Molecularity is a term that refers to the number of particles that are involved in a reaction, or any step of a multi-step mechanism.
Syllabus reference R2.2.8
Reactivity 2.2.8 - The molecularity of an elementary step is the number of reacting particles taking part in that step. (HL)
- Interpret the terms “unimolecular”, “bimolecular” and “termolecular”.
Collision theory
Collision theory is a logical application of the natural laws to chemical reactions. We know that everything is made of particles, and that when a new substance is formed the atoms in the reactant particles are rearranged to form the products.
It follows that for this to happen there must be collisions that break the bonds of the reactants, allowing their atoms to reorganise into different entities.
Hydrogen consists of molecules containing pairs of hydrogen atoms bonded together. Oxygen consists of pairs of oxygen atoms bonded together into oxygen molecules.
However, water consists of molecules containing oxygen atoms bonded to two hydrogen atoms (that are not bonded together).
Collision theory explains that with enough activation energy the bonds in the reactant molecules must break if the atoms are to rearrange. The way they do this is by colliding with sufficient kinetic energy (activation energy).
2H2 + O2 ⇋ 2H2O
Even so, it is difficult to imagine that the process can occur with only one collision; there must be several steps involved.
All of the steps involved when put together must add up to the stoichiometric equation, once any intermediates are cancelled out. We call this process the "mechanism" of the reaction.
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Molecularity
The number of particles involved in collisions during any specific step is called the molecularity.
In reality, collisions involving more than 2 particles are far too unlikely to have any impact on the rate of a chemical reaction.
Hence we can have a two-particle collision, called a bimolecular process, or a 1 particle collision, called a unimolecular process.
However, one an overall rate equation is proposed it is possible that it seems to have three particles involved. This occurs when a fast step must product an intermediate needed for a subsequent slow step.
For example:
Step 1 fast): 2A ⇋ A2
Step 2 (slow): A2 + B → A2B
The rate determining (slow) step needs the intermediate A2, but this is formed in the previous fast step.
In this case the rate equation will show dependence on [A]2, and on [B].
Rate = k[A]2[B]
We say that the reaction is termolecular (molecularity = 3), even though at no stage do three particles actually collide.
Step 1 is bimolecular, and Step 2 is bimolecular.
Worked examples
Q631-01 The following initial
rate data were collected for the reaction:
aA(g) + bB(g) → cC(g) + dD(g)
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|
[A]
|
[B]
|
[C]
|
[D]
|
initial rate/ mol dm-3 s-1
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| Expt 1 |
0.422
|
1.52 x 10-2
|
0
|
0
|
2.72 x 10-5
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| Expt 2 |
0.638
|
1.21 x 10-2
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0
|
0
|
4.93 x 10-5
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| Expt 3 |
0.921
|
1.52 x 10-2
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0
|
0
|
1.29 x 10-4
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Find the rate law that best fits this data.
Answer
|
In experiments 1 and 3 the concentration of B remains the same, so
any effect on the rate is due to concentration of A.
Between expts 1 and 3 the concentration of A changes by a factor
of 0.921/0.422 = 2.18
At the same time the rate changes by a factor of 4.74
If we square the value 2.18 we get 4.75 therefore the reaction is second
order with respect to [A]
Considering expt 1 and 2 the concentration of A increases by
a factor of 0.638/0.422 = 1.51
We would expect this to have a corresponding effect on the rate of a
factor of 1.512 = 2.29
Inspection of the rate change between experiments 1 and 2 reveals an
increase by a factor of 4.93/2.72 = 1.81
During this time the concentration of B decreases by a factor of 1.52/1.21
=1.26
If we divide the expected 2.29 by 1.26 this gives 1.82, which is the
observed rate increase from 1 to 1.
The order is therefore first with respect to [B]
Therefore the full rate expression is: Rate
= k[A]2[B]1
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Q631-02 For the reaction between
hydrogen peroxide and iodide ions in acid solution, represented by the equation:
H2O2 + 3I- + 2H+ → I3- + 2H2O
these kinetic data were gathered:
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|
[H2O2]
|
[I-]
|
[H+]
|
initial rate/ mol dm-3 s-1
|
| Expt 1 |
0.010
|
0.010
|
0.00050
|
1.15 x 10-6
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| Expt 2 |
0.020
|
0.010
|
0.00050
|
2.30 x 10-6
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| Expt 3 |
0.020
|
0.020
|
0.00050
|
4.60 x 10-6
|
| Expt 4 |
0.020
|
0.020
|
0.00100
|
4.60 x 10-6
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What is the rate law for the reaction?
Answer
|
By inspection of expts 1 & 2: ([I-] [H+]
are both constant, therefore any change in the rate is due to [H2O2])
[H2O2] doubles and the rate doubles - therefore
1st order with respect to [H2O2]
By inspection of expts 2 & 3: ([H2O2] [H+]
= constant, therefore any change in the rate is due to [I-])
[I-] doubles and the rate doubles - therefore 1st order with
respect to [I-]
By inspection of expts 3 & 4: ([H2O2] [I-]
= constant, therefore any change in the rate is due to [H+])
[H+] doubles and the rate remains the same - therefore 0th order with
respect to [H+]
The rate expression is therefore: Rate =
k [H2O2]1[I-]1[H+]0
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Q631-03 A certain zero-order
reaction has a value for the rate constant, k = 0.025 mol dm
-3 s
-1
for the disappearance of A. What will be the concentration of A after 15 seconds,
if the initial concentration is 0.50 mol dm
-3?
