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Chapter 15Chapter 15
Rates of ReactionRates of Reaction
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OverviewOverview Reaction RatesReaction Rates
Definition of Reaction RatesDefinition of Reaction Rates Experimental Determination of Rate Experimental Determination of Rate Dependence of Rate on ConcentrationDependence of Rate on Concentration Change of Concentration with TimeChange of Concentration with Time Temperature and Rate; Collision and Transition-State Temperature and Rate; Collision and Transition-State
Theories.Theories. Arrhenius EquationArrhenius Equation
Reaction MechanismsReaction Mechanisms Elementary ReactionsElementary Reactions Rate Law and the MechanismRate Law and the Mechanism CatalysisCatalysis
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Definition of Reaction RateDefinition of Reaction Rate
Reaction rate = increase in concentration Reaction rate = increase in concentration of product of a reaction as a function of of product of a reaction as a function of time or decrease in concentration of time or decrease in concentration of reactant as a function of timereactant as a function of time
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Rate Equation Rate Equation
For the reaction:For the reaction:
A + 2B A + 2B 3C 3C
12
12
change time
A change conc
tt]A[]A[
t]A[
RateA
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Rate EquationRate EquationConcentration vs Reaction Time
A + 2B --> 3C
0.000
0.045
0.090
0 250 500
Time, s
Co
nce
ntr
atio
n, M Init
Rate
Ave.Rate
Inst.Rate
Rates are expressed as positive numbers. For the Rates are expressed as positive numbers. For the reaction in the graph we have:reaction in the graph we have:
t
]A[RA
t
]B[RB
t
]C[RC
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Reaction Rates and Reaction Rates and StoichiometryStoichiometry
A + B A + B C; R C; RCC = R = RAA = R = RBB. .
A + 2B A + 2B 3C; 3C;
For the general reaction: aA + bB For the general reaction: aA + bB cC + dD cC + dD
. .
CBA R31
R21
R
DCBA Rd
aR
c
aR
b
aR
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Calculate the rate of decomposition of HI Calculate the rate of decomposition of HI in the reaction: 2HI(g) in the reaction: 2HI(g) H H22(g) + I(g) + I22(g). (g).
Given: After a reaction time of 100 secs. Given: After a reaction time of 100 secs. the concentration of HI decreased by the concentration of HI decreased by 0.500 M. 0.500 M.
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For the reaction 2A + 3B For the reaction 2A + 3B 4C + 2D; 4C + 2D; determine the rates of B, C and D if the determine the rates of B, C and D if the rate of consumption of A is 0.100 M/s.rate of consumption of A is 0.100 M/s.
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Factors that Affect Reaction RatesFactors that Affect Reaction Rates Nature of reactantsNature of reactants Concentrations of reactantsConcentrations of reactants TemperatureTemperature Presence or absence of a catalyst a Presence or absence of a catalyst a
substance that increases the rate of substance that increases the rate of reaction without being consumed in the reaction without being consumed in the reaction.reaction.
Surface area for heterogeneous reactions. Surface area for heterogeneous reactions.
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Rate Laws and Reaction Rate Laws and Reaction OrderOrder
Rate Law Rate Law – an equation that tells how the – an equation that tells how the reaction rate depends on the concentration of reaction rate depends on the concentration of each reaction.each reaction.
Reaction orderReaction order – the value of the exponents – the value of the exponents of concentration terms in the rate law. of concentration terms in the rate law.
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For the reaction: aA + bB For the reaction: aA + bB cC + dD, the cC + dD, the initial rate of reaction is related to the initial rate of reaction is related to the concentration of reactants. concentration of reactants.
R = k[A]R = k[A]mm[B][B]nn
Varying the initial concentration of one Varying the initial concentration of one reactant at a time produces rates, which reactant at a time produces rates, which will lead to a determination of exponents. will lead to a determination of exponents.
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The rate law describes this dependence: R = The rate law describes this dependence: R = k[A]k[A]mm[B][B]nn where k = rate constant and m and n where k = rate constant and m and n are the orders of A and B respectively. are the orders of A and B respectively.
m = 1 (A varied, B held constant) gives R = k’[A]. m = 1 (A varied, B held constant) gives R = k’[A]. Rate is directly proportional to [A]. Doubling A Rate is directly proportional to [A]. Doubling A doubles Rdoubles R
m = 2 (A varied, B held constant) gives R = k’[A]m = 2 (A varied, B held constant) gives R = k’[A]22. . The rate is proportional to [A]The rate is proportional to [A]22. Doubling A . Doubling A quadruples R.quadruples R.
