Lok Sin Tong Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 14: Rate Equations and Order of Reactions
Chpt.14: p.1
RATE EQUATIONS AND ORDER OF REACTIONS
I.
SIMPLE RATE EQUATIONS
In the general reaction xA + yB  products
experiments show that the reaction rate can be related to the concentrations of individual
reactants by the rate equation or rate law for the reaction under consideration:
Rate k[A]m[B]n
Note : In the rate equation,
<1> rate has the unit of mol dm-3 s-1
<2> [A] and [B] denotes the concentration of the reactants A and B respectively. The units are
mol dm-3.
<3> m and n are constants. The index m is the order of the reaction with respect to A, and
n is the order of the reaction with respect to B. The constants can be positive or negative
and can be integer or a fraction, but the values are usually 0, 1 or 2.
<4> the overall order of the reaction is (m + n).
<5> k is the rate constant for the reaction.
(A) Order of Reaction
The order of a reaction is the sum of the powers of the concentration terms in the experimentally
determined rate equation. This is (m+n) in the above equation.
The order of reaction or the rate expression cannot be determined from the stoichiometric
equation for the reaction. It can only be determined by experiments.
(B) Rate Constant
Rate constant k is a constant whose value depends on the temperature, not the concentration of
reactants. Its value is determined by experiments.
Note
<1> In general, the greater the rate constant, the faster reaction is.
<2> The units of k depends on the order of the reaction.
e.g.
Consider the following rate equation:
rate = k[A]2[B]
then substituting the units of the various quantities,
mol dm-3 s-1 = k (mol dm-3)2(mol dm-3)1
from which the units of k are dm6 mol2 s-1
Order of reaction
Rate equation
Units of rate constant
0
mol dm-3 s-1
Zeroth
rate = k[reactant]
First
rate = k [reactant]1
s-1
Second
rate = k [reactant]2
dm3 mol-1 s-1
Lok Sin Tong Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 14: Rate Equations and Order of Reactions
(C)
Chpt.14: p.2
Simple Rate Equations determined from experimental results
In the following exercises, the order of reaction and rate constant can be determined from a set
of data.
Exercise 1
In a kinetic study, the concentrations of reactant A and B, and the rate of formation of the
product C. were measured at 300K in a number of runs.
Run
1
2
3
4
5
6
7
8
(a)
(b)
(c)
ANSWER.
Concentration of A
(mol dm-3)
0.1
0.0
0.2
0.1
0.3
0.2
0.1
0.3
Concentration of B Initial rate of formation
(mol dm-3)
of C (mol dm-3 s-1)
0.1
0.01 x 10-3
0.2
0.0
0.3
0.18x 10-3
0.3
0.09 x 10-3
0.0
0.0
0.2
0.08 x 10-3
0.2
0.04x10-3
0.1
0.03x10-3
With the information given above, develop a rate law for the reaction.
Calculate the rate constant k.
If 0.6 mol of A and 0.3 mol of B are mixed in a 3 dm3 flask, determine the initial rate of
formation of C.
Lok Sin Tong Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 14: Rate Equations and Order of Reactions
Exercise 2
Nitrogen(II) oxide and hydrogen react at 1000K according to the equation:
2N0(g) + 2H2(g)  N2(g) + 2H20(g)
The following kinetic data are obtained:
Initial concentration / mol dm3
(a)
(b)
ANSWER
Initial rate / mol dm-3s-1.
NO
H2
0.0250
0.100
2.4 x 10.6
0.0250
0.050
1.2 x 10.6
0.0 125
0.100
0.6 x 10.6
Determine the order of the reaction with respect to each reactant.
Determine the rate constant for the reaction at 1000K.
Chpt.14: p.3
Lok Sin Tong Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 14: Rate Equations and Order of Reactions
II.
Chpt.14: p.4
ZEROTH, FIRST AND SECOND ORDER REACTION
Order of reaction with respect to a particular reactant is the power of concentration term of
that reactant in the rate equation.
(A)
Zeroth order reaction
For a zeroth order reaction (with respect to a particular reactant), the rate is constant and does
not change with the concentration of that reactant.
Note
<1> The rate law for a zeroth order reaction is
Rate = k[A]0 = k
Example : The decomposition of ammonia to form H2 and N2 is a
zeroth order reaction with respect to NH3.
2NH3(g)  N2(g) +H2(g)
Rate = k. The reaction is independent of [NH3]
<2>
Since rate = k, thus
log(rate) = log k
A graph of log(rate) versus log[A] gives a straight line with a slope of 0. This shows that
the reaction is zeroth order.
