Ch. 13: Chemical Kinetics
Dr. Namphol Sinkaset
Chem 201: General Chemistry II
I. Chapter Outline
I.
II.
III.
IV.
V.
VI.
VII.
Introduction
The Rate of a Chemical Reaction
Reaction Rate Laws
Integrated Rate Laws
Temperature and Rate
Reaction Mechanisms
Catalysis
I. Introduction
• Some reactions are quick (explosions)
while others are slow (rusting of iron).
• Knowing the rate of a reaction and what
factors influence it allow chemists to
plan accordingly.
• If we understand what contributes to the
rate, we can control the reaction.
I. Introduction
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)
• Balanced equations only give net change.
• Equations tell us nothing about how the
reaction happens.
• One possibility for the above: simultaneous
collision of 6 molecules.
• Unlikely, so reaction must occur in a series of
small steps that leads to final products.
• We study mechanisms later in the chapter.
II. Reaction Rates
• Rates are generally change of
something divided by change in time.
• Reaction rates are no different.
• The rate of a reaction can be written
with respect to any compound in that
reaction.
• However, there can only be one
numerical value for a rate of reaction.
II. Average Rates of Reaction
H2(g) + I2(g) 2HI(g)
II. General Reaction Rates
aA + bB  cC + dD
II. Some Rate Data
• If we plot average
rate data as a
function of time, we
see that the reaction
rate constantly
changes.
• Thus, rate depends
on concentration of
reactants!
III. Rate Laws
• If the rate depends on concentration of
reactants, then we should be able to
write an equation.
• A rate law describes the mathematical
relationship between the concentration
of reactants and how fast the reaction
occurs.
III. A Simple Rate Law
• Consider a decomposition reaction where
A  products
• If the reverse reaction is negligible, then
the rate law is: Rate = k[A]n.
 k is called the rate constant
 n is called the reaction order
III. Reaction Orders
• The reaction order, n, determines how
the rate depends on the concentration
of the reactant. For the previous
reaction, if…
 n = 0, zero order, rate is independent of [A]
 n = 1, first order, rate is directly
proportional to [A]
 n = 2, second order, rate is proportional to
the square of the [A]
III. Reaction Orders and Rate
• The rate law for the
decomposition can
then be either:
 Rate = k[A]0 = k
 Rate = k[A]1
 Rate = k[A]2
• Each will have a
different type of
curve when
graphed.
III. Determining Orders
• Reaction orders can only be determined
by experiment!!
• Reaction orders are not related to the
stoichiometry of a reaction!
• If reaction orders match a reaction’s
stoichiometry, it is just a coincidence.
• Therefore, orders cannot be determined
without experimental data!
III. Sure-fire Method
For the reaction, A  Products, we have the following data:
[A] (M)
Initial Rate (M/s)
0.10
0.015
0.20
0.060
0.40
0.240
III. More Complex Reactions
• What if we have a more complicated
reaction like: aA + bB  cC + dD?
• Writing the general rate law is easy.
Simply include all reactants, each with
its own order.
 Rate = k[A]m[B]n
• If there are more reactants, there are
more terms in the rate law.
III. Example Reaction
• 2H2(g) + 2NO(g)  N2(g) + 2H2O(g)
• After looking at experimental data, the rate
law was found to be Rate = k[H2][NO]2.
• We say the reaction is 1st order in H2, 2nd
order in NO, and 3rd order overall.
• Note that Rate always has units of M/s, so
the units on k will depend on the rate law.
• What are the units of k for the rate law
above?
III. Steps for Finding Rate Law
1) Pick two solutions where one reactant
stays same, but another changes.
2) Write rate law for both w/ as much
information as you have.
3) Ratio the two and solve for an order.
4) Repeat for another pair of solutions.
5) Use any reaction to get value of k.
III. Sample Problem
Determine the complete rate law for the reaction
CHCl3(g) + Cl2(g)  CCl4(g) + HCl(g) using the data
below.
[CHCl3] (M) [Cl2] (M) Initial Rate (M/s)
0.010
0.010
0.0035
0.020
0.020
0.040
0.010
0.020
0.040
0.0069
0.0098
0.027
III. Sample Problem
Sometimes, rate laws can be found by inspection.
