Rate Laws and Order of
Reaction
Half life and chemical kinetics
Page 377 # 1 - 6
381 # 7 - 9
kt ½ = 0.693
Factors that Affect Reaction Rate
1. Temperature
•
•
Collision Theory: When two chemicals react, their
molecules have to collide with each other with sufficient
energy for the reaction to take place.
Kinetic Theory: Increasing temperature means the
molecules move faster.
2. Concentrations of reactants
•
More reactants mean more collisions if enough energy is
present
3. Catalysts
•
Speed up reactions by lowering activation energy
4. Surface area of a solid reactant
•
Bread and Butter theory: more area for reactants to be in
contact
5. Pressure of gaseous reactants or products
•
Increased number of collisions
The Rate Law
The rate law expresses the relationship of the rate of a reaction
to the rate constant and the concentrations of the reactants
raised to some powers.
aA + bB
cC + dD
Rate = k [A]x[B]y
reaction is xth order in A
reaction is yth order in B
reaction is (x +y)th order overall
THE RATE EQUATION
links the rate of reaction to the concentration of reactants
can only be found by doing actual experiments
cannot be found by just looking at the equation
r
k
[]
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
-1
conc. / time
usually mol L s -1
depend on the rate equation
1 litre = 1 dm 3
mol L -1
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
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 equatio n
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 equatio n
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 subst ance 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 equatio n
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 subst ance 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
Rate Laws
•
Rate laws are always determined experimentally.
•
Reaction order is always defined in terms of reactant
(not product) concentrations.
•
The order of a reactant is not related to the
stoichiometric coefficient of the reactant in the balanced
chemical equation.
F2 (g) + 2ClO2 (g)
2FClO2 (g)
rate = k [F2][ClO2] 1
13.2
THE RATE EQUATION
Experimental determination of order
Method 1
Plot a concentration / time graph qnd 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 t o concentration - 1st ORDER.
If the plot is a curve then it must have another order. Try plo tting 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 qnd 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 t o concentration - 1st ORDER.
If the plot is a curve then it must have another order. Try plo tting 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 differen t concentrations of a
reactant but keeping all others constant. Plot a series of conc entration / time graphs
and calculate the initial rate (slope of curve at start) for eac h reaction. From the results
calculate the relationship between concentration and rate and he nce deduce the rate
equation. To find order directly, logarithmic plots are requir ed.
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
units of k:
mol L- 1 sec
-1
1
sec -1
example in the rate equation r = k [A]
in the rate equation r = k [A] [B] 2
2
L mol -1 sec -1
3
L 2 mol-2 sec -1
k will have units of sec -1
k will have units of L
2
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 L -1 s -1
mol L -1
In an experiment between A and B the
initial rate of reaction was found for
various starting concentrations of A and B.
Calculate...
•
•
•
•
•
the
the
the
the
the
individual orders for A and B
overall order of reaction
rate equation
value of the rate constant (k)
units of the rate constant
RATE EQUATION - CALCULATING ORDER OF [A]
[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 - CALCULATING ORDER OF [B]
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 - CALCULATING OVERALL ORDER
[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 - OVERALL EQUATION
[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 - CALCULATING K WITH UNITS
[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 L 2 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
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 L 2mol -2 sec -1
RATE EQUATION QUESTIONS
[A] / mol L-1
No
No 11
Expt 1
Expt 2
Expt 3
CALCULATE
0.25
0.25
0.50
[B] / mol L-1
0.25
0.50
0.25
THE
THE
THE
THE
THE
Rate / mol L-1
s-1
4
8
8
ORDER WITH RESPECT TO A
ORDER WITH RESPECT TO B
OVERALL ORDER OF REACTION
FORMAT OF THE RATE EQUATION
VALUE AND UNITS OF THE RATE CONSTANT
