Chemical Equilibrium

Chemical Equilibrium

I. Dynamic Nature of Equilibrium A. Introduction

Many physical and chemical changes, do not proceed to completion after even a long period of time.

a. A dynamic equilibrium exists when two reversible or opposing changes are continually taking place.

Reactants forward reaction Products

backward reaction

An example of this is walking up an escalator at the same speed as the escalator is moving down. The person appears to be static, but in fact the person is in dynamic equilibrium with escalator.

b. Some examples of irreversible reactions

Some chemical reactions proceed in one direction only and go to completion.

(i) Mg(s) + 2HCl(aq)  MgCl

2

(aq) + H

2

(g) (ii) HCl(aq) + NaOH(aq)  NaCl(aq) + H

2

O(l)

(iii) 4Fe(s) + 3O

2

(g) + 2nH

2

O(l)  2Fe

2

O

3.

nH

2

O(l) Rusting c. Some examples of reversible reactions

Many reactions do not go to completion. These reactions can be made to proceed in either direction. They are described as reversible reactions and can be represented in general as follows:

A + B C + D Forward reaction: A + B  C + D Backward reaction: C + D  A + B Both reactions are proceeding.

(i) CaCO3(s) CaO(s) + CO2(g)

When calcium carbonate is heated in a sealed vessel at a fixed temperature and equilibrium state is reached in which calcium carbonate, calcium oxide, and carbon dioxide are all present.

(ii) CH3COOH(l) + CH3CH2OH(l) CH3COOCH2CH3(l) + H2O(l)

The formation of esters from carboxylic acids and alcohols, for example the formation of ethyl ethanoate from ethanol and ethanoic acid is an equilibrium.

[This reaction can be catalyzed by heating with concentrated sulphuric acid.]

The reaction does not go to completion even if the reaction mixture is heated under reflux for a long time.

This is because some of the ethyl ethanoate reacts with water to give ethanoic acid and ethanol at the same time.

reflux condenser

water in water out

ethanoic acid + ethanol + conc. sulphuric acid + anti-bumping granules

(iii) Industrial preparation of ammonia (Haber Process) N

2

(g) + 3H

2

(g) 2NH

3

(g)

Catalyst: finely divived iron Temperature: 450

^o

C

Pressure: 200 atm

(iv) Ionization of ethanoic acid

CH

3

COOH (aq) CH

3

COO

^-

(aq) + H

⁺

(aq) Weak acids ionize only slightly in water.

B. A Theoretical Approach To Dynamic Equilibrium Let us consider the following reversible reaction:

A + B C + D

Usually, the higher the concentration of the reactants, the faster is the forward reaction. Suppose we mix certain quantities of A and B in a closed system. The rates of the forward and backward reactions will change as follows:

Forward reaction Backward reaction

At start rapid rate

(maximum concentration) zero rate

(no material) As time passes becomes slower as A and B are

consumed

becomes faster as more C and D are formed

At a certain stage rate of forward reaction = rate of backward reaction

At the stage when forward and backward rates become equal, no further change in properties of the mixture can be detected, i.e. the system has reached a state of dynamic equilibrium.

C. Characteristics Of The Equilibrium State

a. A stable state of equilibrium can only be attained in a closed system - one that cannot exchange matter with its surroundings.

b. The equilibrium state can be approached from either direction, that is reactants and products can be interchanged.

c. At equilibrium, the rate of forward reaction is equal to the rate of backward reaction.

d. At equilibrium, both reactants and products are present and their concentrations remain unchanged. Hence, there are no observable changes of the reaction mixture.

Classwork

Equal number of moles of H

2

(g) and I

2

(g) are introduced in a sealed flask at a certain temperature. The following reaction takes place:

H

2

(g) + I

2

(g) 2HI(g)

The reaction mixture is allowed to reach equilibrium.

a. In the diagram below, sketch

(i) the changes in concentration of H

2

, I

2

and HI with time, and

(ii) the changes in the rates of forward reaction and backward reaction with time,

until the reaction mixture reaches equilibrium.

Time Concentration

Time Reaction rate

b. Describe the change in colour intensity of the reaction mixture starting from the very beginning of the reaction to the time when an equilibrium state is reached.