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Determine order of each reactant: Determine order of each reactant:
HCOOH(aq) + BrHCOOH(aq) + Br22(aq) (aq) 2H 2H++(aq) + (aq) +
2Br2Br(aq) + CO(aq) + CO22(g) (g)
R = k[BrR = k[Br22] ]
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The formation of HI gas has the following The formation of HI gas has the following rate law: R = k[Hrate law: R = k[H22][I][I22]. What is the order of ]. What is the order of
each reactant? each reactant?
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ProblemProblem
Determine the reaction orders for the reaction indicated Determine the reaction orders for the reaction indicated from the data provided. from the data provided.
A + 2B + C A + 2B + C Products. Products.
[A]o [B]o [C]o Ro 2.06 3.05 4.00 3.7 0.87 3.05 4.00 0.66 0.50 0.50 0.50 0.013 1.00 0.50 1.00 0.072
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ProblemProblem
Determine the reaction order for each reactant Determine the reaction order for each reactant from the table.from the table.
BrO3BrO3--(aq)+5Br(aq)+5Br(aq)+6H(aq)+6H++(aq)(aq)3Br3Br22(aq)+3H(aq)+3H22O(l)O(l)
[ 3BrO ]o [Br ]o [H+]o Ro
0.10 0.10 0.10 1.2 0.20 0.10 0.10 2.4 0.10 0.30 0.10 3.5 0.20 0.10 0.15 5.4
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First–Order Reaction: First–Order Reaction: Integrated Rate LawIntegrated Rate Law
For a first order reaction, For a first order reaction, Rate = Rate = [A]/[A]/t = k[A]t = k[A]
Linear Graph, In [A] vs timeLinear Graph, In [A] vs time
Equation of the lineEquation of the line
Concentration vs timeConcentration vs time kt]A[
]A[ln
o
o]Aln[kt]Aln[
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ProblemProblem
Calculate the concentration of NCalculate the concentration of N22O O
remaining after its decomposition remaining after its decomposition according to 2Naccording to 2N22O(g) O(g) 2N 2N22(g) + O(g) + O22(g) if (g) if
it’s rate is first order and [Nit’s rate is first order and [N22O]O]oo = 0.20M, k = 0.20M, k
= 3.4 s= 3.4 s11 and T = 780°C. Find its and T = 780°C. Find its concentration after 100 ms.concentration after 100 ms.
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ProblemProblem
When cyclohexane (let's call it C) is When cyclohexane (let's call it C) is heated to 500 heated to 500 ooC, it changes into propene. C, it changes into propene. Using the following data from one Using the following data from one experiment, determine the first order rate experiment, determine the first order rate constant.: constant.:
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Half-Life: First Order ReactionHalf-Life: First Order Reaction
Half-life of First order reactionHalf-life of First order reaction, t, t1/21/2 = 0.693/k. the = 0.693/k. the
time required for the concentration of the reactant to time required for the concentration of the reactant to change to ½ of its initial value. change to ½ of its initial value.
i.e. at ti.e. at t1/21/2 , [A] = ½ [A] , [A] = ½ [A]oo
kt
tk
tkA
A
o
o
/693.0
2
1ln
][
][2/1ln
2/1
2/1
2/1
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ProblemProblem
For the decomposition of NFor the decomposition of N22OO55 at 65 °C, at 65 °C,
the half-life was found to be 130 s. the half-life was found to be 130 s. Determine the rate constant for this Determine the rate constant for this reaction. reaction.
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Second–Order Reactions: Second–Order Reactions: Integrated Rate LawIntegrated Rate Law
Rate law: R = k[A]Rate law: R = k[A]22
Plot of vs. t gives a straight line with Plot of vs. t gives a straight line with a slope of k. a slope of k.
Equation of the Line: Equation of the Line:
Half-life is:Half-life is:
ot ]A[kt
]A[11
t]A[
1
o2/1 ]A[k
1t
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ProblemProblem
At 330°C, the rate constant for the At 330°C, the rate constant for the decomposition of NOdecomposition of NO2 2 is 0.775 L/(mol*s). is 0.775 L/(mol*s).
If the reaction is second-order, what is the If the reaction is second-order, what is the concentration of NOconcentration of NO22 after 2.5x10 after 2.5x1022 s if the s if the
starting of concentration was 0.050 M?starting of concentration was 0.050 M?
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Zero–Order Reactions: Integrated Zero–Order Reactions: Integrated Rate LawRate Law
Rate law: R = k[A]Rate law: R = k[A]00
Plot of [A] vs. t gives a straight line with a Plot of [A] vs. t gives a straight line with a slope of -k. slope of -k.
