REACTION KINETICS
LAUNCH LAB PAGE
558
SECTION 1 – THE REACTION
PROCESS
BY
STUDYING
M ANY
TYP ES
OF
E XP ERIMENTS ,
CHEMISTS
HAVE
FOUND
THAT
CHEMICAL
REACTIONS OCCUR AT WIDELY DIFF ERING RATES.
FOR E XAM PLE , IN THE PRES ENCE OF AI R, I RON
RUSTS
VERY
SLOWLY,
WHEREAS
METHANE
IN
NATURAL GAS BURNS RAPIDLY.
THE SP EED OF A
CHEMICAL REACTION D EPE NDS ON THE E NE RGY
PATHWAY THAT A REACTION FOLLOWS AND THE
CHANGES THAT TAKE PLACE ON THE MOLECULAR
LEVEL WHEN SUBSTANCES INTERACT.
IN THIS
CHAPTER, YOU WILL ST UDY THE FACTORS THAT
AFFECT HOW FAST CHEMICAL REACTIONS TAKE
PLACE.
REACTION MECHANISMS
• If you mix solutions of HCl and NaOH, a quick
neutralization reaction happens
H3O+(aq) + Cl−(aq) + Na+(aq) + OH−(aq) →
2H2O(l) + Na+(aq) + Cl−(aq)
• Reaction is nearly instantaneous – the rate is
only limited by the speed H3O+ and OH− can diffuse
through water to meet each other
• Reactions between
ions of same
charge and
between molecules
are not immediate
• Positive and
negative ions repel
each other
• Electron clouds of
molecules strongly
repel each other at
short distances
• Only ions or
molecules with
very high kinetic
energy overcome
repulsive forces
enough to react
• Colorless hydrogen gas is
made of pairs of hydrogen
atoms bonded together as
diatomic molecules, H2
• Violet-colored iodine
vapor is also diatomic, I2
• Chemical reaction between
two at high temps makes
hydrogen iodide, HI, a
colorless gas
• HI molecules decompose
and re-form H2 and I2
molecules
• Produces violet gas
H2(g) + I2(g) → 2HI(g)
2HI(g) → H2(g) + I2(g)
• These equations only
show which molecular
species disappear as a
result of the reactions
and which species are
made
• They do not show
reaction mechanism →
the step-by-step
sequence of reactions
by which the overall
chemical change occurs
H2(g) + I2(g) → 2HI(g)
2HI(g) → H2(g) + I2(g)
• Experiments can often be designed that
suggest most likely steps in a reaction
mechanism
• Each reaction step is usually simple
• Equation for each step represents the
actual atoms, ions or molecules that
participate
•It was thought the formation of HI was a
simple one-step process
•Involves two molecules, H2 and I2 in forward
reaction and 2 HI molecules in reverse
reaction
•Experiments showed direct reaction between
H2 and I2 never takes place
• Alternative mechanisms for reaction suggested
based on experimental results
• Steps in each reaction had to add up to give
overall equation
• Some species appear in steps but not in overall
net equation
• Intermediates → species that appear in some
steps but not in net equation
2 POSSIBLE MECHANISMS FOR FORMATION OF HI
•This reaction is example of
homongeneous reaction → a
reaction whose reactants and
products exist in a single phase
•In this case, gas
COLLISION THEORY
• For reactions to occur, the atoms/molecules
must collide
• The collisions must also result in interaction
between particles
• Collision theory → set of assumptions
regarding collisions and reactions
• Chemists use theory to understand
observations about chemical reactions
• Let’s consider what happens on molecular scale in one step
of homogeneous reaction system
AB + AB ⇌ A2 + 2B
• According to collision theory, the two AB molecules must
collide in order to react
• Also must collide in the right position with enough energy
to combine valence electrons and disturb the bonds
• If they do, collision results in recombination (A2 and 2B)
•If collision too gentle, two molecules
rebound unchanged (a)
• If molecules aren’t in right position,
they rebound without interacting (b)
• An effective collision must be in right
position and with enough energy (c)
• NOT EVERY COLLISION CREATES A
REACTION
•Chemical reaction produces new
bonds between atoms in colliding
molecules
•Unless collision brings correct
atoms close together in proper
orientation, molecules do not
react
•Ex. If Cl2 collides with oxygen end of
NO, the following reaction may occur:
NO (g) + Cl2 (g) → NOCl (g) + Cl
•This won’t happen unless Cl2 collides
with O of NO
ACTIVATION ENERGY
• Let’s consider reaction for formation of water from
H2 and O2
2H2(g) + O2(g) → 2H2O(l)
• Heat of formation ΔH˚f = −285.8 kJ/mol at 298.15
K
• The free-energy change is also large: ΔG˚ = −237.1 kJ/mol
• Why don’t oxygen and hydrogen combine
spontaneously and immediately to form water?
