Control Chemical Reaction Rates

Rate of Chemical
Reactions
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OVERVIEW
For a chemical reaction to be feasible, it must occur
at a reasonable rate.
Consequently, it is important to be bale to control
the rate of reaction..
Most often, this means making it occur more
rapidly.
When you carry out a reaction in a laboratory, you
want to take place quickly.
A research chemist trying to synthesize anew drug
has the same objective.
Sometimes thought, it is desirable to reduce the
rate of reaction.

Meaning of Reaction Rate
The rate of reaction is a positive quantity that expresses
how the concentration of the reactant or product
changes in time.
To illustrate better, consider the reaction:
N2O5 (𝑔) → 2NO2 (g) +
1
2
O2 (g)
The concentration of Dinitrogen pentoxide decreases
with time; the concentration of Nitrogen dioxide and
Oxygen gas increase.
This is because these species have different coefficients
in the balanced equation, the concentration do not
change at the same rate.

The graph shows the changes
in reactant and product
concentrations with time for
the decomposition reaction of
Dinitrogen pentoxide to
Nitrogen dioxide and Oxygen
gas.
The concentrations of NO2 and
O2 increase with time, whereas
N2O5 decreases.

When one mole of N2O5 decomposes, two moles of
NO2 and one – half mole of O2 are formed:
−∆ N2O5 =
∆ NO2
2
=
∆ O2
1
2
Where Δ[ ] refers to the change in concentration in
moles per liter (or molar concentration, M).
The minus sign in front of N2O5 term is necessary
because it [N2O5] decreases as the reaction takes
place; the numbers on the right (2, ½) are the
coefficients of these species in the balanced
equation.

The rate of reaction can now be defined by dividing by
the change in time, Δt:
rate =
−∆ N2O5
∆t
=
∆ NO2
2∆t
=
∆ O2
1
2
∆t
More generally, for the reaction
𝑎A + 𝑏B → 𝑐C + 𝑑D
Where A, B, C and D represent substances in the gas
phase or in aqueous solution, and a, b, c and d are their
balanced equation, then
rate =
−∆[A]
𝑎∆t
=
−∆[B]
𝑏∆t
=
∆[C]
𝑐∆t
=
∆[D]
𝑑∆t

Let’s have an example:
Molecular nitrogen is disappearing at the rate of
0.1M per minute. What will be the concentration of
ammonia after 1 hour?
N2 + 3H2 → 2NH3
rate =
−∆ N2
∆t
=
−∆[H2]
3∆t
=
∆[NH3]
2∆t
= 0.1𝑀/min
0.1𝑀
min
=
∆[NH3]
2(60min)
∆ NH3 = 0.1𝑀 2 60 = 𝟏𝟐𝑴

Try another example:
1. For the oxidation of Ammonia:
4NH3 + 3O2 → 2N2 + 6H2O
When N2 was found to have a rate of 0.24 M/s, at what rate does water form?
At what rate did ammonia being consumed?
2. Consider the reaction in aqueous solution below:
5Br(aq)
−1
+ BrO3 (aq)
−1
+ 6H(aq)
+1
→ 3Br2 (aq) + 3H2O(l)
If the rate of disappearance of Br–(aq) at a particular moment during the
reaction is 3.5 × 10−4M/s, what is the rate of appearance of Br2(aq) at that
moment?

Measurement of Rate
For the reaction:
N2O5 (𝑔) → 2NO2 (g) +
1
2
O2 (g)
the rate could be determined by measuring:
▪ The absorption of visible light by the NO2 formed;
this species has a reddish – brown color, whereas
N2O5 and O2 are colorless.
▪ The change in pressure that results from the
increase in the number of moles of gas (1 mol
reactant ⟶ 2 ½ mol product).

Reaction Rate and Concentration
The higher the concentration of starting materials, the more
rapidly a reaction takes place.
Pure H2O2, in which its molecular concentration is about 40 M,
is an extremely dangerous substance.
In the presence of trace impurities, it decomposes explosively:
H2O2 (l) → H2O(g) +
1
2
O2 (g)
At a rate too rapid to measure.
The hydrogen peroxide you buy in the drugstore is a dilute
aqueous solution in which [H2O2]= 1M.
At this relatively low concentration, decomposition is so slow
that the solution is stable for several months.

