The reversible chemical reaction for the first 2 questions: A+BC+D has the follo

Question

The reversible chemical reaction for the first 2 questions:

A+BC+D

has the following equilibrium constant:

Kc=[C][D]/[A][B]=1.7

Part A Initially, only A and Bare 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 Value Units Submit Hints My Answers Give Up Review Part Part B What is the final concentration of D at equilibrium if the initial concentrations are Al 1.00 Mand B 2.00 M? Express your answer to two significant figures and include the appropriate units Value Units Submit Hints My Answers Give Up Review Part

Explanation / Answer

Kc=[C][D]/[A][B]=1.7

A)

initially:

[A] = 2

[B] = 2

[C] = 0

[D] = 0

in equilibrium

[A] = 2 - x

[B] = 2 - x

[C] = 0 + x

[D] = 0 + x

substitute in Kc

1.7 = x*x/(2-x)^2

sqrt(1.7) = x/(2-x)

1.3*2 - 1.3x = x

2.3x = 2.6

x = 1.1304

[A] = 2 - 1.1304 = 0.8696

[B] = 2 - 1.1304 = 0.8696

[C] = 0 + x = 1.1304

[D] = 0 + x = 1.1304

[A] = 0.8696 M

B)

similarly:

initially:

[A] = 1

[B] = 2

[C] = 0

[D] = 0

in equilibrium

[A] = 1 - x

[B] = 2 - x

[C] = 0 + x

[D] = 0 + x

substitute in Kc

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

1.7(2-3x+x^2) = x^2

1.7*2 -3*1.7x + 1.7x^2 - x^2 = 0

0.7x^2 - 5.1x + 3.4 = 0

x = 0.742

[A] = 1 - x= 1-0.742 = 0.258

[B] = 2 - x = 2-0.742 = 1.258

[C] = 0 + x = 0.742

[D] = 0 + x = 0.742

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