• When diatomic H2 and O2 approach
each other, the electron clouds repel
each other
• Molecules bounce off and never
actually meet
• For reaction to happen, molecules
must have enough kinetic energy to
actually mix together the valence
electrons
•Bonds of species must be broken in
order for new ones to be formed
•Bond breaking is endothermic
•Bond forming is exothermic
•Initial input of energy needed to
overcome repulsion
•This initial energy activates the
reaction
• Once an exothermic reaction is started,
the energy released is enough to
continue the reaction by activating
other molecules
• Result → reaction rate keeps
increasing
• Limited by time required for reactant
particles to get energy and make
contact
• Activation energy (Ea) → minimum energy
required to transform the reactants into an
activated complex
• Reverse reaction (decomposition of water
molecules) is endothermic
• Water molecules lie at energy level lower than
H2 and O2 levels
• Require higher activation energy before they
decompose
• Represented by Eaʹ
THE ACTIVATED COMPLEX
• When molecules collide, some kinetic
energy converted into internal potential
energy within colliding molecules
• If enough energy converted, molecules
with correct orientation become
activated
• New bonds can now form
• During interval of breaking and reforming
bonds, collision complex is in transition state
• Some kind of partial bonding exists
• Activated complex → a transitional
structure that results from an effective
collision and that remains while old bonds
are breaking and new bonds are forming
• Both forward and
reverse reactions
go through same
activated complex
• Bond broken in
activated complex
for forward
reaction must be
re-formed in
reverse reaction
• Remember speeds (therefore kinetic
energies) of molecules increases as
temperature increases
• Increase in speed → increase in collisions
→ increase in reactions
• Still depends on enough energy to make
activated complex no matter how many
collisions
• Raising temperature gives more molecules
this activation energy, so reactions
increase
•Activated complex is characteristic of
both product and reactant
•May re-form original bonds and
separate back to reactant particles
•May form new bonds and separate into
product particles
•Either are just as likely
REACTION RATE
DEMO PAGE 561
RATE-INFLUENCING FACTORS
•Other than decompositions,
particles must come into contact in
the right position and with enough
energy for activation
•Rate of reaction depends on
collision frequency of reactants and
on collision effectiveness
•5 important factors influence rate of
chemical reaction
1.Nature of reactants
2.Surface area
3.Temperature
4.Concentration
5.Presence of catalysts
1. NATURE OF REACTANTS
•Substances vary greatly in their
ability to react
•Ex. H2 combines strongly with
chlorine under certain conditions
–Under same conditions, hardly reacts
at all with nitrogen
HYDROGEN
AND
CHLORINE
•Na and O2 combine more rapidly
than Fe and O2 under similar
conditions
2. SURFACE AREA
• Gaseous mixtures and dissolved particles can
mix and collide freely
• Reactions with them happen more often and
faster
• Heterogeneous reactions → involve reactants
in two different phases
• Can occur only when two phases are in contact
• Surface area of solid reactant important in
determining rate
• Higher surface area, higher rate of reaction
DEMO PAGE 569
EXAMPLE – ZN AND HCL
• Solid zinc reacts with hydrochloric acid to make zinc
chloride and hydrogen gas
Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)
• Reaction occurs on surface of zinc
• Piece of zinc measuring 1cm on each edge only give
6 cm2 of contact area
• Powdered form – 1,000 times the surface area
EXAMPLE – BURNING COAL
• Big lump of coal burns slowly in air
• Rate of burning increased when you break up
coal into smaller pieces
• If piece of coal is powdered and ignited while
suspended in air → BOOM!