Reaction Rate and Concentration
The dependence of reaction rate on concentration is
readily explained.
Reactions occur as the result of collision between reactant
molecules.
The higher the concentration of molecules, the greater the
number of collisions in unit time and hence the faster the
reaction.
As reactants are consumed, their concentration drop,
collisions occur less frequently, and reaction rate
decreases.

Rate Expression and Rate Constant
The dependence of reaction rate is readily determined for the
decomposition of N2O5.
In the figure on the right shows what happens when reaction
rate is plotted versus [N2O5].
As you would expect, rate increases as concentration increases,
going 0 when [N2O5] = 0 t about 0.00038M/s when [N2O5] =
0.08 M.
Moreover, as you can see, the plot of rate versus concentration
is a straight line through the origin, which means that the rate
must be directly proportional to the concentration:
rate = 𝑘[N2O5]

Rate Expression and Rate Constant
This equation is referred to as the rate expression
for the decomposition of N2O5.
It tells how the rate of reaction
N2O5 (𝑔) → 2NO2 (g) +
1
2
O2 (g)
depends on the concentration of the reactant.
The proportionality constant, k, is called a rate
constant.
It is dependent on the other quantities in the
equation.

Order of Reaction involving a Single Reactant
Rate expressions have been determined by experiment for a large number of
reactions.
For the process
A → 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠
The rate expression has a general form: 𝒓𝒂𝒕𝒆 = 𝒌[𝐀]𝐦
The power to which the concentration of reactant Ais raised in the rate
expression is called the order of the reaction, m.
If m = 0, the reaction is said to be “zero – order.”
If m = 1, the reaction is “first - order”; if m = 2, it is “second order”; and so on.
Ordinarily, the reaction order is integral (0, 1,2, …), but fractional orders, such as
3/2 are possible.

The order of reaction must be determined
experimentally; it cannot be deduced from the
coefficients in the balanced equation.
This must be true because there is only one reaction
order, but there are many ways in which the equation for
the reaction can be balanced.
For example: N2O5(𝑔) → 2NO2(g) +
1
2
O2(g)
To describe the decomposition of N2O5, it could have
been: 2N2O5(𝑔) → 4NO2(g) + O2(g)
The reaction is still first – order no matter how the
equation is written.

One way to find the order of a reaction s to measure the initial rate (i.e., the
rate at t=0) as a function of the concentration of the reactant.
Suppose that we make up two different reaction mixtures differing only in
the concentration of reactant A.
We now measure the rates at the beginning of the reaction, before the
concentration of A has decreased appreciably.
This gives two different initial rates (rate1, rate2) corresponding to two
different starting concentrations of A, [A]1 and [A]2.
From the rate expression, 𝐫𝐚𝐭𝐞𝟐 = 𝒌[𝐀]𝟐
𝒎
and 𝐫𝐚𝐭𝐞𝟏 = 𝒌[𝐀]𝟏
𝒎
Dividing the second by the first gives:
𝐫𝐚𝐭𝐞𝟐
𝐫𝐚𝐭𝐞𝟏
=
[𝐀]𝟐
𝒎
[𝐀]𝟏
𝒎 =
[𝐀]𝟐
[𝐀]𝟏
𝒎

Example:
Acetaldehyde, CH3CHO, occurs naturally in oak and
tobacco leaves, and is present in automobile and diesel
exhaust. The initial rate of decomposition of
acetaldehyde at 600OC
CH3CHO → CH4 + CO
Was measured at a series of concentrations with the
results on the table shown.
Using these data, determine the reaction order (m) in
the equation: 𝑟𝑎𝑡𝑒 = 𝑘[CH3CHO]𝑚
.
Rate [CH3CHO]
0.34 M/s 0.20 M
0.76 M/s 0.30 M
1.4 M/s 0.40 M
2.1 M/s 0.50 M

