Unit 6 - 1
Unit
6
Section
Chemical Equilibrium
6.1
Principles of Chemical Equilibrium
( 1 ) Dynamic equilibrium
Chemical equilibrium is dynamic in nature.
1.
There are five characteristics of dynamic equilibrium :
Dynamic equilibrium involves reversible reactions.
When reactants react to produce products, it is possible that the products themselves may then react
to give the reactants again :
A + B – C + D
reactants
products
To show this, the sign – is used : it means ‘is in equilibrium with’ and implies the rates of the
opposing processes are equal.
Example 1
Consider the reversible reaction :
2 CrO42-(aq) +
yellow
2 H+(aq) – Cr2O72-(aq) +
orange
H2O(l)
The reversibility of the above equilibrium can be demonstrated by the addition of a small amount of
concentrated hydrochloric acid to an aqueous solution of yellow chromate(VI) ions, whose colour can be
seen to change to orange. The solution can be made to go back to yellow by adding an alkali to the
system, which consumes the H+(aq) ions and yellow chromate(VI) ions are formed.
Example 2
Consider the reversible reaction :
Br2(aq) +
orange-brown
H2O(l)
–
HOBr(aq) +
colourless
H+(aq) +
Br-(aq)
Again, the reversibility of the above equilibrium can be readily demonstrated by the addition of
sufficient alkali to the system, which can be seen to turn very pale yellow (almost colourless) as the H+(aq)
ions are consumed and the amount of brown Br2(aq) present are diminished. The brown colour of the
system can be restored by the addition of excess acid back into the system.
2.
At a given temperature the overall properties (such as concentration, amount and partial
pressure of the various species) does not change with time for a system in chemical
equilibrium.
Example :
A +
2B –
C +
3D
Under equilibrium conditions there is no net change in concentration of substances in the system.
If 0.1 mol dm-3 of A and 0.2 mol dm-3 of B turn to products, at the same time 0.1 mol dm-3 of C and 0.3
mol dm-3 of D give back reactants, then the net change is zero.
Unit 6 - 2
3. Although there is no change in reactant or product concentration in a system at dynamic
equilibrium, both forward and backward reactions are still proceeding at equal rates.
Example :
(1)
Consider the approach to equilibrium of the iron(III) / thiocyanate reaction
Fe3+(aq) + NCS-(aq) – FeNCS2+(aq)
blood-red
When iron(III) ions are added to thiocyanate ions, the reaction is rapid at first.
concentration gets less, the forward rate decreases.
As the reactant
(2) At the start of the reaction, there is no product present at all. It begins to be produced by the
forward reaction. As its concentration increases so does the rate of backward reaction
(3) After a particular time interval, the rates of both reactions become equal and the system reaches
equilibrium. This can be shown by including both forward and backward rates on the same graph.
Evidence for the dynamic nature of equilibrium
Since there are no changes in reactant or product concentrations in an equilibrium system, it is fair to
question whether anything is happening at all. For the equilibrium between ethanoic acid, ethanol, ethyl
ethanoate and water, more direct evidence of the dynamic nature of an equilibrium can be obtained as
follows :
CH3COOH(aq) + C2H5OH(aq) – CH3COOC2H5(l) + H2O(l)
(1) Measured amounts of ethanoic acid and ethanol are mixed and left to equilibrate for a long time at
constant temperature.
(2) To check that the system has reached equilibrium :
(i) a small sample is removed and the ethanoic acid concentration is found by titration against
standard sodium hydroxide solution;
(ii) after a period of time, the analysis for acid is repeated;
(iii) if the same value is obtained, it can be assumed that the system is in equilibrium because there
are no concentration changes at equilibrium.
(3) From the known values of the concentrations at the start and the measured equilibrium concentration
of acid, the remaining equilibrium concentrations are calculated.
(4) The equilibrium mixture is now discarded and an exact replica of it is set up in which each acid
molecule contains a ‘labelled’ radioactive isotope, e.g. 14CH3COOH.
(5) After a short while, the mixture is separated by fractional distillation and the radioactivity of the
separated ester is monitored. It is found that the ester now contains ‘labelled’ molecules and this
proves that the forward reaction is still proceeding even in an equilibrium mixture.
Unit 6 - 3
4.
Dynamic equilibrium can be achieved from either the forward or the reverse direction.
Example : The reaction between hydrogen and iodine to form hydrogen iodide
H2(g)
+
I2(g) –
2 HI(g)
(a) Initial amount
/ mol
Equilibrium amount / mol
(b) Initial amount
/ mol
Equilibrium amount / mol
M Bodenstein made a detailed study of this reaction from 1890 to 1900. His method was to
seal in glass bulbs either a mixture of hydrogen and iodine or pure hydrogen iodide, and place the
bulbs in a thermostat bath. After a time interval, he cooled the bulbs rapidly so that chemical
reaction stopped. Then he analyzed the contents of each bulb to find the amounts of hydrogen,
iodine and hydrogen iodide present.
Curve (a) shows a set of results at a certain temperature and pressure. If the reaction between
0.50 mol H2 and 0.50 mol I2 went to completion, 1.00 mol HI would be formed. It appears that,
with 0.78 mol HI formed, no further reaction takes place. The system has reached equilibrium.
Curve (b) shows that the same amount of HI is present in the bulbs when equilibrium is reached
from the other direction, by the dissociation of 1.00 mol of HI.
5.
A stable state of dynamic equilibrium can only be achieved in a closed system (one that cannot
exchange matter with its surroundings).