• This is cause of explosions in coal mines
Coal mine
explosion in
Ukraine
3. TEMPERATURE
•Increase in temperature increases
kinetic energy
•Can result in greater number of
effective collisions
•IF number of effective collisions
increase, reaction rate will increase
MINI LAB PAGE
571
• To be effective, energy of collisions must
be equal to or greater than (≥) the
activation energy
• At higher temperatures, more particles
have enough energy to form activated
complex when collisions occur
• So, rise in temperature produces an
increase in collision energy as well as in
collision frequency
• Decreasing temperature has opposite effect
• Average kinetic energy decreases
• They collide less frequently with less energy
• Produces fewer effective collisions
• Beginning near room temp, reaction rates
of many common reactions roughly double
with each 10K rise in temperature
4. CONCENTRATION
• Pure oxygen has 5 times the concentration of
oxygen molecules in air
• Substance that reacts with oxygen in air reacts
more strongly with pure oxygen
Carbon burns faster in pure
oxygen (a) than in air (b)
because the concentration of
the reacting species, O2, is
greater.
(a)
(b)
• Light produced in pure
oxygen from charcoal is
brighter than in air
• Oxidation of charcoal is
heterogeneous reaction
system (gas and solid)
• Reaction rate depends on
amount of exposed
charcoal AND
concentration of O2
•In homogeneous reaction systems,
reaction rates depend on
concentration of reactants
•If number of effective collisions
increases, rate increases as well
•In general, increase in rate expected
if concentration of one or more
reactants is increased
•In system with only two molecules, only
one collision can occur
•When there are four molecules, there can
be four possible collisions
•Under constant conditions, as
number of molecules in system
increases, so does the total number
of possible collisions
•Lowering concentration should
have the opposite effect
4. PRESENCE OF CATALYSTS
• Some chemical
reactions are slow
• Can be increased by
catalyst → a
substance that changes
the rate of a chemical
reaction without itself
being permanently
used
• Catalysis → the action
of a catalyst
• Catalyst provides
alternative
energy pathway
or reaction
mechanism
• Potential-energy
barrier between
reactants and
products is
lowered
• The reaction rate of the
decomposition of hydrogen peroxide,
H2O2, can be increased by using a
catalyst
• The catalyst used here is manganese
dioxide, MnO2, a black solid
• A 30% H2O2 solution is added
dropwise onto the MnO2 in the
beaker and rapidly decomposes to
O2 and H2O
• Both the oxygen and water appear
as gases because the high heat of
reaction causes the water to
vaporize
• Catalysts do not appear among final
products
• May participate in one step along a reaction
pathway and be regenerated in later step
• Homogeneous catalyst → a catalyst that is
in the same phase as all the reactants and
products in a reaction system
• Heterogeneous catalyst → a catalyst that is
in a different phase from the reactants
RATE LAWS FOR REACTIONS
• Relationship between rate of reaction and
concentration of one reactant is
determined experimentally by keeping
concentrations of other reactants and
temperature constant
• Reaction rate measured for various
concentrations of reactant in question
• Hydrogen gas reacts with nitrogen monoxide
gas at constant volume and at elevated
constant temperature
2H2(g) + 2NO(g) → N2(g) + 2H2O(g)
• 4 mol reactant gases produce 3 mol product
gases
• Result: pressure of system decreases as
