1. CHEMICAL
EQUILIBRIUM

2. CHEMICAL EQUILIBRIUM CONTENTS
Concentration change during a chemical reaction
Dynamic equilibrium
Equilibrium constants
Le Chatelier’s Principle
Haber Process
Equilibrium calculations (ICE)
Effect of temperature on equilibrium constants
Check list

3. CONCENTRATION CHANGE IN A REACTION
As the rate of reaction is dependant on the concentration of reactants... the forward reaction starts off fast but slows as the reactants get less concentrated
FASTEST AT
THE START
THE STEEPER THE GRADIENT, THE FASTER THE REACTION
SLOWS DOWN AS REACTANTS ARE USED UP
TOTAL CONVERSION TO PRODUCTS
In an ordinary reaction; all reactants end up as products; there is 100% conversion

4. EQUILIBRIUM REACTIONS
Initially, there is no backward reaction but, as products form, it speeds up and provided the temperature remains constant there will come a time when the backward and forward reactions are equal and opposite; the reaction has reached equilibrium.
FASTEST AT THE START
NO BACKWARD REACTION
FORWARD REACTION SLOWS DOWN AS REACTANTS ARE USED UP
BACKWARD REACTION
STARTS TO INCREASE
In an equilibrium reaction, not all the reactants end up as products; there is not a 100% conversion.
BUT IT DOESN’T MEAN THE REACTION
IS STUCK IN THE MIDDLE
AT EQUILIBRIUM THE BACKWARD AND FORWARD REACTIONS ARE EQUAL AND OPPOSITE

5. DYNAMIC EQUILIBRIUM IMPORTANT REMINDERS
• a reversible chemical reaction is a dynamic process
• everything may appear stationary but the reactions are moving both ways
• the position of equilibrium can be varied by changing certain conditions
Trying to get up a “down” escalator gives an excellent idea of a non-chemical situation involving dynamic equilibrium.

6. DYNAMIC EQUILIBRIUM IMPORTANT REMINDERS
• a reversible chemical reaction is a dynamic process
• everything may appear stationary but the reactions are moving both ways
• the position of equilibrium can be varied by changing certain conditions
Trying to get up a “down” escalator gives an excellent idea of a non-chemical situation involving dynamic equilibrium.
Summary When a chemical equilibrium is established ...
• both the reactants and the products are present at all times
• the equilibrium can be approached from either side
• the reaction is dynamic - it is moving forwards and backwards
• the concentrations of reactants and products remain constant

7. THE EQUILIBRIUM LAW Simply states
“If the concentrations of all the substances present at equilibrium are raised to the power of the number of moles they appear in the equation, the product of the concentrations of the products divided by the product of the concentrations of the reactants is a constant, provided the temperature remains constant”
There are several forms of the constant; all vary with temperature.
Kc the equilibrium values are expressed as concentrations of mol L-1
Kp the equilibrium values are expressed as partial pressures
The partial pressure expression can be used for reactions involving gases

8. THE EQUILIBRIUM CONSTANT Kc
for an equilibrium reaction of the form...
aA bB cC dD
then (at constant temperature) [C]c . [D]d = a constant, (Kc)
[A]a . [B]b
where [ ] denotes the equilibrium concentration in mol L-1
Kc is known as the Equilibrium Constant and has no units

9. THE EQUILIBRIUM CONSTANT Kc
for an equilibrium reaction of the form...
aA bB cC dD
then (at constant temperature) [C]c . [D]d = a constant, (Kc)
[A]a . [B]b
where [ ] denotes the equilibrium concentration in mol L-1
Kc is known as the Equilibrium Constant and has no units
Example Fe3+(aq) NCS¯(aq) FeNCS2+(aq)
Kc = [ FeNCS2+ ]
[ Fe3+ ] [ NCS¯ ]