Equation of the Line: [A] = kt - [AEquation of the Line: [A] = kt - [Aoo]]
Half-life is: tHalf-life is: t½½ = [A = [Aoo] / 2k] / 2k
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Reaction MechanismsReaction Mechanisms Give insight into sequence of reaction events Give insight into sequence of reaction events
leading to product (leading to product (reaction mechanismreaction mechanism). ). Each of the steps leading to product is called an Each of the steps leading to product is called an
elementary reaction or elementary stepelementary reaction or elementary step. . Consider the reaction of nitrogen dioxide with Consider the reaction of nitrogen dioxide with
carbon dioxide which is second order on NOcarbon dioxide which is second order on NO22::
NONO22(g) + CO(g) (g) + CO(g) NO(g) + CO NO(g) + CO22(g) (g)
Rate = k[NORate = k[NO22]]22. .
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Rate law suggests at least two steps. Rate law suggests at least two steps. A proposed mechanism for this reaction involves A proposed mechanism for this reaction involves
two steps.two steps.
NONO33 is a is a reaction intermediatereaction intermediate = a substance that is = a substance that is
produced and consumed in the reaction so that none produced and consumed in the reaction so that none is detected when the reaction is finished. is detected when the reaction is finished.
Step 1 2NO2(g) NO3(g) + NO(g) Step 2 NO3(g) +CO(g) NO2(g) + CO2(g) Overall NO2 + CO NO + CO2
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The elementary reactions are often The elementary reactions are often described in terms of their molecularity. described in terms of their molecularity. Unimolecular One particle in elementary.Unimolecular One particle in elementary. Bimolecular = 2 particles and Bimolecular = 2 particles and Termolecular = 3 particles Termolecular = 3 particles
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Rate Laws and Reaction Rate Laws and Reaction MechanismsMechanisms
Overall reaction order is often determined by the rate Overall reaction order is often determined by the rate determining step.determining step.
Use rate law of limiting step; No intermediates!Use rate law of limiting step; No intermediates!
2NO2NO22(g) (g) NONO
33(g) + NO(g),(g) + NO(g), RR11 = k = k11[NO[NO
22]]22 SlowSlow
NONO33(g) +CO(g)(g) +CO(g) NONO
22(g) + CO(g) + CO22(g) (g) RR22 = k = k
22[NO[NO33][CO]][CO] FastFast
NONO22 + CO + CO NO + CONO + CO
22 RRobsobs = k[NO = k[NO22]]
22
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ProblemProblem Determine the Rate Law for the mechanism Determine the Rate Law for the mechanism
given belowgiven below
2*[N2O5(g) )g(NO)g(NO 32k 1
1k
] Fast
NO3(g) +NO2(g) )g(O)g(NO)g(NO 22k2
Slow
NO3 + NO )g(NO2 2k3
Fast
2N2O5(g) )g(O)g(NO4 22kobs
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Use steady state approximation. at “equilibrium” rates of forward and reverse reactions are same. Use to eliminate intermediates from rate law equations.
or 32152
11
1]NO][NO[k]ON[k
RR
]NO[]ON[
k
k]NO[
2
52
13
1
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Collision TheoryCollision Theory
Collision theory assumes:Collision theory assumes: Reaction can only occur if collision takes Reaction can only occur if collision takes
place. place. Colliding molecules must have correct Colliding molecules must have correct
orientation and energy. orientation and energy. Collision rate is directing proportional to the Collision rate is directing proportional to the
concentration of colliding particles. concentration of colliding particles.
A + B A + B Products; R Products; Rcc = Z[A][B] = Z[A][B]
2A + B 2A + B Products; R Products; Rcc = Z[A] = Z[A]22[B], etc. [B], etc.
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Only a fraction of the molecules, Only a fraction of the molecules, pp (“ (“stericsteric factor factor”), have correct orientation; multiply collision rate ”), have correct orientation; multiply collision rate by by pp..
Particle must have enough energy. Fraction of Particle must have enough energy. Fraction of those with correct energy follows Boltzmann those with correct energy follows Boltzmann equation where Eequation where Eaa = activation = activation
energy, R = gas constant and T = temp. (Kelvin energy, R = gas constant and T = temp. (Kelvin scale only please). scale only please).
This gives: k = ZpfThis gives: k = Zpf
RT/Eaef
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““Steric Factor”Steric Factor” Molecules must have the correct orientation Molecules must have the correct orientation
before a reaction can take place.before a reaction can take place.
Return to p. 14-14
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Transition State Theory Transition State Theory
Explains the reaction resulting from the collision Explains the reaction resulting from the collision of molecules to form an activated complex. of molecules to form an activated complex.
Activated complex is unstable and can break to Activated complex is unstable and can break to form product.form product.
