4.3   EQUILIBRIUM

 

The equilibrium law and equilibrium constant for a homogeneous system in terms of concentration (Kc) and of partial pressures (Kp). The value of Kc in relation to the extent of reaction. Calculations involving Kc and Kp at constant temperature. Partial pressure derived from mole fractions and total pressure.

(Calculations will not be set which:

(a)  require the solving of quadratic equations using a formula; and

(b)  involve the interconversion of Kc and Kp.)

 

Kinetic – molecular description of equilibrium

 

Some chemical reactions take place in one direction only.

eg.  (s)  +  O2 (g)                         2MgO (s)

Others take place in both directions ie. they are reversible. The conversion from reactants to products is incomplete no matter how long the reaction is allowed to continue. Such reactions will eventually reach a state of equilibrium in which both reactants and products are present.

                        A  +  B   ⇔  C  +  D

If A and B are mixed in a closed vessel molecules combine to form C and D. Molecules of C and D can also combine to reform A and B.

Initially the rate of combination of A and B molecules is greater than the rate of C and D molecules. As the reaction proceeds the concentration of the reactants decreases and eventually become constant. After a time the amount of C and D builds up to a level where the rate of combination of A and B molecules is just equal to the rate of combination of C and D molecules. A state in which the concentrations of reactants and products no longer change is established. The system has reached a state of dynamic equilibrium. The molecules are still reacting but the macroscopic properties are constant.

eg. 

 

 


Concentration

mol dm-3

 

 

 

 

 

 

 


   t0                                            te                    time

 

                                   

The Equilibrium Law

In a series of experiments, all carried out at the same temperature but with different initial concentrations of A and B, the following results were obtained.

                                    A   +   B      ⇔   C   +   D

Experiment

number

Equilibrium concentration in mol dm-3

[A]

[B]

[C]

[D]

1

3.00

2.00

1.00

1.00

2

9.60

10.00

4.00

4.00

3

0.50

3.00

0.50

0.50

4

21.90

1.22

2.11

2.11

Exercise 1

Calculate the following ratio for each of the experiments;

                        [C][D]

                        [A][B]

What conclusion do you come to? The ratio is constant (1/6 or 0.17)

 

At equilibrium:

            Rate of forward reaction  =  Rate of backward reaction

Rate of forward reaction α [A][B]  =  k1 [A][B]

Rate of backward reaction α [C][D]  =  k2 [C][D]

 

 

Therefore          k1 [A][B] =  k2 [C][D]

                        Kc  =   k1  =  [C][D]

                                  k2     [A][B]

Where   Kc is the equilibrium constant (based on concentrations).

            k1 and k2 are the rate constants for the forward and backward reactions. These are constant at constant temperature. Altering the temperature is the only thing that will change their values.

            [  ] represents concentration in mol dm-3.

 

In general, for any system in equilibrium there is a simple relationship between the concentration of the substances present at equilibrium.

            mA  +  nB           ⇔        qC  +  rD

                                    Kc  =  [C]c [D]d

                                             [A]a [B]b

In equilibrium calculations we use the figures from the given equation as our indices.

eg.        H2 (g)  +  I2(g)    ⇔        2HI (g)

Kc  =   [HI]2       =  49

                               [H2] [I2]

For the reaction

            2HI (g)  ⇔  H2 (g)  +  I2(g)

Kc  =  [H2] [I2]   =    1

                                                [HI]2          49

 

Units of Kc

The units of Kc depend on the number of particles on both sides of the equation.

eg.        H2 (g)  +  I2(g)    ⇔        2HI (g)

Kc  =  [HI]2         =        mol2 dm-6              = no units

       [H2] [I2]        mol dm-3 x mol dm-3

 

            N2 (g)  +  3H2 (g) ⇔        2NH3 (g)

                                    Kc  =  [NH3]2                  =        mol2 dm-6              = mol-2 dm6

                                              [N2] [H2]3                            mol dm-3 x mol3 dm-9

 

Exercise 2

Write equilibrium expressions Kc and work out the units for the following reactions;

(a)     2NH3 (g)            ⇔        N2 (g)  +  3H2 (g)

(b)     2NO2 (g)            ⇔        O2 (g)  +  2NO (g)

(c)     NO (g)  +  ½ O2 (g)          ⇔        NO2 (g)

(d)     2SO2 (g)  +  O2 (g)          ⇔        2SO3 (g)

 

Example A

Calculate the equilibrium constant for the following reaction

            CH3COOH   +    C2H5OH                          CH3COOC2H5   +   H2O

If two thirds of the acid is used up when one mole of acid is added to one mole of alcohol at 25oC and allowed to reach equilibrium.

