rate equation chemistry for high school student

15Kin2pp.ppt

1. THE RATE
EQUATION
A guide for A level students
KNOCKHARDY PUBLISHING
2015
SPECIFICATIONS

2. INTRODUCTION
This Powerpointshow is one of several produced to help students understand
selected topics at AS and A2 level Chemistry. It is based on the requirements of
the AQA and OCR specifications but is suitable for other examination boards.
Individual students may use the material at home for revision purposes or it
may be used for classroom teaching if an interactive white board is available.
Accompanying notes on this, and the full range of AS and A2 topics, are
available from the KNOCKHARDY SCIENCE WEBSITE at...
www.knockhardy.org.uk/sci.htm
Navigationis achieved by...
eitherclicking on the grey arrows at the foot of each page
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THE RATE EQUATION
KNOCKHARDY PUBLISHING

3. THE RATE EQUATION
CONTENTS
•Collision theory
•Methods for increasing rate
•Rate changes during a reaction
•The rate equation
•Worked examples
•Graphical methods for determining rate
•Half-life
•Rate determining step

4. Before you start it would be helpful to…
•know how the energy changes during a chemical reaction
•know the basic ideas of Kinetic Theory
•know the importance of catalysts in industrial chemistry
THE RATE EQUATION

5. COLLISION THEORY
Collision theory states that...
•particles must COLLIDEbefore a reaction can take place
•not all collisions lead to a reaction
•reactants must possess at least a minimum amount of energy -ACTIVATION ENERGY
plus
•particles must approach each other in a certain relative way -the STERIC EFFECT
REVISION

6. COLLISION THEORY
Collision theory states that...
•particles must COLLIDEbefore a reaction can take place
•not all collisions lead to a reaction
•reactants must possess at least a minimum amount of energy -ACTIVATION ENERGY
plus
•particles must approach each other in a certain relative way -the STERIC EFFECT
According to collision theory, to increase the rate of reaction you therefore need...
more frequent collisionsincrease particle speed or
have more particles present
more successful collisionsgive particles more energyor
lower the activation energy
REVISION

7. INCREASING THE RATE
•INCREASE THE SURFACE AREA OF SOLIDS
•INCREASE TEMPERATURE
•SHINE LIGHT
•ADD A CATALYST
•INCREASE THE PRESSURE OF ANY GASES
•INCREASE THE CONCENTRATION OF REACTANTS
The following methods may be used to
increase the rate of a chemical reaction
REVISION

8. Reactions are fastest at the start and get slower as the reactants concentration drops.
In a reaction such asA + 2B ——> Cthe concentrations might change as shown
RATE CHANGE DURING A REACTION
Reactants(AandB)
Concentration decreases with time
Product(C)
Concentration increases with time
•the steeper the curve the faster the
rate of the reaction
•reactions start off quickly because of
the greater likelihood of collisions
•reactions slow down with time as
there are fewer reactants to collide
TIME
CONCENTRATION
B
A
C
REVISION

9. Experimental Investigation
•the variation in concentration of a reactant or product is followed with time
•the method depends on the reaction type and the properties of reactants/products
e.g.Extracting a sample from the reaction mixture and analysing it by titration.
-this is often used if an acid is one of the reactants or products
Using a colorimeter or UV / visible spectrophotometer.
Measuring the volume of gas evolved.
Measuring the change in conductivity.
More details of these and other methods can be found in suitable text-books.
MEASURING THE RATE

10. RATE How much concentration changes with time. It is the equivalent of velocity.
MEASURING THE RATE
y
CONCENTRATION
gradient = y
x
x
TIME
•the rate of change of concentration is found from the slope (gradient) of the curve
•the slope at the start of the reaction will give the INITIAL RATE
•the slope gets less (showing the rate is slowing down) as the reaction proceeds
THE SLOPE OF THE GRADIENT OF THE
CURVE GETS LESS AS THE
REACTION SLOWS DOWN
WITH TIME

11. THE RATE EQUATION
Formatlinks the rate of reaction to the concentration of reactants
it can only be found by doing actual experiments
it cannot be found by just looking at the equation
the equation... A + B ——> C + D
mighthave a rate equation like thisr = k [A] [B]
2
rrate of reactionunits of conc. / time usually mol dm
-3
s
-1
krate constant units depend on the rate equation
[ ] concentration units of mol dm
-3
Interpretation
The above rate equation tells you that the rate of reaction is...
proportional to the concentration of reactant A doubling [A] doubles rate
proportional to the square of the concentration of B doubling [B] quadruples (2
2
) rate

