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The reversible chemical reaction A+B⇌C+D has the following equilibrium constant: Kc=[C][D][A][B]=5.9 Initially, only A and B...

The reversible chemical reaction

A+B⇌C+D

has the following equilibrium constant:

Kc=[C][D][A][B]=5.9

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

Part B

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

Solutions

Expert Solution

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

2      2                        0     0

2-x   2-x                      x      x

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

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

x = 1.417

final concentration of A = 2 - x = 2 - 1.417

final concentration of A = 0.583 M

Part B )

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

1      2                        0     0

1-x   2-x                      x      x

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

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

x = 0.882

final concentration of D = 0.882 M

queen_honey_blossom answered 10 months ago
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