reaction proceeds
• Rate of reaction can be determined by
measuring pressure
•Suppose a series of experiments is
done using same initial
concentration of NO but different
initial concentrations of H2
•Initial reaction rate varies directly
with H2 concentration
2H2(g) + 2NO(g) → N2(g) + 2H2O(g)
2H2(g) + 2NO(g) → N2(g) + 2H2O(g)
• Double H2 → double rate
• If R represents reaction rate and [H2] is
concentration of H2 in mol/L,
relationship is shown as follows
• R α [H2]
where α means “is proportional to”
2H 2 (G) + 2NO(G) → N 2 (G) + 2H 2 O(G)
• Suppose same initial [H2] is used BUT initial
[NO] is varied
• Initial rate increase fourfold when [NO] is
doubled
• Increased 9-fold when [NO] tripled
• Reaction rate varies with the square of
concentration of NO
R α [NO]²
• Because R is proportional to [H2] and [NO]²
it is proportional to their product
R ∝ [H2][NO]²
• By introduction of a proportionality
constant, k, the expression becomes an
equality
R = k[H2][NO]²
• Rate law → an equation that relates reaction
rate and concentrations of reactants
• Can use for specific reaction at a given
temperature
• Rise in temp increases rates of most reactions
• Value k usually increases as temp increases
• Relationship between rate and concentration
almost always remains unchanged
USING THE RATE LAW
• General form for rate law is
R = k[A]n[B]m
• R = reaction rate
• K = specific rate constant
• [A] and [B] are concentration of
reactants
• n and m are powers to which
concentrations are raised
•Order → the power to which a
reactant concentration is raised
•Value of “n” is order of reaction with
respect to [A]
•Reaction is “nth order in A”
•Or “mth order in B”
•n and m usually small integers or zero
• Order of 1 means reaction rate (R) is
directly proportional to concentration
of reactant
• Order of 2 means R is directly
proportional to the square of reactant
• Zero means rate does not depend on
concentration, as long as some
reactant is present
• Sum of all reactant orders is called
order of the reaction or overall order
• Equal to n + m
• Orders in rate law may or may not
match coefficients in balanced
equation
• These must be determined from
experimental data
SPECIFIC RATE CONSTANT
•k
•Proportionality constant relating
rate of reaction to concentrations
of reactants
1. Once reaction orders are known, value of k
must be determined from experimental data
2. Value of k is for specific reaction
3. Units of k depend on overall order of reaction
4. Value of k does not change for different
concentrations of reactants or products
5. Value of k is for reaction at specific
temperature – increasing temp increases k
6. Value of k becomes larger with catalyst
SAMPLE PROBLEM
Three experiments that have identical
conditions were performed to measure
the initial rate of the reaction
2HI (g) → H2(g) + I2 (g)
The results for the three experiments in
which only the HI concentration was
varied are as follows:
Experiment [HI] (M)
Rate (M/s)
1
0.015
1.1 x 10-3
2
0.030
4.4 x 10-3
3
0.045
9.9 x 10-3
2HI (g) → H2(g) + I2 (g)
•Write the rate law for the reaction
•Find the value and units of the specific rate
constant
• The general rate law for this reaction
is R= k[HI]n
• we can find the ratio of the reactant
concentrations between experiments 1
and 2, [HI]2÷[HI]1
• Then see how ratio of concentration
affects ratio of rates, R2÷R3
SAMPLE PROBLEM C
3 experiments were performed to measure the initial
rate of reaction
A+ B→ C
Conditions were identical except that the concentration
of reactants varied.
Write the rate law for the reaction.
