About Kinetics Chemists ask three questions when studying chemical reactions: 1. What happens? 2....
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Transcript of About Kinetics Chemists ask three questions when studying chemical reactions: 1. What happens? 2....
Chemical Kinetics
About Kinetics
Chemists ask three questions when studying chemical reactions:1. What happens?2. How fast?3. To what extent?
The first question is addressed by writing balanced chemical reactions, calculating yields, making qualitative observations, etc.
The second question is addressed in an area of chemistry called KINETICS, which will be covered over the next two lectures
The third question will be answered in chapter 19 (Equilibrium)
Importance of Kinetics
Knowing how fast a reaction occurs, or being able to control the speed of the reaction is very important in the real world.
How long after you take that aspirin does your headache go away? How long will the medicine stay in your system? How many days does it take for your milk to spoil in the refrigerator? How long does it take for cement to harden?
If one can predict a reactionβs rate, chance are one can do something to control that rate. Speeding up a process at a large scale can save a company millions of dollars.
Reactions Rates
Different reactions proceed with different rates
The rate of a reaction is defined as the decrease in the concentration of reactant per unit time (or the increase in the concentration of product)
The rate of a reaction depends on several factors, including:reactant concentration (ex. Ozone depletion)temperature (ex. food spoilage)catalysts (ex. catalytic converters; CO + NO CO2 + Β½ N2)surface area (ex. explosives, pharmaceuticals)
Today, we will focus exclusively on the relationship between reaction rates and reactant concentration
Intro
Lets take the reaction: A ---> B.
This reaction tells us that as A is consumed, B is formed at an equal rate. We can express this mathematically in terms of changing concentrations by:
Imagine we have 10 moles of A in 1 L of solution. If we can freeze time for an instant, such that the reaction has not yet begin (t=0), the concentration of A is 10M.
ββ [π΄ ]βπ‘
=β [π΅]β π‘
= 1 mol of A
[A] = 10 M
t = 0
A B
After 10 seconds, 3 moles of B have formed.
[A] = 7 M[B] = 3 M
t = 10
= 1 mol of B
10 more seconds
[A] = 5 M[B] = 5 M
t = 20
20 more seconds
[A] = 4 M[B] = 6 M
40 more seconds
[A] = 3 M[B] = 7 M
t = 40
t = 80
0 10 20 30 40 50 60 70 800
2
4
6
8
10A ---> B
Time (sec)
Co
ncen
trati
on
(M
ol/
L)
A
B
Plotting the data from previous slide
Reactions Follow a Rate Law
The graph in the previous slide shows that the disappearance of A (formation of B) is not linear.
As the reactant concentration decreases, the reaction slows down.
This dependence of rate on concentration suggests that reaction rates follow a rate law, a mathematical expression that ties concentration and rate together
Instantaneous Rates
0 20 40 60 800
2
4
6
8
10 A ---> B
Time (sec)
Con
cen
trati
on
(M
ol/
L)β’ The rate of the reaction is constantly
changing, but we can determine instantaneous rates (reaction rate at a specific time and concentration)
β’ Instantaneous rate at t=0 is the initial rate
β’ We can determine the instantaneous rate by taking the slope of the tangent at the point of interest
β’ Note: a tangent line is linear and ONLY touches the point in question. It does NOT cross the curve
Instantaneous rate of disappearance of A at t=20 sec
πππ‘π20=β8β0π0β50 π
=β .160ππ
Rates and Stochiometry
In the previous example (A---->B), we had 1:1 stoichiometry. Thus, at any given time, the rate of disappearance of A equals the rate of formation of B. If the stoichiometry is NOT 1:1, we have a much different situation, as shown below:
As you can see, 2 moles of N2O5 are consumed for every 4 mole of NO2 and 1 mole of O2 formed. Thus, the disappearance of N2O5 is half as fast as the appearance of NO2, and twice as fast as the appearance of O2.
2π2π5 (π)β4ππ2 (π)+π2(π)
ββ [π 2π5 ]
ππ‘=12
β [ππ2 ]ππ‘
=2β [π2]ππ‘
Example: N2O5(g) ----> 2NO2(g) + Β½ O2(g)
6.4 π₯10β4ππππ
1 .6π₯ 10β 4ππππ
Looking at average ratesaverage rate of disappearance after 10 minutes
average rate of disappearance after 100 minutes
2 .38 π₯10β 4ππππ
0.59 π₯10β 4ππππ
N2O5(g) ----> 2NO2(g) + Β½ O2(g)
fast
slow
GP 1-3
Rate Laws
We see that reducing reactant concentration lowers the reaction rate, but to what extent? What is the mathematical correlation?