Example : Thermal decomposition of calcium carbonate
Unit 6 - 4
(2)
1.
The equilibrium law
Equilibrium constant in terms of concentration ( Kc )
Equilibrium law
If a reversible reaction is allowed to reach equilibrium, then the product of the concentrations of the
products (raised to the appropriate powers) divided by the product of the concentrations of the reactants
(raised to the appropriate powers) has a constant value at a particular temperature. The appropriate
power is the coefficient of that substance in the stoichiometric equation for the reaction.
For a reaction
aP + bQ – cR + dS
the equilibrium constant, expressed in terms of concentrations, Kc, is given by the equilibrium law :
The dimensions of Kc are concentration(c + d – a – b), and the units vary from one equilibrium to
another.
Prove : Assume a chemical equilibrium 2A + 2B – C + D proceeds by 2 steps :
Example 1
Write an expression for Kc and state its unit for each of the following reactions :
(a) 2 HI(g) –
H2(g) +
(b) N2O4(g) –
2 NO2(g)
I2(g)
Kc =
[ H 2( g ) ][ I 2( g ) ]
[ HI ( g ) ] 2
no unit
Example 2
Consider a chemical equilibrium at 700 ℃ involving SO2(g), O2(g) and SO3(g) and represented in
three different ways below :
Kc1
(a)
2 SO2(g) + O2(g) – 2 SO3(g)
(b)
2 SO3(g) – 2 SO2(g) + O2(g)
Kc2
(c)
SO2(g) + 1 O2(g) – SO3(g)
Kc3
2
[ SO3 ( g ) ]
If Kc1 =
= 2.80 x 102 mol-1 dm3, write the expressions and calculate the values for
2
[ SO2 ( g ) ] [O2 ( g ) ]
both Kc2 and Kc3 .
Unit 6 - 5
Example 3
A 10.0 cm3 mixture contains the initial amounts :
ethanol 0.0515 mol; ethanoic acid 0.0525mol; water 0.0167 mol; ethyl ethanoate 0.0314 mol.
The equilibrium amount of ethanoic acid = 0.0255 mol.
CH3COOH + C2H5OH –
Determine the equilibrium constant.
CH3COOC2H5
+
H2O
Initial amount
/ mol
Equilibrium amount /mol
Example 4
In the preparation of ethyl butanoate from the appropriate acid and alcohol, 13.2 g of acid and 6.9 g
of the alcohol were mixed at a certain temperature. This reaction took place in a solvent which
dissolved all the reactants and products. The volume of the reacting mixture at equilibrium was 50 cm3
and the concentration of the ester was found to be 1.8 M.
(a) Write a balanced equation for the above preparation.
(b) Calculate the equilibrium constant, Kc, for the reaction at this temperature.
(c) If a few drops of concentrated hydrochloric acid were added during the preparation, explain how this
would affect
(1) the reaction rate, and
Concentrated hydrochloric acid acts as a catalyst for the esterification reaction.
activation energy for the reaction, reaction rate increases.
As it lowers
(2) the equilibrium position.
Equilibrium position does not change.
equilibrium.
Catalyst would only affect the time to reach
Unit 6 - 6
2.
Equilibrium constant in terms of partial pressure ( Kp )
For gaseous reactions, the concentrations can be given as partial pressures.
equilibrium constant can be written as Kp.
In this case the
Example 1
In the reaction between hydrogen gas and iodine gas :
H2(g) + I2(g) – 2 HI(g)
Kc =
[ HI ( g ) ] 2
[ H 2 ( g ) ][ I 2( g ) ]
Kp =
Since the pressure units cancel out the expression, Kp is dimensionless.
Example 2
In the Haber process for making ammonia :
N2(g) + 3 H2(g) – 2 NH3(g)
Kc =
Kp =
The unit of Kp is P-2, e.g. (k Pa)-2 or atm-2.
Kp can also be expressed in terms of mole fractions of the components in the mixture and the total
pressure :
Example 3
In the equilibrium system at 700 K :
2 SO2(g) + O2(g) – 2 SO3(g)
the partial pressure of the gases in an equilibrium mixture are :
PSO2 = 9.0 kPa, PSO3 = 450 kPa and PO2 = 8.3 kPa.
Calculate Kp for this system.
Unit 6 - 7
Example 4
When 1.00 mol hydrogen and 1.00 mol iodine gases are allowed to reach equilibrium in a 1.00 dm3
flask at 450℃ and 101 kPa, the amount of hydrogen iodide gas at equilibrium is 1.56 mol. Calculate
Kp at 450℃.
Let the total pressure be PT.
PT = 101 kPa
H2(g)
+
I2(g)
–
2 HI(g)
Initial amount
/ mol
Equilibrium amount / mol
Equilibrium pressure / PT
Example 5
Nitrogen and hydrogen are mixed in a molar ration 1 : 3. At equilibrium, at 600℃and 1000 kPa,
the percentage of ammonia in the mixture of gases is 16%. Find Kp at 600℃.
Unit 6 - 8
3.
Determination of equilibrium constants by experiments
Experiment 1 : Finding the Kc of
1.
CH3COOH(aq) + C2H5OH(aq) –
CH3COOC2H5(l) + H2O(l)
To obtain a value of Kc for this system, the following procedure is adopted :
Set up a number of reaction flasks with known initial amounts of each components. The reaction is
catalysed by acid, and so the same volume of concentrated sulphuric acid is added to each flask.
Flask
Moles of acid
1
1
2
2
3
1
4
5
6
7
Flask 1 is a control flask.