Solution:
𝐫𝐚𝐭𝐞𝟐
𝐫𝐚𝐭𝐞𝟏
=
[𝐀]𝟐
𝒎
[𝐀]𝟏
𝒎 =
[𝐀]𝟐
[𝐀]𝟏
𝒎
rate2
rate1
=
0.76
0.34
= 2.2
rate2
rate1
=
CH3CHO 2
CH3CHO 1
𝑚
= (
0.30
0.20
)𝑚 = (1.5)𝑚
2.2 = (1.5)𝑚
𝑙𝑜𝑔2.2 = 𝑙𝑜𝑔(1.5)𝑚
𝑙𝑜𝑔2.2 = 𝒎𝑙𝑜𝑔(1.5)
𝑚 =
𝑙𝑜𝑔2.2
𝑙𝑜𝑔1.5
=
0.3424
0.1761
= 1.9 ≅ 𝟐
The reaction is second order.
Rate [CH3CHO]
0.34 M/s 0.20 M
0.76 M/s 0.30 M
1.4 M/s 0.40 M
2.1 M/s 0.50 M

Order of Reaction with More than 1 Reactants
Many reactions involve more than 1 reactant.
For the reaction of two species A and B,
𝑎A + 𝑏B → 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠
The general form of the rate expression is: 𝑟𝑎𝑡𝑒 = 𝑘[A]𝑚
×
[B]𝑛
In this equation m is referred to as “the order of the reaction
with respect to A.”
Similarly, n is “the order of the reaction with respect to B.”
The overall order of the reaction is the sum of the exponents,
m+n.
If m=1, n=2, then the reaction is first order in A, second in B
and third – order – overall.

When more than one reactant is involved, the order can be
determined by holding the initial concentration of one
reactant constant while varying that of the other reactant.
From rates measured under these conditions, it is possible to
deduce the order of the reaction with respect to the reactant
whose initial concentration is varied.
To see how to do this, consider the reaction between A and B
referred.
Suppose we run two different experiments in which the
initial concentrations of A differ ([A]1, [A]2) but that of B held
constant at [B]. Then,
𝑟𝑎𝑡𝑒1 = 𝑘[A]1
𝑚
× [B]𝑛
𝑟𝑎𝑡𝑒2 = 𝑘[A]2
𝑚
× [B]𝑛

Dividing the second equation by the first,
𝑟𝑎𝑡𝑒2
𝑟𝑎𝑡𝑒1
=
𝑘[A]2
𝑚
× [B]𝑛
𝑘[A]1
𝑚
× [B]𝑛
=
[A]2
𝑚
[A]1
𝑚 =
[A]2
[A]1
𝑚
Knowing the two rates and the ratio of the two
concentrations, we can readily find the value of m.
Example:
Consider the reaction between t-butylbromide and a base
at 55OC: CH3 3CBr + OH−1
→ CH3 3COH + Br−1
A series of experiments is carried out by the results on the
right.
Find the order of reaction with respect to both the
reactant.
Exp [(CH3)3CBr] [OH-1] Rate
M/s
1 0.50 M 0.05 M 0.0
05
2 1.0 M 0.05 M 0.0
1
3 1.50 M 0.05 M 0.0
15
4 1.0 M 0.10 M 0.0
1
5 1.0 M 0.10 M 0.0
1

To solve for m,
Rate ratio:
𝑟𝑎𝑡𝑒3
𝑟𝑎𝑡𝑒1
=
0.015
0.005
= 3
Concentration ratio:
[ CH3 3CBr]3
[ CH3 3CBr]1
𝑚
= (
1.5
0.5
)𝑚
= 3𝑚
3𝑚
= 3
log3𝑚
= log3
mlog3=log3
m=1
To solve for n,
Rate ratio:
𝑟𝑎𝑡𝑒5
𝑟𝑎𝑡𝑒2
=
0.010
0.010
= 1
Concentration ratio:
[OH−]5
[OH−]2
𝑚
=
0.20
0.05
𝑚
= 4𝑚
4m=1
log4m=log1
mlog4=log1
m=0
The reaction is first order with respect to t-butylbromide and zero-order
with respect to OH—1.
Exp [(CH3)3CBr] [OH-1] Rate
M/s
1 0.50 M 0.05 M 0.0
05
2 1.0 M 0.05 M 0.0
1
3 1.50 M 0.05 M 0.0
15
4 1.0 M 0.10 M 0.0
1
5 1.0 M 0.10 M 0.0
1

