15: Packet Principles of Reactivity: Chemical Kinetics
last modified: 06/03/09
CHE 107 Packet 15 - 1
Concept Area I: Terminology
chemical kinetics reaction mechanism rate of reaction instantaneous rate rate equation rate law rate constant order of reaction initial rate half-life, t collision theory activation energy, Ea Arrhenius equation intermediate elementary step molecularity unimolecular bimolecular rate-determining step
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Concept Area II: Collision Theory and Activation Energy
a. You should be able to list the factors that affect the speed of a chemical reaction. b. You should be able to describe the collision theory of reaction rates. c. You should be able to use collision theory to describe the effect of reactant concentration, molecular orientation, temperature, and speed on the reaction rate. d. You should be able to relate activation energy to the rate and thermodynamics of a reaction using reaction coordinate diagrams. e. You should know what a catalyst is and what it does to the rate of a chemical reaction. CHE 107 Packet 15 - 3
Equilibrium tells us how far a reaction goes. Thermodynamics tells us if a reaction will go. Kinetics tells us how fast a reaction will go. Dale Earnhardt, Jr. can race at the Daytona in a car or on a moped.
Both will finish the race (equilibrium & thermodynamics). However, he chooses the car to finish the race in about three hours instead of the moped which might take three days (kinetics).
Remember the diamond?
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Kinetics a measure of the reaction rate using concentration data.
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How can we experimentally measure concentrations?
Top left: by conductivity Top right: by spectrometry Bottom left: by pressure
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What needs to happen for two molecules to react with each other?
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So, how can we increase the number of collisions of reactants?
increase concentration increase temperature increase pressure stir reaction increase surface area decrease volume of container
top image: Kotz & Treichel 5th edition page 623 similar image Kotz, Treichel & Weaver 6th edition page 723 bottom image: Kotz, Treichel & Weaver page 724 CHE 107 Packet 15 - 8
Does the way two molecules hit each other matter for a reaction to occur?
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So, how can we increase the number of collisions of reactants of the correct orientation?
If we increase the total number of collisions, we also increase the number of collisions of the correct orientation!
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Does the speed molecules have when they collide matter for a reaction to occur?
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So, how can we make sure the molecules have enough energy to react when they collide in the proper orientation?
Well, how do we increase the average kinetic energy of something? Another way is to add a catalyst to lower the activation energy of the reaction.
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What is Activation Energy?
Activation energy is the amount of energy required to break the bonds between the reactants. If we don't have enough energy to break the original bonds, we can't form new ones!
If we could lower the activation energy "hill" it would take less energy for our molecules to form product or CHE 107 Packet 15 - 13 "get to the other side".
Activation Energy and Catalysts
Catalysts usually change the amount of activation energy needed for a reaction they change the size of the hill. We normally use catalysts to lower the activation energy of a reaction. Surface catalysts are also sometimes used. These work by increasing the number of collisions with the proper orientation. Let's look at a reaction that occurs in our upper atmosphere: 2 O3 3 O2.
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Ozone
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Ozone with CFC catalyst
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Why do CFC's hurt the ozone layer?
Well, we need ozone to protect us. The high activation energy means that ozone stays around long enough to do its job (protect us from UV light). CFC's come in and react with ozone, destroying it faster than it is supposed to be. Thus, we have a hole in the ozone layer.
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Catalysts
The activation energy hill is usually lowered thus increasing the rate of reaction - by adding a catalyst (although sometimes we add a catalyst to slow-down a reaction). Ozone had a double hump once the CFC catalyst lowered the hill, but this won't always happen. The important part of the ozone graphs is that the activation energy was lowered! Here's a more generic picture:
Note that catalysts can also be used on endothermic reactions!
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Graphically determining H and Ea
80 Ea 80 Ea
50 H 10 reaction H H
50 H 10 reaction
Left reaction is _____________ because _________________. Right reaction is ____________ because _________________.
Units would normally be given for H (either joules or calories). CHE 107 Packet 15 - 19 Here units are omitted for simplicity.
Movie time to summarize!
Rate with 0.3 M HCl Rate with 6.0 M HCl
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Concept Area III: Rates of Reaction and Introduction to Rate Laws
a. You should be able to explain the concept of reaction rate and know how to determine it. b. You should be able to derive instantaneous rates of reaction from experimental information. c. You should be able to interpret a reaction rate in terms of reactant and product concentrations. d. You should know what a rate law equation looks like and be able to define various parts of it. e. You should understand what is meant by the reaction order and be able to give the reaction order for any reactant or for the overall equation.