12. ORDER OF REACTION
Individual orderThe power to which a concentration is raised in the rate equation
Overallorder The sum of all the individual orders in the rate equation.
Order tells you how much the concentration of a reactant affects the rate

13. ORDER OF REACTION
Individual orderThe power to which a concentration is raised in the rate equation
Overallorder The sum of all the individual orders in the rate equation.
e.g.in the rate equation r = k [A] [B]
2
the order with respect to A is 1 1st Order
the order with respect to B is 2 2nd Order
andthe overall order is 3 3rd Order
Value(s)need not be whole numbers
can be zero if the rate is unaffected by how much of a substance is present
Order tells you how much the concentration of a reactant affects the rate

14. ORDER OF REACTION
Individual orderThe power to which a concentration is raised in the rate equation
Overallorder The sum of all the individual orders in the rate equation.
e.g.in the rate equation r = k [A] [B]
2
the order with respect to A is 1 1st Order
the order with respect to B is 2 2nd Order
andthe overall order is 3 3rd Order
Value(s)need not be whole numbers
can be zero if the rate is unaffected by how much of a substance is present
NOTES
The rate equation is derived from experimental evidence not by looking at an equation.
Species appearing in the stoichiometric equation sometimes aren’t in the rate equation.
Substances not in the stoichiometric equation can appear in the rate equation -CATALYSTS
Order tells you how much the concentration of a reactant affects the rate

15. THE RATE EQUATION
Experimental determination of order
Method 1
Plot a concentration / time graph and calculate the rate (gradient) at points on the curve
Plot another graph of the rate (y axis) versus the concentration at that point (x axis)
If it gives a straight line, the rate is directly proportional to concentration -1st ORDER.
If the plot is a curve then it must have another order. Try plotting rate v. (conc.)
2
.
A straight line would mean 2nd ORDER. This method is based on trial and error.

16. THE RATE EQUATION
Experimental determination of order
Method 1
Plot a concentration / time graph and calculate the rate (gradient) at points on the curve
Plot another graph of the rate (y axis) versus the concentration at that point (x axis)
If it gives a straight line, the rate is directly proportional to concentration -1st ORDER.
If the plot is a curve then it must have another order. Try plotting rate v. (conc.)
2
.
A straight line would mean 2nd ORDER. This method is based on trial and error.
Method 2-The initial rates method.
Do a series of experiments (at the same temperature) at different concentrations of a
reactant but keeping all others constant. Plot a series of concentration / time graphs
and calculate the initial rate (slope of curve at start) for each reaction. From the results
calculate the relationship between concentration and rate and hence deduce the rate
equation. To find order directly, logarithmic plots are required.

17. THE RATE CONSTANT (k)
Units The units of k depend on the overall order of reaction.
e.g.if the rate equation is... rate = k [A]
2
the units of k will be dm
3
mol
-1
sec
-1
Divide the rate by as many concentrations as appear in the rate equation.
Overall Order 0 1 2 3
units of k mol dm
-3
sec
-1
sec
-1
dm
3
mol
-1
sec
-1
dm
6
mol
-2
sec
-1
examplein the rate equation r = k [A] k will have units of sec
-1
in the rate equation r = k [A] [B]
2
k will have units of dm
6
mol
-2
sec
-1

18. RATE EQUATION -SAMPLE CALCULATION
In an experiment between A and B the
initial rate of reaction was found for
various starting concentrations of A and B.
Calculate...
•the individual orders for A and B
•the overall order of reaction
•the rate equation
•the value of the rate constant (k)
•the units of the rate constant
0.51 2
1.51 6
0.52 8
1
2
3
[A][B]
Initial
rate (r)
r initial rate of reaction mol dm
-3
s
-1
[ ] concentration mol dm
-3

19. Compare Experiments 1 & 2
[B] same
[A] 3 x bigger
rate 3 x bigger rate [A]
FIRST ORDER with respect to (wrt) A
0.51 2
1.51 6
0.52 8
1
2
3
[A][B]
Initial
rate (r)
CALCULATING ORDER wrt A
Choose any two experiments where...
[A] is changedand, importantly,
[B] is KEPT THE SAME
See how the change in [A] affects the rate
As you can see, tripling [A] has exactly
the same effect on the rate so...
THE ORDER WITH RESPECT TO A = 1
(it is FIRST ORDER)
RATE EQUATION -SAMPLE CALCULATION

