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Chemical KineticsChemical KineticsThe area of chemistry that concerns The area of chemistry that concerns reaction rates and reaction reaction rates and reaction
mechanisms.mechanisms.
Outline: KineticsReaction Rates How we measure rates.
Rate LawsHow the rate depends on amounts of reactants.
Integrated Rate LawsHow to calc amount left or time to reach a given amount.
Half-lifeHow long it takes to react 50% of reactants.
Arrhenius Equation How rate constant changes with T.
MechanismsLink between rate and molecular scale processes.
Factors That Affect Reaction Rates• Concentration of Reactants
– As the concentration of reactants increases, so does the likelihood that reactant molecules will collide.
• Temperature– At higher temperatures, reactant molecules have more kinetic energy, move
faster, and collide more often and with greater energy.• Catalysts
– Speed rxn by changing mechanism.
Reaction RateThe change in concentration of a The change in concentration of a reactant or product per unit of timereactant or product per unit of time
2 1
2 1
[ ] [ ]A at timet A at timetRate
t t
[ ]ARate
t
UNITS FOR RATE
• Mol/L •s
• Mol/L •hr
• Mol/L • day
• Etc…………………….
2NO2NO22(g) (g) 2NO(g) + O 2NO(g) + O22(g)(g)Reaction Rates:
2. Can measure appearance of products
1. Can measure disappearance of reactants
3. Are proportional stoichiometrically
2NO2NO22(g) (g) 2NO(g) + O 2NO(g) + O22(g)(g)Reaction Rates:4. Are equal to
the slope tangent to that point
[NO2]
t
5. Change as the reaction proceeds, if the rate is dependent upon concentration2[ ]
constantNO
t
Products and reactants are stoichiometrically related
• 2 NO2 → 2 NO + O2
• If the rate of disappearance of NO2 is 2.0
mol/L • s what is the rate of appearance for NO and O2 ?
Page 598 – Problem 17
Rate LawsDifferential rate laws express (reveal) the relationship between the concentration of reactants and the rate of the reaction.
Integrated rate laws express (reveal) the relationship between concentration of reactants and time
The differential rate law is usually just called “the rate law.”
• What is the difference between rate and time?
How to write a rate expression
• R = k [A]x [B]y
• X and Y represent the order to which the reactant is raised.
HOW DO YOU DETERMINE THE ORDER
INITIAL RATES METHOD
Writing a (differential) Rate LawWriting a (differential) Rate Law
2 NO(g) + Cl2(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:
ExperimenExperimentt
[NO][NO]
(mol/L)(mol/L)[Cl[Cl22]]
(mol/L)(mol/L)
RateRate
Mol/L·sMol/L·s
11 0.2500.250 0.2500.250 1.43 x 101.43 x 10-6-6
22 0.5000.500 0.2500.250 5.72 x 105.72 x 10-6-6
33 0.2500.250 0.5000.500 2.86 x 102.86 x 10-6-6
44 0.5000.500 0.5000.500 11.4 x 1011.4 x 10-6-6
• Log Rate3 = (Conc3 ) log
• Rate 4 (Conc4 )
ExperimenExperimentt
[NO][NO]
(mol/L)(mol/L)[Cl[Cl22]]
(mol/L)(mol/L)
RateRate
Mol/L·sMol/L·s
11 0.2500.250 0.2500.250 1.43 x 101.43 x 10-6-6
22 0.5000.500 0.2500.250 5.72 x 105.72 x 10-6-6
33 0.2500.250 0.5000.500 2.86 x 102.86 x 10-6-6
44 0.5000.500 0.5000.500 11.4 x 1011.4 x 10-6-6
Writing a Rate LawWriting a Rate LawPart 3 – Determine the overall order for the reaction.
R = k[NO]2[Cl2]
Overall order is the sum of the exponents, or orders, of the reactants
2 + 1 = 3
The reaction is 3rd order
Page 598 – problem 22
Page 598 – problem 24
Page 599 – problems 26
SHORT CUT FOR DETERMINING ORDER
• If the concentration increases and the rate stays the same – zero order
• If the concentration doubles and the rate doubles• – 1st order
• If the concentration doubles and the rate increases by 4 – 2nd order
2What if you have more than one reactant and one is not held constant?