(Note: H

2

(g) and HI(g) are colourless, I

2

(g) is violet in colour.) D. Position Of Equilibrium

a. Intermediate Stage and Position of Equilibrium Let us consider the reversible reaction:

Br2(aq) + H2O(l) H+(aq) + Br-(aq) + HOBr(aq)

brown colourless

(i) Bromine water is yellow/brown, consisting of a mixture of Br2(aq), H2O(l), H+(aq), Br-(aq) and HOBr(aq).

(ii) When excess alkali is added to it, the solution turns very pale yellow (almost colourless), indicating that there is a net reaction from left to right.

(iii) When excess acid is added to the above solution, the yellow/brown colour is restored, showing that there is a net reaction from right to left.

(iv) Any intermediate colour which may be obtained corresponds to an intermediate stage in the reversible reaction.

(v) An intermediate stage in a reversible process at which no observable change is occurring is called A POSITION OF EQUILIBRIUM (or balance point).

b. Shifting position of Equilibrium Consider the equilibrium,

Br2(aq) + H2O(l) H+(aq) + Br-(aq) + HOBr(aq)

brown colourless

(i) When alkali is added,

 the hydrogen ions in the mixture is removed.

 the equilibrium is disturbed and then more Br2 molecules and water molecules will change to H+(aq), Br-(aq) and HOBr(aq), (the colour of the bromine water becomes paler).

 the effect of the disturbance is reduced, the equilibrium position is said to be shifted to the right and the system hence attains a new equilibrium position.

(ii) When acid is added,

 the concentration of the hydrogen ions increases.

 the equilibrium is disturbed. to reduce this effect, more H+(aq), Br-(aq) and HOBr(aq) will changed into Br2 and H2O, (the colour of the bromine water deepens).

 the equilibrium position is said to be shifted to the left.

Classwork

Cr2O72-(aq) + H2O(l) 2CrO42-(aq) + 2H+(aq)

orange yellow

Predict what is the colour of the reaction mixture if (i) excess acid is added,

(ii) excess alkali is added.

II. Equilibrium Law and Equilibrium Constant

A. Equilibrium Constant K

c

a. (i) For a reaction as

a A + b B c C + d D

The letters A, B, C and D represent the chemical substances and a, b, c and d are the stoichiometric coefficients.

(ii) If the concentrations of the various substances at equilibrium are determined experimentally, then the extent to which the reaction proceeds from left to right is given by a simple expression as follows:

K C D

A B

c eqm c

eqm d eqm a

eqm b

 [ ] [ ] [ ] [ ]

 where [A]eqm is the concentration of substance A at equilibrium

(mol dm-3), and similarly for substances B, C and D.

 Kc is called the equilibrium constant for the reaction (the subscript c indicates that the units if concentration are mol dm-3) and has units of (mol dm-3)n; where n = (c + d) - (a + b).

 The above expression is known as the equilibrium law.

b. At a given temperature, Kc is independent of the initial concentrations, i.e. any combination of concentrations of A, B, C and D can be reacted together, yet the same value of Kc is obtained.

c. Two important conclusions arises from this law:

(i) If Kc is large the equilibrium mixture will contain a high proportion of products, that is, the reaction has gone nearly to completion.

If Kc is small, the reaction does not proceed very far at the temperature and the concentration of product is low.

(ii) The addition of more reactants to a system in equilibrium will result in the formation of more products, the system adjusting itself so that the concentrations again satisfy the value for Kc.

Similarly, on addition of more products the equilibrium will move in the opposite direction and the concentration of reactants will increase.

d. The magnitude of K

c

indicates the extent (how far) of a chemical reaction but not the rate (how fast) of the reaction.

e. When stating the value of Kc for a particular reaction, it is important to indicate the equation on which the constant is based.

(i) For example, in the reaction between hydrogen and bromine to form hydrogen bromide, if we write the equation in the form

H2(g) + Br2(g) 2HBr(g)

K HBr g

H g Br g

c eqm

eqm eqm

 [ ( )]

[ ( )] [ ( )]

2

2 2

At 500 K, the value of this constant is about 1012 (ii) The same equilibrium can also be represented by

1

2 H2(g) + 1

2 Br2(g) HBr(g) but then there is a different equilibrium constant

K HBr g

H g Br g

c eqm

eqm eqm

'

/ /

[ ( )]

[ ( )] [ ( )]



2 1 2

2 1 2

It will be seen that K

_c^'

 K

_c

thus K

c'

at 500 K  10

¹²

 10

⁶

f. Units for Kc must be stated clearly.

