Chapter 14 Chemical Kinetics

• Factors that Affect Reaction rates

• Reaction Rates

• Concentration and Rate

• The Change of Concentration with Time

• Temperature and Rate

• Reactions Mechanisms

• Catalysis

Chemical Kinetics

Is the study of the rate at which reactions occur and also gives us information

on how the reaction occurs (the Reaction Mechanism)

14.1 Factors that Affect Reaction Rates

Reaction rates depend on several factors

On a molecular level Reaction rates depend on the

frequency with which molecules collide

Require a quantitative definition of the Reaction rate of a chemical reaction

This is defined in terms of product(s) forming and reactant(s) disappearing

per unit time

N2(g) + 3H2(g) → 2NH3(g)

14.2 Reaction Rates

This is the average rate, it doesn’t gives us an actual rate at a given moment in

time

e.g. Drive the 30km to the Ferries in 30 minutes was I driving 60km/hr all the

way?

Gives information on the rate at a particular moment, for this we plot the

concentration of (product) with time and determine the slope at our time of

interest

Instantaneous Rate

Sample Exercise 14.2

Reaction Rates and Stoichiometry

For the reaction plotted 1 mole of C4H9Cl forms

1mole of C4H9OH. So rate of products formed

is equal to the rate of disappearance of

reactants.

What happens when this is not the case? e.g

2HI(g) →H2(g) + I2(g)

In this case the rate of appearance (formation) of H2 and I2 is equal, but the

rate of HI disappearing is given by

Sample Exercise 14.3

This leads to the generalisation that in a given reaction

aA + bB → cC + dD

How does the rate of reaction change as the concentration of (initial

reactants is changed)?

14.3 Concentration and Rate

For the reaction

NH4+(aq) + NO2

−(aq) N2(g) + 2 H2O(l)

The initial concentrations of both reactants are changed and the observed

rate is measured

The rate law is expressed in the general form

Rate = k[A]m[B]n Where k is the rate constant (depends on temp.)

Big k characterizes fast reaction

m and n are the reaction orders with respect to the concentration of (A and B),

Not stoichiometric coefficients, must be determined experimentally

The exponents in a rate law indicate how the rate is affected by the concentration

of each reactant, usually 0, 1 or 2

If m and n are both = 1 then the rate is first order with respect to [A] and first order

with respect to[B].

So the rate law has an overall reaction order of 1+1 = 2, and the reaction is

second order

Sample Exercise14.5 Determining orders and units

Determining Rate Laws

For the Reaction: A + B → C + D

The experimentally data was tabulated as

Expt [A] M [B] M Init.Rate M s-1

1 0.1 0.1 0.001

2 0.1 0.2 0.002

3 0.2 0.1 0.004

So now we could find the value of k by substituting in our values from, say, expt. 1,

starting with initial concentrations.

The rate of a reaction depends on concentration.

The rate constant k does not, the rate constant is affected by temperature and a

catalyst

Given a rate law we can calculate the rate of reaction using the rate constant and

initial reactant concentrations.

Or given initial concentrations and initial rate we can calculate a rate law and the

rate constant

Sample Exercise 14.6 Determining a rate law from Initial rate data

14.4 The change in Concentration with Time

A rate law is an equation that tells us how a reaction rate depends on reactant

concentrations.

But what if we are interested in how the reactant and product concentrations vary

with time?

From a rate law we can calculate the rate of reaction using the rate constant and

reactant concentrations. We now need an equation that allows us to determine the

concentration of reactants and products at any particular time

First Order Reactions

Since ln[A]t/[A]o = ln[A]t - ln[A]o

We can get this equation in the form y = mx +c by re-arranging

Sample exercise 14.7 Using the Integrated First order rate law

We can test data to see if it fits this equation ….YES? Then first order

Second Order Reactions

Sample Exercise 14.8 Determining Reaction Order from the Integrated Rate Law

Given the concentration of [A] at various times (t), tabulate to include ln[A] and

1/[A]

Plot both vs. t, which gives the linear slope?

Half-Life

The half-life of a reaction t½ is the time required for the concentration of a reactant

to reach half its original value

It is a convenient way of expressing how fast a reaction occurs

First Order Reactions

Half-Life

In contrast to first order reactions the half-life of second order reactions Does

depend on the initial concentrations

Summary of kinetics from an experimental perspective

Second Order Reactions

Rate of reaction can be obtained from the rate law (if it is known)

Without the rate law, determine the rate of a reaction by

The half-life of a second order reaction increases as the reaction progresses

Use the initial rates if they are available (no products to affect the rate)

Determine the order of reaction

OR, Find the graph of the rate data that yields a straight line

OR,Test for constancy of the half-life

Substitute rate data into integrated rate laws to find the law that yields constant

values for k, the rate constantTime (s) [A] M

0 Ao

t1 A1

t2 A2

t3 A3

Why is a reaction first or second order and why does increasing the temperature

speed up a reaction?