20. 0.51 2
1.51 6
0.52 8
1
2
3
[A][B]
Initial
rate (r)
CALCULATING ORDER wrt B
Choose any two experiments where...
[B] is changedand, importantly,
[A] is KEPT THE SAME
See how a change in [B] affects the rate
As you can see, doubling [B] quadruples
the rateso...
THE ORDER WITH RESPECT TO B = 2
It is SECOND ORDER
RATE EQUATION -SAMPLE CALCULATION
Compare Experiments 1 & 3
[A] same
[B] 2 x bigger
rate 4 x bigger rate [B]
2
SECOND ORDER wrt B

21. Compare Experiments 1 & 2
[B] same
[A] 3 x bigger
rate 3 x bigger rate [A]
FIRST ORDER with respect to (wrt) A
0.51 2
1.51 6
0.52 8
1
2
3
[A][B]
Initial
rate (r)
0.51 2
1.51 6
0.52 8
1
2
3
[A][B]
Initial
rate (r)
Compare Experiments 1 & 3
[A] same
[B] 2 x bigger
rate 4 x bigger rate [B]
2
SECOND ORDER wrt B
OVERALL ORDER = THE SUM OF THE INDIVIDUAL ORDERS
= 1 + 2
= 3
RATE EQUATION -SAMPLE CALCULATION

22. 0.51 2
1.51 6
0.52 8
1
2
3
[A][B]
Initial
rate (r)
0.51 2
1.51 6
0.52 8
1
2
3
[A][B]
Initial
rate (r)
rate = k [A] [B]
2
RATE EQUATION -SAMPLE CALCULATION
By combining the two proportionality relationships
you can construct the overall rate equation
Compare Experiments 1 & 3
[A] same
[B] 2 x bigger
rate 4 x bigger rate [B]
2
SECOND ORDER wrt B
Compare Experiments 1 & 2
[B] same
[A] 3 x bigger
rate 3 x bigger rate [A]
FIRST ORDER with respect to (wrt) A

23. 0.51 2
1.51 6
0.52 8
1
2
3
[A][B]
Initial
rate (r)
0.51 2
1.51 6
0.52 8
1
2
3
[A][B]
Initial
rate (r)
Chose one experiment (e.g. Expt. 3) and
substitute its values into the rate equation
k = 8= 4 dm
6
mol
-2
sec
-1
(0.5) (2)
2
rate = k [A] [B]
2
re-arranging k = rate
[A] [B]
2
RATE EQUATION -SAMPLE CALCULATION
Compare Experiments 1 & 3
[A] same
[B] 2 x bigger
rate 4 x bigger rate [B]
2
SECOND ORDER wrt B
Compare Experiments 1 & 2
[B] same
[A] 3 x bigger
rate 3 x bigger rate [A]
FIRST ORDER with respect to (wrt) A

24. 0.51 2
1.51 6
0.52 8
1
2
3
[A][B]
Initial
rate (r)
0.51 2
1.51 6
0.52 8
1
2
3
[A][B]
Initial
rate (r)
SUMMARYRATE EQUATION -SAMPLE CALCULATION
Compare Experiments 1 & 3
[A] same
[B] 2 x bigger
rate 4 x bigger rate [B]
2
SECOND ORDER wrt B
Compare Experiments 1 & 2
[B] same
[A] 3 x bigger
rate 3 x bigger rate [A]
FIRST ORDER with respect to (wrt) A
Chose one experiment (e.g. Expt. 3) and
substitute its values into the rate equation
k = 8= 4 dm
6
mol
-2
sec
-1
(0.5) (2)
2
rate = k [A] [B]
2
re-arranging k = rate
[A] [B]
2

25. RATE EQUATION QUESTIONS
CALCULATE THE ORDER WITH RESPECT TO A
THE ORDER WITH RESPECT TO B
THE OVERALL ORDER OF REACTION
THE FORMAT OF THE RATE EQUATION
THE VALUE AND UNITS OF THE RATE CONSTANT
ANSWER ON NEXT PAGE
[A] / mol dm
-3
[B] / mol dm
-3
Rate / mol dm
-3
s
-1
Expt 1 0.25 0.25 4
Expt 2 0.25 0.50 8
Expt 3 0.50 0.25 8
No 1

26. RATE EQUATION QUESTIONS
Expts 1&2 [A] is constant[B] is doubled Rate is doubled
Therefore rate [B] 1st order wrt B
Explanation: What was done to [B] had exactly the same effect on the rate
[A] / mol dm
-3
[B] / mol dm
-3
Rate / mol dm
-3
s
-1
Expt 1 0.25 0.25 4
Expt 2 0.25 0.50 8
Expt 3 0.50 0.25 8
No 1
ANSWER