The equation that relates the concentration of the reactants to the rate of reaction is called the rate law of the reaction.
We can derive the rate law of a reaction by seeing HOW THE REACTION RATE CHANGES WITH REACTANT CONCENTRATION.
For any reaction aA + bBβ¦β¦. ----> cC + dDβ¦..
In this expression, k is the rate constant, m and n are reaction orders.
πΉπππ=πΒΏ
Lets go back to the previous reaction: Below is a table of data, showing the initial reaction rate as a function of the
starting concentration of N2O5 (g). We perform multiple experiments to collect enough data to determine our rate law.
Our rate law will be in the form: Rate = k [N2O5]m.
Never include products in a rate law !!!!!
Letβs use the data in the table to obtain the value of m. We must pay attention to how the rate is affected by [N2O5]
Reaction Orders and the Method of Initial Rates
N2O5(g) 2NO2(g) + Β½ O2(g)
Experiment [N2O5]o (M) Rate, M/s
1 0.010 .018
2 0.020 .036
3 0.040 .072
Letβs solve for m. We can choose any two sets of data values to set up a system of equations. Weβll choose data from experiments 1 & 2.
Now, we have two equations and two variables (k and m), but the k values can cancel, allowing us to solve for m as shown below:
Rate Laws
0.36ππ β 1
0.18ππ β 1=πΒΏ ΒΏ
2=[2 ]π
π=1
Reaction Orders (example continued)
This means that the reaction is FIRST ORDER WITH RESPECT TO [N2O5]
When we double [N2O5]o, the rate also doubles. When we quadruple [N2O5]o, the rate quadruples.
Physically speaking, a reaction order of 1 means that one molecule of the reactant is involved in the formation of products. Thus, the rate is directly proportional to [N2O5]o by the rate constant, k.
We can write the rate law as:
The overall reaction order is the sum of the individual reaction orders. (1st order reaction). We can easily solve for k by plugging in any corresponding rate and concentration (k = 1.8 s-1)
πΉπππ=πΒΏ
Rate Laws/Reaction Orders
πΉπππ=πΒΏ
β’ Reaction orders of non-elementary reactions must be determined experimentally. You can not assume based on the stoichiometry.
β’ When you have multiple reactants, you must determine the reaction order of each one. β’ To do this, you must vary the concentration of only one reactant
at a time while holding the others fixed.
β’ Letβs attempt to determine the rate law for the reaction below:2NO(g) + O2(g) ---> 2NO2
Example: 2NO(g) + O2(g) ---> 2NO2
Using the data below, determine the rate law of this reaction in the form:
πΉπππ=πΒΏ
Experiment
[NO]o (M) [O2]o (M) Rate (M/s)
1 .0126 .0125 2.82 x 10-2
2 .0252 .0250 1.13 x 10-1
3 .0252 .0125 5.64 x 10-2
β’ This time, we have two reactants. Lets start by determining the value of βmβ. To do so, we hold [O2]o fixed and vary [NO]o. This will show how the rate depends on [NO]o.
β’ In experiments #1 and #3, [O2]o is fixed, so we will use these experiments to find βmβ.
πΉπππ=πΒΏ
Experiment [NO]o (M) [O2]o (M) Rate (M/s)1 .0126 .0125 2.82 x 10-2
2 .0252 .0250 1.13 x 10-1
3 .0252 .0125 5.64 x 10-2
β’ Remember, rate is proportional to [NO] by the power m. The factor of change in the rate is equal to the factor of change of [NO] to the mth power:
factor of rate change factor of change
in [NO]
order
m = 1
β’ The reaction is 1st order with respect to [NO]
( 5.64 π₯10β 22.82 π₯10β 2 )=( .0252.0126 )π
2=2π
πΉπππ=πΒΏ
Run [NO]o (M) [O2]o (M) Rate (M/s)1 .0126 .0125 2.82 x 10-2
2 .0252 .0250 1.13 x 10-1
3 .0252 .0125 5.64 x 10-2
factor of rate change factor of change in [O2]
order
n = 1πΉπππ=πΒΏThe reaction is 1st order with respect to [NO], 1st order with respect to [O2] and 2nd order overall.
