The reversible chemical reaction A+BC+D has the following equilibrium constant:

ID: 890353 • Letter: T

Question

The reversible chemical reaction

A+BC+D

has the following equilibrium constant:

Kc=[C][D][A][B]=3.9

1.)

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.

2.)

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.

1) A + B <===> C + D

Initial 2 2 0 0

change -x -x +x +x

equilbrium (2-x) (2-x) x x

Feed values,

Kc = 3.9 = x^2/(2-x)(2-x)

3.9 = x^2/4 - 4x + x^2

15.6 - 15.6x + 3.9x^2 = x^2

2.9x^2 - 15.6x + 15.6 = 0

x = 1.33 M

So equilibrium concentration of A = 2-1.33 = 0.67 M

2) If initial concentration of [A] = 1.0 M and [B] = 2.0 M

Kc = 3.9 = x^2/(1-x)(2-x)

3.9 = x^2/2 -3x + x^2

7.8 - 11.7x + 3.9x^2 = x^2

2.9x^2 - 11.7x + 7.8 = 0

x = 0.843 M

So the equilibrium concentration of [D] = 0.84 M

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