 

CH3COOH

C2H5OH

CH3COOC2H5

H2O

At start

1

1

0

0

At equilibrium

1/3

1/3

2/3

2/3

[  ]eq

0.33/V

0.33/V

0.67/V

0.67/V

 

Kc  = [CH3COOC2H5]  [ H2O]    =    0.67/V  x  0.67/V          =  4

         [CH3COOH]   [C2H5OH]      0.33/V  x  0.33/V

 

 

 

 

Equilibrium constant in terms of partial pressure, Kp

 

Homogeneous reactions

 

 

Heterogeneous reactions

eg. CaCO3 (s)      ⇔        CaO (s)   +   CO2 (g)

           

Kp  =  (pCaO (s)) (pCO2 (g))

             (pCaCO3 (s))

 

The pressure of solids is taken to be constant.

∴  Kp  =  (pCO2 (g))

 

For heterogeneous reactions Kp involves only the partial pressures of any gases concerned in the equation.

 

NH4Cl (s)           ⇔        NH3 (g)   +   HCl (g)

  Kp   =   (pNH3 (g))(pHCl (g))

 

Exercise 2

For the reaction

3Fe (s)   +  4H2O (g)        ⇔        Fe3O4 (s)   +   4H2 (g)

 

Kp   =  ?

 

 

Knowing the concentration of reactants and products at equilibrium Kc can be calculated.

The larger the value of Kc the higher the proportion of products and the closer the reaction is to completion.

Knowing the value of Kc the concentration of reactants and products at equilibrium can be calculated.

Kc varies with temperature only.

The qualitative effects of changes in temperature and pressure on the position of equilibrium and on the value of the equilibrium constant.

 

Le Chatilier’s Principle

“When a constraint is applied to a system in equilibrium, the system will move in such a way as to remove or cancel out the effect of the constraint.”

(Constraints are changes in concentration, temperature and pressure.)

 

Concentration

                                    A  +  B  ⇔        C  +  D

            Kc  =      [C][D]

                        [A][B]

Kc is a constant and is not affected by changes of concentration. If the concentration of any reactant or product is changed then the equilibrium will move to alter the concentrations of the rest so that the value of Kc remains the same.

Increasing the concentration of any of the products C and D will move the equilibrium to the left to restore the value of Kc (by decreasing [C] and [D] and increasing [A] and [B].

Decreasing the concentration of any of the reactants A and B will also move the equilibrium to the left to restore the value of Kc by increasing [A] and [B] and decreasing {C] and [D].

 

Pressure

This applies only to gaseous reactions. Changing pressure does not affect the value of Kp. An increase in pressure for gasses is equivalent to an increase in concentration. Hence the reaction will be faster and equilibrium will be established more quickly.

eg.        N2 (g)  +  3H2 (g)   ⇔      2NH3 (g)

            2SO2 (g)  +  O2 (g)⇔      2SO3 (g)

In both of these cases an increase in pressure will move the equilibrium to the right (smaller number of molecules) and increase the proportion of products but the value of Kp is not altered.

            2O3 (g)  ⇔        3O2 (g)

            PCl5 (s)  ⇔        PCl3 (s)  +  Cl2 (g)

In these cases increase in pressure moves the equilibrium to the left.

            N2 (g)  +  O2 (g)  ⇔        2NO (g)

Pressure has no effect on the equilibrium position in this case, but does allow the equilibrium to be established more quickly.

 

Temperature

An increase in temperature increases the rate of reaction but does not necessarily increase the yield of product.

eg.        N2 (g)  +  3H2 (g)   ⇔      2NH3 (g)            ΔH -ve

            2SO2 (g)  +  O2 (g) ⇔     2SO3 (g)            ΔH –ve

In both of these cases increase of temperature will move the equilibrium to the left (which absorbs heat as the reaction is endothermic, ΔH +ve, towards the left) and therefore decreases Kp.

