The reversible chemical reaction
A+B⇌C+D
has the following equilibrium constant:
Kc=[C][D][A][B]=7.3
What is the final concentration of D at equilibrium if the initial concentrations are [A] = 1.00 M and [B] = 2.00 M ?
ICE Table:
[A]
[B]
[C]
[D]
initial
1.0
2.0
0
0
change -1x -1x +1x +1x
equilibrium 1.0-1x 2.0-1x +1x +1x
Equilibrium constant expression is
Kc = [C]*[D]/[A]*[B]
7.3 = (1*x)(1*x)/((1-1*x)(2-1*x))
7.3 = (1*x^2)/(2-3*x + 1*x^2)
14.6-21.9*x + 7.3*x^2 = 1*x^2
14.6-21.9*x + 6.3*x^2 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 6.3
b = -21.9
c = 14.6
Roots can be found by
x = {-b + sqrt(b^2-4*a*c)}/2a
x = {-b - sqrt(b^2-4*a*c)}/2a
b^2-4*a*c = 1.117*10^2
roots are :
x = 2.577 and x = 0.8993
x can't be 2.577 as this will make the concentration
negative.so,
x = 0.8993
At equilibrium:
[D] = x = 0.8993 M
Answer: 0.899 M
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