CHEMICAL EQUILIBRIUM Most of the chemistry you have studied so far has been concerned with reactions which are ‘one-way’ or irreversible (represented by the → sign). Any chemical reaction of this type should go to completion, with at least one of the reactants being used up completely. There are many processes which can go either way or reversible (represented by the sign). Chemical reactions of this particular type never go to completion, with both reactants and products co-exist at any time. The following examples are some of the chemical equilibria that we would soon come across in detail : H2O(l) ↔ H2O(g)

phase equilibrium -

+ CH3COO (aq)

Fe3+(aq) + e- ↔ Fe2+(aq)

redox equilibrium

Any reversible reaction can be thought of as comprising a forward and a backward reaction taking place simultaneously in opposite directions : Forward reaction

acid-base equilibrium

rate

CH3COOH(aq) ↔

H+(aq)

Backward reaction

At the beginning

rapid rate

zero rate

As time passes (before equilibrium is reached) At equilibrium

becomes slower as more reactants are used up

becomes faster as more products are formed

rate of forward reaction = rate of backward reaction

time At the stage when forward and backward rates become equal, there is no further change in the amount of reactants and products. The system is said to reach a balanced state or in equilibrium.

Dynamic Equilibrium There are two types of equilibria :

Static equilibrium : children on a see-saw. At the balance point (i.e. the equilibrium position) no movement of the children or the see-saw occurs.

Dynamic equilibrium : boy ascending escalator at the same rate as the escalator descends. At the balance point (i.e. the equilibrium position) the boy and the escalator are moving at the same rate in opposite directions. Thus, the equilibrium position is maintained.

All equilibria in chemistry are dynamic in nature, with both forward and backward reactions continue to proceed at the same rate so that there would be no NET change in amount after equilibrium has been reached. Characteristics of Chemical Equilibrium 1

Dynamic equilibrium can exist only in a closed system (one that does not allow exchange of matter with surroundings).

2

At equilibrium, all reactants and products are present, with amounts (or concentrations) unchanged with time.

3

A state of equilibrium is reached when rate of forward change is equal to rate of backward change.

4

Dynamic equilibrium can be reached by starting from either side of the reversible reaction.

5

Any change in physical condition (e.g. conc., pressure, temp.) would result in a shift of equilibrium position.

Eqm 2

Factors Affecting Equilibrium Position

The direction of shifting in equilibrium position brought by a change in physical condition can be easily predicted by applying the Le Chatelier’s Principle, which states that : If a system in equilibrium is subjected to a change which disturbs the equilibrium, the system will respond in such a way as to counteract the effect of the change. Class Work Consider the following equilibrium :

N2(g) + 3 H2(g) ↔ 2 NH3(g)

∆H = -ve

Determine the change (if any) on the equilibrium position if it is subjected to each of the following changes : (a) adding more hydrogen gas : (b) removing some nitrogen gas : (c) carrying out at a higher pressure : (d) carrying out at a higher temperature : (e) adding more iron powder as catalyst : Class Work Consider the following equilibrium : N2O4(g) ↔ 2 NO2(g) colorless dark brown

∆H = + 57.2 kJ mol-1

Describe and explain any observable change if the above equilibrium is subjected to each of the following changes : (a) adding NO2(g) : (b) decrease the temperature :

(c) increase the pressure :

Class Work Consider the following equilibrium in Contact Process : 2 SO2(g) + O2(g) ↔ 2 SO3(g)

∆H = - ve

Suggest THREE optimum conditions by which the yield of the product could be raised.

conc.

time

Eqm 3 The following table summarizes the effect of concentration, pressure (for gaseous systems), temperature and catalyst on equilibrium position : Change in physical condition increase in concentration of

increase in pressure for gas reactions where

(1) reactant (2) product

The system responds by

Equilibrium position

consuming reactant at a faster rate to give more product consuming product at a faster rate to give more reactant

shifts to right

(1) total no. of moles of gases is the same on both sides (2) total no. of moles of gaseous products less than that of gaseous reactants (3) total no. of moles of gaseous products greater than that of gaseous reactants

no change reducing the applied pressure, hence shifting to the side with less amount of gaseous particles

temperature for

shifts to right shifts to left

(1) endothermic reaction

increase in

shifts to left

(2) exothermic reaction

Presence of catalyst

absorbing the heat to lower the temperature, hence shifting to the side with a higher enthalpy (energy content)

shifts to right

increasing the rates of forward and backward reactions to the same extent

no change

shifts to left

The Equilibrium Law If a reaction in equilibrium is represented by the equation : aA + bB ↔ cC + dD it is found experimentally that

