11/9/2012 1 LECTURE 11: CHEMICAL KINETICS 1. Will the reaction occur? 2. How far will the reaction proceed? 3. How fast will the reaction occur? CHEMICAL REACTIONS C(s, diamond) C(s, graphite) ∆G° rxn = -2.87 kJ/mole Is the reaction favorable? CHEMICAL REACTIONS KINETICS • Studies the rate at which a chemical process occurs. • Besides information about the speed at which reactions occur, kinetics also sheds light on the reaction mechanism (exactly how the reaction occurs).
Transcript
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LECTURE 11: CHEMICAL KINETICS
1. Will the reaction occur? 2. How far will the reaction proceed? 3. How fast will the reaction occur?
CHEMICAL REACTIONS
C(s, diamond) C(s, graphite) ∆G°rxn = -2.87 kJ/mole Is the reaction favorable?
CHEMICAL REACTIONS KINETICS
• Studies the rate at which a chemical process occurs. • Besides information about the speed at which reactions
occur, kinetics also sheds light on the reaction mechanism (exactly how the reaction occurs).
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OUTLINE : KINETICS
Reaction Rates How we measure rates.
Rate Laws How the rate depends on amounts of reactants.
Integrated Rate Laws How to calculate amount left or time to reach a given amount.
Half-life How long it takes to react 50% of reactants.
Arrhenius Equation How rate constant changes with Temperature.
Mechanisms Link between rate and molecular scale processes.
Factors That Influence Reaction Rate
Under a specific set of conditions, every reaction has its own characteristic rate, which depends upon the chemical nature of the reactants.
Four factors can be controlled during the reaction: 1. Concentration - molecules must collide to react 2. Physical state - molecules must mix to collide 3. Temperature - molecules must collide with enough energy to react 4. The use of a catalyst
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 up the reaction by changing
mechanism.
Rates of reactions can be determined by monitoring the change in concentration of either reactants or products as a function of time. [A] vs t
REACTION RATES
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In this reaction, the concentration of butyl chloride, C4H9Cl, was measured at various times, t.
C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)
[C4H9Cl] M
REACTION RATES
The average rate of the reaction over each interval is the change in concentration divided by the change in time:
C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)
Average Rate, M/s
REACTION RATES
• Note that the average rate decreases as the reaction proceeds.
• This is because as the reaction goes forward, there are fewer collisions between reactant molecules.
C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)
REACTION RATES
• A plot of concentration vs. time for this reaction yields a curve like this.
• The slope of a line tangent to the curve at any point is the instantaneous rate at that time.
C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)
REACTION RATES
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• The reaction slows down with time because the concentration of the reactants decreases.
C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)
REACTION RATES
• In this reaction, the ratio of C4H9Cl to C4H9OH is 1:1.
• Thus, the rate of disappearance of C4H9Cl is the same as the rate of appearance of C4H9OH.
C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)
Rate = -[C4H9Cl] t = [C4H9OH]
t
REACTION RATES AND STOICHIOMETRY
• What if the ratio is not 1:1?
H2(g) + I2(g) 2HI(g)
• Only 1/2 HI is made for each H2 used.
REACTION RATES AND STOICHIOMETRY
• To generalize, for the reaction
aA + bB cC + dD
Reactants (decrease) Products (increase)
REACTION RATES AND STOICHIOMETRY
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Sample Problem
PLAN:
Expressing Rate in Terms of Changes in Concentration with Time
PROBLEM: Because it has a nonpolluting product (water vapor), hydrogen gas is used for fuel aboard the space shuttle and in prototype cars with Earth-bound engines:
2H2(g) + O2(g) 2H2O(g)
(a) Express the rate in terms of changes in [H2], [O2], and [H2O] with time.
(b) When [O2] is decreasing at 0.23 mol/L•s, at what rate is [H2O] increasing?
Choose [O2] as a point of reference since its coefficient is 1. For every molecule of O2 which disappears, 2 molecules of H2 disappear and 2 molecules of H2O appear, so [O2] is disappearing at half the rate of change of H2 and H2O.
SOLUTION:
Expressing Rate in Terms of Changes in Concentration with Time
PROBLEM: Because it has a nonpolluting product (water vapor), hydrogen gas is used for fuel aboard the space shuttle and in prototype cars with Earth-bound engines:
2H2(g) + O2(g) 2H2O(g)
- 1 2
[H2] t
= - [O2] t = + [H2O]
t 1 2
0.23 mol/L•s = + [H2O] t
1 2
; = 0.46 mol/L•s [H2O] t
rate = (a)
[O2] t
- = - (b)
Each reaction has its own equation that gives its rate as a function of reactant concentrations.
this is called its Rate Law To determine the rate law we measure the rate at different
starting concentrations.