ANSWER
ANSWER
RATE EQUATION QUESTIONS
[A] / mol L-1
No
No 11
Expt 1
Expt 2
Expt 3
0.25
0.25
0.50
[B] / mol L-1
0.25
0.50
0.25
[B] is doubled
rate " [B]
Rate / mol L-1
s-1
4
8
8
Expts 1&2
[A] is constant
Therefore
Rate is doubled
1st order wrt B
Explanation:
What was done to [B] had exactly the same effect on the rate
ANSWER
ANSWER
RATE EQUATION QUESTIONS
[A] / mol L-1
No
No 11
Expt 1
Expt 2
Expt 3
Expts 1&2
Explanation:
0.25
0.25
0.50
[B] / mol L-1
0.25
0.50
0.25
Rate / mol L-1
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
Expts 1&3
[B] is constant
Therefore
[A] is doubled
rate " [A]
Rate is doubled
1st order wrt A
Explanation:
What was done to [A] had exactly the same effect on the rate
ANSWER
ANSWER
RATE EQUATION QUESTIONS
[A] / mol L-1
No
No 11
Expt 1
Expt 2
Expt 3
Expts 1&2
Explanation:
Expts 1&3
0.25
0.25
0.50
[B] / mol L-1
0.25
0.50
0.25
Rate / mol L-1
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]
ANSWER
ANSWER
RATE EQUATION QUESTIONS
[A] / mol L-1
No
No 11
Expt 1
Expt 2
Expt 3
Expts 1&2
Explanation:
Expts 1&3
0.25
0.25
0.50
[B] / mol L-1
0.25
0.50
0.25
Rate / mol L-1
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
ANSWER
ANSWER
RATE EQUATION QUESTIONS
[A] / mol L-1
No
No 11
Expt 1
Expt 2
Expt 3
Expts 1&2
Explanation:
Expts 1&3
[B] / mol L-1
0.25
0.25
0.50
0.25
0.50
0.25
Rate / mol L-1
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 =
r = 64[A][B]
L mol -1 s -1
RATE EQUATION QUESTIONS
[C] / mol L-1
No
No 22
Expt 1
Expt 2
Expt 3
CALCULATE
0.40
0.20
0.40
THE
THE
THE
THE
THE
[D] / m mol L-1
0.40
0.40
1.20
Rate / mol L-1
s-1
0.16
0.04
1.44
ORDER WITH RESPECT TO C
ORDER WITH RESPECT TO D
OVERALL ORDER OF REACTION
FORMAT OF THE RATE EQUATION
VALUE AND UNITS OF THE RATE CONSTANT
ANSWER
ANSWER
RATE EQUATION QUESTIONS
[C] / mol L-1
No
No 22
Expt 1
Expt 2
Expt 3
0.40
0.20
0.40
[D] / mol L-1
0.40
0.40
1.20
[D] is tripled
rate " [D]2
Rate / mol L-1
s-1
0.16
0.04
1.44
Expts 1&3
[C] is constant
Therefore
Rate is 9 x bigger
2nd order wrt D
Explanation:
Squaring what was done to D affected the rate (3 2 = 9)
ANSWER
ANSWER
RATE EQUATION QUESTIONS
[C] / mol L-1
No
No 22
Expt 1
Expt 2
Expt 3
Expts 1&3
Explanation:
0.40
0.20
0.40
[D] / mol L-1
0.40
0.40
1.20
Rate / mol L-1
s-1
0.16
0.04
1.44
[C] is constant
[D] is tripled
Rate is 9 x bigger
Therefore
rate " [D]2
2nd order wrt D
Squaring what was done to D affected the rate (3 2 = 9)
Expts 1&2
[D] is constant
Therefore
[A] is halved
rate " [C] 2
Explanation:
One half squared = one quarter
Rate is quartered
2nd order wrt C
ANSWER
ANSWER
RATE EQUATION QUESTIONS
[C] / mol L-1
No
No 22
Expt 1
Expt 2
Expt 3
Expts 1&3
Explanation:
Expts 1&2
0.40
0.20
0.40
[D] / mol L-1
0.40
0.40
1.20
Rate / mol L-1
s-1
0.16
0.04
1.44
[C] is constant
[D] is tripled
Rate is 9 x bigger
Therefore
rate " [D]2
2nd order wrt D
Squaring what was done to D affected the rate (3 2 = 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
ANSWER
RATE EQUATION QUESTIONS
[C] / mol L-1
No
No 22
Expt 1
Expt 2
Expt 3
Expts 1&3
Explanation:
Expts 1&2
[D] / mol L-1
0.40
0.20
0.40
Rate / mol L-1
0.40
0.40
1.20
s-1
0.16
0.04
1.44
[C] is constant
[D] is tripled
Rate is 9 x bigger
Therefore
rate " [D]2
2nd order wrt D
Squaring what was done to D affected the rate (3 2 = 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 / 0.2 2 x 0.4 2 = 6.25
Units of k
rate / conc 2 x conc 2
=
L3
mol-3 s -1
Rate is quartered
2nd order wrt C
RATE EQUATION QUESTIONS
[E] / mol L-1
No
No 33
Expt 1
Expt 2
Expt 3
CALCULATE
0.40
0.80
0.80
[F] / mol L-1
0.40
0.80
1.20
THE
THE
THE
THE
THE
Rate / mol L-1
s-1
0.16
0.32
0.32
ORDER WITH RESPECT TO E
ORDER WITH RESPECT TO F
OVERALL ORDER OF REACTION
FORMAT OF THE RATE EQUATION
VALUE AND UNITS OF THE RATE CONSTANT
ANSWER
ANSWER
RATE EQUATION QUESTIONS
[E] / mol L-1
No
No 33
Expt 1
Expt 2
Expt 3
0.40
0.80
0.80
[F] / mol L-1
0.40
0.80
1.20
Rate / mol L-1
s-1
0.16
0.32
0.32
Expts 2&3
[E] is constant
[F] is x 1.5
Rate is UNAFFECTED
Explanation:
Concentration of [F] has no effect on the rate
Rate unchanged
ZERO order wrt F
ANSWER
ANSWER
RATE EQUATION QUESTIONS
[E] / mol L-1
No
No 33
Expt 1
Expt 2
Expt 3
Expts 2&3
0.40
0.80
0.80
[F] / mol L-1
Rate / mol L-1
0.40
0.80
1.20
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
Expts 1&2
[E] is doubled
Therefore
Rate is doubled
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]
Explanation:
[F] is doubled
rate " [E] 2
ANSWER
ANSWER
RATE EQUATION QUESTIONS