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Initial Rate
Reaction Rates
Average Rate
Instantaneous Rate
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Concentration of O3 vs. time during its reaction with C2H4
The slope of a is the instantaneous rate at the beginning of the reaction or the initial rate. The slope of b is the average rate for the entire reaction. The slopes of c and e give average rates over a 10 second time interval. The slope of d is the instantaneous rate at time = 35 seconds.
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Expressing the Rate of a Reaction
So, since a reaction rate is how fast a reaction occurs, or the change in concentration of a species with respect to time: We can...
measure as chemicals disappear:
[O 3 ] [C 2 H 4 ] Rate = - =- t t
measure as chemicals appear:
[O 2 ] [C 2 H 4 O] Rate = + =+ t t
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Expressing the Rate of a Reaction mol:mol ratios How about a reaction where the
aren't all 1:1? Consider: H2(g) + I2(g) 2 HI(g) How would you write the rate in terms of these chemicals appearing and disappearing?
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So, let's generalize: aA + bB cC + dD
Expressing the Rate of a Reaction
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Now, let's try a problem!
Because it has a nonpolluting combustion product (water vapor), hydrogen gas is used for fuel aboard the space shuttle and may some day be used by Earth-bound engines: 2 H2(g) + O2(g) 2 H2O(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/Ls, at what rate is [H2O] increasing?
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2 H2(g) + O2(g) 2 H2O(g)
a. Rate expression a. [O2] is decreasing at 0.23 mol/Ls, [H2O] = ?
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Rate Law Expression
We can also write a rate law or rate equation for a chemical reaction. (Note, only reactants in rate law.) So, let's generalize: aA + bB cC + dD Rate = k[A]m[B]n Can you look at a balanced chemical equation to get the rate law?
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Order of Reactions
aA + bB cC + dD Rate = k[A]m[B]n
When indicating the order of a reaction, we can do two ways: The order can be given with respect to a reactant
The reaction is mth order in A and nth order in B.
The order can be given with respect to the entire reaction
The overall reaction is (m+n)th order.
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Order of Reactions
k k[A]1 zero order independent of concentration 1st order reaction unimolecular depends only on the concentration of A
k[A]2 or 2nd order reaction bimolecular rate is k[A]1[B]1 limited by two molecules colliding k[A]3 etc. 3rd order reaction rate is limited by three molecules colliding
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Is there a limit to the reaction order?
k[A]1 k[A]2 or k[A]1[B]1 k[A]3 etc.
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Your Turn!
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. b. c. d. e. 2 NO(g) + O2(g) 2 NO2(g) rate = k[NO]2[O2] CH3CHO(g) CH4(g) + CO(g) rate = k[CH3CHO]3/2 H2O2(aq) + 3 I(aq) + 2 H+(aq) I3(aq) + 2 H2O(l ) rate = k[H2O2][I] 2 NO(g) + 2 H2(g) 2 N2(g) + 2 H2O(g) rate = k[NO]2[H2] NO(g) + O3(g) 2 NO2(g) + O2(g) rate = k[NO][O3] CHE 107 Packet 15 - 33
Concept Area IV: Deriving Information from Experimental Data
a. You should be able to derive a rate equation and the value of the rate constant from experimental information. b. You should be able to tell the order of a reaction by the units on k.
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Rate Laws are Determined via Experiment
For A B
initial [A] 1.0 2.0 3.0 4.0 Initial Rate (M/s) 0.05 0.10 0.15 0.20
Here when concentration is doubled, tripled and so on, the reaction is double tripled and so on.
Now, we can determine the rate law and the value of k: So, it is both first order with respect to [A] and overall.
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Rate Laws are Determined via Experiment
For CO(g) + NO2(g) CO2(g) + NO(g), use the data below to determine the rate law, the value of the rate constant, and indicate the order of the reaction with respect to each reactant and overall.
Exp. 1 2 3 4 5 [CO] 5.10104 5.10104 5.10104 1.02103 1.53103 [NO2] 3.50105 7.00105 1.75105 3.50105 3.50105 Rate (M/h) 3.4108 6.8108 1.7108 6.8108 10.2108
Let's continue on the next slides to solve this problem:
Your book solves this problem (example 15.3) on pages 710 & 711
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CO(g) + NO2(g) CO2(g) + NO(g)
Exp. 1 2 3 4 5
[CO] 5.10104 5.10104 5.10104 1.02103 1.53103
[NO2] 3.50105 7.00105 1.75105 3.50105 3.50105
Rate (M/h) 3.4108 6.8108 1.7108 6.8108 10.2108
Solving...