EXPERIMENT Initial [ClO2- ]
mol/L 2
Initial [F2 ] mol/L Initial Rate of Increase of [ClO2F]
1 0.010 0.10 2.4 x 10-3
2 0.010 0.40 9.6 x 10-3
3 0.020 0.20 9.6 x 10-3
a. Write the rate law for the reaction 2 ClO2 + → 2 ClO2Fb. Calculate the numerical value of the rate constant and specify the unit.c. In experiment 2, what is the initial rate of decrease of [F2 ].d. Which of the following rate law developed in a? Justify your choice.ClO2 + F2 ↔ ClO2F2
fast
ClO2F2 → ClO2F + F slowClO2 + F → ClO2F fast
or F2 → 2 F slow2 ClO2
+ 2 F → ClO2F fast
UNITS FOR K
Determining Order withConcentration vs. Time data
(the Integrated Rate Law)
.timevs concentrationis linear
.ln( )timevs concentration is linear
1.timevs is linearconcentration
Zero Order:
First Order:
Second Order:
Rate Laws SummaryRate Laws SummaryZero OrderZero Order First OrderFirst Order Second OrderSecond Order
Rate LawRate Law Rate = k Rate = k[A] Rate = k[A]2
Integrated Integrated Rate LawRate Law
[A0 ] – [A] = kt ln[A] = kt
[A]0
1 - 1
[A] [A0 ] = kt
Plot the Plot the produces a produces a straight linestraight line
[A] versus t ln[A] versus t
Relationship Relationship of rate of rate constant to constant to slope of slope of straight linestraight line
Slope = -k Slope = -k Slope = k
Half-LifeHalf-Life
1
[ ]versus t
A
01/ 2
[ ]
2
At
k 1/ 2
0.693t
k 1/ 2
0
1
[ ]t
k A
Solving an Integrated Rate LawTime (s)Time (s) [H[H22OO22] (mol/L)] (mol/L)
00 1.001.00
120120 0.910.91
300300 0.780.78
600600 0.590.59
12001200 0.370.37
18001800 0.220.22
24002400 0.130.13
30003000 0.0820.082
36003600 0.0500.050
Problem: Find the integrated rate law and the value for the rate constant, kA graphing calculator with linear regression analysis greatly simplifies this process!!
Time vs. [HTime vs. [H22OO22]]Time Time (s)(s)
[H[H22OO22]]
00 1.001.00
120120 0.910.91
300300 0.780.78
600600 0.590.59
12001200 0.370.37
18001800 0.220.22
24002400 0.130.13
30003000 0.0820.082
36003600 0.0500.050
Find 2 points use∆Y/∆X = A
Find 2 more points∆Y/∆X = appx AThe slope of both should be close.Pick the plot that is closest to the same numbers
Time vs. ln[HTime vs. ln[H22OO22]]Time (s)Time (s) ln[Hln[H22OO22]]
00 00
120120 -0.0943-0.0943
300300 -0.2485-0.2485
600600 -0.5276-0.5276
12001200 -0.9943-0.9943
18001800 -1.514-1.514
24002400 -2.04-2.04
30003000 -2.501-2.501
36003600 -2.996-2.996
Regression results:
Time vs. 1/[HTime vs. 1/[H22OO22]]Time Time (s)(s)
1/[H1/[H22OO22]]
00 1.001.00
120120 1.09891.0989
300300 1.28211.2821
600600 1.69491.6949
12001200 2.70272.7027
18001800 4.54554.5455
24002400 7.69237.6923
30003000 12.19512.195
36003600 20.00020.000
And the winner is… Time vs. ln[H2O2]
1. As a result, the reaction is 1st order
2. The (differential) rate law is:
2 2[ ]R k H O3. The integrated rate law is:
2 2 2 2 0ln[ ] ln[ ]H O kt H O
4. But…what is the rate constant, k ?
Finding the Rate Constant, kMethod #1: Calculate the slope from the Time vs. ln[H2O2] table. Time (s)Time (s) ln[Hln[H22OO22]]
00 00
120120 -0.0943-0.0943
300300 -0.2485-0.2485
600600 -0.5276-0.5276
12001200 -0.9943-0.9943
18001800 -1.514-1.514
24002400 -2.04-2.04
30003000 -2.501-2.501
36003600 -2.996-2.996
2 2 2 2 0ln[ ] ln[ ]H O kt H O
2 2ln[ ] 2.996
3600
H Oslope
t s
4 18.32 10slope x s
Now remember:
k = -slope
k = 8.32 x 10-4s-1
Determining rxn orderThe decomposition of NO2 at 300°C is described by the equation
NO2 (g) NO (g) + 1/2 O2 (g)
and yields these data:
Time (s) [NO2], M
0.0 0.01000
50.0 0.00787
100.0 0.00649
200.0 0.00481
300.0 0.00380
Graphing ln [NO2] vs. t yields:
Time (s) [NO2], M ln [NO2]
0.0 0.01000 -4.610
50.0 0.00787 -4.845
100.0 0.00649 -5.038
200.0 0.00481 -5.337
300.0 0.00380 -5.573
• The plot is not a straight line, so the process is not first-order in [A].
Determining rxn order
Does not fit:
Second-Order ProcessesA graph of 1/[NO2]
vs. t gives this plot.
Time (s) [NO2], M 1/[NO2]
0.0 0.01000 100
50.0 0.00787 127
100.0 0.00649 154
200.0 0.00481 208
300.0 0.00380 263
• This is a straight line. Therefore, the process is second-order in [NO2].
Finding the Rate Constant, kMethod #2: Obtain k from the linear regresssion analysis.
2 2 2 2 0ln[ ] ln[ ]H O kt H O
4 18.35 10slope a x s
Now remember:
k = -slope
k = 8.35 x 10-4s-1
Regression results:
y = ax + b a = -8.35 x 10-4
b = -.005r2 = 0.99978r = -0.9999
Reaction Mechanism
The reaction mechanism is the series of elementary steps by which a chemical reaction occurs.
