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If 0.25 mol of Br2 and 0.55mol of Cl2 are introduced into a 3.0-L container at...

If 0.25 mol of Br2 and 0.55mol of Cl2 are introduced into a 3.0-L container at 400 K, what will be the equilibrium concentration of BrCl ?? Br2(g)+Cl2(g) 2BrCl at 400 K, Kc= 7.0
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Answer #1
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
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Answer #2
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
2 0
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Answer #3
initial concentration of Br2 =0.25/3 = 0.083
initial concentration of Cl2 = 0.55 mol/ 3.0 L = 0.183 M

Br2 + Cl2 < => 2 BrCl
start
0.25 .. . 0.55
change
-x. . .. . .-x. . . . . .+2x
at equilibrium
0.25-x. .. 0.55-x. . . 2x

Keq = [BrCl]^2 / [Br2][Cl2] = (2x)^2 / (0.25-x)(0.55-x) = 7.0

if we take the square root of both sides of the equation we get

x =0.19

[Br2] = 0.25 - 0.19=0.06 M

[BrCl]= 2x = 0.38M
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