 

Catalysts

Catalysts have no effect on the equilibrium concentrations of reactants and products and therefore no effect on Kc or Kp. Positive catalysts enable equilibrium to be established more quickly.

 

The dissociation of water and Ionic Product Kw. The Brønsted-Lowry theory of acid-base reactions occurring in aqueous solution and the use of the concept of equilibrium to describe proton transfer in acid-base equilibria; Ka and pKa.

 

Ionic Product of Water,Kw

Pure water is a poor conductor of electricity because of self-ionisation.

2H2O (l)            ⇔  H3O+ (aq)    +     OH- (aq)

or more simply

H2O (l)  ⇔  H+ (aq)        +     OH- (aq)

The equilibrium constant Kc is given by ;

                        Kc  =  [H+][OH-]

                                    [H2O]

Only a very few water molecules are ionised so [H2O] can be regarded as constant.

Therefore

                        Kw  =  Kc x  [H2O]  =  [H+] [OH-]

 

Kw is the Ionic Product of Water  and varies with temperature.

At 25oC

                        Kw  =  [H+] [OH-]  =  1 x 10-14 mol2 dm-6

 

In pure water [H+] = [OH-]  =  1 x 10-7 mol dm-3

In acid or alkaline solutions the concentrations of H+ and OH- ions are not equal.

 

Example 1

A solution contains [OH-] of 10-1 mol dm-3.

            Kw  =  [H+] [OH-]  =  1 x 10-14

                     [H+] [10-1]  =  1 x 10-14

               [H+]  =  1 x 10-14  =  1 x 10-13 mol dm-3

                             [10-1]

 

The Brønsted-Lowry theory of acid-base reactions

This theory uses the idea of proton transfer to explain the behaviour of acids and bases in solution.

 

An acid is a substance that donates a proton to a base in solution.

HCl                   H+  +  Cl-

CH3COOH                      CH3COO-  +  H+

 

A base is a substance that accepts a proton in solution

OH-  +  H+                      H2O

NH3  +  H+                     NH4+

 

Conjugate acid-base pairs

The important step in any acid-base reaction is the proton transfer. An acid gives up a proton to a base which accepts it. In the process the acid itself becomes a base, the conjugate base of the acid, and the base becomes its conjugate acid.

eg. When ethanoic acid is added to water some of the acid molecules ionise by losing a proton. Each molecule that ionises forms its conjugate base, the ethanoate ion.

 

CH3COOH          ⇔            CH3COO-         +       H+

ACID                                      CONJUGATE BASE  +  PROTON

 

 

The proton is accepted by water, acting as a base, forming its conjugate acid, the hydronium ion.

H2O          +          H+                  ⇔               H3O+

BASE         +           PROTON                                                CONJUGATE ACID

 

Combination of these two equations gives

CH3COOH        +        H2 O        ⇔         CH3COO-     +                  H3O+

acid 1                     base 2                        base 1                 acid 2

 

There are two conjugate acid-base pairs in the reaction. An acid and its conjugate base differ only by a single proton.

 

 

Exercise 1

Identify the conjugate acid-base pairs in the following reactions.

(a)     NH3 (aq)   +   H2O (aq)                 ⇔        NH4+ (aq)   +   OH- (aq)

(b)     HNO3 (aq)   +   OH- (aq)              ⇔        NO3- (aq)   +   H2O (l)    

(c)     H2O (l)   +   H2O (l)                     ⇔        H3O+ (aq)   +   OH- (aq)

(d)     HSO3- (aq)   +   H2O (l)                ⇔        SO32- (aq)   +   H3O+ (aq)  

 

 

Exercise 2

The following substances act as bases in aqueous solution. Write an equation for each one to show how its conjugate acid is formed.

(a)     OH- (aq)

(b)     HSO3- (aq)

(c)     H2O (l)

(d)     CO32- (aq)

(e)     HCO3-(aq)

 

 

Exercise 3

The following substances behave as acids in some circumstances. For each one write an equation to show the formation of its conjugate base.

(a)     HCOOH (aq)

(b)     H2O (l)

(c)     HCO3- (aq)

(d)     H2S (aq)

(e)     HSO3- (aq)

 

 

The Brønsted-Lowry definitions do not categorise substances as acids or bases. It simply helps us to recognise acid-base behaviour and gives us a way of deciding, in a given situation, whether a substance is acting as an acid or a base.
The Acid Dissociation Constant, Ka

This is a measure of the strength of an acid. Weak acids and bases do not ionise completely in solution.