10. THE EQUILIBRIUM CONSTANT Kc
for an equilibrium reaction of the form...
aA bB cC dD
then (at constant temperature) [C]c . [D]d = a constant, (Kc)
[A]a . [B]b
where [ ] denotes the equilibrium concentration in mol L-1
Kc is known as the Equilibrium Constant and has no units
VALUE OF Kc
AFFECTED by a change of temperature
NOT AFFECTED by a change in concentration of reactants or products
a change of pressure
adding a catalyst

11. LE CHATELIER’S PRINCIPLE
”When a change is applied to a system in dynamic equilibrium, the
system reacts in such a way as to oppose the effect of the change.”
Everyday example
A rose bush grows with increased vigour after it has been pruned.
Chemistry example
If you do something to a reaction that is in a state of equilibrium, the equilibrium position will change to oppose what you have just done

12. FACTORS AFFECTING THE POSITION OF EQUILIBRIUM
CONCENTRATION
The equilibrium constant is not affected by a change in concentration at constant temperature. To maintain the constant, the composition of the equilibrium mixture changes.
If you increase the concentration of a substance, the value of Kc will theoretically be affected. As it must remain constant at a particular temperature, the concentrations of the other species change to keep the constant the same.

13. FACTORS AFFECTING THE POSITION OF EQUILIBRIUM
CONCENTRATION
example CH3CH2OH(l) + CH3COOH(l) CH3COOC2H5(l) H2O(l)
the equilibrium constant Kc = [CH3COOC2H5] [H2O] = (at 298K)
[CH3CH2OH] [CH3COOH]

14. FACTORS AFFECTING THE POSITION OF EQUILIBRIUM
CONCENTRATION
example CH3CH2OH(l) + CH3COOH(l) CH3COOC2H5(l) H2O(l)
the equilibrium constant Kc = [CH3COOC2H5] [H2O] = (at 298K)
[CH3CH2OH] [CH3COOH]
increasing
[CH3CH2OH] - will make the bottom line larger so Kc will be smaller
- to keep it constant, some CH3CH2OH reacts with CH3COOH
- this reduces the value of the bottom line and increases the top
- eventually the value of the constant will be restored

15. FACTORS AFFECTING THE POSITION OF EQUILIBRIUM
CONCENTRATION
example CH3CH2OH(l) + CH3COOH(l) CH3COOC2H5(l) H2O(l)
the equilibrium constant Kc = [CH3COOC2H5] [H2O] = (at 298K)
[CH3CH2OH] [CH3COOH]
increasing
[CH3CH2OH] - will make the bottom line larger so Kc will be smaller
- to keep it constant, some CH3CH2OH reacts with CH3COOH
- this reduces the value of the bottom line and increases the top
- eventually the value of the constant will be restored
decreasing
[H2O] - will make the top line smaller
- some CH3CH2OH reacts with CH3COOH to replace the H2O
- more CH3COOC2H5 is also produced

16. FACTORS AFFECTING THE POSITION OF EQUILIBRIUM
CONCENTRATION
example CH3CH2OH(l) + CH3COOH(l) CH3COOC2H5(l) H2O(l)
the equilibrium constant Kc = [CH3COOC2H5] [H2O] = (at 298K)
[CH3CH2OH] [CH3COOH]
increasing
[CH3CH2OH] - will make the bottom line larger so Kc will be smaller
- to keep it constant, some CH3CH2OH reacts with CH3COOH
- this reduces the value of the bottom line and increases the top
- eventually the value of the constant will be restored
decreasing
[H2O] - will make the top line smaller
- some CH3CH2OH reacts with CH3COOH to replace the H2O
- more CH3COOC2H5 is also produced