CONCENTRATION AND RATES
Compare Experiments 1 and 2: when [NH4
+] doubles, the initial rate doubles.
CONCENTRATION AND RATE
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Likewise, compare Experiments 5 and 6: when [NO2
-] doubles, the initial rate doubles.
CONCENTRATION AND RATE
This equation is called the rate law, and k is the rate constant.
CONCENTRATION AND RATES
• A rate law shows the relationship between the reaction rate and the concentrations of reactants. – For gas-phase reactants use PA instead of [A].
• k is a constant that has a specific value for each reaction. • The value of k is determined experimentally. “Constant” is relative here- k is unique for each rxn k changes with T
RATE LAWS
• Exponents tell the order of the reaction with respect to each reactant.
• This reaction is First-order in [NH4
+] First-order in [NO2
−] • The overall reaction order can be found by adding the
exponents on the reactants in the rate law. • This reaction is second-order overall.
RATE LAWS
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Sample Problem 16.2 Determining Reaction Order from Rate Laws
PROBLEM: For each of the following reactions, use the given rate law to determine the reaction order with respect to each reactant and the overall order:
Using initial rates - Run a series of experiments, each of which starts with a different set of reactant concentrations, and from each obtain an initial rate. See Table 16.2 for data on the reaction
O2(g) + 2NO(g) 2NO2(g) rate = k [O2]m[NO]n
Compare 2 experiments in which the concentration of one reactant varies and the concentration of the other reactant(s) remains constant.
k [O2]2m[NO]2n
k [O2]1m[NO]1n =
rate2
rate1 =
[O2]2m [O2]1m =
6.40 x 10-3 mol/L•s 3.21 x 10-3 mol/L•s
[O2]2 [O2]1
m
= 1.10 x 10-2 mol/L
2.20 x 10-2 mol/L m ; 2 = 2m m = 1
Do similar calculations for the other reactant(s).
Sample Problem 16.3
PLAN:
SOLUTION:
Determining Reaction Orders from Initial Rate Data PROBLEM: Many gaseous reactions occur in a car engine and exhaust
systems. One of these is N O 2 (g ) + C O (g ) N O (g ) + C O 2 (g ) rate = k[NO2]m[CO]n
Use the following data to determine the individual and overall reaction orders: Experiment Initial Rate (mol/L•s) Initial [NO2] (mol/L) Initial [CO] (mol/L)
1 2 3
0.0050 0.080 0.0050
0.10
0.10 0.40 0.10
0.10 0.20
Solve for the order with respect to each reactant using the general rate law using the method described previously.
rate = k [NO2]m[CO]n
First, choose two experiments in which [CO] remains constant and the [NO2] varies.
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Sample Problem 16.3 Determining Reaction Order from Initial Rate Data continued
0.080 0.0050
rate 2 rate 1
[NO2] 2 [NO2] 1
m = k [NO2]m2[CO]n2
k [NO2]m1 [CO]n1 =
0.40 0.10 =
m 16 = 4.0m and m = 2.0
k [NO2]23[CO]n3 k [NO2]21 [CO]n1
[CO] 3 [CO] 1
n = rate 3
rate 1 =
0.0050 0.0050 =
0.20 0.10
n 1 = 2.0n and n = 0
The reaction is 2nd order in NO2.
The reaction is zero order in CO.
rate = k [NO2]2[CO]0 = k [NO2]2 The reaction is second order overall.
SAMPLE PROBLEM: The initial rate of a reaction A + B → C was measured for several different starting concentrations of A and B, and the results are as follows:
Expt. No [A], M [B], M Initial Rate, M/s
1 0.100 0.100 4.0 x 10-5
2 0.100 0.200 4.0 x 10-5
3 0.200 0.100 16.0 x 10-5
Using the data, determine a) rate law for the reaction, b) the rate constant c) rate of the reaction when [A] = 0.50M and [B] = 0.100M
Consider a simple zero order rxn: A B
RATE LAWS
ktAA
kdtAd:Integratet? time after leftis A much How
kAkdtdA- form aldifferenti in Akrate
0
0
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Consider a simple 1st order rxn: A B
How much A is left after time t? Integrate:
Differential form:
RATE LAWS
The integrated form of first order rate law:
Can be rearranged to give:
[A]0 is the initial concentration of A (t = 0). [A]t is the concentration of A at some time, t, during the course of the reaction.