27. RATE EQUATION QUESTIONS
Expts 1&2 [A] is constant[B] is doubled Rate is doubled
Therefore rate [B] 1st order wrt B
Explanation: What was done to [B] had exactly the same effect on the rate
Expts 1&3 [B] is constant[A] is doubled Rate is doubled
Therefore rate [A] 1st order wrt A
Explanation: What was done to [A] had exactly the same effect on the rate
[A] / mol dm
-3
[B] / mol dm
-3
Rate / mol dm
-3
s
-1
Expt 1 0.25 0.25 4
Expt 2 0.25 0.50 8
Expt 3 0.50 0.25 8
No 1
ANSWER

28. RATE EQUATION QUESTIONS
Expts 1&2 [A] is constant[B] is doubled Rate is doubled
Therefore rate [B] 1st order wrt B
Explanation: What was done to [B] had exactly the same effect on the rate
Expts 1&3 [B] is constant[A] is doubled Rate is doubled
Therefore rate [A] 1st order wrt A
Explanation: What was done to [A] had exactly the same effect on the rate
Rate equation isr = k[A][B]
[A] / mol dm
-3
[B] / mol dm
-3
Rate / mol dm
-3
s
-1
Expt 1 0.25 0.25 4
Expt 2 0.25 0.50 8
Expt 3 0.50 0.25 8
No 1
ANSWER

29. RATE EQUATION QUESTIONS
Expts 1&2 [A] is constant[B] is doubled Rate is doubled
Therefore rate [B] 1st order wrt B
Explanation: What was done to [B] had exactly the same effect on the rate
Expts 1&3 [B] is constant[A] is doubled Rate is doubled
Therefore rate [A] 1st order wrt A
Explanation: What was done to [A] had exactly the same effect on the rate
Rate equation isr = k[A][B]
Value of k Substitute numbers from Exp 1 to get value of k
k = rate / [A][B] = 4 / 0.25 x 0.25= 64
[A] / mol dm
-3
[B] / mol dm
-3
Rate / mol dm
-3
s
-1
Expt 1 0.25 0.25 4
Expt 2 0.25 0.50 8
Expt 3 0.50 0.25 8
No 1
ANSWER

30. RATE EQUATION QUESTIONS
Expts 1&2 [A] is constant[B] is doubled Rate is doubled
Therefore rate [B] 1st order wrt B
Explanation: What was done to [B] had exactly the same effect on the rate
Expts 1&3 [B] is constant[A] is doubled Rate is doubled
Therefore rate [A] 1st order wrt A
Explanation: What was done to [A] had exactly the same effect on the rate
Rate equation isr = k[A][B]
Value of k Substitute numbers from Exp 1 to get value of k
k = rate / [A][B] = 4 / 0.25 x 0.25= 64
Units of k rate / conc x conc= dm
3
mol
-1
s
-1
[A] / mol dm
-3
[B] / mol dm
-3
Rate / mol dm
-3
s
-1
Expt 1 0.25 0.25 4
Expt 2 0.25 0.50 8
Expt 3 0.50 0.25 8
No 1
ANSWER

31. RATE EQUATION QUESTIONS
CALCULATE THE ORDER WITH RESPECT TO C
THE ORDER WITH RESPECT TO D
THE OVERALL ORDER OF REACTION
THE FORMAT OF THE RATE EQUATION
THE VALUE AND UNITS OF THE RATE CONSTANT
ANSWER ON NEXT PAGE
[C] / mol dm
-3
[D] / mol dm
-3
Rate / mol dm
-3
s
-1
Expt 1 0.40 0.40 0.16
Expt 2 0.20 0.40 0.04
Expt 3 0.40 1.20 1.44
No 2

32. RATE EQUATION QUESTIONS
[C] / mol dm
-3
[D] / mol dm
-3
Rate / mol dm
-3
s
-1
Expt 1 0.40 0.40 0.16
Expt 2 0.20 0.40 0.04
Expt 3 0.40 1.20 1.44
No 2
Expts 1&3 [C] is constant [D] is tripled Rate is 9 x bigger
Therefore rate [D]
2
2nd order wrt D
Explanation: Squaring what was done to D affected the rate (3
2
= 9)
ANSWER

33. RATE EQUATION QUESTIONS
[C] / mol dm
-3
[D] / mol dm
-3
Rate / mol dm
-3
s
-1
Expt 1 0.40 0.40 0.16
Expt 2 0.20 0.40 0.04
Expt 3 0.40 1.20 1.44
No 2
Expts 1&3 [C] is constant [D] is tripled Rate is 9 x bigger
Therefore rate [D]
2
2nd order wrt D
Explanation: Squaring what was done to D affected the rate (3
2
= 9)
Expts 1&2 [D] is constant [A] is halved Rate is quartered
Therefore rate [C]
2
2nd order wrt C
Explanation: One half squared = one quarter
ANSWER