Moles of
Alcohol
1
1
1
1
-
Moles of ester Moles of water
1
1
1
2
1
1
1
1
Volume of
H2SO4 / cm3
1.0
1.0
1.0
1.0
1.0
1.0
1.0
2.
Allow each system to reach equilibrium by putting them into a thermostatic bath at 298 K for a week
(the reaction is slow). The flasks are stoppered to prevent evaporation losses.
3.
When each system has reached equilibrium, the amount of acid present is found by titrating against
standard sodium hydroxide solution. The control flask shows how much alkali is needed to
neutralize the sulphuric acid which is used as catalyst; this amount is subtracted from each reading.
Results :
Flask
Moles of acid
2
3
4
5
6
7
0.33
1.15
0.66
0.15
0.33
0.45
Moles of
Alcohol
0.33
0.15
Moles of ester Moles of water
Kc
0.67
0.85
0.67
0.85
4.12
4.19
1.15
0.85
0.85
4.19
0.45
1.55
0.55
4.21
Calculations : For flask 2,
CH3COOH(aq) +
Initial amount / mol
1
Equilibrium amount / mol
0.33
C2H5OH(aq) – CH3COOC2H5(l) +
1
0.33
0.67
H2O(l)
0.67
Let V dm3 be the volume of the solution.
Assumption : As the reaction rate of hydrolysis of ester is slow, it is assumed that the hydrolysis of
ethyl ethanoate to give ethanoic acid during titration is negligible.
Unit 6 - 9
Experiment 2 : Finding the Kc of
Fe
3+
(aq)
+
NCS-(aq)
–
2+
FeNCS
(aq)
Being blood red in colour, the FeNCS2+(aq) ions formed in the above reaction absorb certain
frequencies of visible light in the electromagnetic spectrum. When the above reaction is at equilibrium,
the concentration of FeNCS2+(aq) ions can be determined colorimetrically.
1.
The colorimeter can be calibrated by measuring the absorbance of several solutions which contained
known concentrations of FeNCS2+(aq) ions.
To do this a series of standard solutions of FeNCS2+(aq) can be prepared by adding a small
concentration of NCS-(aq) to a large excess of Fe3+(aq) so that essentially all the NCS-(aq) is converted
to FeNCS2+(aq). Under these conditions, the final concentration of FeNCS2+(aq) in these standard
solutions is equal to the concentration of NCS-(aq) solutions used.
A calibration curve is obtained by plotting absorbances against various concentrations of
FeNCS2+(aq).
Standard curve
absorbance
0
[FeNCS2+ (aq)]
2.
In an actual experiment, a standard solution of Fe(NO3)3(aq) and another standard solution of
NaNCS(aq) are required. Mixing of the two solutions dilutes both of them.
3.
The equilibrium concentration of FeNCS2+(aq) in the above mixture is then determined by comparing
its absorbance (colour intensity) with the standard solutions of FeNCS2+(aq) in step 1 by the
colorimetric method.
Calculations :
10 cm3 of 0.021 mol dm-3 Fe(NO3)3(aq) solution was mixed with 10 cm3 of 0.0202 mol dm-3
NaNCS(aq) solution. The equilibrium concentration of FeNCS2+(aq) was found to be 0.0100 mol dm-3.
Initial conc. / mol dm-3
Equilibrium conc. / mol dm-3
Fe3+(aq) + NCS-(aq) –
0.0105
0.0101
FeNCS2+(aq)
0.0100
Unit 6 - 10
(3)
1.
The effect of changes in concentration, pressure and temperature on
equilibria
Effect of concentration change on equilibrium
Under constant temperature, any concentration changes on a system at equilibrium result in
the adjustment of the system without changing the value of equilibrium constant.
Consider the following equilibrium :
A +
B
–
C +
D
If the concentration of reactants (A and /or B) is increased, the position of equilibrium will shift to
the right. More products (C and D) are formed.
If the concentration of reactants (A and /or B) is decreased, the position of equilibrium will shift to
the left. Less products (C and D) are formed.
Explanation of concentration effect :
If the concentration of a reactant in an equilibrium system is suddenly changed, the rate of the
forward reaction alters because rate ∝ [reactant]. Similarly, if a product concentration is suddenly
changed, the rate of the backward reaction alters. In either of these two cases, equilibrium is disrupted
because the rates of the two opposing processes are no longer equal. The concentrations of reactants and
products change until a new equilibrium position is reached.
In both the above cases, the value of Kc remains unchanged although the position of equilibrium is
changing.
Summary of concentration effect :
Changes in concentration
Increase of reactant concentration
Reaction rate
forward rate
Decrease of reactant concentration
forward rate
Increase of product concentration
backward rate
Decrease of product concentration
backward rate
Equilibrium position
shift to
(product side)
shift to
(reactant side)
shift to
(reactant side)
shift to
(product side)
Kc
no change
no change
no change
no change
Example 1
An aqueous solution of bismuth(III) chloride is cloudy because of the hydrolysis
BiCl3(aq) +
colourless
H2O(l) –
BiOCl(s) +
white ppt.
2 HCl(aq)
If a little concentrated hydrochloric acid is added, equilibrium is disrupted and the rate of backward
reaction ___________. The position of equilibrium shifts in the direction that will _________ acid, i.e.
from ___________________. Although the solution cannot absorb all the acid added, the hydrolysis of
bismuth(III) chloride is much reduced and a clear solution results.
