Chemical Equilibria Revision

Chemical Equilibria Revision

Consider the following equilibria

Concentration in mols.dm-3
a) 2 SO3(g) ↔ 2 SO2(g)+ O2(g)
Kc (900oC) = 1.6 x 10-10 mols.dm-3 / [SO3] = 0.10 / [SO2] = 0.10 / [O2] = 0.01
b) N2(g) + 3 H2(g)↔ 2NH3(g)
Kc (650oC) = 0.04 mols-2.dm6 / [N2] = 0.25 / [H2] = 0.10 / [NH3] = 0.15
c) CO(g) + Cl2(g) ↔COCl2(g)
Kc = 1.5 x 104 mols-1.dm3 / [CO] = 0.001 / [Cl2] = 0.0001 / [COCl2] = 0.05

Write an expression for the equilibrium constant in each case.

Using the value for Kc and concentrations given, work out if the reaction is in equilibrium or if not what direction it will shift (R→L, L→R) to reach equilibrium.

Consider the following equilibrium

2H2(g)+S2(g)↔2H2S(g)

In a 12 dm3 flask containing 0.208 moles hydrogen, 1.12 x10-6 moles sulphur, and 0.725 moles of hydrogen sulphide at equilibrium (700oC), what is the value of Kc?

3. 2 SO3(g) ↔ 2 SO2(g)+ O2(g)

Temp (K)Kp (atm)

2980.115

4004.79

5001700

N2O4(g)↔2NO2(g)

Temp (K)Kp (atm)

2984 x 1024

5002.5 x 1010

7003.0 x 104

a) Write expressions for Kp for each reaction.

b) In which reaction does the equilibrium lie furthest to the right?

c) In the above equilibria decide which reaction is exothermic and which is endothermic in the forward direction?

d) If p(SO2) = 0.1 atm and p(O2) = 0.5 atm at equilibrium, what will be the partial pressure of unreacted SO3 present at 400K.

e) If p(N2O4) = 0.1 atm at equilibrium, what is the partial pressure of NO2 formed at 700K?

4. Convert the following [H+] values into pH

a) 1 mol.dm-3

b) 0.01 mol.dm-3

c) 5.0 x 10-3 mol.dm-3

d) 1.55 x 10-5 mol.dm-3

e) 1.0 x 10-7 mol.dm-3

5. Convert the following pH values into [H+]

a) 0.0

b) 0.5

c) 1.9

d) 3.8

e) 8.1

6. Use Ka to work out the pH of a solution of weak acid

a) 0.1 mol.dm-3 ethanoic acid (Ka = 1.7 x 10-5 mol.dm-3)

b) 0.05 mol.dm-3 ethanoic acid (Ka = 1.7 x 10-5 mol.dm-3)

c) 0.01 mol.dm-3 benzoic acid (Ka = 6.3 x 10-5 mol.dm-3)

d) 0.1 mol.dm-3 methanoic acid (Ka = 1.6 x 10-4 mol.dm-3)

e) 0.1 mol.dm-3nitrous acid (HNO2) (Ka = 4.7 x 10-4 mol.dm-3)

7. Use pH and concentration to work out Ka for an acid

a) 0.01 mol.dm-3 HCN with pH 5.15

b) 0.005 mol.dm-3 phenol with pH 6.10

c) 1.0 mol.dm-3 HF with pH 1.66

d) 0.03 mol.dm-3 chloric acid (HClO) with pH 4.5

8. Calculate the pH of the following solution of strong bases using

Kw = 1 x 10-14 mols2.dm-6

a) 0.1 mol.dm-3 NaOH

b) 0.0025 mol.dm-3KOH

c) 1.5 x 10-5 mol.dm-3 LiOH

9. Calculate the pH of buffer solutions containing

a) 1.85 x 10-3 mol.dm-3 H2CO3, 2.09 x 10-2 mol.dm-3HCO3-, Ka = 4.5 x 10-7

b) 0.10 mol.dm-3CH3COOH, 0.30 mol.dm-3CH3COO-, Ka = 1.7 x 10-7mol.dm-3

c) 2.10 x 10-2 mol.dm-3 HCOOH, 4.20 x 10-2 mol.dm-3HCOO-, Ka = 1.8 x 10-4mol.dm-3