Example

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

K SO g

SO g O g

c eqm

eqm eqm

 [ ( )]

[ ( )] [ ( )]

3 2

2 2

2

If a particular equilibrium mixture contains x mol dm-3 SO3, y mol dm-3 SO2, and z mol dm-3 O2

K x

y z

c

 [ ]

[ ] [ ]

mol dm mol dm mol dm

-3

-3 -3

2 2

3 1 2

2 3 2

2

dm

mol mol dm

z y

x z

y

x

_







Classwork

1. What are the units of Kc for the following reactions?

a. N2(g) + 3H2(g) 2NH3(g) ans: mol

^-2

dm

⁶

b. 2NH3(g) N2(g) + 3H2(g) ans: mol

^-2

dm

⁶

c. 2NO(g) + O2(g) 2NO2(g) ans: mol

^-1

dm

³

d. NO(g) + 1

2 O2(g) NO2(g) ans: mol

^-0.5

dm

^1.5

2. The equilibrium constant for the reaction 2NO2(g) N2O4(g)

at 298 K is 200 mol-1 dm3

a. Write an expression for the equilibrium constant for the reaction.

b. If the concentration of N2O4(g) in the equilibrium mixture at 298 K is 210-2 mol dm-3, what is the

concentration of NO2(g)? ans: 0.01 moldm

^-3

c. Calculate the equilibrium constant at 298 K for the reaction, 1

2 N2O4(g) NO2(g)

ans:0.0707 mol

^0.5

dm

^-1.5

3. The equilibrium constants for the synthesis of hydrogen chloride, hydrogen bromide and hydrogen iodide are given

below.

H2(g) + Cl2(g) 2HCl(g) Kc = 1017 H2(g) + Br2(g) 2HBr(g) Kc = 109

H2(g) + I2(g) 2HI(g) Kc = 10

a. What do the values of Kc tell you about the extent of each reaction?

b. Which of these reactions would you regard as virtually complete conversion?

B. Equilibrium constants for Homogeneous equilibrium and heterogeneous equilibrium a. Homogeneous equilibrium

It is an equilibrium system in which all the reactants and products are in the same phase is called a homogeneous equilibrium.

Example: H2(g) + I2(g) 2HI(g)

2

2 2

[ ( )]

[ ( )] [ ( )]

eqm c

eqm eqm

K HI g

H g I g



N

2

(g) + 3H

2

(g) 2NH

3

(g)

2 3

3

2 2

[ ( )]

[ ( )] [ ( )]

eqm c

eqm eqm

K NH g

N g H g



b. Heterogeneous equilibrium

It is an equilibrium system in which two or more phases are present in a heterogeneous equilibrium.

(i) 2NaHCO

3

(s) Na

2

CO

3

(g) + CO

2

(g) + H

2

O(g) For the reaction, the equilibrium constant

2 3 2 2

2 3

[ ( )] [ ( )] [ ( )]

[ ( )]

eqm eqm eqm

c

eqm

Na CO s CO g H O g

K  NaHCO s

Since the densities of solid substances remain the same throughout the reaction. Their molar concentrations

are constants.

Consequently,

2 3 2 2

2 3 '

2 2

[ ( )] [ ( )] [ ( )]

[ ( )]

[ ( )] [ ( )]

eqm eqm eqm

c

eqm

c eqm eqm

Na CO s CO g H O g

K NaHCO s

K CO g H O g





(ii) H

2

O(l) H

⁺

(aq) + OH

^-

(aq)

2

[ ( )] [ ( )]

[ ( )]

eqm eqm

c

eqm

H aq OH aq

K H O l

 



Since the concentration of water is constant, consequently,

2 '

[ ( )] [ ( )]

[ ( )]

[ ( )] [ ( )]

eqm eqm

c

eqm

c eqm eqm

H aq OH aq

K H O l

K H aq OH aq

 

 





C. Calculations involving equilibrium constants Example 1

When 1 mole of hydrogen iodide is allowed to dissociate in a 1.0 dm3 vessel at 440oC, only 0.78 moles of HI are present at equilibrium.

a. What is the equilibrium constant at this temperature for the reaction, 2HI(g) H2(g) + I2(g)?

b. If 2 moles of hydrogen and 1 mole of iodine are mixed together in a 1.0 dm3 vessel at 440oC, how many moles of HI, H2 and I2 will be present at equilibrium?

