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Transcript
CHAPTER SEVEN: REACTION KINETICS
Rates of Reaction
If we have the simple reaction
A
ļ‚® B
Then we can express the rate of reaction either in terms of the disappearance of A or in the production of B:
š‘…š‘Žš‘”š‘’ =
āˆ†[šµ]
āˆ†š‘”
š‘œš‘Ÿ
š‘…š‘Žš‘”š‘’ =
āˆ†[š“]
āˆ†š‘”
where Ī”[A] and Ī”[B] are the change in the molar concentrations and Ī”t is the time for the change to occur.
Question 7.1: What units could the rate be expressed in for the above reaction?
For our simple reaction (A ļ‚® B) we could write down the following equality
āˆ†[šµ]
āˆ†[š“]
= āˆ’
āˆ†š‘”
āˆ†š‘”
as for every molecule of A that disappears, a B molecule ā€œappearsā€. This will not be true however for all reactions.
If we take the following reaction (which is the industrial Haber process)
N2(g) + 3 H2(g) ļ‚® 2 NH3(g)
then the rate of appearance of ammonia molecules will be twice the rate of disappearance of nitrogen molecules, as
for every N2 molecules that reacts, two molecules of NH 3 will appear. The following equality now holds
2āˆ†[š‘2 ]
āˆ†[š‘š»3 ]
=
āˆ†š‘”
āˆ†š‘”
Be careful and note which term has been doubled here. The rate of formation of ammonia is twice that of the rate of
disappearance of N2.
i.e.,
Ī”[NH3] = Ī”[N2] x
2 š‘šš‘œš‘™ š‘š»3
1 š‘šš‘œš‘™ š‘š»3
Therefore it is straightforward to convert the rate of a given reaction into different units if you know the
stoichiometry of the equation.
122
Question 7.2: If the rate of formation of NH3 in the Haber process is given as 1.92 × 10-2 mol NH3 dm-3 s-1, what is
the rate of disappearance of H2?
The major problem that we have when measuring rates of reactions is that they do not remain constant throughout
the time it takes for the reaction to reach completion (or equilibrium).
Question 7.3: If the rate of reaction for A ļ‚® B did remain constant during a reaction what would be the shape of the
following graphs?
[A]
[B]
Time
Time
Instead, the rate of reaction often depends upon the concentration of the reactants. As the concentration of the
reactants changes throughout the course of the reaction, so does the rate. The initial rate (i.e. the reaction rate
during the first few seconds of the reaction) is often the fastest as this is when the concentration of the reactants is
greatest.
For our reaction of A ļ‚® B, the concentration of A might vary like shown below
[A]
Time
The rate of reaction at any given time, (the instantaneous rate), is equal to the negative slope of the tangent to the
concentration curve at that particular time. The slope of the tangent will decrease with time for the graph drawn.
123
Question 7.4: (i) Draw a graph of [B] (y-axis) vs. time for our theoretical equation A ļ‚® B. Indicate on the graph
how you would measure the instantaneous rate at a given time.
(ii) Is the instantaneous rate the same as the average rate of reaction? Explain your answer.
Therefore for chemists to be able to calculate reaction rates they need to measure either the concentration of one of
the reactants or the concentration of one of the products, depending upon which is easier to do experimentally
(remember we can always convert the units of the reaction rate if we know the stoichiometry of the reaction).
Initial Rates of Reaction
As rates of reaction depend upon the extent of reaction, chemists normally talk about the initial reaction rate, i.e. the
rate of reaction right at the start of the reaction (time = 0), before any products are present. This can be calculated as
for at any other time.
[Reactant]
Slope of tangent at t = 0 is equal
to the initial reaction rate
t=0
Time
What experimental factors is the initial reaction rate dependent upon? Firstly it depends upon the temperature at
which the reaction takes place. The vast majority of chemical reactions exhibit an increase in rate when the
temperature is increased. Secondly, the initial rate of a chemical reaction depends upon the initial concentration of
reactants ā€“ the greater the initial concentration the greater the initial rate of reaction.
Finally, it must be
remembered that each and every chemical reaction has a different initial rate of reaction (except by coincidence) as
the rate depends upon the chemical identity of the reactants and the particular reaction they undergo. If we study
one particular chemical reaction at a given temperature, then we can get some insight as to the dependence of the
initial rate of that reaction on the initial concentration.
124
Example 7.1: The following decomposition was studied at a given temperature.