17. FACTORS AFFECTING THE POSITION OF EQUILIBRIUM
SUMMARY
REACTANTS PRODUCTS
INCREASE CONCENTRATION OF A REACTANT
EQUILIBRIUM MOVES TO THE RIGHT
THE EFFECT OF CHANGING THE CONCENTRATION ON THE POSITION OF EQUILIBRIUM
DECREASE CONCENTRATION OF A REACTANT
EQUILIBRIUM MOVES TO THE LEFT
INCREASE CONCENTRATION OF A PRODUCT
DECREASE CONCENTRATION OF A PRODUCT

18. FACTORS AFFECTING THE POSITION OF EQUILIBRIUM
SUMMARY
REACTANTS PRODUCTS
INCREASE CONCENTRATION OF A REACTANT
EQUILIBRIUM MOVES TO THE RIGHT
THE EFFECT OF CHANGING THE CONCENTRATION ON THE POSITION OF EQUILIBRIUM
DECREASE CONCENTRATION OF A REACTANT
EQUILIBRIUM MOVES TO THE LEFT
INCREASE CONCENTRATION OF A PRODUCT
DECREASE CONCENTRATION OF A PRODUCT
Predict the effect of increasing the concentration of O2 on the equilibrium position
2SO2(g) + O2(g) SO3(g)
Predict the effect of decreasing the
concentration of SO3 on the equilibrium position

19. FACTORS AFFECTING THE POSITION OF EQUILIBRIUM
SUMMARY
REACTANTS PRODUCTS
INCREASE CONCENTRATION OF A REACTANT
EQUILIBRIUM MOVES TO THE RIGHT
THE EFFECT OF CHANGING THE CONCENTRATION ON THE POSITION OF EQUILIBRIUM
DECREASE CONCENTRATION OF A REACTANT
EQUILIBRIUM MOVES TO THE LEFT
INCREASE CONCENTRATION OF A PRODUCT
DECREASE CONCENTRATION OF A PRODUCT
Predict the effect of increasing the concentration of O2 on the equilibrium position
2SO2(g) + O2(g) SO3(g) EQUILIBRIUM MOVES TO RHS
Predict the effect of decreasing the
concentration of SO3 on the equilibrium position EQUILIBRIUM MOVES TO RHS

20. FACTORS AFFECTING THE POSITION OF EQUILIBRIUM
PRESSURE
When studying the effect of a change in pressure, we consider the number of gaseous molecules only.
The more particles you have in a given volume, the greater the pressure they exert.
If you apply a greater pressure they will become more crowded (i.e. they are under a greater stress). However, if the system can change it will move to the side with fewer gaseous molecules - it is less crowded.
No change occurs when equal numbers of gaseous molecules appear on both sides.

21. FACTORS AFFECTING THE POSITION OF EQUILIBRIUM
PRESSURE
When studying the effect of a change in pressure, we consider the number of gaseous molecules only.
The more particles you have in a given volume, the greater the pressure they exert.
If you apply a greater pressure they will become more crowded (i.e. they are under a greater stress). However, if the system can change it will move to the side with fewer gaseous molecules - it is less crowded.
No change occurs when equal numbers of gaseous molecules appear on both sides.
INCREASE PRESSURE
MOVES TO THE SIDE WITH FEWER GASEOUS MOLECULES
DECREASE PRESSURE
MOVES TO THE SIDE WITH MORE GASEOUS MOLECULES
THE EFFECT OF PRESSURE ON THE POSITION OF EQUILIBRIUM

22. FACTORS AFFECTING THE POSITION OF EQUILIBRIUM
PRESSURE
When studying the effect of a change in pressure, we consider the number of gaseous molecules only.
The more particles you have in a given volume, the greater the pressure they exert.
If you apply a greater pressure they will become more crowded (i.e. they are under a greater stress). However, if the system can change it will move to the side with fewer gaseous molecules - it is less crowded.
No change occurs when equal numbers of gaseous molecules appear on both sides.
INCREASE PRESSURE
MOVES TO THE SIDE WITH FEWER GASEOUS MOLECULES
DECREASE PRESSURE
MOVES TO THE SIDE WITH MORE GASEOUS MOLECULES
THE EFFECT OF PRESSURE ON THE POSITION OF EQUILIBRIUM
Predict the effect of an increase of pressure on the equilibrium position of..
2SO2(g) + O2(g) SO3(g)
H2(g) + CO2(g) CO(g) + H2O(g)