Solution

a. 2HI(g) H2(g) + I2(g) Initially 1 mol 0 mol 0 mol At eqm 0.78 mol (1-0.78)/2 mol (1-0.78)/2 mol

= 0.11 mol = 0.11 mol

or 0.78 mol dm

^-3

0.11 mol dm

^-3

0.11 mol dm

^-3

2 2

2

-3 -3

-3 2

[ ( )] [ ( )]

[ ( )]

(0.11 mol dm )(0.11 mol dm )

0.0199 (0.78 mol dm )

eqm eqm

c

eqm

c

H g I g

K HI g

K



 

b. 2HI(g) H2(g) + I2(g)

Initially 0 mol 2 mol 1 mol

At eqm 2x mol (2 – x) mol (1 – x) mol

or 2x mol dm

^-3

(2 – x) mol dm

^-3

(1 – x) mol dm

^-3

2 2

2

2

[ ( )] [ ( )]

[ ( )]

(2 )(1 ) 0.0199

(2 )

0.935 or 2.326 (rejected)

eqm eqm

c

eqm

H g I g

K HI g

x x

x x



 





i.e. [HI(g)]

eqm

= 1.87 mol dm

^-3

[H

2

(g)]

eqm

= 1.065 mol dm

^-3

[I

2

(g)]

eqm

= 0.065 mol dm

^-3

Example 2

A student mixed 10 cm

³

of 2.010

^-3

M Fe(NO

3

)

3

(aq) with 10 cm

³

of 2.010

^-3

M KSCN(aq) Fe

³⁺

(aq) + SCN

^-

(aq) [Fe(SCN)]

²⁺

(aq)

When the system reaches the equilibrium, the concentration of [Fe(SCN)]

²⁺

(aq) is 1.410

^-4

M. Determine the

Let x mol of H2 and x mol of I2 react to give 2x mol of HI.

equilibrium constant (K

c

)of the reaction.

Solution

Fe

³⁺

(aq) + SCN

^-

(aq) [Fe(SCN)]

²⁺

(aq) Initially 1.010

^-3

mol dm

^-3

1.010

^-3

mol dm

^-3

0 mol dm

^-3

At eqm 1.010

^-3

- 1.4010

^-4

= 8.610

^-4

mol dm

^-3

1.4010

^-4

mol dm

^-3

= 8.610

^-4

mol dm

^-3

2 3

4 -3

4 -3 4 -3

-1 3

[ ( )] ( )

[ ( )] [ ( )]

1.4 10 mol dm

(8.6 10 mol dm )(8.6 10 mol dm ) 189.29mol dm

eqm c

eqm eqm

c

c

Fe SCN aq

K Fe aq SCN aq

K K



 



 



 

 

 Classwork

1. In the determination of the equilibrium constant (K

c

) of:

Fe

²⁺

(aq) + Ag

⁺

(aq) Fe

³⁺

(aq) + Ag(s)

100 cm

³

of 0.100 M AgNO

3

(aq) and 100 cm

³

of 0.100 M FeSO

4

(aq) are mixed in a dry conical flask. The mixture is then allowed to stand overnight and filtered. The concentration of Ag

⁺

(aq) is found by titration. 25.00 cm

³

of the filtrate with 0.050 M KSCN(aq) and 6.10 cm

³

of the KSCN(aq) is required for complete reaction.

a. Calculate the equilibrium concentrations of Ag

⁺

(aq), Fe

²⁺

(aq) and Fe

³⁺

(aq).

b. Calculate the equilibrium constant (K

c

) c. What is the significance of

(i) using a dry conical flask?

(ii) allowing the mixture to stand overnight?

Ans: [Ag

⁺

] = 0.0122 M; [Fe

²⁺

] = 0.0122 M; [Fe

³⁺

] = 0.0378 M; Kc = 262.5 mol

^-1

dm

³

2. At 400 K, 0.250 mole of PCl

3

(g) and 0.009 mol of PCl

5

(g) were mixed in a 1 dm

³

flask. After the system was left overnight, an equilibrium was established and 0.002 mole of chlorine gas was found in the flask. Determine the equilibrium constant (K

c

) of the reaction:

PCl

5

(g) PCl

3

(g) + Cl

2

(g) Ans: 0.072 mol dm

^-3

3. For the equilibrium

CH3COOH(l) + CH3CH2OH(l) CH3COOCH2CH3(l) + H2O(l)

When 1 mole of ethanoic acid and 2.05 moles of ethanol are heated to equilibrium, it is found that 0.15 mole of ethanoic acid is present.

a. How many moles of ethyl ethanoate is present.

b. How many moles of water are formed?

c. Calculate Kc.