Unit 6 - 11
Example 2 : Consider the formation of ammonia
N2(g) +
3 H2(g) –
2 NH3(g)
If hydrogen gas is added at constant volume, the following effects will be observed :
1. The forward rate increases with increased [H2].
2. Thus the forward rate becomes greater than the backward rate.
3. The concentration of ammonia increases while that of nitrogen decreases until the two
rates become equal again.
4. Kc remains unchanged.
5. The reaction quotient Qc is smaller than Kc. There is a shift of equilibrium position to
the right until Qc = Kc.
Qc
=
Rate of
Reaction
Time
Example 3 : Consider the formation of ammonia
N2(g) +
3 H2(g) –
2 NH3(g)
If ammonia is removed at constant volume, the following effects will be observed :
1. The backward rate decreases with decreased [NH3].
2. Thus the backward rate becomes less than the forward rate.
3. The concentration of ammonia builds up again at the expense of nitrogen and hydrogen
until the two rates become equal again.
4. Kc remains unchanged.
5. The reaction quotient Qc is smaller than Kc. There is a shift of equilibrium position to
the right until Qc = Kc.
Qc
=
Rate of
Reaction
Time
Unit 6 - 12
2.
Effect of pressure change on equilibrium
Changes in pressure result in the adjustment of the system without changing the value of
equilibrium constant, as in the case of concentration change.
Pressure changes can only affect those equilibria that involve gaseous reactants or products
The effect of increased pressure on an equilibrium system can be divided into three cases :
Case
Rates of forward and
Position of
Kp or Kc
backward reactions
equilibrium
Number of gaseous molecules are Rates of both forward and
No change
No change
the same on both sides of equation. backward reactions increase to
the same extent.
Example :
H2(g) + I(g) – 2 HI(g)
Number of gaseous reactant
Rates of both forward and
Shift to the right No change
molecules is greater than that of
backward reactions increase, but (product side)
gaseous product molecules.
the forward rate increases more.
Example :
2 SO2(g) + O2(g) – 2 SO3(g)
Number of gaseous product
molecules is greater than that of
gaseous reactant molecules.
Rates of both forward and
Shift to the left
backward reactions increase, but (reactant side)
the backward rate increases more.
No change
Example :
N2O4(g) – 2 NO2(g)
A change in the volume of container holding a gaseous system also affects the position of
equilibrium. When the volume of container decreases, the concentrations and partial pressures of
reactants and products are increased. The gaseous system reduces its own pressure to minimize
disturbance to the system. The equilibrium position is shifted to the side having smaller number of
gaseous molecules.
Example 1 : Investigation of the effect of pressure change on the following reaction
– 2 NO2(g)
N2O4(g)
pale yellow
brown
When the plunger is pushed in rapidly from position 1 to 2, (a) the colour momentarily deepens, then
(b) goes paler after a short while.
Explain while there is a colour change in (a) and (b).
(a) As volume is reduced, the concentration of NO2(g) increases and thus the colour will be deepened.
(b) Increase in pressure of the gas mixture will cause the reaction to go in the direction from NO2 to
N2O4. It is because rates of both forward and backward reactions increase, but the backward rate
increases more. The colour in the syringe will fade as more pale yellow N2O4 molecules are
formed.
Unit 6 - 13
Example 2
N2(g) +
3 H2(g) –
2 NH3(g)
If the pressure of an equilibrium mixture of nitrogen, hydrogen and ammonia is increased, the
equilibrium shifts in the direction that tends to decrease the pressure. It does this by decreasing the total
number of molecules present, i.e., equilibrium position shifts to the right (product side).
Let PT be the total pressure of the system.
be expressed in terms of their mole fractions :
Thus
Kp
Partial pressures of nitrogen, hydrogen and ammonia can
=
If PT is increased, as Kp is constant at constant temperature, the mole fraction factor must increase to
cancel out the fall in the pressure factor (1 / P2T). Therefore :
1. Kp remains unchanged.
2. The reaction quotient Qp is smaller than Kp. There is a shift of equilibrium position to
the right until Qp = Kp.
Qp
3.
4.
=
Amount of NH3 increases.
Amount of H2 and N2 decrease.
Example 3 : The addition of an inert gas at constant volume
If argon is added at constant volume to an equilibrium mixture of nitrogen, hydrogen and ammonia
at constant volume,
N2(g) + 3 H2(g) – 2 NH3(g)
1.
2.
3.
PT increases, but
each partial pressure remains constant, and therefore
there is no effect on the position of equilibrium.
Example 4 : The addition of an inert gas at constant pressure
If argon is added at constant pressure to an equilibrium mixture of nitrogen, hydrogen and ammonia
at constant pressure, there is a dilution effect as the volume of mixture increases.
N2(g) + 3 H2(g) – 2 NH3(g)
Therefore :
1.
2.
3.
Each partial pressure decreases.
Kp remains unchanged.
The reaction quotient Qp is larger than Kp.
the left until Qp = Kp.
Qp
4.
5.
=
Amount of N2 and H2 increase.
Amount of NH3 decreases.
There is a shift of equilibrium position to
Unit 6 - 14
3.
Effect of temperature change on equilibrium
As equilibrium constant is derived from the rate constants which are temperature dependent, thus the
value of equilibrium constant, Kc or Kp, changes with temperature.
Endothermic reaction
For an equilibrium system whose forward reaction is endothermic, an increase in temperature will
shift the equilibrium position in a direction to consume more heat, i.e., shifts to the right.