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Potential Energy DiagramPotential Energy Diagram
Endothermic reactionsEndothermic reactions In endothermic reactions heat energy is taken in In endothermic reactions heat energy is taken in
from the surroundings and turned into potential from the surroundings and turned into potential energy in the products. As a result, the enthalpy energy in the products. As a result, the enthalpy of the products is greater than the enthalpy of of the products is greater than the enthalpy of the reactants.the reactants.
The potential energy (enthalpy) diagram for an The potential energy (enthalpy) diagram for an endothermic reaction is shown on the next slideendothermic reaction is shown on the next slide
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Potential Energy DiagramPotential Energy Diagram
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Exothermic reactionsExothermic reactions In exothermic reactions potential energy in the In exothermic reactions potential energy in the
reactants is turned into heat energy and given reactants is turned into heat energy and given off to the surroundings. As a result, the enthalpy off to the surroundings. As a result, the enthalpy of the products is less than the enthalpy of the of the products is less than the enthalpy of the reactants.reactants.
The potential energy (enthalpy) diagram for an The potential energy (enthalpy) diagram for an exothermic reaction is shown on the next slideexothermic reaction is shown on the next slide
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Potential Energy DiagramPotential Energy Diagram
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Reaction Rates and Temperature: Reaction Rates and Temperature: The Arrhenius EquationThe Arrhenius Equation
Rate (rate constant) increases exponentially with Rate (rate constant) increases exponentially with temperature. temperature.
Collision theory indicates collisions every 10Collision theory indicates collisions every 1099s – s – 10101010s at 25°C and 1 atm.s at 25°C and 1 atm.i.e. only a small fraction of the colliding molecules i.e. only a small fraction of the colliding molecules actually react.actually react.
RTEaef /
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The Arrhenius EquationThe Arrhenius Equation
RT
EAlnkln a
Summary: where A = frequency factor.Summary: where A = frequency factor.
Linear form: . Linear form: .
Plot ln k vs. 1/T; the slope gives EPlot ln k vs. 1/T; the slope gives Eaa/R./R.
Two point equation sometimes used also: Two point equation sometimes used also:
RTE
expAk a
Arrhenius Plot
1/T, K
0.00300 0.00305 0.00310 0.00315
ln k
-8
-7
-6
-5
21
a
1
2T
1
T
1
R
E
k
kln
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Rate constant increases when TRate constant increases when T22>T>T11
k, sk, s11 Temp., °CTemp., °C Temp., KTemp., K4.8x104.8x1044 45.045.0 318.15318.158.8x108.8x1044 50.050.0 323.15323.151.6x101.6x1033 55.055.0 328.15328.152.8x102.8x1033 60.060.0 333.15333.15
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ProblemProblem
Determine the activation energy for the Determine the activation energy for the decomposition of Ndecomposition of N22OO55 from the from the
temperature dependence of the rate temperature dependence of the rate constant.constant.
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ProblemProblem
Determine the rate constant at 35°C for Determine the rate constant at 35°C for the hydrolysis of sucrose, given that at the hydrolysis of sucrose, given that at 37°C it is 0.91mL/(mol*sec). The 37°C it is 0.91mL/(mol*sec). The activation energy of this reaction is 108 activation energy of this reaction is 108 kJ/mol.kJ/mol.
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CatalysisCatalysis CatalystsCatalysts a substance that increases the rate of a substance that increases the rate of
a reaction without being consumed in the a reaction without being consumed in the reaction. reaction.
Catalyst provides an alternative pathway from Catalyst provides an alternative pathway from reactant to product which has a rate determining reactant to product which has a rate determining step that has a lower activation energy than that step that has a lower activation energy than that of the original pathway. of the original pathway.
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Potential Energy Diagram with Potential Energy Diagram with CatalystCatalyst
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Homogeneous and Homogeneous and Heterogeneous CatalystsHeterogeneous Catalysts
Homogeneous catalystHomogeneous catalyst: catalyst : catalyst existing in the same phase as the existing in the same phase as the reactants.reactants.
Heterogeneous catalysisHeterogeneous catalysis: catalyst : catalyst existing in a different phase than the existing in a different phase than the reactants.reactants.
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The catalytic hydrogenation of ethylene is an The catalytic hydrogenation of ethylene is an example of a heterogeneous catalysis reaction:example of a heterogeneous catalysis reaction:
)g(CHCH)g(H)g(CHCHPt
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ENZYMESENZYMES (biological (biological catalysts) catalysts)
They are proteins (large They are proteins (large organic molecules that are organic molecules that are composed of amino acids). composed of amino acids).
Slotlike active sites. The Slotlike active sites. The molecule fits into this slot molecule fits into this slot and reaction proceeds. and reaction proceeds. Poisons can block active Poisons can block active site or reduce activity by site or reduce activity by distorting the active site.distorting the active site.