β’ Now we can find βnβ by varying [O2]o and holding [NO]o fixed. We can use experiments #2 and #3 for this. This will show how the rate depends on [O2]o.
( 5.64 π₯10β 21.13 π₯10β 1 )=( .0125.025 )π
( 12 )=(12 )π
k = 179 M-1s-1
Pay Attention to the Units of k, As They Change with Overall Reaction Order
The rate constant, k, is the constant of proportionality between rate and concentration. Higher values of k = faster reactions
It is important to note that the units of k depend on the overall reaction order. Ex:
Rate is always in units of molarity per unit time (sec, hr, etc). Concentrations are always expressed as molarities (M or mol/L). Thus, we have:
Recall for a 1st order reaction:
For zth order (z = any integer)
πΉπππ=πΒΏ
ππ
=π (π ) (π )βπ=1ππ
=πβ1π β1 Units of k for a 2nd order reaction
π=π β1 Units of k for a 1st order reaction
π=πβ(π§ β1 )π β 1
Group Example
Determine the relative (m & n) and overall (m+n) reaction order of the reaction below. Then, derive the rate law and determine the value of k.
Nπ2+πΆπβππ+πΆπ2
πΉπππ=πΒΏ
Experiment [NO2]o (M) [CO]o (M) Rate (M/s)1 .0300 .200 1 x 105
2 .0900 .200 9 x 105
3 .300 .0400 1 x 107
4 .300 .0800 1 x 107
β’ Tripling [NO2] causes the rate to increase nine-fold. This means that the rate is squarely proportional to [NO2], so the reaction is second order with respect to NO2 (m=2). Doubling [CO] does nothing. Thus, the rate does not depend on [CO], and is zero order with respect to CO (m=0). Overall 2nd order.
πΉπππ=πΒΏ k = 1.11 x 108 M-1s-1
The rate of a first order reaction, AB, can be described by the following expression:
First Order Dependence
βπ[ π΄ ]ππ‘
=π[ π΄ ]
π [ π΄ ][π΄ ]
=βπππ‘
β«ΒΏΒΏ
ΒΏ
ln [ π΄]π‘β ln ΒΏ
ππ[π¨ ]π=βππ+π₯π§ ΒΏ
Time-dependent equation for first-order reaction!
Determining the Overall Rate Order of A Reaction Graphically
This equation is in y = mx + b form. Therefore, for any 1st order reaction, the plot of the natural log of [A]t vs time will be linear. The slope of the line will be βk.
π₯π§ ΒΏy axis m
(slope)
bx axis
Plotting 1st Order Reactions
π₯π§ ΒΏb
time values on x-axis
slope = -kunits: s-
1
ln [A]t on y-axis
GP 4
The rate of a second order reaction, AB, can be described by the following expression:
Second Order Dependence
βπ[ π΄ ]ππ‘
=π ΒΏ
π [ π΄ ]ΒΏΒΏ
β«ΒΏΒΏ
ΒΏ
β1ΒΏΒΏ Time-dependent
equation for second order reaction!
1ΒΏΒΏ
Determining the Overall Rate Order of A Reaction Graphically
A second-order reaction depends on the concentration of [A] to the 2nd power. For the reaction: A ----> B
Therefore, for any 2nd order reaction, the plot of the inverse of [A]t vs time will be linear. The slope of the line will be k.
πΉπππ=πΒΏπΒΏΒΏy
mβ’x
b
Plotting a 2nd Order Reaction
πΒΏΒΏ
b
slope = kunits = M-
1 s-1
time values on x-axis
1/[A]t on y-axis
GP 5, 6
Determining Overall Rate Order From Plotting Time-Dependent Data
β’ We can determine if a process is first or second order by plotting the data against both equations. Which ever fitting method yields a linear plot gives the overall order.
π₯π§ ΒΏ πΒΏΒΏ
not linear:NOT 1st order
linear!2nd order
Important!!! Natural logs (ln) are exponential terms! To solve for the value of a
term within a natural log function, use the exponential function on your calculator (ex)
example : ln ( x )=0.3 solve for x
e ln (x)=e0.3
e ln ( x )=x
Ch 18: Kinetics Pt. 2 Temperature Dependence of Rate
Constants
Temperature and Rate
The rates of most chemical reactions increase with temperature.
