The reversible chemical reaction A+B⇌C+D has the following equilibrium constant: Kc=[C][D][A][B]=7.5 Part A 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
Kc=[C][D][A][B]=7.5
[A] = [B] = 2 M
Find [A] when in equilibrium
in the equilibrium
[A] = [B] = 2-x
[C] = [D] = +x
substitute in K
Kc=[C][D][A][B]=7.5
(x)^2 /(2-x)^2 = 7.5
x/(2-x) = sqrt(7.5)
x = (2-x) * 2.74
x = 5.477 -2.74x
3.7x = 5.477
x = 1.48
substitute in [A]
[A] = 2-x = 2-1.48 = 0.52
B)
Find D when
[A] = 1 M
[B] = 2 M
then in equilibrium
[A] = 1 -x
[B] = 2 -x
[C] = [D] = x
Substitute in K
Kc=[C][D][A][B]=7.5
7.5 = (x^2)/((2 -x)( 1 -x))
solve for x
7.5(2-3x+x^2) =x^2
7.5(2-3x+x^2) - x^2 =
6.5x^2-22.5x+15 = 0
x = 0.90 or 2.56
x = 0.90 since 2.5 is impossible
then
[D] = x = 0.90
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