23. FACTORS AFFECTING THE POSITION OF EQUILIBRIUM
PRESSURE
When studying the effect of a change in pressure, we consider the number of gaseous molecules only.
The more particles you have in a given volume, the greater the pressure they exert.
If you apply a greater pressure they will become more crowded (i.e. they are under a greater stress). However, if the system can change it will move to the side with fewer gaseous molecules - it is less crowded.
No change occurs when equal numbers of gaseous molecules appear on both sides.
INCREASE PRESSURE
MOVES TO THE SIDE WITH FEWER GASEOUS MOLECULES
DECREASE PRESSURE
MOVES TO THE SIDE WITH MORE GASEOUS MOLECULES
THE EFFECT OF PRESSURE ON THE POSITION OF EQUILIBRIUM
Predict the effect of an increase of pressure on the equilibrium position of..
2SO2(g) + O2(g) SO3(g) MOVES TO RHS :- fewer gaseous molecules
H2(g) + CO2(g) CO(g) + H2O(g) NO CHANGE:- equal numbers on both sides

24. FACTORS AFFECTING THE POSITION OF EQUILIBRIUM
TEMPERATURE
• temperature is the only thing that can change the value of the equilibrium constant.
• altering the temperature affects the rate of both backward and forward reactions
• it alters the rates to different extents
• the equilibrium thus moves producing a new equilibrium constant.
• the direction of movement depends on the sign of the enthalpy change.

25. FACTORS AFFECTING THE POSITION OF EQUILIBRIUM
TEMPERATURE
• temperature is the only thing that can change the value of the equilibrium constant.
• altering the temperature affects the rate of both backward and forward reactions
• it alters the rates to different extents
• the equilibrium thus moves producing a new equilibrium constant.
• the direction of movement depends on the sign of the enthalpy change.
REACTION TYPE DH
INCREASE TEMP
DECREASE TEMP
EXOTHERMIC -
TO THE LEFT
TO THE RIGHT
ENDOTHERMIC +

26. FACTORS AFFECTING THE POSITION OF EQUILIBRIUM
TEMPERATURE
• temperature is the only thing that can change the value of the equilibrium constant.
• altering the temperature affects the rate of both backward and forward reactions
• it alters the rates to different extents
• the equilibrium thus moves producing a new equilibrium constant.
• the direction of movement depends on the sign of the enthalpy change.
REACTION TYPE DH
INCREASE TEMP
DECREASE TEMP
EXOTHERMIC -
TO THE LEFT
TO THE RIGHT
ENDOTHERMIC +
Predict the effect of a temperature increase on the equilibrium position of...
H2(g) + CO2(g) CO(g) + H2O(g) DH = kJ mol-1
2SO2(g) + O2(g) SO3(g) DH = - ve

27. FACTORS AFFECTING THE POSITION OF EQUILIBRIUM
TEMPERATURE
• temperature is the only thing that can change the value of the equilibrium constant.
• altering the temperature affects the rate of both backward and forward reactions
• it alters the rates to different extents
• the equilibrium thus moves producing a new equilibrium constant.
• the direction of movement depends on the sign of the enthalpy change.
REACTION TYPE DH
INCREASE TEMP
DECREASE TEMP
EXOTHERMIC -
TO THE LEFT
TO THE RIGHT
ENDOTHERMIC +
Predict the effect of a temperature increase on the equilibrium position of...
H2(g) + CO2(g) CO(g) + H2O(g) DH = kJ mol-1 moves to the RHS
2SO2(g) + O2(g) SO3(g) DH = - ve moves to the LHS