An decrease in temperature will shift the equilibrium position in a direction to produce more heat,
i.e., shifts to the left.
Exothermic reaction
For an equilibrium system whose forward reaction is exothermic, an increase in temperature will
shift the equilibrium position in a direction to consume more heat, i.e., shifts to the left.
An decrease in temperature will shift the equilibrium position in a direction to produce more heat,
i.e., shifts to the right.
Explanation of temperature effect
If the temperature of an equilibrium system is increased, the rates of both the forward and the
backward reaction increase. However, the two do not increase by the same extent because their
activation energies are different. The two rates therefore become unequal and equilibrium is disrupted.
Endothermic reaction
Change in temperature
Increase in temperature
Ea 〉bEa
Decrease in temperature
Exothermic reaction
Increase in temperature
f
b
Ea 〉fEa
Reaction rate
Decrease in temperature
The reactant and product concentrations adjust until a new equilibrium position is reached.
The new equilibrium is governed by the value of new equilibrium constant.
Therefore, a change in temperature results in the adjustment of the system to a new
equilibrium constant.
Summary of temperature effect
Endothermic reaction
∆H (
)
Exothermic reaction
∆H (
)
Change in temperature
Increase in temperature ( )
Decrease in temperature ( )
Increase in temperature ( )
Decrease in temperature ( )
Position of equilibrium
Shift to
( )
Shift to
( )
Shift to
( )
Shift to
( )
Kc or Kp
(
(
(
(
)
)
)
)
Unit 6 - 15
Example 1 : Investigation of the effect of pressure change on the following reaction
increase in temperature
N2O4(g) – 2 NO2(g)
∆H = + 58.0 kJ mol-1
decrease in temperature
A gas syringe of nitrogen is immersed in a freezing mixture for a few minutes. The temperature,
colour and volume of the gas are recorded. The experiment is repeated with the gas immersed in water
at room temperature and 100℃.
Temperature
change
Increase
Position of
equilibrium
shifts to right
Observation and reason
1. Gases become darker brown as more NO2 is formed.
Reason :
The rates of both forward and backward reactions
increase as temperature increases. The rate of the
forward reaction is increased by a large extent than the
rate of the backward reaction in endothermic reaction.
Equilibrium position shifts to right.
2. The plunger moves out.
Reason :
Volume of gases increases because the total number of
gas molecules increases.
Decrease
shift to left
1. Gases become pale yellow as more N2O4 is formed.
Reason :
The rates of both forward and backward reactions
decrease as temperature decreases. The rate of the
forward reaction is decreased by a large extent than the
rate of the backward reaction in endothermic reaction.
Equilibrium position shifts to left.
2. The plunger moves in.
Reason :
Volume of gases decreases because the total number of
gas molecules decreases.
Unit 6 - 16
Example 2
The formation of ammonia is exothermic :
N2(g) +
3 H2(g) –
2 NH3(g)
∆H = - 92.0 kJ mol-1
If the temperature is raised, the system can absorb heat by the dissociation of ammonia into
hydrogen and nitrogen. The equilibrium constant for the formation of ammonia is decreased.
In an equilibrium system at 500 K, Kp = 3.55 x 10-2 atm-2 and the rates of the forward and backward
reaction are equal. When the temperature is increased to 700 K, the rates of both reactions are increased.
The rate of the backward reaction is increased by a large extent than the rate of the forward reaction
because bEa > fEa . The backward rate is now greater than the forward rate and so [NH3] decreases
while [N2] and [H2] increase until the two rates become equal again
The new equilibrium constant gets smaller.
Kp = 7.76 x 10-5 atm-2 at 700 K.
The overall effect of an increase in temperature on the equilibrium system is therefore :
1. An increase in both forward and backward rates.
2. Backward rate is increased by a large extent than the rate of the forward reaction.
3. A decrease in the value of Kp.
4. A shift in equilibrium position to the left.
5. A decrease in [NH3] and an increase in [N2] and [H2].
Example 3
The formation of water-gas is endothermic :
C(s) +
Kp
H2O(g) –
H2(g) + CO(g)
∆H = + 131 kJ mol-1
=
A temperature increase leads to a greater increase in forward rate than in backward rate because
Ea > bEa . The concentration of the reactants therefore tends to decrease while that of the products
increases until the two rates are equal again.
f
The overall effect is summarized below :
1. An increase in both forward and backward rates.
2. Forward rate is increased by a large extent than the rate of the backward reaction.
3. An increase in the value of Kp.
4. A shift in equilibrium position to the right.
5. A decrease in [H2O] and an increase in [H2] and [CO].
Unit 6 - 17
Relation of temperature and the value of equilibrium constant
From thermodynamic derivation, the actual relationship between equilibrium constant and enthalpy
change of reaction is given by the van’t Hoff’s equation :
∆H
ln K = constant RT
By measuring K at a number of different temperatures, a plot of ln K against 1 / T can be used to
find the enthalpy change for the reaction concerned. The plot should give a straight line of slope
-∆H / R.
Example 1
(a) At 283 K and 1.33 kPa , two gases A and B are in equilibrium :
A(g) –
2 B(g)
The mole fraction of gas A is 0.772.
Calculate the equilibrium constant Kp at this temperature.
(b) The enthalpy change for the reaction is + 52.45 kJ mol-1, calculate Kp at 333K.
Example 2
The following table gives the equilibrium constant, Kp, at several temperatures for the dissociation of
ethanoic acid dimers in the gas phase :
T/K
Kp / N m-2
298
72.9
303
121.1
(a) Show graphically that the data obey the equation
enthalpy change for the dissociation of acid dimers.
