Transcript THE RATE EQUATION A guide for A level students KNOCKHARDY PUBLISHING SPECIFICATIONS KNOCKHARDY PUBLISHING THE RATE EQUATION INTRODUCTION This Powerpoint show is one of several produced to.

THE RATE
EQUATION
A guide for A level students
2015
KNOCKHARDY PUBLISHING
SPECIFICATIONS
KNOCKHARDY PUBLISHING
THE RATE EQUATION
INTRODUCTION
This Powerpoint show 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
Navigation is achieved by...
either
or
clicking on the grey arrows at the foot of each page
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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
THE RATE EQUATION
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
REVISION
COLLISION THEORY
Collision theory states that...
• particles must COLLIDE before 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
COLLISION THEORY
Collision theory states that...
• particles must COLLIDE before 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 collisions
increase particle speed
have more particles present
more successful collisions give particles more energy or
lower the activation energy
or
REVISION
INCREASING THE RATE
The following methods may be used to
increase the rate of a chemical reaction
• INCREASE THE SURFACE AREA OF SOLIDS
• INCREASE TEMPERATURE
• SHINE LIGHT
• ADD A CATALYST
• INCREASE THE PRESSURE OF ANY GASES
• INCREASE THE CONCENTRATION OF REACTANTS
REVISION
RATE CHANGE DURING A REACTION
Reactions are fastest at the start and get slower as the reactants concentration drops.
In a reaction such as
A + 2B ——> C the concentrations might change as shown
Reactants (A and B)
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
CONCENTRATION
Concentration decreases with time
C
A
B
TIME
MEASURING THE RATE
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
RATE
How much concentration changes with time. It is the equivalent of velocity.
CONCENTRATION
THE SLOPE OF THE GRADIENT OF THE
CURVE GETS LESS AS THE
REACTION SLOWS DOWN
WITH TIME
y
x
gradient = y
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 RATE EQUATION
Format
r
k
[]
links 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
might have a rate equation like this
r = k [A] [B]2
rate of reaction
rate constant
concentration
units of
units
units of
conc. / time usually mol dm-3 s-1
depend on the rate equation
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 (22) rate
ORDER OF REACTION
Order tells you how much the concentration of a reactant affects the rate
Individual order
The power to which a concentration is raised in the rate equation
Overall order
The sum of all the individual orders in the rate equation.
ORDER OF REACTION
Order tells you how much the concentration of a reactant affects the rate
Individual order
The power to which a concentration is raised in the rate equation
Overall order
The sum of all the individual orders in the rate equation.
Value(s)
e.g.
in the rate equation
and
the order with respect to A is
the order with respect to B is
the overall order is
r = k [A] [B]2
1
2
3
1st Order
2nd Order
3rd Order
need not be whole numbers
can be zero if the rate is unaffected by how much of a substance is present
ORDER OF REACTION
Order tells you how much the concentration of a reactant affects the rate
Individual order
The power to which a concentration is raised in the rate equation
Overall order
The sum of all the individual orders in the rate equation.
Value(s)
e.g.
in the rate equation
and
the order with respect to A is
the order with respect to B is
the overall order is
r = k [A] [B]2
1
2
3
1st Order
2nd Order
3rd Order
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
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.
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.
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 dm3 mol-1 sec-1
Divide the rate by as many concentrations as appear in the rate equation.
Overall Order
0
units of k
mol dm-3 sec-1
1
sec-1
example in the rate equation r = k [A]
in the rate equation r = k [A] [B]2
2
dm3 mol-1 sec-1
3
dm6 mol-2 sec-1
k will have units of sec-1
k will have units of dm6 mol-2 sec-1
RATE EQUATION - SAMPLE CALCULATION
r
[]
[A]
[B]
Initial
rate (r)
1
0.5
1
2
2
1.5
1
6
3
0.5
2
8
initial rate of reaction
concentration
mol dm-3 s-1
mol dm-3
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
RATE EQUATION - SAMPLE CALCULATION
[A]
[B]
Initial
rate (r)
1
0.5
1
2
2
1.5
1
6
3
0.5
2
8
Compare Experiments 1 & 2
[B]
same
[A]
3 x bigger
rate
3 x bigger  rate  [A]
FIRST ORDER with respect to (wrt) A
CALCULATING ORDER wrt A
Choose any two experiments where...