28. FACTORS AFFECTING THE POSITION OF EQUILIBRIUM
CATALYSTS
Catalysts work by providing an alternative reaction pathway involving a lower activation energy. Ea
MAXWELL-BOLTZMANN DISTRIBUTION OF MOLECULAR ENERGY
EXTRA MOLECULES WITH SUFFICIENT ENERGY TO OVERCOME THE ENERGY BARRIER
MOLECULAR ENERGY
NUMBER OF MOLECUES WITH
A PARTICULAR ENERGY

29. FACTORS AFFECTING THE POSITION OF EQUILIBRIUM
CATALYSTS
An increase in temperature is used to speed up chemical reactions but it can have an undesired effect when the reaction is reversible and exothermic.
In this case you get to the equilibrium position quicker but with a reduced yield because the increased temperature moves the equilibrium to the left.
In many industrial processes a compromise temperature is used (see Haber and Contact Processes). To reduce the problem one must look for a way of increasing the rate of a reaction without decreasing the yield i.e. with a catalyst.

30. FACTORS AFFECTING THE POSITION OF EQUILIBRIUM
CATALYSTS
An increase in temperature is used to speed up chemical reactions but it can have an undesired effect when the reaction is reversible and exothermic.
In this case you get to the equilibrium position quicker but with a reduced yield because the increased temperature moves the equilibrium to the left.
In many industrial processes a compromise temperature is used (see Haber and Contact Processes). To reduce the problem one must look for a way of increasing the rate of a reaction without decreasing the yield i.e. with a catalyst.
Adding a catalyst DOES NOT AFFECT THE POSITION OF EQUILIBRIUM. However, it does increase the rate of attainment of equilibrium. This is especially important in reversible, exothermic industrial reactions such as the Haber or Contact Processes where economic factors are paramount.

31. HABER PROCESS N2(g) + 3H2(g) 2NH3(g) : DH = - 92 kJ mol-1
Conditions Pressure kPa (200 atmospheres)
Temperature °C
Catalyst iron

32. HABER PROCESS N2(g) + 3H2(g) 2NH3(g) : DH = - 92 kJ mol-1
Conditions Pressure kPa (200 atmospheres)
Temperature °C
Catalyst iron
Equilibrium theory favours
low temperature exothermic reaction - higher yield at lower temperature
high pressure decrease in number of gaseous molecules

33. HABER PROCESS N2(g) + 3H2(g) 2NH3(g) : DH = - 92 kJ mol-1
Conditions Pressure kPa (200 atmospheres)
Temperature °C
Catalyst iron
Equilibrium theory favours
low temperature exothermic reaction - higher yield at lower temperature
high pressure decrease in number of gaseous molecules
Kinetic theory favours
high temperature greater average energy + more frequent collisions
high pressure more frequent collisions for gaseous molecules
catalyst lower activation energy

34. HABER PROCESS N2(g) + 3H2(g) 2NH3(g) : DH = - 92 kJ mol-1
Conditions Pressure kPa (200 atmospheres)
Temperature °C
Catalyst iron
Equilibrium theory favours
low temperature exothermic reaction - higher yield at lower temperature
high pressure decrease in number of gaseous molecules
Kinetic theory favours
high temperature greater average energy + more frequent collisions
high pressure more frequent collisions for gaseous molecules
catalyst lower activation energy
Compromise conditions
Which is better? A low yield in a shorter time or
a high yield over a longer period.
The conditions used are a compromise with the catalyst
enabling the rate to be kept up, even at a lower temperature.