The sum of the elementary steps must give the overall balanced equation for the reaction The mechanism must agree with the experimentally determined rate law
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 The experimental rate law must agree with the rate-determining agree with the rate-determining step step
Identifying the Rate-Determining StepFor the reaction:
2H2(g) + 2NO(g) N2(g) + 2H2O(g)The experimental rate law is:
R = k[NO]2[H2]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 IntermediatesFor the reaction:
2H2(g) + 2NO(g) N2(g) + 2H2O(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
Collision ModelKey Idea: Molecules must collide to react.However, only a small fraction of collisions produces a reaction. Why?
The Collision Model
Furthermore, molecules must collide with the correct orientation and with enough energy to cause bond breakage and formation.
Activation Energy• In other words, there is a minimum amount of energy
required for reaction: the activation energy, Ea.
• Just as a ball cannot get over a hill if it does not roll up the hill with enough energy, a reaction cannot occur unless the molecules possess sufficient energy to get over the activation energy barrier.
Collision ModelCollisions must have Collisions must have sufficient sufficient energyenergy to produce the reaction to produce the reaction (must equal or exceed the (must equal or exceed the activation energy).activation energy).
Colliding particles must be Colliding particles must be correctly correctly orientedoriented to one another in order to to one another in order to produce a reaction.produce a reaction.
1.1.
2.2.
Factors Affecting RateFactors Affecting RateIncreasing temperature always increases the rate of a reaction.
Particles collide more frequently Particles collide more energeticallyIncreasing surface area increases
the rate of a reaction
Increasing Concentration USUALLY increases the rate of a reaction
Presence of Catalysts, which lower the activation energy by providing alternate pathways
Endothermic ReactionsEndothermic Reactions
Exothermic ReactionsExothermic Reactions
Maxwell–Boltzmann Distributions
• As the temperature increases, the curve flattens and broadens.
• Thus at higher temperatures, a larger population of molecules has higher energy.
The Arrhenius Equation
/aE RTk Ae kk = rate constant at = rate constant at temperature Ttemperature T AA = frequency factor = frequency factor EEaa = activation energy = activation energy RR = Gas constant, 8.31451 = Gas constant, 8.31451 J/K·molJ/K·mol
The Arrhenius Equation, Rearranged1
ln( ) ln( )aEk AR T
Simplifies solving for Ea
-Ea / R is the slope when (1/T) is plotted against ln(k) ln(A) is the y-intercept Linear regression analysis of a table of (1/T) vs. ln(k) can quickly yield a slope Ea = -R(slope)
CatalysisCatalysis•Catalyst: A substance that speeds up a reaction : A substance that speeds up a reaction without being consumedwithout being consumed
•Enzyme: A large molecule (usually a protein) that : A large molecule (usually a protein) that catalyzes biological reactions.catalyzes biological reactions.
•Homogeneous catalyst: Present in the same phase as : Present in the same phase as the reacting molecules.the reacting molecules.
•Heterogeneous catalyst: Present in a different phase : Present in a different phase than the reacting moleculesthan the reacting molecules..
Lowering of Activation Energy Lowering of Activation Energy by a Catalystby a Catalyst
Catalysts
One way a catalyst can speed up a reaction is by holding the reactants together and helping bonds to break.
Enzymes• Enzymes are catalysts
in biological systems.• The substrate fits into
the active site of the enzyme much like a key fits into a lock.
Catalysts Increase the Number of Catalysts Increase the Number of Effective CollisionsEffective Collisions
Heterogeneous CatalysisHeterogeneous Catalysis
Step #1: Step #1: Adsorption and Adsorption and
activation of activation of the reactants.the reactants.
Carbon monoxide and nitrogen
monoxide adsorbed on a
platinum surface
Heterogeneous CatalysisHeterogeneous Catalysis
Step #2: Step #2:
Migration of the Migration of the adsorbed adsorbed
reactants on the reactants on the surface.surface.
Carbon monoxide and nitrogen
monoxide arranged prior to
reacting
Heterogeneous CatalysisHeterogeneous Catalysis
Step #3: Step #3:
Reaction of the Reaction of the adsorbed adsorbed
substances.substances.
Carbon dioxide and nitrogen form
from previous molecules
Heterogeneous CatalysisHeterogeneous Catalysis
Step #4: Step #4:
Escape, or Escape, or desorption, of desorption, of the productsthe products..
Carbon dioxide and nitrogen gases escape
(desorb) from the platinum surface
Slow Initial Step
• The rate law for this reaction is found experimentally to be
Rate = k [NO2]2
• CO is necessary for this reaction to occur, but the rate of the reaction does not depend on its concentration.
• This suggests the reaction occurs in two steps.
NO2 (g) + CO (g) NO (g) + CO2 (g)
Slow Initial Step• A proposed mechanism for this reaction is
Step 1: NO2 + NO2 NO3 + NO (slow)
Step 2: NO3 + CO NO2 + CO2 (fast)
• The NO3 intermediate is consumed in the second step.
• As CO is not involved in the slow, rate-determining step, it does not appear in the rate law.
Fast Initial Step• A proposed mechanism is
Step 1 is an equilibrium- it includes the forward and reverse reactions.