[A] is changed
and, 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
CALCULATING ORDER wrt B
Choose any two experiments where...
[B] is changed
and, importantly,
[A] is KEPT THE SAME
See how a change in [B] affects the rate
As you can see, doubling [B] quadruples
the rate so...
THE ORDER WITH RESPECT TO B = 2
It is SECOND ORDER
[A]
[B]
rate
[A]
[B]
Initial
rate (r)
1
0.5
1
2
2
1.5
1
6
3
0.5
2
8
Compare Experiments 1 & 3
same
2 x bigger
4 x bigger
 rate  [B]2
SECOND ORDER wrt B
RATE EQUATION - SAMPLE CALCULATION
[A]
[B]
Initial
rate (r)
1
0.5
1
2
2
1.5
1
3
0.5
2
[A]
[B]
Initial
rate (r)
1
0.5
1
2
6
2
1.5
1
6
8
3
0.5
2
8
Compare Experiments 1 & 2
[B]
same
[A]
3 x bigger
rate
3 x bigger  rate  [A]
[A]
[B]
rate
FIRST ORDER with respect to (wrt) A
OVERALL ORDER
Compare Experiments 1 & 3
same
2 x bigger
4 x bigger
 rate  [B]2
SECOND ORDER wrt B
= THE SUM OF THE INDIVIDUAL ORDERS
= 1 + 2
=
3
RATE EQUATION - SAMPLE CALCULATION
[A]
[B]
Initial
rate (r)
1
0.5
1
2
2
1.5
1
3
0.5
2
[A]
[B]
Initial
rate (r)
1
0.5
1
2
6
2
1.5
1
6
8
3
0.5
2
8
Compare Experiments 1 & 2
[B]
same
[A]
3 x bigger
rate
3 x bigger  rate  [A]
[A]
[B]
rate
FIRST ORDER with respect to (wrt) A
 rate = k [A] [B]2
Compare Experiments 1 & 3
same
2 x bigger
4 x bigger
 rate  [B]2
SECOND ORDER wrt B
By combining the two proportionality relationships
you can construct the overall rate equation
RATE EQUATION - SAMPLE CALCULATION
[A]
[B]
Initial
rate (r)
1
0.5
1
2
2
1.5
1
3
0.5
2
[A]
[B]
Initial
rate (r)
1
0.5
1
2
6
2
1.5
1
6
8
3
0.5
2
8
Compare Experiments 1 & 2
[B]
same
[A]
3 x bigger
rate
3 x bigger  rate  [A]
[A]
[B]
rate
FIRST ORDER with respect to (wrt) A
 rate = k [A] [B]2
re-arranging
k = rate
[A] [B]2
Compare Experiments 1 & 3
same
2 x bigger
4 x bigger
 rate  [B]2
SECOND ORDER wrt B
Chose one experiment (e.g. Expt. 3) and
substitute its values into the rate equation
k =
8
(0.5) (2)2
= 4 dm6 mol-2 sec-1
RATE EQUATION - SAMPLE CALCULATION
[A]
[B]
Initial
rate (r)
1
0.5
1
2
2
1.5
1
3
0.5
2
[A]
[B]
Initial
rate (r)
1
0.5
1
2
6
2
1.5
1
6
8
3
0.5
2
8
Compare Experiments 1 & 2
[B]
same
[A]
3 x bigger
rate
3 x bigger  rate  [A]
[A]
[B]
rate
FIRST ORDER with respect to (wrt) A
 rate = k [A] [B]2
re-arranging
SUMMARY
k = rate
[A] [B]2
Compare Experiments 1 & 3
same
2 x bigger
4 x bigger
 rate  [B]2
SECOND ORDER wrt B
Chose one experiment (e.g. Expt. 3) and
substitute its values into the rate equation
k =
8
(0.5) (2)2
= 4 dm6 mol-2 sec-1
RATE EQUATION QUESTIONS
[A] / mol dm-3
No 1
Expt 1
Expt 2
Expt 3
CALCULATE
0.25
0.25
0.50
[B] / mol dm-3
0.25
0.50
0.25
Rate / mol dm-3 s-1
4
8
8
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
RATE EQUATION QUESTIONS
[A] / mol dm-3
No 1
Expt 1
Expt 2
Expt 3
Expts 1&2
Explanation:
0.25
0.25
0.50
[B] / mol dm-3
0.25
0.50
0.25
ANSWER
Rate / mol dm-3 s-1
4
8
8
[A] is constant
[B] is doubled
Rate is doubled
Therefore
rate  [B]
1st order wrt B
What was done to [B] had exactly the same effect on the rate
RATE EQUATION QUESTIONS
[A] / mol dm-3
No 1
Expt 1
Expt 2
Expt 3
Expts 1&2
Explanation:
Expts 1&3
Explanation:
0.25
0.25
0.50
