Chapter 12 Chemical Kinetics
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Transcript of Chapter 12 Chemical Kinetics
Prentice Hall © 2003 Chapter 14
Chapter 12Chapter 12Chemical KineticsChemical Kinetics
CHEMISTRY The Central Science
9th Edition
Originated by: David P. White
Prentice Hall © 2003 Chapter 14
• Kinetics is the study of how fast chemical reactions occur.
• There are 4 important factors which affect rates of reactions:– reactant concentration,
– temperature,
– action of catalysts, and
– surface area.
• Goal: to understand chemical reactions at the molecular level.
Factors that Affect Reaction Factors that Affect Reaction RatesRates
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• Speed of a reaction is measured by the change in concentration with time.
• For a reaction A B
• Suppose A reacts to form B. Let us begin with 1.00 mol A.
Reaction RatesReaction Rates
€
=Δ moles of B( )
Δt€
Average rate =change in number of moles of B
change in time
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Reaction RatesReaction Rates
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– At t = 0 (time zero) there is 1.00 mol A (100 red spheres) and no B present.
– At t = 20 min, there is 0.54 mol A and 0.46 mol B.
– At t = 40 min, there is 0.30 mol A and 0.70 mol B.
– Calculating,
Reaction RatesReaction Rates
€
Average rate =Δ moles of B( )
Δt
€
=moles of B at t =10( ) − moles of B at t = 0( )
10 min − 0 min
€
=0.26 mol − 0 mol
10 min − 0 min= 0.026 mol/min
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• For the reaction A B there are two ways of measuring rate:– the speed at which the products appear (i.e. change in moles of
B per unit time), or
– the speed at which the reactants disappear (i.e. the change in moles of A per unit time).
Reaction RatesReaction Rates
€
Average rate with respect to A = −Δ moles of A( )
Δt
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Change of Rate with Time• For the reaction A B there are two ways of• Most useful units for rates are to look at molarity. Since
volume is constant, molarity and moles are directly proportional.
• Consider:
C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)
Reaction RatesReaction Rates
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Change of Rate with Time
C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)– We can calculate the average rate in terms of the disappearance
of C4H9Cl.
– The units for average rate are mol/L·s or M/s.
– The average rate decreases with time.
– We plot [C4H9Cl] versus time.
– The rate at any instant in time (instantaneous rate) is the slope of the tangent to the curve.
– Instantaneous rate is different from average rate.
– We usually call the instantaneous rate the rate.
Reaction RatesReaction Rates
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Reaction Rate and Stoichiometry• For the reaction
C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)
we know
• In general for
aA + bB cC + dD
Reaction RatesReaction Rates
€
Rate = −Δ C4H9Cl[ ]
Δt=
Δ C4H9OH[ ]Δt
€
Rate = −1
a
Δ A[ ]Δt
= −1
b
Δ B[ ]Δt
=1
c
Δ C[ ]Δt
=1
d
Δ D[ ]Δt
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• In general rates increase as concentrations increase.
NH4+(aq) + NO2
-(aq) N2(g) + 2H2O(l)
Concentration and RateConcentration and Rate
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• For the reaction
NH4+(aq) + NO2
-(aq) N2(g) + 2H2O(l)
we note – as [NH4
+] doubles with [NO2-] constant the rate doubles,
– as [NO2-] doubles with [NH4
+] constant, the rate doubles,
– We conclude rate [NH4+][NO2
-].
• Rate law:
• The constant k is the rate constant.
Concentration and RateConcentration and Rate
€
Rate = k[NH4+][NO2
−]
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Exponents in the Rate Law• For a general reaction with rate law
we say the reaction is mth order in reactant 1 and nth order in reactant 2.
• The overall order of reaction is m + n + ….• A reaction can be zeroth order if m, n, … are zero.• Note the values of the exponents (orders) have to be
determined experimentally. They are not simply related to stoichiometry.