[C]c [D]d [A]a [B]b

= KC

where KC , the equilibrium constant, is a constant value at a given temperature. Class Work A gaseous mixture of 2.0 mole H2(g) and 1.0 mole I2(g) is allowed to react at a particular temperature and pressure. Upon equilibrium, the amount of HI(g) is found to be 1.8 mole. (a)

Calculate the equilibrium constant, Kc, for the reaction H2(g) + I2(g) ↔ 2 HI(g)

(b) What would be the amount of HI(g) formed if 1.0 mole of H2(g) is allowed to react with 1.0 mole of I2(g) at the same temperature and pressure ?

Eqm 4 (c) Calculate the amount of H2(g) required to react with 1.0 mole of I2(g) so that half the amount of I2(g) would be converted into HI(g) at the same temperature and pressure.

Notice that : 1

the law is only valid after equilibrium has been reached, i.e. [A], [B], [C] and [D] refer to the respective concentrations AT equilibrium

2

it is essential to relate any numerical value for Kc to the particular equation concerned : 2 SO2 + O2 ↔ 2 SO3

Kc = 2.8 × 102

SO2 + ½ O2 ↔ SO3

Kc =

2.8 × 10 2

3

the unit for Kc are different, depending on the relative amount of reactants and products

4

the convention to express equilibrium law is to write product terms in the numerator while putting reactant terms in the denominator, in doing so providing a parallel relationship between the magnitude of Kc and the yield : large Kc means that at equilibrium the reaction almost goes to completion to the product side (i.e. higher yield)

5

the magnitude of Kc tells us nothing about the RATE of a reaction; i.e. an equilibrium reaction with a large Kc value may not even occur if it has a reaction rate too slow

Homogeneous and Heterogeneous Equilibria An equilibrium can be classified, according to the phase of the reacting species involved, into : 1

2

homogeneous - reacting species are in the same phase Fe3+(aq) + SCN-(aq) ↔ Fe(SCN)2+(aq)

aqueous state

N2(g) + 3 H2(g) ↔ 2 NH3(g)

gaseous state

heterogeneous - at least one reacting species (reactant or product) are in a different phase (usually solid state) from the rest Fe2+(aq) + Ag+(aq) ↔ Fe3+(aq) + Ag(s)

For homogeneous equilibria in gaseous phase, equilibrium constants in terms of partial pressure, Kp , is a more convenient quantity instead of those in terms of concentration, Kc

Kp =

PC ⋅ PD PA ⋅ PB

where PX represents the partial pressure of X . Class Work For the same reaction in the class work on p.3, give the expression for the equilibrium constant, Kp and calculate its value.

Eqm 5

Class Work At 1100 K, Kp = 0.13 atm-1 for the system

2 SO2(g) + O2(g) ↔ 2 SO3(g) If 2.0 mol of SO2 and 2.0 mol of O2 are mixed and allowed to react, what must the total pressure be to give a 90% yield of SO3 ?

For heterogeneous equilibria, the equilibrium law can usually be simplified by considering the fact that concentration of solids or liquids are effectively constant, which can then be incorporated into the equilibrium constant to yield another constant : e.g.

CaCO3(s) ↔ CaO(s) + CO2(g)

K =

[CaO( s )] ⋅ [CO 2 ( g )] [CaCO 3 ( s )]

K ⋅[CaCO 3 ( s )] = [CO2(g)] [CaO( s )] K’ = [CO2(g)] Class Work Solid ammonium hydrogen sulphide dissociates into hydrogen sulphide and ammonia as represented by the following equation : NH4HS(s) ↔ H2S(g) + NH3(g) At 295 K, when solid ammonium hydrogen sulphide is introduced into an evacuated vessel, the total pressure at equilibrium of the gas mixture is found to be 0.54 atm. (i) Calculate the equilibrium constant, Kp , for the reaction.