308
182.7
313
302.0
∆H
, where ∆H is the
RT
Calculate the value of ∆H.
ln K = constant -
(b) From the result in (a) estimate the energy of a hydrogen bond.
Unit 6 - 18
4.
Effect of catalyst on equilibrium
A catalyst is a substance that changes the rate of a reaction but is itself left chemically unaltered at
the end of the process. When a positive catalyst is added to an equilibrium system, it speeds up the rate
of both the forward and backward reactions. Since the decrease in activation energy for both forward
and backward reactions is by the same extent, the increase in rate is also the same for both forward and
backward reactions. A catalyst cannot alter the value of the equilibrium constant, it can only change
the time to reach equilibrium.
Potential
Energy
rate of
reaction
Reaction coordinate
time
Example
The graphs below give the percentage of Z in the equilibrium mixture at constant pressure and
constant temperature for the reversible reaction :
x X(g) + y Y(g) – z Z(g)
(a) Suggest how you would increase the yield of Z.
The yield of Z would be increased by conditions of low temperature and high pressure.
(b) In practice, the conditions chosen for the synthesis of Z are 723 K and 250 atm. Justify this choice.
723 K is chosen so that the % yield of Z is not too low whilst the time to reach equilibrium is not too
long. 250 atm is not too high a pressure used; too high a pressure used would mean high cost of
plant construction and maintenance.
(c) A catalyst has been used for the above reaction to give the results shown. What would happen if
the catalyst were poisoned ? Draw a new graph for the reaction at 873 K to illustrate your answer.
The time to reach equilibrium would be longer but the yield of Z is unaffected.
(d) Suppose the equilibrium constant for the above reaction is K, the rate constant for the forward
reaction is k1, and the rate constant for backward reaction is k-1.. Express K in terms of k1 and k-1.
Unit 6 - 19
5.
Calculations on equilibrium composition involving changes in concentration / pressure
Example 1
Ethanoic acid and ethanol react to form ethyl ethanoate and water. A solution was prepared by
mixing 1.0 mol CH3COOH and 2.0 mol C2H5OH. At the temperature of the experiment, Kc = 4.00.
Calculate the equilibrium amounts of ethanoic acid, ethanol, ethyl ethanoate and water.
CH3COOH(aq) +
C2H5OH(aq) –
CH3COOC2H5(l) +
H2O(l)
Initial amount
/ mol
Equilibrium amount /mol
Example 2
Consider the equilibrium for the synthesis of ammonia at constant pressure :
N2(g) + 3 H2(g) – 2 NH3(g)
A mixture of 2.00 mol of N2(g) and 6.00 mol of H2(g) was placed in a vessel of volume 10.0 dm3 and
heated to 600 K. When the system had reached equilibrium, it was found that 2.00 mol of NH3 was
present.
(a) Calculate the equilibrium constant, Kc, of this reaction at 600 K.
N2(g) + 3 H2(g) – 2 NH3(g)
Equilibrium amount /mol
(b) When the volume of this vessel was increased and the system allowed to come to a new equilibrium
at the same temperature, 4.50 mol of H2(g) was found to be present. Calculate the new volume.
N2(g) + 3 H2(g) – 2 NH3(g)
Equilibrium amount /mol
(c) Another 10.0 dm3 vessel contains 10.0 mol of N2(g)at the same temperature. How many moles of
H2(g) must be added to this vessel in order that, at equilibrium, half of the N2(g) will be converted to
NH3(g) ?
N2(g) + 3 H2(g) – 2 NH3(g)
Equilibrium amount /mol
Unit 6 - 20
Example 3
In a closed container of 1 dm3 at 1100 K, equilibria among the gases hydrogen, carbon monoxide,
carbon dioxide and steam are studied and the following results are recorded :
Experiment
1
2
3
H2
14
23
32
Partial pressures at equilibrium / k N m-2
CO
CO2
35
65
27
23
19
31
H2O
35
26
69
(a) Write an equation with carbon monoxide on the product side to show the equilibrium established in
the container.
(b) Calculate the equilibrium constant of the above reaction at 1100K.
(c) Calculate the total number of moles of the gases at equilibrium in experiment 1.
( R = 8.31 J K-1 mol-1 )
(d) The equilibrium constant for the above reaction at 1300 K is 2.80.
equilibrium ? Explain your answer.
Example 4
At 1100 K, Kp = 0.13 atm-1 for the system :
2 SO2(g) + O2(g) –
What is the effect of heat on the
2 SO3(g)
If 2.00 mol of SO2(g) and 2.00 mol of O2(g) are mixed and allowed to react, what must be the total
pressure be to give a 90% yield of SO3(g) ?
2 SO2(g) +
initial amount / mol
equilibrium amount / mol
O2(g) –
2 SO3(g)
Unit 6 - 21
6.
Haber Process
Manufacture of ammonia by Haber process
Ammonia is manufactured by direct combination of nitrogen and hydrogen :
N(g) + 3H2(g) – 2NH3(g) ∆H = - 92.0 kJ mol-1
The flow diagram of the Haber process is given below :
Nitrogen is produced by the fractional distillation of liquid air.
Hydrogen is obtained from methane, naphtha, or other hydrocarbons by steam reforming :
Purification is necessary because compounds of oxygen and sulphur poison the catalyst in the
synthesis of ammonia and have to be removed. The purified nitrogen and hydrogen are mixed in a
ration of 1 : 3 respectively by volume, compressed to 200-1000 atmospheres and heated in the heat
exchanger by the hot product gases.
