Question

In: Chemistry

1) The reversible chemical reaction A(aq)+B(aq)⇌C(aq)+D(aq) has the following equilibrium constant: K=[C][D][A][B]=5.2 a) Initially, only A...

1) The reversible chemical reaction

A(aq)+B(aq)⇌C(aq)+D(aq)

has the following equilibrium constant:

K=[C][D][A][B]=5.2

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?

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

Expert Solution

a)

A + B ßà C + D

2 2 0 0

2-x 2-x x x

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

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

x/(2-x) = sqrt (5.2)

x/(2-x) = 2.28

x = 4.56 – 2.28*x

3.28*x = 4.56

x = 1.39 M

[A] = 2-x = 2 – 1.39 = 0.61 M

Answer: 0.61 M

b)

A + B ßà C + D

1 2 0 0

1-x 2-x x x

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

5.2 = x^2 / (2-3x+x^2)

5.2*x^2 -15.6*x +10.4 = x^2

4.2*x^2 – 15.6*x + 10.4 = 0

solving above quadratic equation we get,

x = 2.84 and x=0.87

x can’t be 2.84 as it will make concentration of A and B negative

so,

x = 0.87 M

[D] = 0.87 M

Answer: 0.87 M

queen_honey_blossom answered 1 year ago

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