Order of Reaction: Your Turn
1. Consider the following hypothetical reaction:
X(g)⟶Y(g)
The disappearance of X is monitored at timed
intervals shown in the table on the right. Assume
that the temperature and volume are kept constant.
Calculate the order of reaction with respect to X.
2. The peroxisulfate ion reacts with the iodide ion
in aqueous solution according to the following
equation: S2O8(𝑎𝑞)
−2
+ 3I(𝑎𝑞)
−1
→ 2SO4(𝑎𝑞)
−2
+ I3(aq)
−1
What is the order of reaction with respect to [S2O8
—
2] and [I—1]?
Expt. [X] Rate
1 0.600M 0
2 0.515M 0.00043 M/s
3 0.425M 0.00018 M/s
4 0.355M 0.00010 M/s
5 0.330M 0.00006 M/s
Expt [S2O8
—2]
(M)
[I—1]
(M)
Rate
(M/min)
1 0.0200 0.0155 1.15x10-4
2 0.0250 0.0200 1.85x10-4
3 0.0300 0.0020 2.22x10-4
4 0.0300 0.0275 3.06x10-4

Models for Reaction Rate – Collision Model;
Activation Energy
For every reaction, there is a certain minimum energy that molecules
must possess for collision to be effective.
This is referred to as activation energy.
It has a symbol Ea, and is expressed in kJ/mol.
The collision model of reaction rates just developed can be made
quantitative.
The rate constant for a reaction, k, is the product of three factors:
k = p x Z x f
Where:
▪ p, called a steric factor, considers the fact that only certain
orientations of colliding molecules are likely to lead to reaction.
▪ Z, is the collision frequency, which gives the number of molecular
collisions occurring in unit time at unit concentrations of reactants.
▪ f, the fraction of collisions in which the energy of the colliding
molecules is equal to or greater than Ea. (f=e—Ea/RT)

Models for Reaction Rate – Transition-State
Model; Activation Energy Diagrams
The idea of the activated complex was developed by
among others, Henry Erying at Princeton in the 1930s.
It forms the basis of the transition – state model for
reaction rate, which assumes that the activated complex
- is in equilibrium, at low concentrations, with the
reactants.
- May either decompose to products by “climbing” over
the energy barrier, or alternatively, revert back to the
reactants.
The transition- state model is generally somewhat more
accurate than the collision model.
It explains why the activation energy is ordinarily much
smaller than the bond enthalpies in the reactant
molecules.

Reaction Rate and Temperature
The rates of most reactions increase as the
temperature rises.
The effect of temperature on reaction rate can be
explained in terms of kinetic theory.
Raising the temperature greatly increases the fraction
of molecules having very high speeds and hence high
kinetic energies most likely to react when they collide.
The higher the temperature, the larger the fraction of
molecules that can provide the activation of energy
required for the reaction.
The increase of fraction of effective collision causes
reaction rate to increase with temperature.

Catalysis
A catalyst is a substance that increases the rate of a
reaction without being consumed by it.
It does this by changing the reaction path to one with a
lower activation energy.
The catalyzed path consists of two or more steps in the
catalyzed reaction.
A heterogeneous catalyst is one that is in a different phase
from the reaction mixture.
Most commonly, the catalyst is a solid that increases the rate
of a gas – phase or liquid – phase reaction. An example is a
decomposition of Nitrous oxide on gold:
N2O
Au
N2 +
1
2
O2

Catalysis
A homogeneous catalyst is one that is present in the
same phase as the reactants.
It speeds up the reaction by forming a reactive
intermediate that decomposes to give products.
In this way, the catalyst provides an alternative process
of lower activation energy.
Many reactions that take place slowly under ordinary
conditions occur readily in living organisms in the
presence of catalysts called enzymes.
Enzymes are protein molecules of high molar mass.

Catalysis
Enzymes, like all other catalysts, lower the activation
energy for reaction.
They can be enormously effective; it is not uncommon
for the rate constant to increase by a factor of 1012 or
more.
However, from a commercial standpoint, enzymes have
drawbacks.
A particular enzyme operates best over a narrow range
of temperature.
An increase in temperature frequently deactivates an
enzyme by causing molecules to “unfold,” changing its
characteristic shape.

Reaction Mechanisms
A reaction mechanism is a description of a path, or
sequence of steps, by which the reaction occurs at the
molecular level.
In the simplest case, only a single step is involved.
This is a collision between two reactant molecules.
Reaction mechanisms frequently change with
temperature and sometimes with pressure.
The nature of the rate expression and hence the
reaction order depends on the mechanism by which the
reaction takes place.

Control Chemical Reaction Rates

Control Chemical Reaction Rates

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