d) 1.5 x 10-3 mol.dm-3 NH4+, 8.0 x 10-2 mol.dm-3NH3, Ka = 4.5 x 10-7 mol.dm-3

10. CO2(aq)+H2O(l)↔HCO3-(aq)+H+(aq)

This equilibrium is important for maintaining the pH of blood.

Ka = 4.5 x 10-7 mols.dm-3

a) Calculate the concentration of CO2(aq) in a sample of blood where [HCO3-] = 2.5 x 10-2 mols.dm-3 and the pH = 7.4

b) Explain what is meant by a buffer system using the above equilibrium as an example.

11.

a) Explain how an aq. Solution containing ethanoic acid and sodium ethanoate resists changes in pH when contaminated with small amounts of acid or alkali.

b) What is the pH equal to when [ethanoic acid] = [ethanoate]

c) Calculate the mass of sodium ethanoate to be dissolved in 1 dm3 of 0.10 mol.dm-3 ethanoic acid solution so that the pH = 5.5, Ka = 1.8 x 10-5 mol.dm3.

d) Name a suitable indicator, giving reasons, which you would use for titrating ethanoic acid with NaOH.

e) State with reasons whether the pH of an aq. solution of NH4Cl would be greater or less than 7.

f) State with reasons whether the pH of an aq. solution of CH3COONa would be greater or less than 7.

Mark Scheme

1.a)Kc= [SO2]2 [O2]

————————

[SO3]2

=(0.1)2 X 0.01 / (0.1)2

=0.01 mols.dm-3

This value is > than Kc for this reaction (1.6 x 10-10mols.dm-3) so reaction will need to shift to left to reach equilibrium.

b)Kc= [NH3]2

————————

[H2]3[N2]

=(0.25)2 / (0.1)3 x 0.25

=90.0 mols-2.dm6

This value is > than Kc for this reaction (0.04 mols-2 .dm6) so reaction will need to shift to left to reach equilibrium.

c)Kc= [COCl2]

————————

[CO][Cl2]

=0.05 / 0.001 x 0.0001

=5 x 105 mols-1 .dm3

This value is > than Kc for this reaction (1.5 x 104 mols-1 .dm3) so reaction will need to shift to left to reach equilibrium.

[H2]=0.208 / 12= 1.73 x 10-2 mols.dm-3

[S2]=1.12 x 10 -6 / 12 = 9.33 x 10-8 mols.dm-3

[H2S]=0.725 / 12= 6.04 x 10-2 mols.dm-3

Kc= [H2S]2

————————

[H2]2[S2]

=(6.04 x 10-2 )2 / (1.73 x 10-2)2 x 9.33 x 10-8

=1.31 x 108 mol-1.dm3

3.a)Kp= p(SO2)2 x p(O2)

————————

p(SO3)2

Kp= p(NO2)2

————

p(N2O4)

b)N2O4(g)↔2NO2(g)

This reaction lies further to the right as it has a much larger Kp value.

c)2 SO3(g) ↔ 2 SO2(g)+ O2(g)

Reaction is endothermic as Kp rises with temperature.

N2O4(g)↔2NO2(g)

Reaction is exothermic as Kp falls with temperature

d) Kp= p(SO2)2 x p(O2)=(0.1)2 x 0.5 / p(SO3)2

————————

p(SO3)2

p(SO3) = √ (0.005/4.79)

= 0.032 atm

e)Kp= p(NO2)2= p(NO2)2 / 0.1 ————

p(N2O4)

p(NO2) = √ (3.0 x 104 x 0.1)

= 54.8 atm.