[B] / mol dm-3
0.25
0.50
0.25
ANSWER
Rate / mol dm-3 s-1
4
8
8
[A] is constant
[B] is doubled
Rate is doubled
Therefore
rate  [B]
1st order wrt B
What was done to [B] had exactly the same effect on the rate
[B] is constant
[A] is doubled
Rate is doubled
Therefore
rate  [A]
1st order wrt A
What was done to [A] had exactly the same effect on the rate
RATE EQUATION QUESTIONS
[A] / mol dm-3
No 1
Expt 1
Expt 2
Expt 3
Expts 1&2
Explanation:
Expts 1&3
0.25
0.25
0.50
[B] / mol dm-3
0.25
0.50
0.25
ANSWER
Rate / mol dm-3 s-1
4
8
8
[A] is constant
[B] is doubled
Rate is doubled
Therefore
rate  [B]
1st order wrt B
What was done to [B] had exactly the same effect on the rate
Explanation:
[B] is constant
[A] is doubled
Rate is doubled
Therefore
rate  [A]
1st order wrt A
What was done to [A] had exactly the same effect on the rate
Rate equation is
r = k[A][B]
RATE EQUATION QUESTIONS
[A] / mol dm-3
No 1
Expt 1
Expt 2
Expt 3
Expts 1&2
Explanation:
Expts 1&3
0.25
0.25
0.50
[B] / mol dm-3
0.25
0.50
0.25
ANSWER
Rate / mol dm-3 s-1
4
8
8
[A] is constant
[B] is doubled
Rate is doubled
Therefore
rate  [B]
1st order wrt B
What was done to [B] had exactly the same effect on the rate
Explanation:
[B] is constant
[A] is doubled
Rate is doubled
Therefore
rate  [A]
1st order wrt A
What was done to [A] had exactly the same effect on the rate
Rate equation is
r = 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
RATE EQUATION QUESTIONS
[A] / mol dm-3
No 1
Expt 1
Expt 2
Expt 3
Expts 1&2
Explanation:
Expts 1&3
0.25
0.25
0.50
[B] / mol dm-3
0.25
0.50
0.25
ANSWER
Rate / mol dm-3 s-1
4
8
8
[A] is constant
[B] is doubled
Rate is doubled
Therefore
rate  [B]
1st order wrt B
What was done to [B] had exactly the same effect on the rate
Explanation:
[B] is constant
[A] is doubled
Rate is doubled
Therefore
rate  [A]
1st order wrt A
What was done to [A] had exactly the same effect on the rate
Rate equation is
r = 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 = dm3 mol-1 s-1
RATE EQUATION QUESTIONS
[C] / mol dm-3
No 2
Expt 1
Expt 2
Expt 3
CALCULATE
0.40
0.20
0.40
[D] / mol dm-3
0.40
0.40
1.20
Rate / mol dm-3 s-1
0.16
0.04
1.44
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
RATE EQUATION QUESTIONS
[C] / mol dm-3
No 2
Expt 1
Expt 2
Expt 3
Expts 1&3
Explanation:
0.40
0.20
0.40
[D] / mol dm-3
0.40
0.40
1.20
ANSWER
Rate / mol dm-3 s-1
0.16
0.04
1.44
[C] is constant
[D] is tripled
Rate is 9 x bigger
2
Therefore
rate  [D]
2nd order wrt D
Squaring what was done to D affected the rate (32 = 9)
ANSWER
RATE EQUATION QUESTIONS
[C] / mol dm-3
No 2
Expt 1
Expt 2
Expt 3
Expts 1&3
Explanation:
Expts 1&2
Explanation:
0.40
0.20
0.40
[D] / mol dm-3
0.40
0.40
1.20
Rate / mol dm-3 s-1
0.16
0.04
1.44
[C] is constant
[D] is tripled
Rate is 9 x bigger
2
Therefore
rate  [D]
2nd order wrt D
Squaring what was done to D affected the rate (32 = 9)
[D] is constant
[A] is halved
Therefore
rate  [C] 2
One half squared = one quarter
Rate is quartered
2nd order wrt C
ANSWER
RATE EQUATION QUESTIONS
[C] / mol dm-3
No 2
Expt 1
Expt 2
Expt 3
Expts 1&3
Explanation:
Expts 1&2
0.40
0.20