The gaseous mixture is then passes over a heated catalyst in the catalytic chamber or ammonia
converter. The catalyst is finely divided iron promoted by aluminium oxide and potassium oxide which
is heated electrically to 450℃-500℃. As the reaction between nitrogen and hydrogen forming ammonia
is exothermic, further heating is unnecessary.
N(g) +
3H2(g)
– 2NH3(g)
∆H = - 92.0 kJ mol-1
The hot gases leaving the converter contains 10-15% ammonia pass through the heat exchanger and
cooled. The ammonia formed in the process is then absorbed in water or liquefied under pressure. The
uncombined nitrogen and hydrogen are recycled.
Unit 6 - 22
Equilibrium and kinetic considerations in Haber process
The reaction, N(g) + 3H2(g) – 2NH3(g) , is reversible and exothermic. The formation of
ammonia at different conditions of temperatures and pressures are shown in figure :
Choice of pressure conditions
Theoretically, a higher yield of ammonia would be obtained on increasing the pressure. In addition
to increasing the yield, the reaction should attain equilibrium at a faster rate as the pressure is increased.
This is because with an increase in pressure, the probability of collisions between gaseous particles also
increases due to a decrease in the volume. However, with increasing pressure, the cost of industrial
plant becomes more expensive, a compromise is reached between a high yield at high cost and a lower
yield at lower cost. Most industrial plants choose to operate at the pressure range 200-1000 atmospheres,
depending on the scale of the manufacturing plant.
Choice of temperature conditions
Since the synthesis of ammonia is an exothermic process, a lower temperature will lead to a larger
equilibrium constant and a higher yield of ammonia. However, a decrease in the temperature will
decrease the rate of formation of ammonia, that is the equilibrium will be attained in a longer time; this
makes the process uneconomical. A compromise has again to be reached between high yield in a long
time or a lower yield in shorter time. The operating temperature is usually between 450℃-500℃
which gives a fair amount of ammonia at a reasonable rate.
Choice of Catalyst
A catalyst is added to increase the rate of reaction. This catalyst is chemically unchanged at the end
of the reaction, so it does not contribute to the enthalpy change of the system, and does not change the
position of equilibrium. Thus the equilibrium concentrations of NH3(g), H2(g) and N2(g) remained
unchanged with the addition of catalyst, which only changes the time to reach equilibrium. The catalyst
of Haber process is mainly finely divided iron; promotors consisting of Al2O3, K2O, ZnO, and V2O5 have
also been added to this catalyst. They increase the efficiency of the iron by providing more active sites
for this heterogeneous catalyst. The catalysts are easily subjected to poisoning by contaminants CO,
CO2 and H2S in the reactant gases. Thus the purity requirement is particularly high. Catalyst life is of
the order of five years.
Unit 6 - 23
(4)
1.
Partition coefficient
Partition coefficient of a non-volatile solute distributed between two immiscible liquids
When a solute is added to a pair of immiscible liquids, it may dissolve in both liquids. In this case
the solute will distribute itself between the two solvents. It may well be more soluble in one solvent
than the other.
Provided that there is insufficient solute to saturate either of the two solvents and the solute is in the
same molecular state in both solvents, then a dynamic equilibrium exists between the solute in two
immiscible solvents :
solute in solvent 1 – solute in solvent 2
solvent 2
solvent 2
rate of
dissolution
solvent 1
solvent 1
time
It is found that the ratio of the solute concentrations in the two solvents is always the same.
If C1 and C2 are the concentrations of the solute in solvent 1 and solvent 2 respectively, then :
The constant Kd is called the partition coefficient or distribution coefficient. Like other equilibrium
constants it is temperature-dependent.
Example 1 : Determination of partition coefficient of iodine between water and tetrachloromethane
Iodine distributes itself between water and tetrachloromethane. A known mass of iodine (0.9656 g) is
added to 50 cm3 of water and 50 cm3 tetrachloromethane in a separating funnel.
The mixture is shaken well and allowed to stand.
The tetrachloromethane layer is run out of the bottom of the separating funnel and a portion (25 cm3)
of the aqueous layer is titrated with standard thiosulphate solution. 4.40 cm3 of 0.0100 mol dm-3
thiosulphate solution is required for complete reaction.
Find the partition coefficient of iodine between water and CCl4 . (Mr of I2 = 254)
Unit 6 - 24
Example 2 : Determination of the partition coefficient of ethanoic acid between water and
2-methylpropan-1-ol
Procedures
Some (a) 25 cm3 of 0.2 M ethanoic acid was delivered by a measuring cylinder into a 100 cm3
separating funnel. This is followed by about (a) 25 cm3 of 2-methylpropan-1-ol. The funnel is
stoppered and the mixture is shaken well for 1.5 min (b) and is left for equilibration. The room
temperature is also recorded.
After a while about 20 cm3 of the lower aqueous layer was run into a beaker and 10.0 cm3 was
pipetted into a clean conical flask. This solution was titrated with 0.10 M NaOH solution, using
phenolphthalein as indicator.
The junction layer (c) was discarded from the separating funnel, then 10.0 cm3 of the alcohol solution
was pipetted (d) into another clean conical flask. A small volume of distilled water (e) was added into
the alcohol solution and the mixture was again titrated with 0.10 M NaOH using the same indicator. The
mixture was shaken well after each addition of NaOH (e).
Results
Expt.