(ii) If the vessel initially contains hydrogen sulphide gas at a pressure of 0.335 atm, what is the partial pressure of ammonia when the system is allowed to reach equilibrium at the same temperature ?

Experimental Determination of Equilibrium Constant General procedure : 1

Mix known volumes of known concentrations of the reactants for the reaction

2

Allow the reaction to equilibrate in a closed system and at a constant temperature

3

‘Freeze’ the equilibrium so that the equilibrium position would not shift significantly during the subsequent titration

4

Determine the equilibrium concentration of one of the reacting species by employing a chemical (via titration) or physical method (e.g. colorimetry)

5

The equilibrium concentrations of the rest of reacting species can be obtained by calculations

Eqm 6

Colorimetric method in determining Kc of the reaction : Fe3+(aq) + SCN-(aq) ↔ Fe(SCN)2+(aq) 1

The concentration of Fe(SCN)2+, which is blood red in colour, can be determined either by visual comparison or by using a colorimeter.

2

Calibrate a colorimeter by measuring the absorbance (i.e. colour intensity) of several solutions which contain known concentrations of Fe(SCN)2+(aq).

3

By mixing comparable volumes of standard Fe(NO3)3(aq) and KSCN(aq) solutions and measure the [Fe(SCN)2+(aq)] formed with the calibrated colorimeter, [Fe3+(aq)], [SCN-(aq)] and [Fe(SCN)2+(aq)] at equilibrium can be found.

Factors Affecting Equilibrium Constant The numerical value of the equilibrium constant remains unchanged at one particular temperature. This means that Kc and Kp are unaffected by catalysts or by changes in pressure and concentration. The equilibrium constant does, however, vary with temperature (i.e. temperature-dependent), as illustrated by the following Kp values at different temperatures for two important reactions : N2(g) + 3H2(g) ↔ 2NH3(g) T/K 400 500 600

Kp =

( p NH3 ) 2 p N 2 ( pH2 )3

N2O4(g) ↔ 2 NO2(g)

atm -2

Kp =

1.0 × 102 1.6 × 10-1 3.1 × 10-3

( p NO2 ) 2 p N 2O 4

atm

5.1 × 10 1.5 × 103 1.4 × 104

Explain whether these two reactions are exothermic or endothermic.

The relationship between K and T is best illustrated by the following equation : ln K = constant -

∆H RT

The close resemblance of this and Arrhenius equation reveals a parallel relationship between K and k : while the values of equilibrium constants tell us about the yield of a reaction from an energetic point of view (∆H); the values of rate constants tell us about the rate of a reaction from a kinetic point of view (Ea). Notice that : 1

changes in concentration and pressure result in a shift in equilibrium position in order to preserve the Kc value

2

changes in temperature result in a shift in equilibrium position because Kc has been adjusted to a new value

Applying the Principles of Reaction Rates and Equilibria to Industrial Processes optimum condition to increase the rate Haber Process N2(g) + 3H2(g) ↔ 2NH3(g) Contact Process 2 SO2(g) + O2(g) ↔ 2 SO3(g)

optimum condition to increase the yield

real condition in industry

Predicting Equilibrium Shift by Reaction Quotient

Eqm 7

In order to predict the direction of shifts in a chemical equilibrium, we can compare the reaction quotient, Q, of the reaction with the equilibrium constant. The reaction quotient is defined in the same way as the equilibrium constant except that the concentrations of reactants and products can be taken at any moment of the reaction (i.e. not necessarily at the eqm). For the hypothetical reaction: aA + bB ↔ cC + dD the expression of reaction quotient in terms of concentrations is:

[C]c [D]d Qc = [A]a [B]b There are three possible cases when the reaction quotient is compared with the equilibrium constant of the reaction: 1

Q=K

system is at equilibrium

2

Q>K

system is not at eqm as concentrations of products are greater than their eqm values eqm shift to LHS (reactant side) until eqm is reached

3

Q