TOPIC 16
CHEMICAL KINETICS
16.1
Rate Expression and Reaction Mechanism
By: Merinda Sautel Alameda Int’l Jr/Sr High School Lakewood, CO [email protected]
ESSENTIAL IDEA
Rate expressions can only be determined empirically and these limit possible reaction mechanisms. In particular cases, such as a linear chain of elementary
reactions, no equilibria and only one significant
activation barrier, the rate equation is equivalent to the slowest step of the reaction.
NATURE OF SCIENCE (2.7)
Principle of Occam’s razor – newer theories need to remain as simple as possible while maximizing explanatory power.
The low probability of three molecule collisions means
stepwise reaction mechanisms are more likely.
INTERNATIONAL-MINDEDNESS
The first catalyst used in industry was for the production of sulfuric acid. Sulfuric acid
production closely mirrored a country’s
economic health for a long time. What are some current indicators of a country’s economic
health?
THEORY OF KNOWLEDGE
Reaction mechanism can be supported by indirect evidence. What is the role of
empirical evidence in scientific theories? Can
we ever be certain in science?
UNDERSTANDING/KEY IDEA 16.1.A
The order of a reaction can be either integer or fractional in
nature. The order of a reaction can
describe, with respect to a reactant,
the number of particles taking part
in the rate-determining step.
Distinguish between the terms rate
constant, overall order of reaction
and order of reaction with respect
to a particular reactant.
DEFINITIONS
The order of reaction with respect to a
particular reactant is the power to which its concentration is raised in the rate equation.
The overall order for the reaction is the sum of all the individual orders for all reactants.
The rate constant (k) is a constant for a particular reaction at a specified
temperature.
RATE LAW EXPRESSION
A + B PRODUCTS
The rate law can be written as follows:
R = k [A]
m[B]
nwhere R is the rate
k is the rate constant A and B are reactants
m and n are the experimentally found
orders of the reaction with respect to
each reactant
2H 2 + 2NO 2H 2 O + N 2
R = k [NO]
2[H
2]
Notice that the coefficients in the equation
are not the exponents in the rate expression.
Please do not confuse this with equilibrium.
This is a 2
ndorder rxn with respect to NO and a 1
storder rxn with respect to H
2.
The overall order is found by adding the
individual orders 2 + 1 = 3.
UNDERSTANDING/KEY IDEA 16.1.B
The value of the rate constant (k) is affected by temperature and its
units are determined from the
overall order of the reaction.
UNITS OF THE RATE CONSTANT (k)
Zero order Rate = k k = units of rate
= mol dm
-3s
-1
First order Rate = k[A] k = units of rate/units of conc
= (mol dm
-3s
-1)/(mol dm
-3) = s
-1
Second order Rate = k[A]
2k = units of rate/(units of conc)
2
= (mol dm
-3s
-1)/(mol dm
-3)
2= mol
-1dm
3s
-1
Third order Rate = k[A]
3k = units of rate/(units of conc)
3
= (mol dm
-3s
-1)/(mol dm
-3)
3= mol
-2dm
6s
-1UNDERSTANDING/KEY IDEA 16.1.C
Rate equations can only be
determined experimentally.
APPLICATION/SKILLS
Be able to deduce the rate
expression for an equation from experimental data and solve
problems involving the rate
expression.
DETERMINING THE EXPONENTS IN A RATE LAW
This must be done experimentally. NEVER assume the exponents equal the coefficients in the balanced equation.
Factor by which Factor by which Order of conc. is changed rate changes Reaction
2 No change 0
3 No change 0
2 2 = 21 1
3 3 = 31 1
4 4 = 41 1
2 4 = 22 2
3 9 = 32 2
4 16 = 42 2
2 8 = 23 3
3 27 = 33 3
4 64 = 43 3
INITIAL RATES METHOD
2 NO(g) + Cl
2(g) 2 NOCl(g)
Problem - Write the rate law, determine the value of the rate constant, k, and the overall order for the following reaction:
Experiment [NO]
(mol/dm3)
[Cl2] (mol/dm3)
Rate mol/dm3·s
1 0.250 0.250 1.43 x 10-6
2 0.500 0.250 5.72 x 10-6
3 0.250 0.500 2.86 x 10-6
4 0.500 0.500 11.4 x 10-6
OBSERVATIONS
When given initial concentration data, you want to look at each reactant
separately.
Try to find 2 reactions where your
identified reactant is changing and the other reactant(s) are constant.
For example in rxns 1 and 2, [NO] is
changing and [Cl
2] is constant.
Part 1 – Determine the values for the exponents in the rate law:
Experiment [NO]
(mol/dm3)
[Cl2] (mol/dm3)
Rate mol/dm3·s
1 0.250 0.250 1.43 x 10-6
2 0.500 0.250 5.72 x 10-6
3 0.250 0.500 2.86 x 10-6
4 0.500 0.500 1.14 x 10-5
In experiment 1 and 2, [Cl2] is constant while [NO] doubles.
R = k[NO]
x[Cl
2]
yThe rate quadruples, so the reaction is second order with respect to [NO]
R = k[NO]
2[Cl
2]
yPart 1 – Determine the values for the exponents in the rate law:
Experiment [NO]
(mol/dm3)
[Cl2] (mol/dm3)
Rate mol/dm3·s
1 0.250 0.250 1.43 x 10-6
2 0.500 0.250 5.72 x 10-6
3 0.250 0.500 2.86 x 10-6
4 0.500 0.500 1.14 x 10-5
R = k[NO]
2[Cl
2]
yIn experiment 2 and 4, [NO] is constant while [Cl2] doubles.
