The reversible chemical reaction A + B C + D has the following equilibrium const

Question

The reversible chemical reaction A + B C + D has the following equilibrium constant: K_c = [C] [D]/[A] [B] = 1.5 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

part A)

A    +    B   -------------> C    +    D

2           2                       0            0

2-x       2-x                     x             x

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

1.5 = x^2 / (2-x)^2

x = 1.101

concentration of A at equilibrium = 2 - 1.101 = 0.90

[A] = 0.90 M

part B)

A    +    B   -------------> C    +    D

1           2                       0            0

1-x       2-x                     x             x

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

1.5 = x^2 / (1-x)(2-x)

x = 0.725

equilibrium concentration of D = 0.725

[D] = 0.72 M

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