Question

The reversible chemical reaction

A+B⇌C+D

has the following equilibrium constant:

Kc=[C][D][A][B]=3.8

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 ?

Answer 1

A+B⇌C+D

K_eq = {[C]*[D]}/{[A]*[B]}

Initially, [A] = [B] = 2 M & [C] = [D] = 0 M

Let at equilibrium, [A] = [B] = (2-x)M & [C] = [D] = x M

Thus,3.8 = x²/(2-x)²

or, 1.949 = x/(2-x)

or, x = 1.322 M

Thus, at equilibrium, [A] = [B] = 2 - x = 0.678 M

2)

Initially, [A] = 1 M , [B] = 2 M & [C] = [D] = 0 M

Let at equilibrium, [A] = (1-x) , [B] = (2-x)M & [C] = [D] = x M

Thus,3.8 = x²/{(2-x)*(1-x)}

or, 3.8*(x² - 3x + 2) = x²

or, 2.8x² - 11.4x + 7.6 = 0

or, x = 0.84 M

Thus, at equilibrium , [D] = x = 0.84 M