Answer
|
For a zeroth order reaction, Rate = k[A]0
From this the concentration does not affect the rate. Therefore Rate
= k = 0.025 mol dm-3 s-1.
After 15 seconds the number of moles used up = 15 x 0.025 = 0.375 mol
dm-3
If the initial concentration = 0.5 mol dm-3
Then the mol dm-3 remaining = 0.5 - 0.375 = 0.125
mol dm-3
|
Q631-04 The following initial
rate data were collected for the reaction:
aA(g) + bB(g) → cC(g) + dD(g)
| |
[A] |
[B] |
initial rate/ mol dm-3 s-1 |
| Expt 1 |
0.42 |
1.5 x 10-2 |
2.7 x 10-5 |
| Expt 2 |
0.42 |
3.0 x 10-2 |
5.4 x 10-5 |
| Expt 3 |
0.84 |
1.5 x 10-2 |
1.1 x 10-4 |
Find the orders of reaction with respect to [A] and [B], and hence the rate
expression.
Answer
|
Firstly, remember that the stoichiometry of the equation has NO EFFECT
on the rate equation.
By inspection of experiments 1 and 2, you can see that the concentration
of B doubles while [A] remains constant. Meanwhile, the initial rate doubles. Thus [B] is directly proportional to the initial rate, therefore the
order with respect to [B] is one.
By inspection of experiments 1 and 3, you can see that the concentration
of A doubles while [B] remains constant. However, the initial rate increases by a factor of four. Thus [A] has a squared effect on the initial
rate, therefore the order with respect to [A] is 1.
The rate equation is therefore: Rate
= k [A]2[B]1
|
Q631-05 The following initial
rate data were collected for the reaction:
aA(g) + bB(g) → cC(g) + dD(g)
| |
[A] |
[B] |
initial rate/ mol dm-3 s-1 |
| Expt 1 |
0.42 |
1.5 x 10-2 |
2.7 x 10-5 |
| Expt 2 |
0.42 |
3.0 x 10-2 |
5.4 x 10-5 |
| Expt 3 |
0.84 |
1.5 x 10-2 |
2.7 x 10-5 |
Find the orders of reaction with respect to [A] and [B], and hence the rate
expression.
Answer
|
Firstly, remember that the stoichiometry of the equation has NO EFFECT
on the rate equation.
By inspection of experiments 1 and 2, you can see that the concentration
of B doubles while [A] remains constant. The initial rate meanwhile doubles. Hence, [B] is proportional to the initial rate, therefore the
order with respect to [B] is one.
By inspection of experiments 1 and 3, you can see that the concentration
of A doubles while [B] remains constant. However, the initial rate remains
unchanged. Thus [A] has no effect on the initial
rate, therefore the order with respect to [A] is 0.
The rate equation is therefore: Rate
= k [A]0[B]1
|
Q631-06 The following initial
rate data were collected for the reaction:
2NO(g) + 2H2(g) → N2(g) + 2H2O(g)
| |
[NO] |
[H2] |
initial rate/ mol N2 dm-3 s-1 |
| Expt 1 |
0.1 |
0.1 |
2.53 x 10-6 |
| Expt 2 |
0.1 |
0.2 |
5.05 x 10-6 |
| Expt 3 |
0.2 |
0.1 |
1.01 x 10-5 |
Find the orders of reaction with respect to [NO] and [H2], and
hence the rate expression.
Answer
|
Firstly, remember that the stoichiometry of the equation has NO EFFECT
on the rate equation.
By inspection of experiments 1 and 2, you can see that the [H2] doubles while [NO] remains constant. Meanwhile, the initial rate also doubles. Thus [H2] is directly proportional to the initial rate, therefore the
order with respect to [H2] is one.
By inspection of experiments 1 and 3, you can see that the concentration
of [NO] doubles while [H2] remains constant. Meanwhile, the initial rate increases by a factor of four. Thus [NO] has a squared effect on the initial
rate, therefore the order with respect to [NO] is 1.
The rate equation is therefore: Rate
= k [H2]1[NO]2
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Q631-07 The following data were
obtained for the reaction of nitrogen monoxide gas, NO(g) with oxygen gas to form
nitrogen dioxide gas, NO
2(g) at 25ºC.
| |
[NO] |
[O2] |
initial rate/ mol dm-3 s-1 |
| Expt 1 |
0.50 |
0.2 |
2.7 x 10-5 |
| Expt 2 |
0.50 |
0.4 |
5.4 x 10-5 |
| Expt 3 |
1.00 |
0.2 |
5.4 x 10-5 |
Find the orders of reaction with respect to [NO] and [O2], and hence
the rate expression.