<3>
A plot of [A] versus time gives a straight line of slope -k.
(B)
First order reaction
For a first order reaction (with respect to a particular reactant), the rate is directly proportional to
the concentration of that reactant.
Note
<1> The rate law for a first order reaction is
Rate = k[A]
A plot of the rate against (A) gives a straight line of slope k.
<2> Example : Decomposition of N2O5, to form NO2 and O2
2N2O5(g)  4NO2(g) + O2(g)
rate = k[N2O5]
<3> Since
log(rate) = log k + log[A]
A graph of log(rate) versus log(A] gives a straight line with s slope of 1. This shows that
the reaction is first order.
<4> A plot of ln [A] versus time gives a straight line of slope -k
<5> Radioactive decay is a first order reaction which has a constant half-lfe.
Lok Sin Tong Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 14: Rate Equations and Order of Reactions
(C)
Chpt.14: p.5
Second order reaction
For a second order reaction (with respect to a particular reactant), the rate is directly
proportional to square of the concentration of that reactant.
Note
<1> The rate law for a second order reaction is
Rate = k[A]2
A plot of the rate against [A]2 gives a straight line is produced.
<2> Example :
Decomposition of NO2 to form NO and O2
2NO2(g)  2NO(g) + O2(g)
rate = k[NO2]2
<3> Since
log(rate) = log k + 2log[A]
A graph of log (rate) versus log[M] gives a straight line with slope of 2. This shows that
the reaction is second order.
<4> A plot of
1 / [A] versus time gives a straight line graph of slope k.
Lok Sin Tong Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 14: Rate Equations and Order of Reactions
(D)
Comparison of Rate laws
Chpt.14: p.6
Lok Sin Tong Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 14: Rate Equations and Order of Reactions
(E)
Chpt.14: p.7
Rate-Concentration graph
The diagram below compares the type of rate-concentration graph obtained for a particular
reactant when the reaction is zeroth order, first order and second order with respect to that
reactant.
Exercise 4
Consider the reaction
A+BX+Y
which is first order with respect to A and second order with respect to B.
(a) Write an expression of the reaction.
(b) Draw and label suitable straight line graphs that would allow the rate constant to be
determined from a series of experiments in which
(1) [A], is kept constant, but [B] is varied.
(2) [B] is kept constant, but [A] is varied.
ANSWER
Lok Sin Tong Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 14: Rate Equations and Order of Reactions
Exercise 7
The following kinetic results are obtained for the decomposition of NO2
2NO2(g)  2NO(g) + O2(g)
at 300 C.
If the rate expression is rate = k [NO2]x
Plot a suitable graph to determine
(a) the order of the reaction, and
(b)
ANSWER
the rate constant.
Chpt.14: p.8
Lok Sin Tong Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 14: Rate Equations and Order of Reactions
Chpt.14: p.9
Exercise 8
The results of an experiment to investigate the rate of the reaction
H2O2(aq) + 2H+(aq) + 2I-(aq)  I2(aq) + 2H2O(l)
at room temperature are given in the table below:
Run
Volume of 1.0 M
H2SO4 (cm3)
Volume of
0.1 M KI (cm3)
Water
(cm’)
0.1M H202
(cm’)
1/t (s-1)
1
10.0
5.0
20.0
5.0,
0.013
2
10.0
10.0
15.0
5.0
0.022
3
10.0
15.0
10.0
5.0
0.032
4
10.0
20.0
5.0
5.0
0.044
(a)
(b)
(c)
ANSWER
How would you determine t for each sample, where t is the time for a small fixed
quantity of I2 to be formed?
From the data given in the table, determine the order of reaction with respect to I-.
Suggest how you would modify the experiment to determine the order of the reaction
with respect to H+(aq) ion.
Lok Sin Tong Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 14: Rate Equations and Order of Reactions
Chpt.14: p.10
III. RADIOACTIVE DECAY
Nuclear decays are typical of first order kinetics.
The integrated form of a first order reaction (nuclear decay) may be expressed as
Nt = N0 e-kt
and ln N0 / Nt = kt
where N0 and Nt respectively represent the initial number and number at time t of the
radioisotope, and
k is the decay constant.
Note
<1> Since ln[Nt] = ln[N0] - kt
a plot of ln[Nt] against time gives a straight line of slope -k and an intercept of ln[N0].
<2> The number of radioactive isotope at time t follows an exponential decay. which has a
constant half-life.
(A)
Half—life of First order reaction
In first order reaction, half—life is the time it takes for the concentration of reactant to decrease
to half the original concentration.