[E] / mol L-1
No
No 33
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 L-1
0.40
0.80
1.20
Rate / mol L-1
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]
1st 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
ANSWER
RATE EQUATION QUESTIONS
[E] / mol dm-3
No
No 33
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]
1st 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
Determine the rate law and calculate the rate constant for
the following reaction from the following data:
S2O82- (aq) + 3I- (aq)
2SO42- (aq) + I3- (aq)
Experiment
[S 2 O 82-]
[I-]
Initial Rate
(M/s)
1
0.08
0.034
2.2 x 10-4
2
0.08
0.017
1.1 x 10 -4
3
0.16
0.017
2.2 x 10-4
rate = k [S2O82-]x[I-]y
y=1
x=1
rate = k [S2O82-][I-]
Double [I-], rate doubles (experiment 1 & 2)
Double [S2O82-], rate doubles (experiment 2 & 3)
2.2 x 10-4 M/s
rate
k=
=
= 0.08/M•s
2[S2O8 ][I ] (0.08 M)(0.034 M)
GRAPHICAL DETERMINATION OF RATE
The variation in rate can be investigated by measuring the chang e in concentration of
one of the reactants or products, plotting a graph and then find ing 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 SLOPE
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
SLOPE = y / x
GRAPHICAL DETERMINATION OF RATE
The variation in rate can be investigated by measuring the chang e 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 L-1
SLOPE
- 1.60 mol L-1
=
1520 min
rate
= - 1.05 x 10-3 mol L-1
The rate is negative because
the reaction is slowing down
RATE CALCULATION
The rate of reaction at
any moment can be
found from the SLOPE
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
SLOPE = y / x
GRAPHICAL DETERMINATION OF RATE
The variation in rate can be investigated by measuring the chang e in concentration of
one of the reactants or products, plotting a graph and then find ing 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
The half-life, t½ , is the time required for the concentration of a
reactant to decrease to half of its initial concentration.
t½ = t when [A] = [A]0/2
ln
t½ =
[A]0
[A]0/2
k
Ln 2 0.693
=
=
k
k
What is the half-life of N2O5 if it decomposes with a rate
constant of 5.7 x 10-4 s-1?
0.693
t½ = Ln 2 =
= 1200 s = 20 minutes
-4
-1
k
5.7 x 10 s
How do you know decomposition is first order?
units of k (s-1)
FIRST ORDER REACTIONS AND HALF LIFE
A useful relationship
k t ½ = log e 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
13.3
ORDER OF REACTION – GRAPHICAL DETERMINATION
The order of reaction can be found by measuring the rate at diff erent times during the
reaction and plotting the rate against either concentration or t ime. 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 diff erent times during the
reaction and plotting the rate against either concentration or t ime. The shape of the
curve provides an indication of the order.
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 diff erent times during the
reaction and plotting the rate against either concentration or t ime. The shape of the
curve provides an indication of the order.
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 diff erent times during the
reaction and plotting the rate against either concentration or t ime. The shape of the
curve provides an indication of the order.
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 diff erent times during the
reaction and plotting the rate against either concentration or t ime. The shape of the
curve provides an indication of the order.
PLOTTING RATE AGAINST CONCENTRATION
SECOND ORDER – the rate is
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 diff erent times during the
reaction and plotting the rate against either concentration or t ime. The shape of the
curve provides an indication of the order.
PLOTTING RATE AGAINST CONCENTRATION
SECOND ORDER – the rate is
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
Run # Initial [A]
([A]0)
Initial [B]
([B]0)
Initial Rate (v 0)
1
1.00 M
1.00 M
1.25 x 10-2 M/s
2
1.00 M
2.00 M
2.5 x 10-2 M/s
3
2.00 M
2.00 M
2.5 x 10-2 M/s
What is the order with respect to A?
0
What is the order with respect to B?
1
What is the overall order of the
reaction?
1
[NO(g)] (mol dm-3)
[Cl2(g)] (mol dm-3)
Initial Rate
(mol dm-3 s-1)
0.250
0.250
1.43 x 10-6
0.250
0.500
2.86 x 10-6
0.500
0.500
1.14 x 10-5
What is the order with respect to Cl2?
1
What is the order with respect to NO?
2
What is the overall order of the
reaction?
3
© 2003 JONATHAN HOPTON & KNOCKHARDY PUBLISHING