The rate doubles, so the reaction is first order with respect to [Cl2].
R = k[NO]
2[Cl
2]
Part 2 – Determine the value for k, the rate constant, by using any set of experimental data:
Experiment [NO]
(mol/dm3)
[Cl2] (mol/dm3)
Rate mol/dm3·s
1 0.250 0.250 1.43 x 10-6
R = k[NO]
2[Cl
2]
2
1.43 10
6mol 0.250 mol 0.250 mol
x k
L s L L
6 3 2
5
3 3 2
1.43 10
9.15 10 0.250
x mol L L
k x
L s mol mol s
APPLICATION/SKILLS
Be able to sketch, identify and
analyze graphical representations for zero, first and second order
reactions.
GUIDANCE
Be familiar with concentration vs time and rate vs concentration
graphs.
concentration time
concentration rate
ZERO-ORDER REACTION
Rate = k[A]0
concentration time
concentration rate
FIRST-ORDER REACTION
Rate = k[A]
HALF-LIFE
If a reactant has a constant half-life, then the reaction must be first order with respect to that reactant.
This is only true for 1
storder reactions.
The shorter the half-life, the faster the reaction.
Half-life (t
1/2) is the time it takes for half
the concentration to decrease.
concentration time
concentration rate
SECOND-ORDER REACTION
The concentration – time graph starts out steeper than the first order and levels off more than first order.
Rate = k[A]2
concentration time
concentration rate
SUMMARY
UNDERSTANDING/KEY IDEA 16.1.D
Reactions may occur by more than one step and the slowest step
determines the rate of reaction.
(rate determining step/RDS)
APPLICATION/SKILLS
Evaluate proposed reaction
mechanisms to be consistent with
kinetic and stoichiometric data.
Reaction Mechanism
The reaction mechanism is the series of
elementary steps by which a chemical reaction occurs. It is a theory about the sequence of
events in the progression of reactants to products.
The sum of the elementary steps must give
the overall balanced equation for the reaction
Rate-Determining Step
In a multi-step reaction, the slowest step is the rate-determining step. It therefore determines the rate of the reaction.
The experimental rate law must agree
with the rate-determining step.
Identifying the Rate-Determining Step
For the reaction:
2H
2(g) + 2NO(g) N
2(g) + 2H
2O(g) The experimental rate law is:
R = k[NO]
2[H
2]
Which step in the reaction mechanism is the rate-determining (slowest) step?
Step #1 H2(g) + 2NO(g) N2O(g) + H2O(g) Step #2 N2O(g) + H2(g) N2(g) + H2O(g)
Step #1 agrees with the experimental rate law
Identifying Intermediates
For the reaction:
2H
2(g) + 2NO(g) N
2(g) + 2H
2O(g)
Which species in the reaction mechanism are intermediates (do not show up in the final, balanced equation?)
Step #1 H2(g) + 2NO(g) N2O(g) + H2O(g) Step #2 N2O(g) + H2(g) N2(g) + H2O(g)
2H2(g) + 2NO(g) N2(g) + 2H2O(g)
N2O(g) is an intermediate
GUIDANCE
Be able to use potential energy
level profiles to illustrate multi-step reactions; showing the higher E
ain the rate-determining step in the
profile.
www.chemguide.co.uk
The activation energy for the overall reaction is equal to the activation energy of the rate determining step.
GUIDANCE
You should be familiar with reactions where the rate-
determining step is not always the
first step.
www.chemistry.msu.edu
Notice the second step is the rate determining
step since it has a higher activation energy.
When the rate determining step is not the first step in the mechanism, it is a bit more
complicated.
The reactant concentrations depend upon
earlier steps so these earlier steps must be
taken into account.
2NO(g) + O 2 (g) 2NO 2 (g)
Step #1 NO(g) + NO(g) N2O2(g) fast
Step #2 N2O2(g) + O2(g) 2NO2(g) RDS (slow) Overall 2NO(g) + O2(g) 2NO2(g)
So the rate equation is taken from step 2
Rate = k[N2O2][O2] but N2O2 is an intermediate whose
concentration is based on step 1 which is NO(g) + NO(g).
We then substitute back into the rate equation getting
Rate = k[NO]2[O2] which matches the coefficients in the overall balanced equation thus meeting the requirements for the
mechanism.
UNDERSTANDING/KEY IDEA 16.1.E
The molecularity of an elementary step is the number of reactant
particles taking part in that step.
The term molecularity is used in reference to an elementary step to indicate the number of reactant species involved.
Unimolecular – one reactant species
Bimolecular – two reactant species
Termolecular – three reactant species
The probability of more than two particles colliding at the same time with sufficient energy and correct
orientation is extremely low.
UNDERSTANDING/KEY IDEA 16.1.F
Catalysts alter a reaction
mechanism, introducing a step with
lower activation energy.
GUIDANCE
Know that catalysts are involved in
the rate-determining step.
Catalysts alter the reaction
mechanism by providing an alternate pathway for the rate-determining step with a lower activation energy.
Duch.sd57.bc.ca
Citations
International Baccalaureate Organization. Chemistry Guide, First assessment 2016. Updated 2015.
Brown, Catrin, and Mike Ford. Higher Level Chemistry. 2nd ed.
N.p.: Pearson Baccalaureate, 2014. Print.
ISBN 978 1 447 95975 5 eBook 978 1 447 95976 2
Most of the information found in this power point comes directly from this textbook.
The power point has been made to directly complement the Higher Level Chemistry textbook by Brown and Ford and is used for direct instructional purposes only.