0.40
[D] / mol dm-3
0.40
0.40
1.20
Rate / mol dm-3 s-1
0.16
0.04
1.44
[C] is constant
[D] is tripled
Rate is 9 x bigger
2
Therefore
rate  [D]
2nd order wrt D
Squaring what was done to D affected the rate (32 = 9)
Explanation:
[D] is constant
[A] is halved
Therefore
rate  [C] 2
One half squared = one quarter
Rate equation is
r = k[C]2[D]2
Rate is quartered
2nd order wrt C
ANSWER
RATE EQUATION QUESTIONS
[C] / mol dm-3
No 2
Expt 1
Expt 2
Expt 3
Expts 1&3
Explanation:
Expts 1&2
[D] / mol dm-3
0.40
0.20
0.40
Rate / mol dm-3 s-1
0.40
0.40
1.20
0.16
0.04
1.44
[C] is constant
[D] is tripled
Rate is 9 x bigger
2
Therefore
rate  [D]
2nd order wrt D
Squaring what was done to D affected the rate (32 = 9)
Explanation:
[D] is constant
[A] is halved
Therefore
rate  [C] 2
One half squared = one quarter
Rate equation is
r = 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 / 0.22 x 0.42 = 6.25
Units of k
rate / conc2 x conc2
=
dm9 mol-3 s-1
Rate is quartered
2nd order wrt C
RATE EQUATION QUESTIONS
[E] / mol dm-3
No 3
Expt 1
Expt 2
Expt 3
CALCULATE
0.40
0.80
0.80
[F] / mol dm-3
0.40
0.80
1.20
Rate / mol dm-3 s-1
0.16
0.32
0.32
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
ANSWER ON NEXT PAGE
ANSWER
RATE EQUATION QUESTIONS
[E] / mol dm-3
No 3
Expt 1
Expt 2
Expt 3
Expts 2&3
Explanation:
0.40
0.80
0.80
[F] / mol dm-3
0.40
0.80
1.20
Rate / mol dm-3 s-1
0.16
0.32
0.32
[E] is constant
[F] is x 1.5
Rate is UNAFFECTED
Concentration of [F] has no effect on the rate
Rate unchanged
ZERO order wrt F
ANSWER
RATE EQUATION QUESTIONS
[E] / mol dm-3
No 3
Expt 1
Expt 2
Expt 3
Expts 2&3
Explanation:
Expts 1&2
Explanation:
0.40
0.80
0.80
[F] / mol dm-3
0.40
0.80
1.20
Rate / mol dm-3 s-1
0.16
0.32
0.32
[E] is constant
[F] is x 1.5
Rate is UNAFFECTED
Concentration of [F] has no effect on the rate
Rate unchanged
ZERO order wrt F
[E] is doubled
[F] is doubled
Rate is doubled
Therefore
rate  [E] 2
2nd order wrt E
Although both concentrations have been doubled, we know [F]
has no effect. The change must be all due to [E]
ANSWER
RATE EQUATION QUESTIONS
[E] / mol dm-3
No 3
Expt 1
Expt 2
Expt 3
Expts 2&3
Explanation:
Expts 1&2
Explanation:
Rate equation is
0.40
0.80
0.80
[F] / mol dm-3
0.40
0.80
1.20
Rate / mol dm-3 s-1
0.16
0.32
0.32
[E] is constant
[F] is x 1.5
Rate is UNAFFECTED
Concentration of [F] has no effect on the rate
Rate unchanged
ZERO order wrt F
[E] is doubled
[F] is doubled
Rate is doubled
Therefore
rate  [E] 2
2nd order wrt E
Although both concentrations have been doubled, we know [F]
has no effect. The change must be all due to [E]
r = k[E]
ANSWER
RATE EQUATION QUESTIONS
[E] / mol dm-3
No 3
Expt 1
Expt 2
Expt 3
Expts 2&3
Explanation:
Expts 1&2
Explanation:
0.40
0.80
0.80
[F] / mol dm-3
0.40
0.80
1.20
Rate / mol dm-3 s-1
0.16
0.32
0.32
[E] is constant
[F] is x 1.5
Rate is UNAFFECTED
Concentration of [F] has no effect on the rate
Rate unchanged
ZERO order wrt F
[E] is doubled
[F] is doubled
Rate is doubled
Therefore
rate  [E] 2
2nd order wrt E
Although both concentrations have been doubled, we know [F]
has no effect. The change must be all due to [E]
