a. At some temperature, K = 244 for the gas phase reaction What is the concentra

Question

a.

At some temperature, K = 244 for the gas phase reaction

What is the concentration of HF in an equilibrium mixture established by adding 3.51 mol each of H2 and F2 to a 1.00 L container at this temperature?

[HF] =  M

What would be the equilibrium concentration of HF if 5.60 mol HF were removed from the above equilibrium mixture?

[HF] =  M

b.

What is the equilibrium concentration of CN1- ion in a solution prepared by mixing 45.0 mL of 0.250 M HCN and 45.0 ml of 0.250 M NH3?

[CN1-] =  M

Explanation / Answer

First, let us define the equilibrium constant for any species:

The equilibrium constant will relate product and reactants distribution. It is similar to a ratio

The equilibrium is given by

rReactants -> pProducts

Keq = [products]^p / [reactants]^r

For a specific case:

aA + bB = cC + dD

Keq = [C]^c * [D]^d / ([A]^a * [B]^b)

a)

K = [HF]^2 /([H2][F2])

initially

[H2] = 3.51

[F2] = 3.51

[HF] = 0

in equilbirium

[H2] = 3.51 - x

[F2] = 3.51 - x

[HF] = 0 + 2x

244 = (2x) ^2 / (3.51 - x)(3.51 - x)

sqrt(244) = 2x/ (3.51 - x)

15.62*3.51 - 15.62x = 2x

17.62x = 15.62*3.51

x = (15.62*3.51)/(17.62)

x = 3.111

[H2] = 3.51 - 3.111= 0.399

[F2] = 3.51 - 3.111 = 0.399

[HF] = 0 + 2*3.111= 6.222

then, if we add

[HF] = 6.222+5.6 = 11.822

there will be reverse reaction, since now htere is excess HF

2HF <-> H2 + F2

K = 1/244

initially

[H2] = 3.51 - 3.111= 0.399

[F2] = 3.51 - 3.111 = 0.399

[HF] = 11.822

in equilbirium

[H2] = 0.399 + x

[F2] = 0.399 + x

[HF] = 11.822 - 2x

substitute in K

1/244 = (0.399 + x)(0.399 + x) / (11.822 - 2x)^2

sqrt(1/244) = (0.399 +x) /(11.822 - 2x)

0.0640*11.822 - 2*0.0640x = 0.399+x

(1+2*0.0640)x = 0.0640*11.822-0.399

x = 0.357608 / (1+2*0.0640) =

x = 0.317028

[HF] = 11.822 - 2x = 11.822-2*0.317028

[HF] = 11.1879 M

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