14.5 Temperature and Rate

The Collision Model, explains what is happening on a molecular level

In a gas phase reaction there may be 1030 collisions per second, only a small

number of collisions lead to reaction

The Molecules must also have sufficient

energy to react (breaking bonds)

The arrangement of atoms at the top of the energy barrier is the transition state

The Activation energy is the minimum amount of energy required to initiate a

chemical reaction, (form the transition state)

Arrhenius developed a relationship between the Activation Energy and the rate

constant k

Determining the Activation Energy

Determining k at several different temperatures and plotting ln k vs. 1/T the Ea can

be found from the slope Remember to plot T in Kelvin as the gas constant R has

units of J / mol-K !

Sample exercises 14.10 and 14.11

The Lower the Activation Energy, the faster the reaction

The higher the temperature the greater the proportion of molecules with KE ≥ Ea

14.6 Reaction Mechanisms

The number of molecules participating as reactants in an elementary process is

defined by the molecularity

A reaction mechanism tells us the sequence of events that describes the process

of forming products from reactants

The reaction may occur in a series of steps

The rate law can not generally be deduced from the overall balanced equation for

a reaction as it depends on the rate law of the mechanism steps (Slowest)

Rate Laws for Elementary Reactions are determined by molecular proportions

For this elementary reaction (step 1 in a 2 step mechanism)

H2 + ICl → HI + HCl

The rate Law is given as

Sample exercise 14.12 and 14.13

Multi-step Mechanisms

A balanced chemical equation often occurs in a multi-step mechanism through a

sequence of Elementary reactions

The sequence of Elementary reactions produces intermediates

One step is slower

Rate determining step

A valid mechanism must meet three criteria

Sample exercise 14.14

Two types of Mechanisms

Mechanisms with a slow initial step

2NO2(g) + F2(g) → 2NO2F(g

Sample exercise 14.14

Mechanisms with a fast initial step

If the slow step is not the first step, an intermediate is a reactant in the rate

determining step

Example:

2NO(g) + Br2(g) → 2NOBr(g)

Has an experimentally determined Rate law of Rate = k[NO]2[Br2]

The proposed mechanism is:

NO(g) + Br2(g) NOBr2(g)

NOBr2(g) + NO(g) → 2NOBr(g)

Because step 2 is the slow step the overall rate law is governed by that step

Rate = k2[NOBr2][NO]

But NOBr2 is an unknown and unstable intermediate

Because the forward and reverse reactions occur faster than step 2 an equilibrium

is established

In any dynamic equilibrium the rate of the forward reaction is equal to the rate of

the reverse reaction, so

When a fast step precedes a slow step, we can solve for the concentration of an

intermediate by assuming that an equilibrium is established in the fast step

Sample exercise 14.15

14.7 Catalysis

A catalyst increases the rate of a reaction (by lowering the Activation Energy),

without being consumed in the reaction

Homogeneous Catalyst

Heterogeneous Catalyst

The reaction I3- (aq) + 2N3

- (aq) → 3I- (aq) + 3 N2 (g) is catalyzed by CS2 (aq)

The mechanism is:

CS2 + N3- → S2CN3

- SLOW

2S2CN3- + I3

- → 2CS2 + 3N2 + 3I- FAST

What is the rate law?

Heterogeneous Catalysts

Exist in a different Phase to the reactants

For example catalytic hydrogentation of alkenes with a Ni catalyst and Catalytic

Converters.

Enzyme Catalysis

Substrate (yellow) binds to

enzyme (purple) in an active site

Each enzyme catalyses a

specific reaction in the same

way a key fits a given lock

The product is produced and the

enzyme is unchanged.

An enzyme can increase a

reaction rate up to 1018 times!

The product(s) leaving the active

site is the rate determining step,

once the products have left the

site can be filled by another

substrate molecule

At A, reduction Catalyst

Pt and Rh, rips the N out of the molecule, N + N gives N2 de-adsorbed

2NO => N2 + O2 or 2NO2 => N2 + 2O2

At B, oxidation Catalyst

Pt and Pd oxidize CO and hydrocarbons to CO2 with remaining O2 in exhaust gas

2CO + O2 => 2CO2