34. RATE EQUATION QUESTIONS
[C] / mol dm
-3
[D] / mol dm
-3
Rate / mol dm
-3
s
-1
Expt 1 0.40 0.40 0.16
Expt 2 0.20 0.40 0.04
Expt 3 0.40 1.20 1.44
No 2
Expts 1&3 [C] is constant [D] is tripled Rate is 9 x bigger
Therefore rate [D]
2
2nd order wrt D
Explanation: Squaring what was done to D affected the rate (3
2
= 9)
Expts 1&2 [D] is constant [A] is halved Rate is quartered
Therefore rate [C]
2
2nd order wrt C
Explanation: One half squared = one quarter
Rate equation isr = k[C]
2
[D]
2
ANSWER

35. RATE EQUATION QUESTIONS
[C] / mol dm
-3
[D] / mol dm
-3
Rate / mol dm
-3
s
-1
Expt 1 0.40 0.40 0.16
Expt 2 0.20 0.40 0.04
Expt 3 0.40 1.20 1.44
No 2
Expts 1&3 [C] is constant [D] is tripled Rate is 9 x bigger
Therefore rate [D]
2
2nd order wrt D
Explanation: Squaring what was done to D affected the rate (3
2
= 9)
Expts 1&2 [D] is constant [A] is halved Rate is quartered
Therefore rate [C]
2
2nd order wrt C
Explanation: One half squared = one quarter
Rate equation isr = k[C]
2
[D]
2
Value of k Substitute numbers from Exp 2 to get value of k
k = rate / [C]
2
[D]
2
= 0.04 / 0.2
2
x 0.4
2
= 6.25
Units of k rate / conc
2
x conc
2
= dm
9
mol
-3
s
-1
ANSWER

36. RATE EQUATION QUESTIONS
CALCULATE THE ORDER WITH RESPECT TO E
THE ORDER WITH RESPECT TO F
THE OVERALL ORDER OF REACTION
THE FORMAT OF THE RATE EQUATION
THE VALUE AND UNITS OF THE RATE CONSTANT
No 3
ANSWER ON NEXT PAGE
[E] / mol dm
-3
[F] / mol dm
-3
Rate / mol dm
-3
s
-1
Expt 1 0.40 0.40 0.16
Expt 2 0.80 0.80 0.32
Expt 3 0.80 1.20 0.32

37. RATE EQUATION QUESTIONS
[E] / mol dm
-3
[F] / mol dm
-3
Rate / mol dm
-3
s
-1
Expt 1 0.40 0.40 0.16
Expt 2 0.80 0.80 0.32
Expt 3 0.80 1.20 0.32
No 3
Expts 2&3 [E] is constant [F] is x 1.5 Rate unchanged
Rate is UNAFFECTED ZERO order wrt F
Explanation: Concentration of [F] has no effect on the rate
ANSWER

38. RATE EQUATION QUESTIONS
[E] / mol dm
-3
[F] / mol dm
-3
Rate / mol dm
-3
s
-1
Expt 1 0.40 0.40 0.16
Expt 2 0.80 0.80 0.32
Expt 3 0.80 1.20 0.32
No 3
Expts 2&3 [E] is constant [F] is x 1.5 Rate unchanged
Rate is UNAFFECTED ZERO order wrt F
Explanation: Concentration of [F] has no effect on the rate
Expts 1&2 [E] is doubled [F] is doubled Rate is doubled
Therefore rate [E]
2
2nd order wrt E
Explanation: Although both concentrations have been doubled, we know [F]
has no effect. The change must be all due to [E]
ANSWER

39. RATE EQUATION QUESTIONS
[E] / mol dm
-3
[F] / mol dm
-3
Rate / mol dm
-3
s
-1
Expt 1 0.40 0.40 0.16
Expt 2 0.80 0.80 0.32
Expt 3 0.80 1.20 0.32
No 3
Expts 2&3 [E] is constant [F] is x 1.5 Rate unchanged
Rate is UNAFFECTED ZERO order wrt F
Explanation: Concentration of [F] has no effect on the rate
Expts 1&2 [E] is doubled [F] is doubled Rate is doubled
Therefore rate [E]
2
2nd order wrt E
Explanation: Although both concentrations have been doubled, we know [F]
has no effect. The change must be all due to [E]
Rate equation isr = k[E]
ANSWER

