Question

The reversible chemical reaction

A+B⇌C+D

has the following equilibrium constant:

*K*c=[C][D]/[A][B]=2.5

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

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

Answer #1

The "REVERSIBLE" chemical reaction-

A + B <--> C + D

Kc = [C][D] / [A][B] = 2.5

1.)

Concentration at the start and after equilibrium-

[A] = 2.00 M --> (2.00 - X) M

[B] = 2.00 M --> (2.00 - X) M

[C] = 0.00 M --> X M

[D] = 0.00 M --> X M

2.5 = X^2 / (2.00 - X)^2

take the square root of both sides-

1.581= X / (2.00 - X)

X = 3.1622 - 1.581X

3.1622 = 2.581 X

X = 1.225

at equilibrium-

**[A] = 0.775 M**

[B] = 0.775 M

[C] = 1.225 M

[D] = 1.225 M

2.)

Concentration at the start and after equilibrium-

[A] = 1.00 M --> (1.00 - X) M

[B] = 2.00 M --> (2.00 - X) M

[C] = 0.00 M --> X M

[D] = 0.00 M --> X M

2.5 = X^2 / (1.00 - X)*(2.00 - X)

2.5(2.00 -3.00X + X^2) = X^2

2.5 X^2 - 7.5 X + 5.00 = X^2

1.5 X^2 - 7.5 X = -5

X = 1.288

at equilibrium-

[A] = -0.288 M

[B] = 0.712 M

[C] = 1.288 M

**[D] = 1.288 M**

**Infact it is not possible.**

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