INTEGRATED RATE LAWS
Manipulating this equation produces…
…which is in the form y = mx + b
INTEGRATED RATE LAWS
If a reaction is first-order, a plot of ln [A]t vs. t will yield a straight line with a slope of -k.
So, use graphs to determine reaction order.
FIRST ORDER PROCESSES
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Methyl methanoate is hydrolyzed when dissolved in excess hydrochloric acid at 298 K. The ester’s concentration was 0.01 mol/li at the start of the reaction, but 8.09 x 10-3 after 21 min. What is the value of the first order rate constant k1?
What is the concentration of methyl methanoate after 30
minutes?
SAMPLE PROBLEM
Consider the process in which methyl isonitrile is converted to acetonitrile.
CH3NC CH3CN
How do we know this is a first order reaction?
FIRST ORDER PROCESSES
This data was collected for this reaction at 198.9°C.
CH3NC CH3CN
Does rate=k[CH3NC] for all time intervals?
FIRST ORDER PROCESSES
• When ln P is plotted as a function of time, a straight line results. – The process is first-order. – k is the negative slope: 5.1 10-5 s-1.
FIRST ORDER PROCESSES
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Similarly, integrating the rate law for a process that is second-order in reactant A:
also in the form y = mx + b
Rearrange, integrate:
SECOND ORDER PROCESSES
So if a process is second-order in A, a plot of 1/[A] vs. t will yield a straight line with a slope of k.
If a reaction is first-order, a plot of ln [A]t vs. t will yield a straight line with a slope of -k.
First order:
SECOND ORDER PROCESSES
SAMPLE PROBLEM
The dimerization of methyl viologen, an active ingredient in weed killers is second order with respect to methyl viologen, MV. Calculate the value of the rate constant, k if the initial concentration of MV was 0.001 mol/L and the concentration dropped to 4 x 10-4 mol/L after 0.02 s.
SAMPLE PROBLEM
Consider a second order reaction which consumes 15% of the initial material after 12 min and 23 s. If the initial concentration was 0.001 mol/L calculate the rate constant k.
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The 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
DETERMINING ORDER OF REACTION
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].
Does not fit:
DETERMINING ORDER OF REACTION
A 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].
SECOND ORDER PROCESSES
The second type of second order reactions is represented by:
A + B products
SECOND ORDER PROCESSES
kt]B][A[]A][B[ln
]A[]B[1
]B][A[kdt
]B[ddt
]A[drate
o
0
00
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For some reactions like: CH3COCl(aq) + H2O(l) CH3COOH(aq) + HCl(aq)
SECOND ORDER PROCESSES
reaction. order first pseudo called also
]OH[k'k where ]COClCH[krate:is rate reactionthus ,negligible
is consumed amount excess, inalways is reactant one if but]OH][COClCH['krate
23
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SAMPLE PROBLEM: Determining order of reaction from Rate Law
For each of the following reactions, use the given rate law to determine the reaction order with respect to each reactant and the overall order:
A 1:1 reaction occurs between A and B, and is second order. The initial concentrations of A and B are 0.1 mol/L and 0.2 mol/L, respectively. What is the rate constant, k if the concentration of A after 0.5 hr is 0.05 mol/L?
• Half-life is defined as the time required for one-half of a reactant to react.
• Because [A] at t1/2 is one-half of the original [A],
[A]t = 0.5 [A]0.
HALF-LIFE
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For a first-order process, set [A]t=0.5 [A]0 in integrated rate equation:
NOTE: For a first-order process, the half-life does not depend on [A]0.
HALF LIFE
For a second-order process, set [A]t=0.5 [A]0 in 2nd order equation.
HALF-LIFE SECOND ORDER
SAMPLE PROBLEM
The half-life of 60Co is 10.5 min. If we start with 100g of 60Co, how much remains after 42 minutes?
SUMMARY OF RATE EQUATIONS: Order Differential Form Integrated Form Half-Life Units of the Rate
Constants
0
1
2
2 -
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DETERMINATION OF REACTION ORDER
Order (m) ∆[A] by a factor of: Effect on rate Zero (0) 2, 4, 15, ½, etc. None
1st (1) 2 2X 3 3X
2nd (2)
2 4X 3 9X ½ ¼X
Reaction Orders For the reaction: A →B, the rate law is:
rate = k[A]m
Determining Reaction Orders from Initial Rate Data
PROBLEM: Many gaseous reactions occur in a car engine and exhaust systems. One of these is
N O 2 (g ) + C O (g ) N O (g ) + C O 2 (g ) rate = k[NO2]m[CO]n
Use the following data to determine the individual and overall reaction orders:
SAMPLE PROBLEM: Consider the following reaction: hydrogen peroxide decomposes in the presence of excess cerous ion (Ce+3). The following data were obtained:
Using the differential method, determine the order of reaction with respect to OCI-.