40. RATE EQUATION QUESTIONS
[E] / mol dm
-3
[F] / mol dm
-3
Rate / mol dm
-3
s
-1
Expt 1 0.40 0.40 0.16
Expt 2 0.80 0.80 0.32
Expt 3 0.80 1.20 0.32
No 3
Expts 2&3 [E] is constant [F] is x 1.5 Rate unchanged
Rate is UNAFFECTED ZERO order wrt F
Explanation: Concentration of [F] has no effect on the rate
Expts 1&2 [E] is doubled [F] is doubled Rate is doubled
Therefore rate [E]
2
2nd order wrt E
Explanation: Although both concentrations have been doubled, we know [F]
has no effect. The change must be all due to [E]
Rate equation isr = k[E]
Value of k Substitute numbers from Exp 1 to get value of k
k = rate / [E] = 0.16 / 0.4= 0.40
Units of k rate / conc = s
-1
ANSWER

41. GRAPHICAL DETERMINATION OF RATE
RATE CALCULATION
The rate of reaction at
any moment can be
found from the gradient
of the tangentat that
point. The steeper the
gradient, the faster the
rate of reaction
Place a rule on the
outside of the curve and
draw a line as shown on
the graph.
y
x
gradient = y / x
In the reaction…
A(aq) + B(aq) ——> C(aq) + D(aq)
the concentration of B was measured
every 200 minutes.The reaction is
obviously very slow!
The variation in rate can be investigated by measuring the change in concentration of
one of the reactants or products, plotting a graph and then finding the gradients of the
curve at different concentrations.

42. GRAPHICAL DETERMINATION OF RATE
RATE CALCULATION
The rate of reaction at
any moment can be
found from the gradient
of the tangentat that
point. The steeper the
gradient, the faster the
rate of reaction
Place a rule on the
outside of the curve and
draw a line as shown on
the graph.
y
x
gradient = y / x
concentration = 1.20 mol dm
-3
gradient = -1.60 mol dm
-3
1520 min
rate = -1.05 x 10
-3
mol dm
-
3
The rate is negative because
the reaction is slowing down
The variation in rate can be investigated by measuring the change in concentration of
one reactants or product, plotting a graph and then finding the gradients of tangents to
the curve at different concentrations.

43. GRAPHICAL DETERMINATION OF RATE
RATE CALCULATION
The rate of reaction at
any moment can be
found from the gradient
of the tangentat that
point. The steeper the
gradient, the faster the
rate of reaction
Place a rule on the
outside of the curve and
draw a line as shown on
the graph.
y
x
gradient = y / x
The gradients of tangents at several other
concentrations are calculated.
Notice how the gradient gets less as the
reaction proceeds, showing that the reaction
is slowing down.
The tangent at the start of the reaction is used
to calculate the initial rateof the reaction.
The variation in rate can be investigated by measuring the change in concentration of
one of the reactants or products, plotting a graph and then finding the gradients of the
curve at different concentrations.

44. FIRST ORDER REACTIONS AND HALF LIFE
One characteristic of a FIRST ORDER REACTION
is that it is similar to radioactive decay. It has a
half-life that is independent of the concentration.
It should take the same time to drop to one half of
the original concentration as it does to drop from
one half to one quarter of the original.
The concentration of a reactant
falls as the reaction proceeds

45. The concentration of reactant A
falls as the reaction proceeds
The concentration drops from
4 to 2 in 17 minutes
FIRST ORDER REACTIONS AND HALF LIFE

46. The concentration drops from
4 to 2 in 17 minutes
2 to 1 in a further 17 minutes
FIRST ORDER REACTIONS AND HALF LIFE
The concentration of reactant A
falls as the reaction proceeds

47. The concentration drops from
4 to 2 in 17 minutes
2 to 1 in a further 17 minutes
1 to 0.5 in a further 17 minutes
FIRST ORDER REACTIONS AND HALF LIFE
The concentration of reactant A
falls as the reaction proceeds

48. The concentration drops from
4 to 2 in 17 minutes
2 to 1 in a further 17 minutes
1 to 0.5 in a further 17 minutes
FIRST ORDER REACTIONS AND HALF LIFE
The concentration of reactant A
falls as the reaction proceeds

49. A useful relationship
k t
½= log
e2
= 0.693
where t
½= the half life
FIRST ORDER REACTIONS AND HALF LIFE
Half life= 17 minutes
k t
½= 0.693
k = 0.693
t
½
k = 0.693 = 0.041 min
-1
17

50. ORDER OF REACTION –GRAPHICAL DETERMINATION
The order of reaction can be found by measuring the rate at different times during the
reaction and plotting the rate against either concentration or time. The shape of the
curve provides an indication of the order.