2 N2O5(g) ļ‚® 4 NO2(g) + O2(g)
(7.1)
The following initial rates of reaction were measured for a range of initial concentrations of N 2O5(g), [N2O5]o.
[N2O5]o
0.010
0.020
0.040
0.060
0.080
Initial reaction rate/mol dm-3 s-1
5.20 × 10-5
1.04 × 10-4
2.08 × 10-4
3.12 × 10-4
4.16 × 10-4
(i) Confirm that the initial rate of reaction is proportional to [N2O5]o.
(ii) Calculate the initial rate of reaction at this temperature when [N2O5]o = 0.062 mol dm-3.
Solution:
125
Question 7.5: Determine the relationship between the initial rate of reaction and the reactant concentration
for the following reaction at 300°C. Show your reasoning clearly.
2 NO2(g) ļ‚® 2 NO(g) + O2(g)
[NO2]o mol dm-3
3.0 × 10-2
4.5 × 10-2
6.0 × 10-2
8.4 × 10-2
1.0 × 10-1
(7.2)
Initial rate mol dm-3 h-1
2.5
5.6
10
19
28
The relationship between the initial rate of reaction and the initial concentration of a reactant is called the ORDER
of the reaction with respect to that particular reactant. For example for the reaction
2 N2O5(g) ļ‚® 4 NO2(g) + O2(g)
it was found that the initial rate was proportional to [N 2O5]o
We can write this as
Rate ļ” [N2O5]1
The order then is the value of the exponent that the concentration of the reactant is raised to.
The reaction is thus 1st (first) order with respect to N2O5. Hopefully you would have calculated that for the reaction
2 NO2(g) ļ‚® 2 NO(g) + O2(g)
nd
the reaction rate is 2 (second) order with respect to [NO2]o (Question 7.5).
Question 7.6: Some reactions can be zero order with respect to a particular reactant. Draw a graph showing the
behaviour of the initial reaction rate versus the concentration of a zero order reactant.
Rate Laws
When we relate the order of a reactant to the rate of a reaction, we are producing a mathematical relationship which
explains the experimental data that have been measured. Thus we found that for equation 7.1
Rate ļ” [N2O5]
and for equation 7.2
Rate ļ” [NO2]2
These mathematical expressions to describe rates of reaction are known as Rate Laws.
Most reactions of course, involve more than one reactant and the rate law for these equations must include the order
for each of the reactants.
126
For example, take the following reaction
2 NO(g) + O2(g) ļ‚® 2 NO2(g)
(7.3)
Experimental data proves the following relationships hold.
Rate ļ” [NO]2
Rate ļ” [O2]
Thus the reaction is 1st order with respect to O2(g) but 2nd order with respect to NO(g). The overall rate can thus be
expressed as
Rate ļ” [NO]2 [O2]
The overall order is three, the sum of the individual reactant orders.
For reaction 7.3, the order for each reactant is equal to its stoichiometric factor in the equation
2 NO (g) + 1 O2 (g)
Rate ļ” [NO]2
ļ‚® 2 NO2 (g)
Rate ļ” [O2]1
This is by no means always true. For instance the reaction between carbon monoxide and nitrogen dioxide
NO2(g) + CO(g) ļ‚® CO2(g) + NO(g)
is found to obey the following rate law
Rate ļ” [NO2]2
The rate laws arise from the reaction mechanism i.e. how the molecules actually break and form bonds. The orders
for each reactant must be determined experimentally.
However, in many reactions which have a relatively straightforward reaction mechanism the order for each reactant
is often the same as its stoichiometric factor in the balanced chemical equation. For example,
H2 (g) + I2 (g) ļ‚® 2 HI (g)
The rate law is found to be
Rate ļ” [H2] [I2]
Elementary Reactions are those which exhibit this relationship between the stoichiometric factor and the order for
the rate law.
NOTE: One must never assume that a reaction is an elementary reaction. One must instead derive the rate law from
experimentation.
127
To fully express a rate law we must include the constant of proportionality. Thus for equation 7.1, as
Rate ļ” [N2O5]
and we can express the rate law for this reaction as
Rate = k [N2O5]
where k is the constant of proportionality. This value is known as the rate constant.
The rate constant, k, is a true constant for a particular reaction, at a particular temperature
The rate constant does have units. However these depend upon the exact form of the rate law (in particular the
overall order of the rate law).
For the 1st order reaction in equation 7.1
Rate
=
k [N2O5]
-3 -1
=
k × (mol dm-3)
=
s-1
(mol dm s )
ļœk
For the second order equation 7.2
Rate = k × [NO2]2
k = (mol dm-3 s-1)/(mol2 dm-6) = mol-1 dm3 s-1
Question 7.7: Determine the units of the rate constant for a reaction which has an overall order of three.