35. IMPORTANT USES OF AMMONIA AND ITS COMPOUNDS
HABER PROCESS
IMPORTANT USES OF AMMONIA AND ITS COMPOUNDS
MAKING
FERTILISERS 80% of the ammonia produced goes to make fertilisers such as
ammonium nitrate (NITRAM) and ammonium sulphate
NH HNO3 ——> NH4NO3
2NH H2SO4 ——> (NH4)2SO4
NITRIC ACID ammonia can be oxidised to nitric acid
nitric acid is used to manufacture... fertilisers (ammonium nitrate)
explosives (TNT)
polyamide polymers (NYLON)

36. CALCULATIONS INVOLVING Kc
construct the balanced equation, including state symbols (aq), (g) etc.
determine the number of moles of each species at equilibrium
divide moles by volume (in L) to get the equilibrium concentrations in mol L-1
(If no volume is quoted, use a V; it will probably cancel out)
from the equation constructed in the first step, write out an expression for Kc.
substitute values from third step and calculate the value of Kc with any units
Example 1
One mole of ethanoic acid reacts with one mole of ethanol at 298K. When equilibrium is reached it is found that two thirds of the acid has reacted. Calculate the value of Kc.

37. CALCULATIONS INVOLVING Kc
construct the balanced equation, including state symbols (aq), (g) etc.
determine the number of moles of each species at equilibrium
divide moles by volume (in L) to get the equilibrium concentrations in mol L1
(If no volume is quoted, use a V; it will probably cancel out)
from the equation constructed in the first step, write out an expression for Kc.
substitute values from third step and calculate the value of Kc with any units
Example 1
One mole of ethanoic acid reacts with one mole of ethanol at 298K. When equilibrium is reached it is found that two thirds of the acid has reacted. Calculate the value of Kc.
CH3COOH(l) C2H5OH(l) CH3COOC2H5(l) H2O(l)

38. CALCULATIONS INVOLVING Kc
construct the balanced equation, including state symbols (aq), (g) etc.
determine the number of moles of each species at equilibrium
divide moles by volume (in L) to get the equilibrium concentrations in mol L-1
(If no volume is quoted, use a V; it will probably cancel out)
from the equation constructed in the first step, write out an expression for Kc.
substitute values from third step and calculate the value of Kc with any units
Example 1
One mole of ethanoic acid reacts with one mole of ethanol at 298K. When equilibrium is reached it is found that two thirds of the acid has reacted. Calculate the value of Kc.
CH3COOH(l) C2H5OH(l) CH3COOC2H5(l) H2O(l)
moles (Initially)
moles (Change) / / / /3
moles (at Equilibrium) / / / /3

39. CALCULATIONS INVOLVING Kc
construct the balanced equation, including state symbols (aq), (g) etc.
determine the number of moles of each species at equilibrium
divide moles by volume (in L) to get the equilibrium concentrations in mol L-1
(If no volume is quoted, use a V; it will probably cancel out)
from the equation constructed in the first step, write out an expression for Kc.
substitute values from third step and calculate the value of Kc with any units
Example 1
One mole of ethanoic acid reacts with one mole of ethanol at 298K. When equilibrium is reached it is found that two thirds of the acid has reacted. Calculate the value of Kc.
CH3COOH(l) C2H5OH(l) CH3COOC2H5(l) H2O(l)
moles (initially)
moles (Change) / / / /3
moles (at Equilibrium) / / / /3
equilibrium concs /3 / V /3 / V /3 / V /3 / V
V = volume (L) of the equilibrium mixture

40. CALCULATIONS INVOLVING Kc
construct the balanced equation, including state symbols (aq), (g) etc.
determine the number of moles of each species at equilibrium
divide moles by volume (in L) to get the equilibrium concentrations in mol L1
(If no volume is quoted, use a V; it will probably cancel out)
from the equation constructed in the first step, write out an expression for Kc.
substitute values from third step and calculate the value of Kc with any units
Example 1
One mole of ethanoic acid reacts with one mole of ethanol at 298K. When equilibrium is reached it is found that two thirds of the acid has reacted. Calculate the value of Kc.
CH3COOH(l) C2H5OH(l) CH3COOC2H5(l) H2O(l)
moles (Initially)
moles (Change) / / / /3
moles (at Equilibrium) / / / /3
equilibrium concs /3 / V /3 / V /3 / V /3 / V
V = volume (L) of the equilibrium mixture
Kc = [CH3COOC2H5] [H2O]
[CH3COOH] [C2H5OH]