51. ORDER OF REACTION –GRAPHICAL DETERMINATION
The order of reaction can be found by measuring the rate at different times during the
reaction and plotting the rate against either concentration or time. The shape of the
curve provides an indication of the order.
PLOTTING RATE AGAINST CONCENTRATION
RATE OF REACTION / mol dm
-
3
s
-
1
CONCENTRATION / mol dm
-3

52. ORDER OF REACTION –GRAPHICAL DETERMINATION
The order of reaction can be found by measuring the rate at different times during the
reaction and plotting the rate against either concentration or time. The shape of the
curve provides an indication of the order.
PLOTTING RATE AGAINST CONCENTRATION
RATE OF REACTION / mol dm
-
3
s
-
1
CONCENTRATION / mol dm
-3
ZERO ORDER –the rate does not
depend on the concentration. The
line is parallel to the x axis.

53. ORDER OF REACTION –GRAPHICAL DETERMINATION
The order of reaction can be found by measuring the rate at different times during the
reaction and plotting the rate against either concentration or time. The shape of the
curve provides an indication of the order.
PLOTTING RATE AGAINST CONCENTRATION
RATE OF REACTION / mol dm
-
3
s
-
1
CONCENTRATION / mol dm
-3
ZERO ORDER –the rate does not
depend on the concentration. The
line is parallel to the x axis.
FIRST ORDER–the rate is
proportional to the concentration
so you get a straight line of fixed
gradient. The gradient of the line
equals the rate constant for the
reaction.

54. ORDER OF REACTION –GRAPHICAL DETERMINATION
The order of reaction can be found by measuring the rate at different times during the
reaction and plotting the rate against either concentration or time. The shape of the
curve provides an indication of the order.
PLOTTING RATE AGAINST CONCENTRATION
RATE OF REACTION / mol dm
-
3
s
-
1
CONCENTRATION / mol dm
-3
ZERO ORDER –the rate does not
depend on the concentration. The
line is parallel to the x axis.
FIRST ORDER–the rate is
proportional to the concentration
so you get a straight line of fixed
gradient. The gradient of the line
equals the rate constant for the
reaction.
SECOND ORDER –the rate is
proportional to the square of the
concentration. You get an
upwardly sloping curve.

55. ORDER OF REACTION –GRAPHICAL DETERMINATION
The order of reaction can be found by measuring the rate at different times during the
reaction and plotting the rate against either concentration or time. The shape of the
curve provides an indication of the order.
PLOTTING RATE AGAINST CONCENTRATION
RATE OF REACTION / mol dm
-
3
s
-
1
CONCENTRATION / mol dm
-3
ZERO ORDER –the rate does not
depend on the concentration. The
line is parallel to the x axis.
FIRST ORDER–the rate is
proportional to the concentration
so you get a straight line of fixed
gradient. The gradient of the line
equals the rate constant for the
reaction.
SECOND ORDER –the rate is
proportional to the square of the
concentration. You get an
upwardly sloping curve.

56. ORDER OF REACTION –GRAPHICAL DETERMINATION
The order of reaction can be found by measuring the rate at different times during the
reaction and plotting the rate against either concentration or time. The shape of the
curve provides an indication of the order.
PLOTTING RATE AGAINST TIME
RATE OF REACTION / mol dm
-
3
s
-
1
TIME / s
ZERO ORDER
A straight line showing a constant decline in concentration.
FIRST ORDER
A slightly sloping curve which drops with a constant half-life.
SECOND ORDER
The curve declines steeply at first then levels out.

57. ORDER OF REACTION
GRAPHICAL
DETERMINATION
Calculate the rate of reaction at
1.0, 0.75, 0.5 and 0.25 mol dm
-3
Plot a graph of rate v [A]
Calculate the time it takes
for [A] to go from...
1.00 to 0.50 mol dm
-3
0.50 to 0.25 mol dm
-3
Deduce from the graph
that the order wrt A is 1
Calculate the value and
units of the rate constant, k

58. RATE DETERMINING STEP
Many reactions consist of a series of separate stages.
Each step has its own rate and rate constant.
The overall rate of a multi-step process is governed by the slowest
step(like a production line where overall output can be held up by
a slow worker).
This step is known as theRATE DETERMINING STEP .
If there is more than one step, the rate equation may not contain
all the reactants in its format.

59. RATE DETERMINING STEP
THE REACTION BETWEEN PROPANONE AND IODINE
Iodine and propanone CH
3COCH
3+ I
2 CH
3COCH
2I + HI
react in the presence of acid
The rate equation is... r = k [CH
3COCH
3] [H
+
]
Why do H
+
ions appear in
the rate equation?
Why does I
2not appear
in the rate equation?

