Question

The reversible chemical reaction

A+B⇌C+D

has the following equilibrium constant:

*K*c=[C][D][A][B]=6.7

A: initially, only A and B are present, each at 2.00 *M*.
What is the final concentration of A once equilibrium is
reached?

B:What is the final concentration of D at equilibrium if the
initial concentrations are [A] = 1.00 *M* and [B] = 2.00
*M* ?

Please include all steps!

Answer #1

A + B⇌C + D

I 2 2 0 0

C -x -x +x +x

E 2-x 2-x +x +x

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

6.7 = x*x/(2-x)(2-x)

6.7 = (x/2-x)^2

6.7 = x/2-x

2.53 = x/2-x

2.53*(2-x) = x

x = 1.43

[A] = 2-x = 2-1.43 = 0.57M

B.

A + B ⇌ C + D

I 1 2 0 0

C -x -x +x +x

E 1-x 2-x +x +x

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

6.7 = x*x/(1-x)(2-x)

6.7 = x*x/(1-x)(2-x)

6.7*(1-x)(2-x) = x^2

x = 0.89

[A] = 1-x = 1-0.89 = 0.11M

[B] = 2-x = 2-0.89 = 1.11M

[D] = x = 0.89M

[C] = x = 0.89M

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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
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Express your answer to two significant figures and...

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