Ans: a. 0.85 mol b. 0.85 mol c. 4.0

Since the volume of the solution is doubled, ie, the initial concentrations are reduced by half.

4. For the reaction

H2(g) + CO2(g) H2O(g) + CO(g)

Kc is 0.771 at 750oC. If 1 mole of H2 and 1 mole of CO2 are mixed in a 1.0 dm3 container at 750oC. What are the concentrations of all substances at equilibrium?

Ans: [H2O]=[CO]=0.468 mol dm-3, [H2]=[CO2]=0.532 mol dm-3

5. For the equilibrium

PCl5(g) PCl3(g) + Cl2(g)

Kc is 4.1610-2 mol dm-3 at 250oC. In a given system at 250oC, the equilibrium concentrations are [PCl

5

]=1 mol dm-3 and [PCl

3

] = [Cl

2

] = 0.204 mol dm-3. If the pressure of the system is reduced by half, what are the new equilibrium concentrations?

Ans: [PCl5] = 0.4635 mol dm-3; [PCl3] = [Cl2] = 0.1385 mol dm-3

6. 1 mole of ethanoic acid, CH3COOH, is heated with 1 mole of ethanol, CH3CH2OH, at 25oC. After equilibrium is reached, the excess acid requires 66.6 cm3 of 5 M NaOH solution for complete neutralization.

a. Calculate the equilibrium constant for the esterification reaction. Ans: 4.0

b. When 0.5 mole of ethanol is added to 1.0 mole of ethanoic acid at 25oC, how much ester will be formed at

equilibrium? Ans: 0.423 mol

D. Determination of equilibrium constant from experiments

The equilibrium between ethanoic acid, ethanol, ethyl ethanoate and water

CH3COOH(l) + C2H5OH(l) CH3COOC2H5(l) + H2O(l) at 373 K Data for the reaction

Initial no. of moles of ethanoic acid

Initial no. of moles of ethanol

No. of moles of ethyl ethanoate at

equilibrium

Kc

1.00 0.18 0.171 3.9

1.00 0.33 0.293 3.3

1.00 0.50 0.414 3.4

1.00 1.00 0.667 4.0

1.00 2.00 0.858 4.6

1.00 8.00 0.966 3.9

Procedures

a. The experiment was carried out by sealing known amounts of the acid and alcohol in a number of glass tubes which were then heated to a temperature of 373 K (100oC) until equilibrium was attained.

b. The tubes were then rapidly cooled to "freeze" the equilibrium.

c. The amount of acid left determined by titration with a standard solution of alkali. The assumption is made that the equilibrium position was not shifted significantly during the analysis.

d. From the amount of acid left, the amount of acid used, the amounts of ester and water produced, and the amount of alcohol remained can be calculated.

Results and Analysis

(i) Using the last set of data as an example.

Since 0.966 mole of ethyl ethanoate is formed,

number of moles of ethanoic acid left = 1.00 – 0.966 = 0.034 mol number of moles of ethanol remained = 8.00 – 0.966 = 7.034 mol number of moles of water formed = 0.966 mol

If V is the total volume of the liquid mixture in dm3 we have:

H3COOH(l) + C2H5OH(l) CH3COOC2H5(l) + H2O(l)

Initial conc./ mol dm-3

1 00 . V

8 00 . V

0 V

0 V

Eqm conc./ mol dm-3

0 034 . V

7 034 . V

0 966 . V

0 966 . V

Kc =

[ ] [ ]

[ ] [ ]

CH COOC H H O

CH COOH C H OH

eqm eqm

eqm eqm

3 2 5 2

3 2 5

= ( .

)( . ) ( .

)( . ) 0 966 0 966 0 034 7 034 V V

V V =

( . )( . ) ( . 0 966 0 966 )( . ) .

0 034 7 034  3 90

Notice that the equilibrium constant for this reaction has no units.

(ii) If we take the value of Kc to be 4.0, then for an equimolar ratio of ethanoic acid and ethanol there is approximately 67% conversion into ethyl ethanoate and water (see the fourth set of the data).