Determination of the Rate Law
In the previous section we noted that the order of each reactant could be determined experimentally by measuring
the initial rate of reaction over a range of initial concentrations. If we do this for each reactant then it is possible to
determine the overall order of the reaction and consequently the rate constant.
Example 7.2: The initial rate of formation of a substance X in the hypothetical reaction
A+B+Cļ‚®X+Y
was studied at a constant temperature for a variety of different initial concentrations of A, B, and C. The following
data were determined:
Experiment
1
2
3
4
Initial concentration mol dm-3
A
0.010
0.020
0.020
0.010
B
0.010
0.010
0.010
0.020
C
0.010
0.010
0.020
0.010
128
Initial rate
mol (X) dm-3 s-1
× 105
5.0
10.0
20.0
20.0
Determine the rate law, the order of each reactant and the overall order.
Solution:
129
Question 7.8: CO2, H2 and N2 can react at very high temperatures to form an amino acid (the building blocks of
DNA). This reaction is an elementary one and can be represented by the following equation
x CO2(g) + y H2(g) + z N2(g) ļ‚® amino acid
Determine the rate law given the following data:
Experiment
1
2
3
4
[CO2]
1.0
1.0
1.0
2.0
Initial concentrations
mol dm-3 × 101
[H2]
1.0
1.0
2.0
1.0
Hint: Read the headings in the table carefully.
130
Initial reaction rate × 105
mol (amino acid) dm-3 s-1
[N2]
1.0
3.0
1.0
3.0
5.0
15
160
60
Change of Concentration with Time
From the rate law, we are able to calculate the rate of a reaction from the rate constant and the reactant
concentrations. We can also derive an equation so that we can calculate the concentration of the reactants and
products at any given time during the reaction.
Zero-Order Reactions
In a zero-order reaction, the rate is independent of the concentration of the reacting species.
For a reaction of the type
A ļ‚® products
the rate law is given by:
Rate = - (Ī”[A]/Ī”t) = k[A]o = k
By integrating this differential rate law, we obtain an equation that relates the concentration of A at the start of the
reaction, [A]o , to the concentration at any other time t, [A]t
[A]t =
- kt
+ [A]o
First-Order Reactions
A first-order reactionā€™s rate depends on the concentration of a single reactant raised to the first power.
For a reaction of the type
A ļ‚® products
the rate law may be first order:
Rate = - (Ī”[A]/Ī”t) = k[A]
Integrating the above equation leads to:
ln[A]t - ln[A]o =
ln[A]t =
- kt or ln([A]t /[A]o) = - kt
- kt
+
ln[A]o
The above equation has the form of the general equation for a straight line, y = mx + c
For a first-order reaction, a graph of ln[A]t vs time gives a straight line with slope of -k and the y-intercept of ln[A]o .
Second-Order Reactions
A second-order reactionā€™s rate depends on the concentration of a single reactant raised to the second power.
Letā€™s consider reactions of the type
A ļ‚® products and A + B ļ‚® products
that are second order in A:
Rate = - (Ī”[A]/Ī”t) = k[A]2
Integrating the above equation leads to:
1/[A]t = kt + 1/[A]o
The above equation is also in the form of the general equation for a straight line, y = mx + c
If the reaction is second order, a plot of 1/[A] t vs t yields a straight line with slope = k and the y-intercept = 1/[A]o
131
Example 7.3: The following data were obtained for the gas-phase decomposition of nitrogen dioxide at 300°C,
NO2(g) ļ‚® NO(g) + ½ O2(g)
Time (s)
[NO2] (M)
0.0
0.01000
50.0
0.00787
100.0
0.00649
200.0
0.00481
300.0
0.00380
Is the reaction first or second order?
Solution:
132
Half-Life
The half-life of a reaction, t½ , is the time require for the concentration of the reactant to reach one-half of its initial
value.
For a first-order reaction:
t½ = 0.693/k
whereas, for a second-order reaction:
t½ = 1/k [A]o
Example 7.4: From the following data for the first-order gas-phase isomerization of CH3NC at 215°C, calculate the
first-order rate constant and the half-life for the reaction.
Time (s)
Pressure CH3N (kPa)
0
0.661
2000
0.441
5000
0.237
8000
0.126
12000
0.0549
15000
0.0295
Solution:
133