br2 --> 2Br
I.........0.063.....0.012
C..........-x.......+2x
E........0.063-x...0.012+2x

Substitute the E line into the Keq expression and solve for x, then evaluate 0.063-x and 0.012 + 2x.

Answer 2

But wasn't it supposed to be [Br]^2/[Br2] to calculate the Qc. Therefore we will get
.....Br2...>2Br
I...0.063....0.012
C...+x........-2x
E...0.063+x....0.012-2x

Answer 3

I think that when Qc>Kc so the system wil shift to the left..

Answer 4

well, am I every embarrassed to make a simple freshman mistake. You are absolutely right and I am way way wrong.
Yes, Keq = (Br^-)^2/(Br2)
(0.012)^2/(0.063) = 0.00228 so Qc>K, the rxn will proceed to the left so the equilibrium should be written as
........Br2 ==> 2Br^-
I.....0.063.....0.012
C......+x........-2x
E.....0.063+x....0.012-2x

And go from there. Twenty lashes with a wet noodle for me. :-)

Answer 5

Ur hell