Let's start with [CO]. Find two experiments where [NO2] doesn't change.
Now, let's use the formula xy=z where x is the ratio the concentration changed, y is the order with respect to that species and z is the ratio the rate changed. So, for [CO] we have: So, what order do we have with respect to CO?
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CO(g) + NO2(g) CO2(g) + NO(g)
Exp. 1 2 3 4 5
[CO] 5.10104 5.10104 5.10104 1.02103 1.53103
[NO2] 3.50105 7.00105 1.75105 3.50105 3.50105
Rate (M/h) 3.4108 6.8108 1.7108 6.8108 10.2108
Solving...
Now, let's do the same for [NO2]. Find two experiments where [CO] doesn't change.
So, for [NO2] we have: So, what order do we have with respect to NO2?
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CO(g) + NO2(g) CO2(g) + NO(g)
Exp. 1 2 3 4 5
[CO] 5.10104 5.10104 5.10104 1.02103 1.53103
[NO2] 3.50105 7.00105 1.75105 3.50105 3.50105
Rate (M/h) 3.4108 6.8108 1.7108 6.8108 10.2108
Solving...
So, we can now write our rate law since we know first order with respect to both CO and NO2: Now, let's figure out the value of the rate constant by plugging in the values from any trial:
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Units on k
By now you should have noticed that the units on k depend on the order of the reaction. So, the units on k change! Therefore, if you are given k you can tell the order of the reaction by looking at the units!
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Units on k
0 1 2 rate = k so units on k are M/s rate = k[A]1 since [A]'s units are M we'll find that the units on k are 1/s or s1 rate = k[A]2 here the units on k are 1/Ms or or rate = M1s1 k[A]1[B]1
So, we can generalize k's units to: M(1 order)s1.
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While we're talking about k
k is constant for a specific temperature. A good rule of thumb is that the rate doubles for every 10 C increase, so k also approximately doubles for every 10 C increase.
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Concept Area V: Integrated Rate Laws
a. You should be able to describe and use the relationships between concentration reactant and time for zero, first and second-order reactions. b. You should be able to apply graphical methods for determining reaction order and the rate constant from experimental data. c. Use the concept of half-life, especially for first-order reactions.
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Integrated Rate Laws 1 order
st
So far we've seen that and
[A] Rate = - t Rate = k[A]
Let's set these two equal to each other: Ooo... we could use calculus to integrate over time to get: ln [A]t = -kt
[A]0
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[A] - = k[A] t
Integrated Rate Laws 1 order
st
Why is this useful? Well, let's rearrange our little equation:
[A ]t ln = -kt [A ]0 ln[A ]t - ln[A]0 = -kt
Remember the equation of a line? Look similar?
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ln[A ]t = -kt + ln[A]0
Kotz, Treichel & Weaver page 718
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Integrated Rate Laws 2 order
nd
So far we've seen that and
[A] Rate = - t
Rate = k[A]2
Let's set these two equal to each other: Wow! We could use calculus to integrate over time to get: 1 - 1 = kt
[A]t [A]0
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[A] - = k[A]2 t
Integrated Rate Laws 2 order
nd
Why is this useful? Well, let's rearrange our little equation:
1 1 - = kt [A]t [A]0 1 1 = kt + [A]t [A]0
Remember the equation of a line? y = mx + b Look similar?
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Kotz, Treichel & Weaver page 718
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Integrated Rate Laws 0 order
th
So far we've seen that and
[A] Rate = - t Rate = k
Let's set these two equal to each other: OK, we could use calculus to integrate over [A]0 - [A]t = kt time to get:
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[A] - =k t
Integrated Rate Laws 0 order
th
Why is this useful? Well, let's rearrange our little equation:
[A]0 - [A]t = kt [A]t = -kt + [A]0
Remember the equation of a line? y = mx + b Look similar?
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Kotz, Treichel & Weaver page 717
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Integrated rate laws graphical summary:
1/[A]t = kt + 1/[A]0 ln[A]t = -kt + ln[A]0 [A]t = -kt + [A]0
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Given the following data, determine the order of the reaction.
Time (min) 0 10 20 30 40 50 60 [N2O5] 0.0165 0.0124 0.0093 0.0071 0.0053 0.0039 0.0029 ln [N2O5] 4.104 4.390 4.68 4.95 5.24 5.55 5.84 1/ [N2O5] 60.6 80.6 110 140 190 260 340
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So how can this be used?
