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:

Kc=[C][D][A][B]=3.6

A) 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.

B) What is the final concentration of D at equilibrium if the initial concentrations are [A] = 1.00M and[B] = 2.00M ?Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

Initially the conc of [A] = [B] = 2

at equilibrium ; [A] = [B] = 2-x   [C] = [D] = x

Kc = 3.6 = x X x / (2-x)^2

1.897 = x / (2-x)

3.794 - 1.897x = x

3.794 = 2.897 x

x = 1.309

Final concentration of A = 0.691 M

b)

initial concentration = A = 1

B= 2

At equilbirium :

[A] = 1-x

[B] = 2-x

[C] = x = [D]

Kc = 3.6 = x2 / (1-x) (2-x)

3.6 = x2 / 2-3x +x2

7.2 - 10.8x + 3.6x2 = x2

7.2 -10.8x + 2.6x2 = 0

x = 0.834

so Conc of D = 0.834

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.