1
2
3
4
5
Eqm. conc. of acid in
alcohol layer / M
0.050
0.059
0.077
0.100
0.115
Eqm. conc. of acid in
water layer / M
0.152
0.182
0.242
0.300
0.358
Partition coefficient of acid
between water and alcohol
3.04
3.08
3.14
3.00
3.11
Questions
(a) Why is the initial measurement of the volumes of distilled water and alcohol not necessary to be
highly accurate ?
The exact volumes of distilled water and alcohol would not affect the partition coefficient of
ethanoic acid between water and alcohol.
(b) What is the purpose of shaking the mixture ? What precautions must be undertaken during this
process ?
By gently shaking the two liquids together, the area of contact will be increased and the chances of
transferring ethanoic acid molecules will be improved. This ensures the early attainment of the
partition equilibrium. The precautions are :
(i) The alcohol is inflammable. When shaking the mixture all flames in the vicinity must be put
off.
(ii) The alcohol is volatile. During the shaking the vapour pressure of alcohol caused by
absorption of heat from the hands has to be released occasionally by reversing the funnel and
opening the tap.
(iii) Do not shake too hard as a hard shaking can break up the layers into tiny droplets that can take
a very long time to collect back together into separate layers.
(c) What happens to the junction layer ? Why is it necessary to discard it ?
2-methylpropan-1-ol is slightly soluble. At the boundary interface an emulsion is formed, which
contains both alcohol and water, and is unsuitable for any kind if titrimetric analysis.
(d) Why is a small volume of water added in the titration of the alcohol solution and why vigorous
shaking is of paramount importance after each addition of NaOH ?
The acid can be draw n from the alcohol to the water added and be readily neutralized. Vigorous
shaking allows the fast equilibration of the acid between the alcohol and aqueous layer.
(e) Determine on a graph the value of the partition coefficient from the above results.
Unit 6 - 25
2.
Solvent Extraction
At the end of an organic synthesis when the product is in a large volume of aqueous solution, it is
often impractical to remove the water by evaporation or boiling. The organic product may be too
volatile or unstable to heat.
In these cases one common technique is to find a second solvent which is very volatile, immiscible
with water and will easily dissolve the organic substance. By shaking the two solvents together, the
organic product will distribute itself between the two layers according to the distribution law. The
second solvent is then run off using a tap funnel and allowed to evaporate (or distil off) leaving the
organic product.
Ethoxyethane (ether) is a good solvent for many organic compounds and is immiscible with water.
When an aqueous solution of a product is shaken with ethoxyethane in a separating funnel, a large
fraction of product will pass into the ether layer if the partition coefficient is high. The ether layer is
separated from the aqueous layer and dried. With its low boiling point, ethoxyethane is easily distilled
off to leave the product.
Figure : Solvent extraction using separating funnel
It is more efficient to use the ether in portions for repeated extractions than to use it all in one
operation.
Example 1
The product of an organic synthesis, 5.00 g of X, is obtained in solution in 1.00 dm3 of water. The
partition coefficient of X between ethoxyethane and water is 40.0 at room temperature. Calculate the
mass of X that can be extracted from the aqueous solution by
(a) 50.0 cm3 of ethoxyethane, and
(b) two successive portions of 25.0 cm3 of ethoxyethane.
Unit 6 - 26
Example 2
At room temperature, the partition coefficient of an organic compound X between ether and water is
3.10.
(a) Calculate the mass of X extracted into the ether layer when 100 cm3 of water containing 10.0 g of
X is shaken with 100 cm3 of ether.
(b)
Calculate the mass of X which would be removed from the aqueous layer in each of TWO further
extraction using 100 cm3 of ether each time.
(c)
Compare the total mass of X extracted in these three successive extractions with that obtained
from a single extraction using 300 cm3 of ether. Comment on the difference.
Example 3
The distribution coefficient of a compound X between two immiscible liquids D and E is 12.
more soluble in D. What mass of X can be extracted from 50 cm3 of a solution E which initially
contains 4 g of X when shaken with 100 cm3 of D ?
X is
Unit 6 - 27
3. Chromatography
Chromatography is a technique used to separate the components of a mixture. It is used in
qualitative analysis to identify separated components of a mixture and also to determine the purity of a
substance.
Chromatography depends on the partition of the components of the mixture between a stationary
phase and a mobile phase.
In paper chromatography the stationary phase is water adsorbed on paper. The mobile phase is
usually an organic liquid. Paper chromatographic method consists essentially of two steps :
1. Distribution of the components between the two phases.
2. Separation of the components in the stationary phase by a continuous flow of the mobile phase.
The distribution of a component between the two phases is given by the partition coefficient Kd.
Kd
=
A component of a mixture with a high value for Kd remains largely dissolved in the mobile phase
and thus passes over the stationary phase rapidly.
A component with small value for Kd remains largely dissolved in the stationary phase. As the
moving phase passes over it, this component moves slowly along the stationary phase.
There are two types of paper chromatography. In ascending paper chromatography, a paper strip is
suspended from a glass rod in a tank. The strip dips into a small amount of solvent at the bottom of the
tank. In descending paper chromatography, the solvent is contained in a trough at the top of the tank.
The paper chromatogram is dried and developed. The various components of a colour mixture can
be visually identified. If they are colourless, the paper has to be developed. This can be done by
spraying with a reagent, by ultra-violet radiation, by heat or by exposure to a gas such as iodine vapour.
Each component has a characteristic Rf value.
is given by :
Rf =
The Rf value is related to its partition coefficient and