The reversible chemical reaction A +B C+D has the following equilibrium constant

Question

The reversible chemical reaction A +B C+D has the following equilibrium constant K_c = [C][D]/[A][B] = 7.2 Initially, only A and B are present, each at 2.00 M. What is the final concentration of A once equilibrium is reached? Express your answer to two significant figures and include the appropriate units. [A] = Value Units What is the final concentration of D at equilibrium if the initial concentrations are [A] = 1.00 M and [B] = 2.00 M? Express your answer to two significant figures and include the appropriate units. [D] = Value Units

Explanation / Answer

Part A

The "REVERSIBLE" chemical reaction-
A + B <--> C + D
Kc = [C][D] / [A][B] = 7.2
Concentration at the start and after equilibrium-

. . . . . A . . .+ . . .B <===> . C . .+ . . .D
initial : 2.00 . . . . 2.00
react : . .-n . . . . . ..-n . . . . . .+n . . . . .+n
final : . 2.00-n . . .2.00-n . . . . .n . . . . . .n

Kc= [A][B] / [C][D]
7.2 = (2.00-n)(2.00-n) / ( n*n)
n = 0.53

the final concentration of A = 2.00 - 0.53 = 1.47 M

Part -B

The reversible chemical reaction
A+BC+D
1...2...0...0
1-x..2-x..x..x
so
K = 7.2= x^2/((1-x)*(2-x))

This equation will produce two solutions for x, 2.66 and 0.890. The first solution will be eliminated, since it implies that the equilibrium concentrations of A and B are negative.

This means that the equilibrium concentration of D will be

[D]=0.890 M

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