Consider the weak acid HA.

            HA        +           H2O      ⇔        H3O+     +           A-

 

If the solution is dilute the equilibrium constant Kc is given by;

                        Kc  =      [H3O+] [A-]

                                    [HA] [H2O]

the value of [H2O] is large and effectively constant.

Therefore

[H3O+] [A-]  =  Kc [H2O]  =  Ka

                                         [HA]

Ka is called the Acid Dissociation Constant and is a constant at constant temperature.

All Ka values are very small (eg. Methanoic acid Ka = 1.6 x 10-4) and it is usually more convenient to use pKa values where

pKa  =  -log10 Ka

 

Note: the smaller the value of pKa, the larger the value of Ka and the stronger the acid.

 

Examples

1.        Calculate the pKa of ethanoic acid if Ka = 1.8 x 10-5 mol dm-3.

pKa  = - log10Ka  =  - log10 (1.8 x 10-5 )  =  4.75

           

2.       Calculate the pH of a 0.1M ethanoic acid solution if Ka = 1.8 x 10-5 mol dm-3

CH3COOH          ⇔        CH3COO-     +     H+

Ka  =    [CH3COO-] [H+]  =  1.8 x 10-5

            [CH3COOH]

 

If the acid is only slightly dissociated  [CH3COOH] = 0.1 mol dm-3 and [CH3COO-] =  [H+]

Ka  =  [H+]2  =  1.8 x 10-5

         [0.1]

[H+]  =  √(1.8 x 10-6)  =  1.34 x 10-3

 

pH  =  -log10[H+]  =  -log10 [1.34 x 10-3]  =  2.87

 

3.       Calculate the Ka of methanoic acid if a 0.01M solution has a pH of 2.9.

pH  =  -log10[H+]  = 2.9

[H+]  =  10-2.9  =  1.26 x 10-3

HCOOH   ⇔      HCOO-     +     H+

[HCOO-]  =  [H+]  =  1.26 x 10-3

[HCOOH]  =  0.01 - 1.26 x 10-3  =  8.74 x 10-3

 

Ka  =    [HCOO-] [H+]  =  (1.26 x 10-3)2  =  1.82 x 10-4

            [HCOOH]          8.74 x 10-3

 

 

 

Definition of pH. Calculations involving pH and concentration for strong acids and bases. Ka and concentration of weak acids.

 

pH

The pH scale describes the degree of acidity or alkalinity of a solution.

In a neutral solution [H+]  =  [OH-]  = 10-7 mol dm-3.

In acid solution [H+] >[OH-]  and in alkaline solution [H+] <[OH-] .

 

Sørensen (a Danish biochemist who worked for Carlsberg on the problems connected with brewing beer, in which the control of acidity is important)  defined pH as ;

                                    pH  =  -log10 [H+]

 

For a neutral solution [H+] = 10-7 mol dm-3  pH  =  -log10 [10-7]   =  7

Solutions with a pH < 7 are acidic.

eg. 0.1M HCl solution.      [H+] = 10-1 mol dm-3

                                    pH  =  -log10 [H+]  =  -log10 [10-1] 

                                    pH  =  1

 

Example 1

What is the pH of a solution with [H+]  =  6.28 x 10-6 mol dm-3 ?

 

pH  =  -log10 [H+]  =  -log10 [6.28 x 10-6]   =  5.20

 

 

The pH of alkaline solutions can be found using the ionic product of water Kw.

 

Example 2

What is the pH of a 0.01M NaOH solution?

 

[OH-]  =  0.01  =  10-2 mol dm-3

Kw  =  [H+] [OH-]  =  1 x 10-14  =  [H+] [10-2] 

[H+]  =  1 x 10-14 / 10-2  =   1 x 10-12

pH = -log10 [H+]  =  -log10 [1 x 10-12]  =  12

 

 

Experimental determination of titration curves. Use of titration curves in choice of an indicator.

 

Acid – Base Titrations

Titration is used to determine the concentration of either an acid or a base. The aim of a titration is to determine the volume of one solution needed to react exactly with a known volume of another solution. The end-point is marked by the change in colour of an indicator.