A is a graph using ___ order. B is a graph using ___ order. C is a graph using ___ order.
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Half-Lives
A half-life is the time it takes for the sample to disappear. At t, [A]t = 1 or [A]t = 1 [A]0
[A]0 2 2
For first order reactions:
[A]t 1 ln = -kt ; ln = -kt [A]0 2 ln 0.5 0.693 t = - ; t = + k k
So, if you know t, you know k and vice versa.
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Notice that the amount decayed is less and less, but the time is the same. According to Dr. D.W. Margerum of Purdue University, after 7 halflives, most of the data is gone. After 10, it is useless.
Kotz, Treichel & Weaver page 720
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Half-Life Problem
Cyclopropane is the smallest cyclic hydrocarbon. Because its 60 bond angles reduce orbital overlap, its bonds are weak. As a result, it is thermally unstable and rearranges to propene at 1000C via the first-order reaction shown below. The rate constant is 9.2 s1.
CH2 H2C CH2 (g)
H3C
CH
CH2 (g)
a. What is the half-life of the reaction? b. How long does it take for the concentration of cylcopropane to reach one-quarter of the initial value?
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a.
What is the half-life of the reaction?
"Easy" Half-Life Problem
a.
How long does it take for the concentration of cylcopropane to reach one-quarter of the initial value?
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Another Half Life Problem
The following data are for the decomposition of hydrogen peroxide in dilute sodium hydroxide at 20 C.
[H2O2] Minutes 8.98102 4.49102 2.25102 1.13102 0 488 976 1464
What is the observed half life? What is the order of the reaction? What is the value of the rate constant?
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Last Comment on Half Lives
If time that it takes for each concentration to drop to one half of its value is not constant, it is not first order. What order is it?
It is zero order if, the rate of disappearance of the species is essentially constant with the same average value. Of course, that value is actually the value of the rate constant as well. It is second order if, (1/[X])/t is essentially constant with the same average value. Of course, that is actually the value of the rate constant as well.
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Summary
Order Half-life
0 1 2 Kotz, Treichel & Weaver page 719 [R]0/2k (ln 2)/k 1/(k[R]0)
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Concept Area VI: The Arrhenius Equation
a. You should be able to describe the effect of temperature on the reaction rate using the Arrhenius equation. b. You should be able to use the Arrhenius equation to calculate the activation energy from experimental data.
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Collision Theory Revisited
So, we know that lots of factors are involved for molecules to react together (speed, orientation, concentration, etc.). How can we describe all of this mathematically?
Where A is the frequency and orientation of collisions, R is the gas constant (8.314103kJ/molK), T is the temperature, and Ea is the activation energy (in kJ). The whole term eEa/RT is the fraction of molecules with the minimum amount of energy to react.
The Arrhenius equation comes from the work done by the Swedish chemist Svante Arrhenius in 1889.
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Collision Theory Revisited
Now, this equation may not look very useful. Let's try to rearrange it a bit! - E a / RT k = Ae Why did we do this?
k = e - Ea / RT A k ln = ln e - Ea / RT A Ea ln k - ln A = RT Ea 1 ln k = + ln A R T
Well remember the equation for a line? y = mx + b Does our formula look familiar now?
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Collision Theory Revisited
So, we've got an equation for a line. Now what?
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Concept Area VII: Reaction Mechanisms and Rate Laws
a. You should understand the concept of a reaction mechanism. b. You should be able to describe the elementary steps of a mechanism and give their molecularity. c. You should be able to recognize the ratedetermining step in a mechanism and identify any reaction intermediates. d. You should understand how a catalyst can change the mechanism of a reaction. e. You should now fully understand and be able to label a reaction coordinate diagram.
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Reaction Mechanisms Steps in the Overall Reaction
Can you tell how a car works just by looking at it? No! You must look under the hood at the engine. Reactions are similar. We can't tell how they work just by looking at a balanced chemical reaction. We have to look at the various steps it took to get there!
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Reaction Mechanisms Steps in the Overall Reaction
Consider the following reaction: 2A +B E + F The details "under the hood" might be: 1. A + B C 2. C + A D 3. D E + F When these three reactions are added together, we get our overall reaction. Notice, we've produced C and D which are not written in the overall reaction because they are only intermediates in the reaction!
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Mechanism Terminology
reaction intermediate a substance that is formed a...