Question

Part A

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

Express your answer to two significant figures and include the appropriate units.

The reversible chemical reaction

A(aq)+B(aq)⇌C(aq)+D(aq)

Part B

What is the final concentration of Dat equilibrium if the initial concentrations are [A] = 1.00 mol L−1 and [B] = 2.00 mol L−1 ?

Express your answer to two significant figures and include the appropriate units.

has the following equilibrium constant:

K=[C][D][A][B]=2.5

Answer 1

A)

A(aq)+B(aq)⇌C(aq)+D(aq)

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])

substitite

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

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

solve for x

2*sqrt(Kc) - x*sqrt(Kc) = x

(1+sqrt(Kc)) *x = 2*sqrt(Kc)

x = 2*sqrt(Kc) / (1+sqrt(Kc))

You will need the value of Kc:

then, get [A]

[A] = 2-x

[A] = 2- 2*sqrt(Kc) / (1+sqrt(Kc))

if Kc = 2.5

[A] = 2- 2*sqrt(2.5) / (1+sqrt(2.5)) = 0.77485 M

B)

A(aq)+B(aq)⇌C(aq)+D(aq)

initially:

[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])

substitite

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

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

0.4x^2 = (1-3x+x^2)

0.6x^2 -3x + 1 = 0

x = 0.359127

[A] = 1-x = 1-0.359127 = 0.6408 M