4.a) 0.00

b) 2.00

c) 2.30

d) 4.81

e) 7.00

5. a) 1.0 mol.dm-3

b) 0.32 mol.dm-3

c) 1.26 x 10-2 mol.dm-3

d) 1.59 x 10-4 mol.dm-3

e) 7.94 x 10-9 mol.dm-3

6. Ka = [H+] [A-] / [HA]

[H+] = [A-][H+]2 = Ka x [HA]

[H+] = √ (Ka x [HA])

a) [H+] = √ (1.7 x 10-5 x 0.1) = 1.3 x 10-3mol.dm-1pH = 2.88

b) [H+] = √ (1.7 x 10-5 x 0.05) = 9.23 x 10-4 mol.dm-1pH = 3.03

c) [H+] = √ (6.3 x 10-5 x 0.01) = 7.94 x 10-4 mol.dm-1pH = 3.10

d) [H+] = √ (1.6 x 10-4 x 0.1) = 4.0 x 10-3mol.dm-1pH = 2.40

e) [H+] = √ (4.7 x 10-4 x 0.1) = 6.86 x 10-3 mol.dm-1pH = 2.16

7.Ka = [H+]2 / [HA]

a) [H+]= 7.09 x 10-6 mol.dm-3,Ka = 5.01 x 10 -9 mol.dm-3

b)[H+]= 7.94 x 10-7 mol.dm-3,Ka = 1.26 x 10 -10 mol.dm-3

c) [H+]= 0.022 mol.dm-3,Ka = 4.79 x 10 -9 mol.dm-4

d) [H+]= 3.16 x 10-5 mol.dm-3,Ka = 3.33 x 10 -8 mol.dm-3

8.[H+] = Kw / [OH-]

a) pH = 13

b) pH = 10.4

c) pH = 9.2

9.[H+] = Ka x ( [HA] / [A-])

a) [H+] = 3.98 x 10-8mols.dm-3pH = 7.40

b) [H+] = 6.67 x 10-8mols.dm-3pH = 7.25

c) [H+] = 9.00 x 10-5mols.dm-3pH = 4.04

d) [H+] = 8.44 x 10-9mols.dm-3pH = 8.07

10.[CO2] = [HCO3-] x [H+] / Ka

= (2.5 x 10-2) x (3.98 x 10-8) / 4.5 x 10-7

= 2.21 x 10-3 mols.dm-3

11.a)CH3COOH↔CH3COO-+ H+

If a small amount of alkali is added it reacts with the H+ to form water so the pH does not increase. The equilibrium shifts to the right to replace the H+ removed.

If a small amount of acid is added it reacts with the ethanoate to form ethanoic acid so the pH does not decrease.

b) When [CH3COO-] = [CH3COOH] , [H+] = Ka x 1 so pH = pKa

The pKa of a weak acid is the pH at which 50% of the acid is dissociated and so is a good measure of the strength of the acid.

c) Ka = [H+] [CH3COO-] / [CH3COOH] = 3.16 x 106 x [CH3COO-] / 0.10

[CH3COO-] = (1.8 x 10-5) x 0.1 / 3.16 x 10-6

= 0.57 mols.dm-3

CH3COONa = 82 g.mol-1

Mass of sodium ethanoate required = 82 x 0.57 = 46.7g

d) For a weak acid and strong base use phenolphathalein, as the end point will be alkaline. This fits with when phenolphathalein changes colour (straight part of the titration curve).

e) NH4+ ↔H++ NH3

The ammonium ion is a weak acid (salt of strong acid, weak base) and will dissociate according to the above equilibrium to give a slightly acidic solution.

f) CH3COO- +H2O↔CH3COOH+ OH-

The ethanoate ion is a weak base (salt of strong base, weak acid) and will dissociate according to the equilibrium above to give a slightly alkaline solution.