REACTION MECHANISM: A reaction mechanism is a sequence of molecular events, or reaction steps, that defines the pathway from reactants to products.
OH-(aq) + CH3Cl(aq) CH3OH(aq) + Cl-(aq)
REACTION MECHANISM: Reaction is broken into steps with intermediates being formed. “some reactions occur in one step, but most occur in in multiple steps.” Each step is called an elementary step, and the number of molecules involved in each step defines the molecularity of the step. Molecularity: is the number of reactants.
REACTION MECHANISM: UNIMOLECULAR REACTIONS: A single reactant is involved.
O3* O2 + O
BIMOLECULAR REACTIONS: elementary steps that involves two reactants (most common) HI + HI activated complex H2 + I2 TERMOLECULAR REACTIONS: Three reactants are involved (rare, due to probability of orientation and energy both being correct.) O(g) + O2(g) + N2(g) O3(g) + “energetic” N2(g)
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13.5
Reaction Mechanisms
The overall progress of a chemical reaction can be represented at the molecular level by a series of simple elementary steps or elementary reactions.
The sequence of elementary steps that leads to product formation is the reaction mechanism.
Intermediates are species that appear in a reaction mechanism but not in the overall balanced equation. An intermediate is always formed in an early elementary step and consumed in a later elementary step.
The rate-determining step is the slowest step in the sequence of steps leading to product formation.
The reaction rate and rate law can be determined from the stoichiometry (coefficients) of the slowest elementary step in the mechanism
*Know the difference between an elementary reaction and the net reaction !
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The Raschig process for the preparation of hydrazine (N2H4)
“Cancel intermediates and “add steps” to give overall RXN.” 2NH3(g) + OCl-(aq) N2H4(aq) + Cl-(aq) + H2O(l)
The overall rate law, mechanism, and the total order can’t be predicted from the stoichiometry, only by experiment.
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The following is only true for individual steps:
The rate law of an elementary step is given by the product of a rate constant and the conc. of the reactants in the step.
Step Molecularity rate law
A Product(s) uni rate = k[A]
A + B Product(s) bi rate = k[A][B]
A + A Product(s) bi rate = k[A]2
2A + B Product(s) ter rate = k[A]2[B]
The overall mechanism must match the observed rate law.
Usually one STEP is assumed to be the rate determining step. 74
Example:
Overall RXN: 2NO2(g) + F2(g) 2NO2F(g)
Observed Experimental rate law: rate = k[NO2][F2]
Question: Why does this rule out a single step reaction?
Answer: rate law for single step process would be: rate = k[NO2]2[F2]
“Let’s try to work out a mechanism that matches the observed rate law.”
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Example:
Overall RXN: 2NO2(g) + F2(g) 2NO2F(g)
Observed Experimental rate law: rate = k[NO2][F2]
slow: NO2(g) + F2(g) NO2F(g) + F(g) ‡ k1
fast: NO2(g) + F(g) ‡ NO2F(g) k2
The rate law is dependent upon the slow step.
rate = k[NO2][F2]
2NO2(g) + F2(g) 2NO2F(g)
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Let’s try making the 2nd reaction slow and the first fast
Overall RXN: 2NO2(g) + F2(g) 2NO2F(g)
Observed Experimental rate law: rate = k[NO2][F2]
fast: NO2(g) + F2(g) NO2F(g) + F(g) ‡ k1
slow: NO2(g) + F(g) ‡ NO2F(g) k2
The rate law is dependent upon the slow step.
2NO2(g) + F2(g) 2NO2F(g)
rate = k[NO2][F ‡]
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SAMPLE PROBLEM: The experimental rate law for the reaction between NO2 and CO to produce NO and CO2 is rate = k[NO2]2. The reaction is believed to occur via two steps: STEP 1 NO2 + NO2 NO3 + NO STEP 2 NO3 + CO NO2 + CO2 What is the overall reaction? What is/are the intermediate/s? What can you say about the relative rates of reaction 1 and 2?
SAMPLE PROBLEM: The following substitution reaction has a first order rate law: (Co(CN)5(H2O)2- + I- Co(CN)5I3- + H2O Rate = k[Co(CN)5(H2O) 2-] Suggest a mechanism in accord with the rate law.