In radioactive decay, which is a typical first order reaction,
Half—life is the time it takes for the radioactivity of a sample (or number of radioactive
isotopes) to decrease to half the original activity (or half the original number of radioactive
isotopes.
Since ln N0 / Nt = kt, and after a half-life, the number of radioactive isotope is half the original
number, the half-life is found by putting N0 = 2Nt
ln 2N / N = kti/2
or
ln2 = kt1/2
t1/2 = ln 2 / k = 0.693 / k
Note
The equation shows that the half-life does not depend on concentration. Half-life can be
determined if k is known. In fact,
The half—life is constant for radioactive decays.
Lok Sin Tong Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 14: Rate Equations and Order of Reactions
Chpt.14: p.11
Exercise 9
The half-life of Krypton-85 is 10.6 years. The number of half- lives, x for 85Kr to decay to
1/100 of its initial value is given by (1/2)x = 1/100.
How long will it take for 99% of the 85Kr to disintegrate?
ANSWER
Exercise 10
The rate constant for the decay of 222Rn is 0.1810 day-1
(a) What is the half-life for 222Pn expressed in days?
(b) What fraction of a sample of 222Rn decays in a period of exactly one week?
ANSWER
Exercise 11
Plutonium-239, an isotope produced by nuclear reactors, has a half-life of 24000 years.
Given that the world’s electrical power reactors are producing about 20000 kg of 239Pu each
year, calculate how long it would take for one year’s total production of 239Pu to decay to 1.0
kg.
ANSWER
Lok Sin Tong Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 14: Rate Equations and Order of Reactions
IV.
Chpt.14: p.12
CARBON—14 DATING
Radioactive carbon-14 is produced in the upper atmosphere by the bombardment of nitrogen
atoms with neutrons formed by cosmic radiation:
14
1
14
1
7N + 0n  6C + 1H
Note
<1> This radioactive carbon is distributed over the earth’s surface in the form of carbon
dioxide (CO2) in the air and hydrogencarbonate ions (HCO3) in the sea. It is therefore
incorporated into growing plants as they photosynthesize.
<2> Carbon-14 decomposes through  particle production.
14
 7N14 + -1e0
6C
(A)
<3>
Over the years, the production and decomposition of carbon-14 maintains a constant
amount of carbon-14 in the atmosphere.
<4>
Radioactive carbon-14 decays with a half-life of 5570 years. If it is assumed that the
level of carbon-14 has remained constant over centuries, the age of articles of
histological and archaeological interest can be calculated.
Determination of the age of an archaeological specimen using Carbon-14 dating
Example : A piece of wood from ancient time has a count rate of 10 per minute. The count
rate form the same mass of a similar piece of modern wood is 15 per minute.
Determine the age of the ancient wood.
Calculation
Since radioactive decay reactions are first order,
ln N0/Nt = kt
where
N0 =
number of C-14 atoms in the archaeological specimen at ancient times
Nt = number of C-14 atoms in the archaeological specimen at present
k = rate constant for the radioactive decay reaction (decay constant)
t =
age of the archaeological specimen.
At time equals to half-life of C-14, t1/2
Nt = 1/2 N0 and t1/2 = ln 2 / k = 0.693 / k
Knowing the half-life of C-14, we have
k = 0.693
t1/2
= 1.24 x 104 year –1
Lok Sin Tong Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 14: Rate Equations and Order of Reactions
Chpt.14: p.13
If the number of C-14 atoms is proportional to the number of counts, then
ln 15/10 = 1.24x10-4 t
t = 3270 years
n
Alternatively, 15 x (15/10) = 10, where n=number of half- lives passed.
n =
0.585
t =
0.585x 5570
= 3258 years.
Exercise 13
The radioactive isotope carbon-14 is formed by the bombardment on neutrons nitrogen,
Carbon-14 in turn disintegrates by the emission of  particles, with a half-life of about 5570
years.
(a) Write nuclear equations (in terms of atomic numbers and atomic mass numbers) to show the
formation of carbon—14 and its decay.
(b)If 32 counts per minute per gram of carbon is emitted by living organisms, while a fossilized bone
gives 8 counts per minute of carbon, determine the age of the fossil.
ANSWER
Lok Sin Tong Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 14: Rate Equations and Order of Reactions
Chpt.14: p.14
Exercise 14
A radioactive element decays with a rate constant of 2.0x104s1.
How long will it take for 0.5 g of the substance to decay to 0.10 g?
ANSWER
Exercise 15
If the half-life of a radioactive element is 150s, what percentage of the isotope will remain after
600s?
ANSWER
Exercise 16
If a radioisotope loses 95% of its activity in 110 minutes, what is its half-life?
ANSWER