Determine the rate law for the reaction 2NO(g) +
2H2(g)  N2(g) + H2O(g) using the data below.
[NO] (M) [H2] (M) Initial Rate (M/s)
0.10
0.10
0.00123
0.10
0.20
0.00246
0.20
0.10
0.00492
IV. Concentration and Time
• Study and elucidation of rate laws allow
the prediction of when a reaction will end.
• An integrated rate law for a chemical
reaction is a relationship between the
concentrations of reactants and time.
• Integrated rate laws depend on the order
of the reaction; thus, we examine each
separately.
• We will only consider reactions with one
reactant.
IV. 1st Order Integrated Rate Law
IV. 1st Order Integrated Rate Law
• Notice this equation is in y = mx + b form.
• A plot of ln[A] vs. t for a 1st order reaction
yields a straight line with m = -k and b =
ln[A]0.
IV. 2nd Order Integrated Rate Law
IV. 2nd Order Integrated Rate Law
• Again, this equation is in y = mx + b form.
• A plot of 1/[A] vs. t yields a straight line
with slope equal to k and y-intercept equal
to 1/[A]0.
IV. Zero Order Integrated Rate Law
IV. Zero Order Integrated Rate Law
• Yet again in y = mx + b form!
• Plot of [A] vs. t results in a straight line
with slope equal to -k and b = [A]0.
IV. Reaction Half Lives
• The half-life, t1/2, of a reaction is the
time required for the concentration of a
reactant to decrease to half its initial
value.
• Half life equations depend on the order
of the reaction.
IV. 1st Order Reaction Half Life
IV. 1st Order Reaction Half Life
• Notice that the half life doesn’t depend on
reactant concentration!
• Unique for 1st order.
• The half life for a 1st order reaction is
constant.
IV. 1st Order Half Lives
IV. 2nd Order Reaction Half Life
IV. 2nd Order Reaction Half Life
• For 2nd order, the half life depends on initial
concentration.
• As concentration decreases, half life gets
longer and longer.
IV. Zero Order Reaction Half Life
IV. Zero Order Reaction Half Life
• We see that for zero order reactions, the half
life depends on concentration as well.
V. Temperature and Rate
• In general, rates of reaction are highly
sensitive to temperature.
• If Rate = k[A]n, where does the
temperature factor in?
• It’s in the constant k!
• Generally, increasing temperature
increases k.
V. The Arrhenius Equation
• Note that R is the gas constant, and T is
temperature in kelvin.
V. Parameters of Arrhenius Eqn.
• We can describe the physical meanings of
the aspects of the Arrhenius equation by
considering a specific reaction.
V. Activation Energy
• To get to product state, reactant must go through
high-energy activated complex, or transition state.
• Even though reaction is exo overall, it must go
through an endo step.
• Higher Ea means slower reaction.
V. Frequency Factor
• The frequency factor represents the number of
approaches to the activation barrier per unit
time.
• For this reaction, it represents how often the
NC part of the molecule vibrates.
• Note that not all approaches result in reaction
due to not having enough energy.
• A frequency factor of 109/s means that there
are 109 vibrations per second of the NC group.
V. Exponential Factor
• The exponential factor is a number
between 0 and 1 that represents the
fraction of molecules that successfully
react upon approach.
• An exponential factor of 10-7 means that
1 out of every 107 molecules has
enough energy to cross the energy
barrier.
V. Exponential Factor & Temp
• Since exponential factor = e-Ea/RT,
temperature has a huge influence.
• As T  0, the factor goes to 0, and as T
 ∞, the factor goes to 1.
• Thus, higher temperatures mean more
successful approaches because the
molecules have more energy to
overcome the activation barrier.
V. Finding A and Ea
V. Arrhenius Plots
• If we have kinetic data at various
temperatures, we can plot ln k vs. 1/T.
• We should get a straight line with m = -Ea/R
and b = ln A.
V. Two-Point Form
V. Sample Problem
• The decomposition of HI has rate
constants of k = 0.079 1/M·s at 508 °C
and k = 0.24 1/M·s at 540 °C. What is
the activation energy of this reaction in
kJ/mole?