REACTION MECHANISM and RATE LAWS
Determine rate law by experiment. Devise a mechanism. Predict the rate law for the mechanism If the predicted rate laws do not agree, retry devising a mechanism. If predicted and experimental rate laws agree, look for additional supporting evidence.
Nitrogen oxide is reduced by hydrogen to give water and nitrogen, 2 H2(g) + 2 NO(g) → N2(g) + 2 H2O(g)
SAMPLE PROBLEM
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One possible mechanism to account for this reaction is: 2 NO(g) → N2O2(g)
N2O2(g) + H2(g) → N2O(g) + H2O(g)
N2O(g) + H2(g) → N2(g) + H2O(g)
What is the molecularity of each of the three steps?
Show that the sum of these elementary
steps is the net reaction.
A MODEL FOR KINETICS
Generally reactions occur more rapidly at higher temperatures than at lower temperatures.
The rate generally doubles for every 10 K rise in
temperature. It’s an exponential increase!
TRANSITION STATE THEORY
• energy barrier must be overcome • reaction energy diagram [humpy diagrams] • transition state energy--max of rxn. E diagram • activated complex--deformed molecules in their transition
state, formed at the Ets--unstable, can go either way! • Activation energy, E*, Ea--energy a reacting molecule must
absorb from its environment in order to react.
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uncatalyzed catalyzed
COLLISION THEORY
Assumes molecules must collide in order to react! EFFECTIVE COLLISIONS ARE hindered by
concentration, temperature and geometry of reactants and
transition states
THE EFFECT OF TEMPERATURE OF REACTION RATE: ARRHENIUS EQUATION
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A is the frequency factor units of L/(mol • s) & depends on the frequency of
collisions and the fraction of these that have the correct geometry--# of effective collisions
e-E*/RT is always less than 1 and is the fraction of
molecules having the minimum energy required for reaction
*Notice in the equation: As "Ea" increases, "k" gets smaller and thus, the rate would decrease.
Also, notice that as "T" goes up, "k" increases
and so the rate would also increase.
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Taking this equation, plot 1/T vs. ln [k], and get a straight line. From the straight line, find the slope and then find the activation energy. slope = - Ea R so ... Ea = - (R) (slope)
Used to calculate
-value of activation energy from temperature dependence of the rate constant
-rate constant for a given temp - if the E*
[also known as Ea] and A factor are known.
Points to remember!!
1. Ea is smaller; k is greater; the reaction is faster. 2. Ea is greater; k is smaller; the reaction is slower.
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SAMPLE PROBLEM The colorless gas dinitrogen tetroxide decomposes to
the brown gas NO2 in a first order reaction with a value of:
k = 4.5 x 103/s at 274K. If k is 1.00 x 104/s at 283K, what is the energy of
activation?
SAMPLE PROBLEM Calculate the activation energy for the following set of data:
T (°C) k (l/mol- s) 3 1.4 x 10-3
13 2.9 x 10-3
24 6.2 x 10-3
33 1.2 x 10-2
Exercise 12.8
The gas-phase reaction between methane and diatomic sulfur is given by the equation: CH4 (g) + 2S2 (g) CS2 (g) + 2H2S (g)
At 550° C the rate constant for this reaction is 1.1 M-1 s-1 and at 625° C the rate constant is 6.4 M-1 s-1. Using these values, calculate Ea for this reaction.
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COMPLEX REACTIONS – REVERSIBLE REACTIONS
2H2 (g) + O2 (g) 2H2O (g)
SAMPLE PROBLEM:
For the reaction A + B C + D, various initial rate measurements were run using A and B only and C and D only. From the data below calculate the equilibrium constant for the reaction:
Rate, (M/s) [A] [B] 1.081 x 10-5 0.660 1.23 6.577 x 10-5 4.01 1.23 1.312 x 10-4 4.01 2.25 Rate, (M/s) [C] [D] 7.805 x 10-7 2.88 0.995 1.290 x 10-6 2.88 1.65 4.528 x 10-7 1.01 1.65
THE COLLISION THEORY OF REACTION RATES
• Particles must collide. • Only two particles may collide at one time. • Proper orientation of colliding molecules so that
atoms can come in contact with each other to become products.
The collision must occur with enough energy to overcome the electron/electron repulsion of the valence shell electrons of the reacting species and must have enough energy to transform translational energy into vibrational energy in order to penetrate into each other so that the electrons can rearrange and form new bonds.
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This new collision product is at the peak of the activation energy hump and is called
the activated complex or the transition state. At this
point, the activated complex can still either fall to
reactants or to products.
With all of these criteria met, the reaction may proceed in the forward direction.