41. CALCULATIONS INVOLVING Kc
construct the balanced equation, including state symbols (aq), (g) etc.
determine the number of moles of each species at equilibrium
divide moles by volume (in dm3) to get the equilibrium concentrations in mol dm-3
(If no volume is quoted, use a V; it will probably cancel out)
from the equation constructed in the first step, write out an expression for Kc.
substitute values from third step and calculate the value of Kc with any units
Example 1
One mole of ethanoic acid reacts with one mole of ethanol at 298K. When equilibrium is reached it is found that two thirds of the acid has reacted. Calculate the value of Kc.
CH3COOH(l) C2H5OH(l) CH3COOC2H5(l) H2O(l)
moles (initially)
moles (change) / / / /3
moles (at Equilibrium) / / / /3
equilibrium concs /3 / V /3 / V /3 / V /3 / V
V = volume (dm3) of the equilibrium mixture
Kc = [CH3COOC2H5] [H2O] = 2/3 / V . 2/3 / V = 4
[CH3COOH] [C2H5OH] 1/3 / V . 1/3 / V

42. CALCULATIONS INVOLVING Kc
Example 2
Consider the equilibrium P Q R S (all species are aqueous)
One mole of P and one mole of Q are mixed. Once equilibrium has been achieved 0.6 moles of P are present. How many moles of Q, R and S are present at equilibrium ?
P Q R S
Initial moles
Change (2 x 0.4)
at Equilibrium 0· ·2 0· ·4
Explanation • if 0.6 mol of P remain of the original 1 mol, 0.4 mol have reacted
• the equation states that 2 moles of Q react with every 1 mol of P
• this means that 0.8 (2 x 0.4) mol of Q have reacted, leaving 0.2 mol
• one mol of R and S are produced from every mol of P that reacts
• this means 0.4 mol of R and 0.4 mol of S are present at equilibrium

43. EFFECT OF TEMPERATURE ON Kc
If temperature changes the magnitude of Kc changes as well.
Le Chatelier’s Principle predicts that an increase in temperature favours an endothermic process i.e. a process that absorbs heat tends to minimise a temperature increase.
If the forward reaction is endothermic and increase in temperature will results in a larger value of Kc.
Example: N2O4 (g)  2 NO2(g) ΔH = kJ mol-1
Kc = 5 x 10-4 at 0oC and 0.36 at 100oC.
The concentration of NO2 relative to that of N2O4 increases as the temperature increases.
If the forward reaction if is exothermic, an increase in temperature leads to a decrease in the value of Kc.
Example: N2(g) + 3 H2(g)  2 NH3(g) ΔH = kJ mol-1
Kc = 5 x 108 at room temperature and 0.5 at 400oC
The concentration of NH3 decreases relative to that of N2 and H2 as the temperature increases.

44. What should you be able to do?
REVISION CHECK
What should you be able to do?
Understand how concentration changes during chemical reactions
Recall the the characteristics of dynamic equilibrium
Construct an expression for the equilibrium constant Kc
Recall the factors which can affect the position of equilibrium
Apply Le Chatelier’s Principle to predict changes in the position of equilibrium
Recall and explain the conditions used in the Haber process
Recall the importance of ammonia and its compounds
Recall the definitions of and differences between weak acids and bases
Recall the properties of hydrochloric acid and its role in salt formation
CAN YOU DO ALL OF THESE? YES NO

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47. © 2003 JONATHAN HOPTON & KNOCKHARDY PUBLISHING
AN INTRODUCTION TO
CHEMICAL
EQUILIBRIUM
THE END
© 2003 JONATHAN HOPTON & KNOCKHARDY PUBLISHING