However for a given quantity of ethanoic acid (in this case one mole) a large excess of ethanol , increases the amounts of ethyl ethanoate and water produced (approaching 97% conversion of ethanoic acid in the last set of results).

This illustrates an important point, namely that the position of equilibrium can be altered by varying the

relative concentrations of the participating substances, but the equilibrium constant itself remains

constant unless the temperature is altered.

III. The Effect of Changes in Concentration and Temperature on Chemical Equilibrium A. Factors Affecting Position of Equilibrium: Le Chatelier's Principle

 Le Chatelier's Principle states that if a system at equilibrium is subjected to a change, the position of equilibrium will shift in such a way that reduces the effect of the change.

 Le Chatelier's Principle enable us to predict the effect of an altered factor on equilibrium position without having to consider the effects on the forward and backward rates.

For example, when the concentration of a reactant in an equilibrium mixture is increased, the position of equilibrium will shift in such a way so as to decrease the concentration of that reactant. This is achieved by shifting the equilibrium position to the right.

a. Effect of Concentration Change on Equilibrium Position (i) Examples

1. Consider the equilibrium below,

Cr2O72-(aq) + H2O(l) 2CrO42-(aq) + 2H+(aq)

orange yellow

a. What will happen when acid is added into this system?

b. What will happen when alkali is added into this system?

c. What will happen when water is added into this system?

2.

a. For the reaction H

2

(g) + I

2

(g) 2HI(g), the following data are determined at 490

^o

C, [H

2

(g)] = 0.22 mol dm

^-3

; [I

2

(g)]=0.22 mol dm

^-3

and [HI(g)]=1.56 mol dm

^-3

Calculate the equilibrium constant (K

c

) at 490

^o

C.

b. If an additional 0.200 mol dm

^-3

of H

2

(g) is added to the above equilibrium mixture while keeping volume and temperature constant, what will happen? Calculate the equilibrium concentrations of all species when equilibrium is reached.

(ii) Reaction Quotient Q

c

The reaction quotient for a reversible reaction a A + b B c C + d D

[ ] [ ] [ ] [ ]

c d

c a b

Q C D

 A B

When comparing Q

c

with K

c

of a reaction, we can predict the shifting direction of the equilibrium position.

Comparison of Q

c

and K

c

Shifting direction of equilibrium position

Q

c

< K

c

To the right / product side

i.e. more products are formed

Q

c

> K

c

To the left / reactant side

i.e. more reactants are formed Q

c

= K

c

No change in equilibrium position

i.e. the concentrations of both reactants and products remain constant

Classwork

1. The equilibrium constant, K

c

, for the reaction Fe

³⁺

(aq) + SCN

^-

(aq) Fe(SCN)

²⁺

(aq) Is 128.4 mol

^-1

dm

³

at a given temperature.

If a 2.0 dm

³

flask contains 0.0320 mol of Fe

³⁺

(aq), 0.0600 mol of SCN

^-

(aq) and 0.1020 mol of FeSCN

²⁺

(aq).

Calculate the reaction quotient to show that the system is not at equilibrium. To which direction will the equilibrium position proceed to reach equilibrium?

Ans: Q

c

= 106.25 mol

^-1

dm

³

; the equilibrium position should be shifted to the right.

2. At 713 K, the equilibrium constant for the reaction is 49.5.

H

2

(g) + I

2

(g) 2HI(g)

A 10 dm

³

vessel is filled with 0.40 mol HI and allowed to decomposed at this temperature until equilibrium is reached.

a. What will be the concentrations of H

2

, I

2

and HI at equilibrium?

b. If 0.10 mol of HI is added to the equilibrium mixture in the vessel at the temperature, calculate the reaction quotient and hence state the direction of shift in the equilibrium position.

c. Calculate the concentrations of H

2

, I

2

and HI at the new equilibrium state at the same temperature.

d. Based on your calculation in (c), state whether all the HI added can be removed by the system to re- establish a new equilibrium or not.