Concentration and RateConcentration and Rate
€
Rate = k[reactant 1]m[reactant 2]n
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Using Initial Rates to Determines Rate Laws
• A reaction is zero order in a reactant if the change in concentration of that reactant produces no effect.
• A reaction is first order if doubling the concentration causes the rate to double.
• A reacting is nth order if doubling the concentration causes an 2n increase in rate.
• Note that the rate constant does not depend on concentration.
Concentration and RateConcentration and Rate
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Example 1Example 1
[A] [B] Rate (M/s)
0.10 0.10 1.0 x 10-2
0.20 0.10 2.0 x 10-2
0.10 0.20 4.0 x 10-2
Rate = k [A]1 [B]2
k = rate/[A]1 [B]2 = 0.010/ (0.10) (0.10)2 = 10 M-2 s-1
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Example 1Example 1
Showing work for previous slide:
€
rate2
rate1=
k2[A]x[B]y
k1[A]x[B]y=
0.020
0.010=
k2(0.20)x (0.10)y
k1(0.10)x (0.10)y
€ 2 = 2x; x = 1
€
rate3
rate1=
k3[A]x[B]y
k1[A]x[B]y=
0.040
0.010=
k3(0.10)x (0.20)y
k1(0.10)x (0.10)y
4 = 2y; y = 2
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Example 2Example 2
[A] [B] [C] Rate (M/s)
0.10 0.10 0.10 1.0 x 10-2
0.20 0.10 0.10 2.0 x 10-2
0.10 0.20 0.10 4.0 x 10-2
0.20 0.20 0.20 6.4 x 10-1
What is the rate expression, including k with units?
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First Order Reactions• Goal: convert rate law into a convenient equation to give
concentrations as a function of time.• For a first order reaction, the rate doubles as the
concentration of a reactant doubles.
The Change of The Change of Concentration with TimeConcentration with Time
€
Rate = −Δ[A]
Δt= k[A]
€
ln A[ ]t− ln A[ ]0
= −kt
€
lnA[ ]t
A[ ]0
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟= −kt
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First Order Reactions
• A plot of ln[A]t versus t is a straight line with slope -k and intercept ln[A]0.
• In the above we use the natural logarithm, ln, which is log to the base e.
The Change of The Change of Concentration with TimeConcentration with Time
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First Order Reactions
The Change of The Change of Concentration with TimeConcentration with Time
€
ln A[ ]t= −kt + ln A[ ]0
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Second Order Reactions• For a second order reaction with just one reactant
• A plot of 1/[A]t versus t is a straight line with slope k and intercept 1/[A]0
• For a second order reaction, a plot of ln[A]t vs. t is not linear.
The Change of The Change of Concentration with TimeConcentration with Time
€
1
A[ ]t
= kt +1
A[ ]0
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Second Order Reactions
The Change of The Change of Concentration with TimeConcentration with Time
€
1
A[ ]t
= kt +1
A[ ]0
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Half-Life• Half-life is the time taken for the concentration of a
reactant to drop to half its original value.
• For a first order process, half life, t½ is the time taken for [A]0 to reach ½[A]0.
• Mathematically,
The Change of The Change of Concentration with TimeConcentration with Time
€
t 12
= −ln 1
2( )k
=0.693
k
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Half-Life• For a second order reaction, half-life depends in the
initial concentration:
The Change of The Change of Concentration with TimeConcentration with Time
€
t 12
= −1
k A[ ]0
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The Collision Model• Most reactions speed up as temperature increases. (E.g.
food spoils when not refrigerated.)• When two light sticks are placed in water: one at room
temperature and one in ice, the one at room temperature is brighter than the one in ice.
• The chemical reaction responsible for chemiluminescence is dependent on temperature: the higher the temperature, the faster the reaction and the brighter the light.
Temperature and RateTemperature and Rate
Temperature and RateTemperature and Rate
The Collision Model
• As temperature increases, the rate increases.
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The Collision Model• Since the rate law has no temperature term in it, the rate
constant must depend on temperature.