Find the value and units of the specific rate constant
RATE LAWS AND REACTION PATHWAY
• The form of the rate law depends on the
reaction mechanism
• For reaction with single step, reaction rate
of that step is proportional to the product
of the reactant concentrations, each of
which is raised to its stoichiometric
coefficient
(wait a minute, you will understand)
EXAMPLE
• 1 molecule of gas A collides with 1 molecule of gas
B to form molecules of C
A + B → 2C
• One particle of each reactant is involved in each
collision
• Doubling concentration of either reactant will
double collision frequency
• Also doubles reaction rate for this step
A+B→
2C
•The rate for this step is directly
proportional to the concentration
of A and B
R = k[A][B]
• Now suppose reaction is reversible
• In reverse, 2 molecules of C must decompose
to form 1 molecule of A and 1 molecule of B
2C →
A + B
• Reaction rate for the reverse is directly
proportional to [C] x [C]
R = k[C]²
• Power of the molar concentration of
each reactant is same as the coefficient
in balanced equation
• Relationship is true ONLY if reaction is
one-step
• If reaction is in sequence of steps, rate
law is determined from slowest step
because it has the lowest rate
• Rate-determining step → the slowestrate step for the chemical reaction
•Consider reaction of nitrogen dioxide and
carbon monoxide
NO2(g) + CO(g) → NO(g) + CO2(g)
•Reaction is believed to be in two steps
•Step 1: NO2 + NO2 → NO3 + NO slow
•Step 2: NO3 + CO → NO2 + CO2 fast
STEP 1: NO 2 + NO 2 → NO 3 + NO (SLOW)
STEP 2: NO 3 + CO → NO 2 + CO 2 (FAST)
• Step 1 – two molecules NO2 collide,
forming intermediate NO3
• NO3 then collides with one molecule
CO and reacts quickly to make one
molecule NO2 and one CO2
• First step is slower, so it is the ratedetermining step
•Rate law from rate-determining step
R = k[NO2]²
•Tells us 2 molecules NO2 are reactants in
slower, rate-determining step
•CO not in rate law because it reacts
AFTER the rate-determining step
GENERAL FORM OF RATE LAW
R = k[A]n[B]m….
• R → reaction rate
• k → rate constant
• [A], [B] → molar concentrations of reactants
• n, m, → respective powers reactants are raised
• Must be determined from experimental data
SAMPLE PROBLEM D
• Nitrogen dioxide and fluorine react in the gas
phase according to the following equation.
2NO2(g) + F2(g) → 2NO2F(g)
• A proposed mechanism for this reaction follows.
• Step 1: NO2 + F2 → NO2F + F slow
• Step 2: F + NO2 → NO2F fast
• Identify the rate-determining step and write an
acceptable rate law.
• If we combine these two steps, the
intermediate, F, cancels out and we are
left with the original equation.
• The first step is the slower step, and is
considered the rate-determining step.
• We can write the rate law from this ratedetermining step.
R = k [NO2][F2]
SAMPLE PROBLEM E
• A reaction involving reactants X and Y was
found to occur by a one-step mechanism:
X + 2Y → XY2
• Write the rate law for this reaction, and then
determine the effect of each of the following on
the reaction rate:
• a. doubling the concentration of X
• b. doubling the concentration of Y
• c. using one-third the concentration of Y
• Because the equation represents a single-step mechanism,
the rate law can be written from the equation (otherwise,
it could not be)
• The rate will vary directly with the concentration of X,
which has an implied coefficient of 1 in the equation, and
will vary directly with the square of the concentration of Y,
which has the coefficient of 2: R = k[X][Y]²
• a. Doubling the concentration of X will double the rate (R =
k[2X][Y]²)
• b. Doubling the concentration of Y will increase the rate
fourfold (R = k[X][2Y]²)
• c. Using one-third the concentration of Y will reduce the
rate to one-ninth of its original value (R = k[X][ 1Y]²)
PRACTICE PROBLEM 1
• The rate of a reaction involving L, M,
and N is found to double if the
concentration of L is doubled, to increase
eightfold if the concentration of M is
doubled, and to double if the
concentration of N is doubled. Write the
rate law for this reaction.
• R = k[L][M]³[N]
PRACTICE PROBLEM 2
• At temperatures below 498 K, the following
reaction takes place.
NO2(g) + CO(g) → CO2(g) + NO(g)
• Doubling the concentration of NO2 quadruples the
rate of CO2 being formed if the CO concentration is
held constant. However, doubling the concentration
of CO has no effect on the rate of CO2 formation.
Write a rate-law expression for this reaction.
• R = k[NO2]²
CHEM LAB PAGE
584