Bromine gas is allowed to reach equilibrium according to the equation.
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 .
...............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.
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
I think that when Qc>Kc so the system wil shift to the left..
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. :-)
Ur hell
To calculate the concentrations of Br2 and Br at equilibrium, we can use the equilibrium constant expression (Kc) and the given initial concentrations.
The equation for the equilibrium reaction is: Br2 ⇌ 2Br
The equilibrium constant expression in terms of concentrations is:
Kc = [Br]² / [Br2]
Given information:
Initial concentration of Br2 = 0.063 M
Initial concentration of Br = 0.012 M
Equilibrium constant (Kc) = 0.0011
Let's assume the equilibrium concentration of Br2 is x M. Since we have the stoichiometric ratio 2:1 between Br2 and Br, the equilibrium concentration of Br will be 2x M.
Now, using the given Kc and initial concentrations, we can set up the equation:
Kc = [Br]² / [Br2]
0.0011 = (2x)² / x
0.0011 = 4x² / x
Rearranging the equation:
0.0011 = 4x
We can now solve this equation for x:
x = 0.0011 / 4
x = 0.000275 M
The equilibrium concentration of Br2 is 0.000275 M, and the equilibrium concentration of Br is 2(0.000275) = 0.00055 M.
Therefore, at equilibrium, the concentrations are:
[Br2] = 0.000275 M
[Br] = 0.00055 M