60. RATE DETERMINING STEP
THE REACTION BETWEEN PROPANONE AND IODINE
Iodine and propanone CH
3COCH
3+ I
2 CH
3COCH
2I + HI
react in the presence of acid
The rate equation is... r = k [CH
3COCH
3] [H
+
]
Why do H
+
ions appear in The reaction is catalysed by acid
the rate equation? [H
+
] affects the rate but is unchanged overall
Why does I
2not appear The rate determining step doesn’t involve I
2
in the rate equation?

61. RATE DETERMINING STEP
THE REACTION BETWEEN PROPANONE AND IODINE
Iodine and propanone CH
3COCH
3+ I
2 CH
3COCH
2I + HI
react in the presence of acid
The rate equation is... r = k [CH
3COCH
3] [H
+
]
Why do H
+
ions appear in The reaction is catalysed by acid
the rate equation? [H
+
] affects the rate but is unchanged overall
Why does I
2not appear The rate determining step doesn’t involve I
2
in the rate equation?
The slowest step of any multi-step reaction is known as the rate determining step and it
is the species involved in this step that are found in the overall rate equation.
Catalysts appear in the rate equation because they affect the rate but they do not
appear in the stoichiometric equation because they remain chemically unchanged

62. RATE DETERMINING STEP
HYDROLYSIS OF HALOALKANES
Haloalkanes (general formula RX) are RX + OH
-
ROH + X
-
hydrolysed by hydroxide ion to give alcohols.
With many haloalkanes the rate equation is...r = k [RX][OH
-
]SECOND ORDER
This is because both the RX and OH
-
must
collide for a reaction to take place in ONE STEP

63. RATE DETERMINING STEP
HYDROLYSIS OF HALOALKANES
Haloalkanes (general formula RX) are RX + OH
-
ROH + X
-
hydrolysed by hydroxide ion to give alcohols.
With many haloalkanes the rate equation is...r = k [RX][OH
-
]SECOND ORDER
This is because both the RX and OH
-
must
collide for a reaction to take place in ONE STEP
but with others it only depends on [RX]... r = k [RX] FIRST ORDER
The reaction has taken place in TWO STEPS...
-the first involves breaking an R-X bond i) RX R
+
+ X
-
Slow
-the second step involves the two ions joiningii)R
+
+ OH
-
ROH Fast
The first step is slower as it involves bond breaking and energy has to be put in.
The first order mechanism is favoured by tertiary haloalkanes because the hydroxide ion is hindered
in its approach by alkyl groups if the mechanism involves the hydroxide ion and haloalkane colliding.

64. RATE DETERMINING STEP
The reaction H
2O
2+ 2H
3O
+
+ 2I¯ I
2+ 4H
2O takes place in 3 steps
Step 1H
2O
2+ I¯ IO¯ + H
2O SLOW
Step 2IO¯ + H
3O
+
HIO + H
2O FAST
Step 3HIO + H
3O
+
+ I¯ I
2+ 2H
2O FAST
The rate determining step is STEP 1 as it is the slowest

65. RATE DETERMINING STEP
The reaction H
2O
2+ 2H
3O
+
+ 2I¯ I
2+ 4H
2O takes place in 3 steps
Step 1H
2O
2+ I¯ IO¯ + H
2O SLOW
Step 2IO¯ + H
3O
+
HIO + H
2O FAST
Step 3HIO + H
3O
+
+ I¯ I
2+ 2H
2O FAST
The rate determining step is STEP 1 as it is the slowest
The reaction 2N
2O
5 4NO
2+ O
2 takes place in 3 steps
Step 1N
2O
5 NO
2+ NO
3 SLOW
Step 2NO
2+ NO
3 NO + NO
2+ O
2 FAST
Step 3NO + NO
3 2NO
2
from another Step 1FAST
The rate determining step is STEP 1 rate = k [N
2O
5]

66. OTHER TOPICS
Autocatalysis
A small number of reactions appear to speed up, rather than slow down, for a time.
This is because one of the products is acting as a catalyst and as more product is
formed the reaction gets faster. One of the best known examples is the catalytic
properties of Mn
2+
(aq) on the decomposition of MnO
4¯(aq). You will notice it in a
titration of KMnO
4with either hydrogen peroxide or ethanedioic (oxalic) acid.
Molecularity The number of individual particles of the reacting species taking
part in the rate determining step of a reaction
e.g. A + 2B C + D molecularity is 3-one Aand two B’sneed to collide
A 2B however has a molecularity of 1-only one Ais involved

67. THE RATE
EQUATION
The End
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