Rate equation is
r = 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
GRAPHICAL DETERMINATION OF RATE
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.
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!
RATE CALCULATION
The rate of reaction at
any moment can be
found from the gradient
of the tangent at 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
GRAPHICAL DETERMINATION OF RATE
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.
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
RATE CALCULATION
The rate of reaction at
any moment can be
found from the gradient
of the tangent at 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
GRAPHICAL DETERMINATION OF RATE
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.
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 rate of the reaction.
RATE CALCULATION
The rate of reaction at
any moment can be
found from the gradient
of the tangent at 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
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
FIRST ORDER REACTIONS AND HALF LIFE
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
The concentration of reactant A
falls as the reaction proceeds
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
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
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
A useful relationship
k t½ = loge 2
= 0.693
where t½ = the half life
Half life
= 17 minutes
k t½ = 0.693
k
= 0.693
t½
k
= 0.693 = 0.041 min-1
17
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.
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.
RATE OF REACTION / mol dm-3 s-1
PLOTTING RATE AGAINST CONCENTRATION
CONCENTRATION / mol dm-3
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.
RATE OF REACTION / mol dm-3 s-1
PLOTTING RATE AGAINST CONCENTRATION
ZERO ORDER – the rate does not
depend on the concentration. The
line is parallel to the x axis.
CONCENTRATION / mol dm-3
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.
RATE OF REACTION / mol dm-3 s-1
PLOTTING RATE AGAINST CONCENTRATION
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.
CONCENTRATION / mol dm-3
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
SECOND ORDER – the rate is
RATE OF REACTION / mol dm-3 s-1
proportional to the square of the
concentration. You get an
upwardly sloping curve.
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.
CONCENTRATION / mol dm-3
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
SECOND ORDER – the rate is
RATE OF REACTION / mol dm-3 s-1
proportional to the square of the
concentration. You get an
upwardly sloping curve.
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.
CONCENTRATION / mol dm-3
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
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.
TIME / s
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
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 the RATE DETERMINING STEP.
If there is more than one step, the rate equation may not contain
all the reactants in its format.
