In: Chemistry
The reversible chemical reaction
A+B⇌C+D
has the following equilibrium constant:
Kc=[C][D][A][B]=5.9
Initially, only A and B are present, each at 2.00 M. What is the final concentration of A once equilibrium is reached?
Part B
What is the final concentration of D at equilibrium if the initial concentrations are [A] = 1.00 M and [B] = 2.00 M ?
A + B <-------------> C + D
2 2 0 0
2-x 2-x x x
Kc = [C][D] / [A][B]
5.9 = x^2 / (2 -x)^2
x = 1.417
final concentration of A = 2 - x = 2 - 1.417
final concentration of A = 0.583 M
Part B )
A + B <-------------> C + D
1 2 0 0
1-x 2-x x x
Kc = [C][D] / [A][B]
5.9 = x^2 / (1-x)(2-x)
x = 0.882
final concentration of D = 0.882 M