Ans: a. [H

2

(g)]

eqm

= 0.00443 mol dm

^-3

; [I

2

(g)]

eqm

= 0.00443 mol dm

^-3

; [HI(g)]

eqm

= 0.0311 mol dm

^-3

b. Q

c

= 86.07; the equilibrium position shifts to the left.

c. [H

2

(g)]

eqm

= 0.00553 mol dm

^-3

; [I

2

(g)]

eqm

= 0.00553 mol dm

^-3

; [HI(g)]

eqm

= 0.0389 mol dm

^-3

d. No.

b. Effect of Pressure Change on Equilibrium Position (For the equilibriums involving gaseous components)

(i) For the equilibrium between dinitrogen tetroxide (N2O4) and nitrogen dioxide (2NO2)

N2O4(g) 2NO2(g)

pale yellow dark brown

N2O4 2NO2

Gas syringe

plunger

(ii) Suppose we increase the pressure of the system. The colour darkens immediately (due to decreased volume and hence increased concentration of gas). But quickly, the colour lightens, showing that the concentration of NO2 molecules must be decreasing.

N₂O₄ 2NO₂

Equilibrium position shifts to the left

(iii) If the pressure of the equilibrium mixture is reduced (by pulling the plunger out), the reverse change takes place.

The equilibrium position moves to the right to produce more gas molecules so as to increase the pressure.

Thus the colour of the mixture lightens immediately at first (due to increased volume and hence decreased concentration of gas) and quickly darkens.

N₂O₄ 2NO₂

Equilibrium position shifts to the right

Classwork

1. Predict the direction of the shift of the equilibrium position of the following reaction in response to each of the following changes in conditions:

As4O6(s) + 6C(s) As4(g) + 6CO(g) a. Addition of CO(g)

b. Removal of gaseous arsenic (As4(g))

2. Predict the shift in equilibrium position that will occur for each of the following processes when the volume is reduced:

a. P4(s) + 6Cl2(g) 4PCl3(l) b. PCl3(g) + Cl2(g) PCl5(g)

c. PCl3(g) + 3NH3(g) P(NH2)3(g) + 3HCl(g)

Note: It is important to realize that although the changes discussed may alter equilibrium position, they do not alter the equilibrium constant. According to Le Chatelier's principle, provided the temperature is kept constant the equilibrium shifts in such a direction as to reduce the change and the new equilibrium concentrations still satisfy the original equilibrium constant.

c. Effect of Temperature Change on Equilibrium

(i) Increase in temperature always increases rate of a reaction.

(ii) For a system to be in equilibrium, the forward and backward rates have to be the same.

If an equilibrium mixture is heated or cooled, the forward and backward rates will change at once. But since the forward and backward reactions are different chemical reactions, a change in temperature will not affect the rates to the same extent.

Therefore, when the temperature of an equilibrium system is changed,

 the equilibrium position will shift

 the Kc value of the system will change

(iii) 1. reactants products H= +ve (endothermic)

 equilibrium position shifts to the right when temperature increases

 Kc value increases when temperature increases 2. reactants products H= -ve (exothermic)

 equilibrium position shifts to the left when temperature increases

 Kc value increases when temperature decreases (iv) For the Haber Process

N

2

(g) + 3H

2

(g) 2NH

3

(g) H = -92.6 kJmol

^-1

Temperature (K)

K

c

(mol

^-2

dm

⁶

)

298 4.110

⁸

373 2.310

⁵

473 440

573 7.3

673 0.41

773 0.05

873 0.0095

973 0.0026

(v) For the equilibrium between N

2

O

4

and NO

2

N2O4(g) 2NO2(g) H=+58 kJ mol-1 Temperature (K) K

c

(mol dm

^-3

)

298 0.0045

323 0.027

343 0.098

373 0.500

An increase in temperature will change the equilibrium to shift to the right and the value of K is increased, as the concentration of NO2(g) is increased whereas the concentration of N2O4(g) is decreased.

(vi)

Reaction type Temperature change K

c

value Change in equilibrium position

Exothermic forward reaction

Increased Decrease To the left

Decreased Increase To the right

Endothermic forward reaction

Increased Increase To the right

Decreased Decrease To the left

Example

Haber process is an important industrial process to manufacture ammonia with the use of nitrogen and hydrogen.

Ammonia has numerous uses like making fertilizers and explosives. The reaction between nitrogen and hydrogen is a reversible reaction. It takes place with release of thermal energy.

N

2

(g) + 3H

2

(g) 2NH

3

(g) H = -92.6 kJmol

^-1

a. Based on your knowledge about “chemical equilibrium”, predict the necessary conditions to increase the yield of ammonia in the Haber process.

b. The actual operating conditions of the Haber process are a temperature of about 450

^o

C, a pressure of about 400 atm and the presence of a catalyst (e.g. iron). Justify the conditions used.

