For the equilibrium Br2( g )+Cl2( g )?2BrCl( g ) at 400 K, K c = 7.0. 1) If 0.25
ID: 804870 • Letter: F
Question
For the equilibrium
Br2(g)+Cl2(g)?2BrCl(g)
at 400 K, Kc = 7.0.
1) If 0.25 mol of Br2 and 0.55 mol of Cl2 are introduced into a 3.0-L container at 400 K, what will be the equilibrium concentration of Br2? Express your answer to two significant figures and include the appropriate units.
2) If 0.25 mol of Br2 and 0.55 mol of Cl2 are introduced into a 3.0-L container at 400 K, what will be the equilibrium concentration of Cl2?
3) If 0.25 mol of Br2 and 0.55 mol of Cl2 are introduced into a 3.0-L container at 400 K, what will be the equilibrium concentration of BrCl?
Please include units for all answers :)
Explanation / Answer
Br2 + Cl2 <-> 2BrCl
I .083 .183 0
C -x -x +2x
E .083-x .183-x 2x
note that you have 3-L so to get M you must divide 0.25mol Br2/3-L=.083 and do the same for Cl2
Kc=7.0
7.0=(2x)^2/[(.083-x)(.183-x)]
7.0=4x^2/(x^2-.266x+.0153)
4x^2=7.0(x^2-.266x+.0153)
4x^2=7x^2-1.8666x+.1069
0=3x^2-1.8666x+.1069
a=3, b=-1.8666, c=.1069
-(-1.866)+/-squareroot(1.8666^2-4(3)(....
X= ----------------------------------------...
2(3)
x= .558, 0.0638
input both for x if you get a neg then its not valid.....
Br2=1.9x10^-2
Cl2=0.12M
BrCl=0.13M