• Consider the first order reaction CH3NC CH3CN. – As temperature increases from 190 C to 250 C the rate
constant increases from 2.52 10-5 s-1 to 3.16 10-3 s-1.
• The temperature effect is quite dramatic. Why?• Observations: rates of reactions are affected by
concentration and temperature.
Temperature and RateTemperature and Rate
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The Collision Model• Goal: develop a model that explains why rates of
reactions increase as concentration and temperature increases.
• The collision model: in order for molecules to react they must collide.
• The greater the number of collisions the faster the rate.• The more molecules present, the greater the probability
of collision and the faster the rate.
Temperature and RateTemperature and Rate
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The Collision Model• The higher the temperature, the more energy available to
the molecules and the faster the rate.• Complication: not all collisions lead to products. In fact,
only a small fraction of collisions lead to product.
The Orientation Factor• In order for reaction to occur the reactant molecules must
collide in the correct orientation and with enough energy to form products.
Temperature and RateTemperature and Rate
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The Orientation Factor• Consider:
Cl + NOCl NO + Cl2
• There are two possible ways that Cl atoms and NOCl molecules can collide; one is effective and one is not.
Temperature and RateTemperature and Rate
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The Orientation Factor
Temperature and RateTemperature and Rate
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Activation Energy• Arrhenius: molecules must posses a minimum amount of
energy to react. Why?– In order to form products, bonds must be broken in the
reactants.
– Bond breakage requires energy.
• Activation energy, Ea, is the minimum energy required to initiate a chemical reaction.
Temperature and RateTemperature and Rate
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Activation Energy• Consider the rearrangement of methyl isonitrile:
– In H3C-NC, the C-NC bond bends until the C-N bond breaks and the NC portion is perpendicular to the H3C portion. This structure is called the activated complex or transition state.
– The energy required for the above twist and break is the activation energy, Ea.
– Once the C-N bond is broken, the NC portion can continue to rotate forming a C-CN bond.
Temperature and RateTemperature and Rate
H3C N CC
NH3C H3C C N
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Activation Energy• The change in energy for the reaction is the difference in
energy between CH3NC and CH3CN.
• The activation energy is the difference in energy between reactants, CH3NC and transition state.
• The rate depends on Ea.
• Notice that if a forward reaction is exothermic (CH3NC CH3CN), then the reverse reaction is endothermic (CH3CN CH3NC).
Temperature and RateTemperature and Rate
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Activation Energy• How does a methyl isonitrile molecule gain enough
energy to overcome the activation energy barrier?• From kinetic molecular theory, we know that as
temperature increases, the total kinetic energy increases.• We can show the fraction of molecules, f, with energy
equal to or greater than Ea is
where R is the gas constant (8.314 J/mol·K).
Temperature and RateTemperature and Rate
€
f = e−Ea
RT
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Activation Energy
Temperature and RateTemperature and Rate
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The Arrhenius Equation• Arrhenius discovered most reaction-rate data obeyed the
Arrhenius equation:
– k is the rate constant, Ea is the activation energy, R is the gas constant (8.314 J/K-mol) and T is the temperature in K.
– A is called the frequency factor.
– A is a measure of the probability of a favorable collision.
– Both A and Ea are specific to a given reaction.
Temperature and RateTemperature and Rate
€
k = Ae−Ea
RT
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Determining the Activation Energy
• If we have a lot of data, we can determine Ea and A graphically by rearranging the Arrhenius equation:
• From the above equation, a plot of ln k versus 1/T will have slope of –Ea/R and intercept of ln A.
Temperature and RateTemperature and Rate
€
lnk = −Ea
RT+ ln A
Temperature and RateTemperature and Rate
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Determining the Activation Energy• If we do not have a lot of data, then we recognize
Temperature and RateTemperature and Rate
€
lnk1 = −Ea
RT1
+ ln A and lnk2 = −Ea
RT2
+ ln A
€
lnk1 − lnk2 = −Ea
RT1
+ ln A ⎛
⎝ ⎜
⎞
⎠ ⎟− −
Ea
RT2
+ ln A ⎛
⎝ ⎜
⎞
⎠ ⎟
€
lnk1
k2
=Ea
R
1
T2
−1
T1
⎛
⎝ ⎜
⎞
⎠ ⎟
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• The balanced chemical equation provides information about the beginning and end of reaction.