Answer

a. (i) Since the reaction is exothermic, a lower temperature will shift the equilibrium to the right-hand side and hence increase the yield of ammonia.

(ii) There are totally 4 molecules on the left-hand side but 2 molecules on the right-hand side. A higher pressure will shift the equilibrium to the right-hand side and more ammonia is produced.

(iii)Increase the concentration of N

2

and H

2

or removing the NH

3

will shift the equilibrium to the right-hand side.

b. (i) Using high pressure, i.e. 200 atm to shift the equilibrium to the right-hand side and also increase the rate of reaction.

(ii) The use of catalysts shortens the time for the reaction to reach equilibrium.

(iii) Using of high temperature, i.e. 450

^o

C is to increase the rate of manufacture of ammonia, although shifting the equilibrium to the left and decreasing the equilibrium constant.

Classwork

1. Consider the following equilibrium system of the reaction.

2SO

2

(g) + O

2

(g) 2SO

3

(g) H = -196 kJmol

^-1

a. Is the forward reaction an exothermic or endothermic change?

b. Predict the yield of SO

3

when (i) the temperature is raised.

(ii) the temperature is lowered.

Ans: a. Exothermic b. (i) decrease (ii) increase

2. Consider the following equilibrium system in which the forward reaction is endothermic:

Co

²⁺

(aq) + 4Cl

^-

(aq) CoCl

42-

(aq) H = +ve

Pink blue

At relatively low concentrations of chloride ions, Cl

^-

(aq), the equilibrium lies far to the left, and the solution is pink. If there is a high concentration of Cl

^-

(aq), the equilibrium lies to the right, and the solution is blue. The equilibrium is also sensitive to temperature change. Deduce the colour of the mixture if

a. the temperature of the equilibrium mixture is increased, b. the temperature of the equilibrium mixture is decreased.

Ans: a. The colour of the mixture becomes blue.

b. The colour of the mixture becomes pink.

d. Effect of Catalyst on Equilibrium Position

The function of a catalyst is to alter the rate of a chemical reaction. It works by providing an alternative path which requires a different activation energy. In a reversible reaction, a positive catalyst increase the rates of the forward and backward reactions to the same extent, so that the position of equilibrium remains unchanged. But the addition of a positive catalyst will enable the same position of equilibrium to be reached in a short time.

e. Summary of the effects of changes of various factors on the equilibrium a A(g) + b B(g) c C(g) + d D(g)

Factor Equilibrium position Equilibrium constant

Increase in concentration of A or B Shifts to right No change

Increase in concentration of C or D Shifts to left No change

Increase in pressure by reducing the volume of the container

Shifts to right if (c+d) < (a+b) Shifts to left if (c+d) > (a+b)

No change of a+b = c+d

No change

Increase in temperature

Shifts to right if the forward reaction is endothermic

Shifts to left if the forward reaction is exothermic

Endothermic reaction:

increase

Exothermic reaction:

decrease

Addition of a catalyst No change No change

B. Limitations of Le Chaterlier’s Principle

For the equilibrium between N

2

O

4

and NO

2

in a closed container, N2O4(g) 2NO2(g) H= +58 kJ mol-1

an increase in temperature will increase the total pressure insider the container.

By Le Chaterlier’s Principle, increase in temperature will shift the equilibrium to the right. However, increase in pressure will shift the equilibrium to the left.

The principle offers no way of deciding which direction of equilibrium shift will predominate in this case.

IV. Applications of Chemical Equilibrium in Industries A. Industrial production of nitric acid

3NO

2

(g) + H

2

O(l) 2HNO

3

(aq) + NO(g)

In this process, NO(g) is removed from the system by further reaction with oxygen gas. The removal of NO(g) causes the system to shift to the right.

B. Haber Process

N

2

(g) + 3H

2

(g) 2NH

3

(g) H = -92.6 kJmol

^-1

To optimize the production of ammonia, the NH

3

(g) produced is removed once it is formed in order to shift the equilibrium to the right. Besides, a low temperature favours the formation of NH

3

(g) because the forward reaction is exothermic.

However, too low in temperature will increase the time to reach equilibrium. The optimal temperature of the

process is usually 450

^o

C.