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] = 5.4 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. 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.

Explanation / Answer

Initially

[A] = 2

[B] = 2

[C] = 0

[D] = 0

in equilibrium

[A] = 2-x

[B] = 2-x

[C] = 0+x

[D] = 0+x

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

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

solve for x

sqrt(5.4) = x/(2-x)

2.3237900 = x/(2-x)

(2-x) = 1/(2.3237900 )x

2-x = 0.43x

1.43x = 2

x = 2/1.43 = 1.398

[A] = 2-x = 2-1.398= 0.602 M

2)

IF

[A] = 1

[B] = 2

[C] = 0

[D] = 0

in equilibrium

[A] = 1-x

[B] = 2-x

[C] = 0+x

[D] = 0+x

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

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

solve for x

5.4(2-3x+x^2) = x^2

2-3x+x^2 = (1/5.4)x^2

2-3x+x^2 = 0.18518x^2

(1-0.18518)x^2 -3x +2 =0

x = 0.9824

then

[D] = 0+x= 0.9824

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