RATE DETERMINING STEP
THE REACTION BETWEEN PROPANONE AND IODINE
Iodine and propanone
react in the presence of acid
CH3COCH3 + I2
The rate equation is...
r = k [CH3COCH3] [H+]
Why do H+ ions appear in
the rate equation?
Why does I2 not appear
in the rate equation?
CH3COCH2I + HI
RATE DETERMINING STEP
THE REACTION BETWEEN PROPANONE AND IODINE
Iodine and propanone
react in the presence of acid
CH3COCH3 + I2
CH3COCH2I + HI
The rate equation is...
r = k [CH3COCH3] [H+]
Why do H+ ions appear in
the rate equation?
The reaction is catalysed by acid
[H+] affects the rate but is unchanged overall
Why does I2 not appear
in the rate equation?
The rate determining step doesn’t involve I2
RATE DETERMINING STEP
THE REACTION BETWEEN PROPANONE AND IODINE
Iodine and propanone
react in the presence of acid
CH3COCH3 + I2
CH3COCH2I + HI
The rate equation is...
r = k [CH3COCH3] [H+]
Why do H+ ions appear in
the rate equation?
The reaction is catalysed by acid
[H+] affects the rate but is unchanged overall
Why does I2 not appear
in the rate equation?
The rate determining step doesn’t involve I2
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
RATE DETERMINING STEP
HYDROLYSIS OF HALOALKANES
Haloalkanes (general formula RX) are
hydrolysed by hydroxide ion to give alcohols.
RX +
OH-
With many haloalkanes the rate equation is...
r = k [RX][OH-]
This is because both the RX and OH- must
collide for a reaction to take place in ONE STEP
ROH +
X-
SECOND ORDER
RATE DETERMINING STEP
HYDROLYSIS OF HALOALKANES
Haloalkanes (general formula RX) are
hydrolysed by hydroxide ion to give alcohols.
RX +
OH-
ROH +
X-
With many haloalkanes the rate equation is...
r = k [RX][OH-]
This is because both the RX and OH- must
collide for a reaction to take place in ONE STEP
SECOND ORDER
but with others it only depends on [RX]...
FIRST ORDER
The reaction has taken place in TWO STEPS...
- the first involves breaking an R-X bond
- the second step involves the two ions joining
r = k [RX]
i)
ii)
RX
R+ + XR+ + OHROH
Slow
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.
RATE DETERMINING STEP
The reaction
H2O2 + 2H3O+ + 2I¯
Step 1
H2O2 + I¯
Step 2
IO¯ + H3O+
Step 3
HIO + H3O+ + I¯
I2 + 4H2O
IO¯ + H2O
HIO
+
H2O
I2 + 2H2O
The rate determining step is STEP 1 as it is the slowest
takes place in 3 steps
SLOW
FAST
FAST
RATE DETERMINING STEP
The reaction
H2O2 + 2H3O+ + 2I¯
Step 1
H2O2 + I¯
Step 2
IO¯ + H3O+
Step 3
HIO + H3O+ + I¯
I2 + 4H2O
IO¯ + H2O
HIO
+
takes place in 3 steps
SLOW
H2O
FAST
I2 + 2H2O
FAST
The rate determining step is STEP 1 as it is the slowest
The reaction
2N2O5
4NO2 + O2
takes place in 3 steps
Step 1
N2O5
NO2 + NO3
Step 2
NO2 + NO3
NO + NO2 + O2
FAST
Step 3
NO +
2NO2
FAST
NO3
The rate determining step is STEP 1
SLOW
from another Step 1
rate = k [N2O5]
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 Mn2+(aq) on the decomposition of MnO4¯(aq). You will notice it in a
titration of KMnO4 with either hydrogen peroxide or ethanedioic (oxalic) acid.
Molecularity
e.g.
The number of individual particles of the reacting species taking
part in the rate determining step of a reaction
A + 2B
A
C+D
2B
molecularity is 3 - one A and two B’s need to collide
however has a molecularity of 1 - only one A is involved
THE RATE
EQUATION
The End
© 2015 JONATHAN HOPTON & KNOCKHARDY PUBLISHING