Answer
|
Firstly, remember that the stoichiometry of the equation has NO EFFECT
on the rate equation.
By inspection of experiments 1 and 2, you can see that the [O2] doubles while [NO] remains constant. Meanwhile, the initial rate doubles. Thus [O2] has a proportional effect on the initial rate, therefore the
order with respect to [O2] is one.
By inspection of experiments 1 and 3, you can see that [NO] doubles while [O2] remains constant. Meanwhile, the initial rate remains
also doubles. Thus [NO] has a directly proportional effect on the initial
rate, therefore the order with respect to [NO] is 1.
The rate equation is therefore: Rate
= k [NO]1[O2]1
|
Q631-08 The following data were
obtained for the reaction of compounds A and B at constant temperature.
| |
initial [A]/mol dm-3 |
initial [B]/mol dm-3 |
initial rate/ mol dm-3 s-1 |
| Expt 1 |
0.15 |
0.24 |
0.45 x 10-5 |
| Expt 2 |
0.30 |
0.24 |
0.90 x 10-5 |
| Expt 3 |
0.60 |
0.48 |
7.20 x 10-5 |
Find the orders of reaction with respect to [A] and [B], and hence the rate
expression.
Answer
Firstly, remember that the stoichiometry of the equation has NO EFFECT
on the rate equation.
In experiments 1 & 2, the concentration of B is constant. Any effect on the initial rate is due to changing the concentration of A. The concentration of A doubles and at the same time the initial rate doubles. The concentration of A is directly proportional to the rate. The order with respect to [A] = 1.
Between experiments 1 & 3 the concentration of A increases by a factor of four. This should engender a similar four-fold increase in the rate. However, in fact, the rate changes from 0.45 x 10-5 to 7.20 x 10-5, representing a 16-fold increase. This can only be due to the change in the concentration of B, which has doubled. Hence, doubling [B] causes an four-fold increase in the rate. The order with respect to [B] = 1.
The rate equation is therefore: Rate
= k [A]1[B]2 |
Q631-09 The following data were
obtained for the reaction of compounds P and Q at constant temperature. The reaction
can be represented by the rate expression:
Rate = [P]2[Q]
| |
initial [P]/mol dm-3 |
initial [Q]/mol dm-3 |
initial rate/ mol dm-3 s-1 |
| Expt 1 |
0.20 |
0.30 |
4.80 x 10-3 |
| Expt 2 |
0.10 |
0.10 |
|
| Expt 3 |
0.40 |
|
9.60 x 10-3 |
| Expt 4 |
|
0.60 |
19.2 x 10-3 |
Find the concentration of Q in experiment 3.
Answer
The order with respect to [P] is 1. Hence, the doubling of [P] between experiments 1 and 3 should cause a four-fold increase in the rate. But the rate changes from 4.80 x 10-3 to 9.60 x 10-3, which represents only a two-fold increase.
The effect of [Q] on the rate is one of direct proportionality. Hence [Q] must have halved from Expt 1 to 3 to reduce the rate by a factor of 1.
Hence, the value of [Q] in experiment 3 = 0.15 mol dm-3 |
Q631-10 The bromination of acetone
that occurs in acid solution is represented by this equation.
CH3COCH3(aq) + Br2(aq) → CH3COCH2Br(aq) + H+(aq) + Br-(aq)
These kinetic data were obtained for given reaction concentrations.
|
Initial Concentrations,
/ mol dm-3
|
Initial Rate of disappearance of Br2,
/ mol dm-3 s-1
|
| [CH3COCH3] |
[Br2]
|
[H+]
|
|
|
0.30
|
0.050
|
0.050
|
5.7 x 10-5
|
|
0.30
|
0.10
|
0.050
|
5.7 x 10-5
|
|
0.30
|
0.10
|
0.10
|
1.1 x 10-4
|
|
0.40
|
0.050
|
0.20
|
3.1 x 10-4
|
Based on these data, what is the rate equation?
Answer
|
In experiments 1 & 2 [CH3COCH3] and [H+]
are constant:
Doubling [Br2] has no effect on the rate, therefore the order is 0 wrt
[Br2]
In experiments 2 & 3 [CH3COCH3] and [Br2]
are constant:
Doubling [H+] doubles the rate, therefore the order wrt [H+] is 1
In experiments 3 & 4 nothing is constant but we already know that
the order is 0 wrt [Br2]:
Doubling [H+] doubles the rate, therefore we expect an effect on the
rate because of the [H+] to be doubled. The rate should change from
1.2 x 10-4 to 2.4 x 10-4
However the new rate is 3.1 x 10-4, the extra increase must
be due to the increased concentration of [CH3COCH3]
The concentration of [CH3COCH3] increases by
a factor of 4/3.
Multiplying this factor by the value expected of the rate, 2.4 x 10-4,
gives 3.2 which is the rate in experiment 4.
The order wrt [CH3COCH3] is 1.
Therefore the rate expression is: Rate =
k [CH3COCH3]1[Br2]0[H+]1
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