• The reaction mechanism gives the path of the reaction.• Mechanisms provide a very detailed picture of which
bonds are broken and formed during the course of a reaction.
Elementary Steps• Elementary step: any process that occurs in a single step.
Reaction MechanismsReaction Mechanisms
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Elementary Steps• Molecularity: the number of molecules present in an
elementary step.– Unimolecular: one molecule in the elementary step,
– Bimolecular: two molecules in the elementary step, and
– Termolecular: three molecules in the elementary step.
• It is not common to see termolecular processes (statistically improbable).
Reaction MechanismsReaction Mechanisms
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Multistep Mechanisms• Some reaction proceed through more than one step:
NO2(g) + NO2(g) NO3(g) + NO(g)
NO3(g) + CO(g) NO2(g) + CO2(g)
• Notice that if we add the above steps, we get the overall reaction:
NO2(g) + CO(g) NO(g) + CO2(g)
Reaction MechanismsReaction Mechanisms
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Multistep Mechanisms• If a reaction proceeds via several elementary steps, then
the elementary steps must add to give the balanced chemical equation.
• Intermediate: a species which appears in an elementary step which is not a reactant or product.
Reaction MechanismsReaction Mechanisms
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Rate Laws for Elementary Steps• The rate law of an elementary step is determined by its
molecularity:– Unimolecular processes are first order,
– Bimolecular processes are second order, and
– Termolecular processes are third order.
Rate Laws for Multistep Mechanisms• Rate-determining step: is the slowest of the elementary
steps.
Reaction MechanismsReaction Mechanisms
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Rate Laws for Elementary Steps
Reaction MechanismsReaction Mechanisms
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Rate Laws for Multistep Mechanisms• Therefore, the rate-determining step governs the overall
rate law for the reaction.
Mechanisms with an Initial Fast Step• It is possible for an intermediate to be a reactant.• Consider
2NO(g) + Br2(g) 2NOBr(g)
Reaction MechanismsReaction Mechanisms
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Mechanisms with an Initial Fast Step
2NO(g) + Br2(g) 2NOBr(g)
• The experimentally determined rate law is
Rate = k[NO]2[Br2]
• Consider the following mechanism
Reaction MechanismsReaction Mechanisms
NO(g) + Br2(g) NOBr2(g)k1
k-1
NOBr2(g) + NO(g) 2NOBr(g)k2
Step 1:
Step 2:
(fast)
(slow)
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Mechanisms with an Initial Fast Step• The rate law is (based on Step 2):
Rate = k2[NOBr2][NO]
• The rate law should not depend on the concentration of an intermediate (intermediates are usually unstable).
• Assume NOBr2 is unstable, so we express the concentration of NOBr2 in terms of NOBr and Br2 assuming there is an equilibrium in step 1 we have
Reaction MechanismsReaction Mechanisms
€
[NOBr2] =k1
k−1
[NO][Br2]
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Mechanisms with an Initial Fast Step• By definition of equilibrium:
• Therefore, the overall rate law becomes
• Note the final rate law is consistent with the experimentally observed rate law.
Reaction MechanismsReaction Mechanisms
€
k1[NO][Br2] = k−1[NOBr2]
€
Rate = k2
k1
k−1
[NO][Br2][NO] = k2
k1
k−1
[NO]2[Br2]
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• A catalyst changes the rate of a chemical reaction.• There are two types of catalyst:
– homogeneous, and
– heterogeneous.
• Chlorine atoms are catalysts for the destruction of ozone.
Homogeneous Catalysis• The catalyst and reaction is in one phase.
CatalysisCatalysis
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End of Chapter 14End of Chapter 14Chemical KineticsChemical Kinetics