How is this temperature dependence reflected in the rate expression?
Rates increase with temperature because rate constants increase with temperature. An example is the 1st order reaction:
CH3NC ---> CH3CN
ππππ=π[πͺ π―ππ΅πͺ ]
Variation in k with temperature.
Collision Model
The collision model makes sense of this. This model is based on kinetic theory
Weβve seen that the thermal energy of a molecule is converted to kinetic energy in order to facilitate motion, and weβve seen that the velocity of a molecule increases with T.
The central idea of the collision model is that molecules must collide to react. The more collisions per second, the faster the reaction goes.
This model also rationalizes the concentration dependence. The more molecules present, the more collisions you have.
π π
Activation Energy
Of course, there is more to a chemical reaction than just collisions of molecules.
Molecules must have some minimum energy in order to overcome the energy barrier of reaction
Upon colliding, the kinetic energy of molecules is used to stretch, bend, and break bonds in order to cause a reaction.
But if the molecules donβt have enough kinetic energy, they simply bounce off one another.
This minimum energy that molecules must have is called the activation energy,(Ea)
π π ππ
E too low
Activation Energy Needed to Reach Transition State
β’ Lets go back to the reaction CH3NC ---> CH3CN
β’ The molecule passes through a transition state, a high-energy intermediate in which the CN bond gets rotated 90o.
β’ This is an unstable configuration and requires an activation energy to be reached.
β’ Once this transition state is reached, the process is energetically downhill. The transition state is always the highest energy point in the reaction pathway. As drawn, this reaction is exothermic because the energy of the products is less than the energy of the reactants.
Transition state
nitrogen
carbon
ΞH
En
thal
py
Arrhenius noted that reaction-rate data depended on three aspects(1) the fraction of molecules possessing an energy Ea or greater(2) the number of collisions per second(3) the fraction of molecules oriented in the right way for a reaction to
occur
These factors are incorporated into the Arrhenius Equation
This equation relates k to temperature. A is the frequency factor, which is related to criteria 2) and 3) above, and the fraction of particles possessing an energy of Ea or greater is given by
eβE aRT
π€=ππβπ¬π
πΉπ»
Rearranging the Arrhenius Equation
Taking the natural log of both sides, the Arrhenius equation becomes:
π₯π§π=βπ¬π
πΉπ»+π₯π§ π¨
β’ As you see, we once again have an equation in y=mx+b form.
y=ln kπ (π ππππ )=β
πΈπ
π
π₯=1π
Plotting ln k vs 1/T yields a linear plot with slope βEa/R and a y-intercept of ln A
Showing Changes in k Graphically
From the data below, determine the activation energy of the 1st order reaction
Temperature, oC K (s-1)189.7 2.52 x 10-5
198.9 5.25 x 10-5
230.3 6.30 x 10-4
251.2 3.16 x 10-3
π₯π§π=βπ¬π
πΉπ»+π₯π§ π¨
Plot ln k vs 1/T to get the slope. We know R, so we can calculate Ea. T must be in Kelvin.
1/T (oK-1) ln K
0.002161 -10.5887
0.002119 -9.8547
0.001987 -7.36979
0.001908 -5.75718
0.0018 0.0019 0.002 0.0021 0.0022
-12
-10
-8
-6
-4
-2
0
f(x) = β 19038.464147378 x + 30.516686342063
1/Tln
k
π ππππ=β19038=βπΈπ
π =β
πΈπ
(8.314 π½πππ πΎ
)
π¬π=π .ππ ππππ π±πππ
Calculating Changes in k Mathematically
The Arrhenius equation can be rearranged once again to compare k values at different temperatures
Remember:
lnπ2π1
=πΈπ
π (π 2βπ1
π1π2)
πln (π₯)=π₯
Example
The reaction above has an activation energy of 43.5 kJ/mol. The reaction occurs at 298 K with a rate constant of 110 s-1. What will the rate constant be if we increase the temperature to 308 K?
lnπ2π1
=πΈπ
π (π 2βπ1
π1π2)
lnπ2110
= 43500 π½πππβ 1
8.314 π½ πππβ1πΎβ 1 ( 308β29891784 )ln
π2110
=.57
π( ππ π2
110)=π.57π2110
=1.77
π2=194.7 π β1
Reaction Mechanisms
Consider the following reaction:
In looking at this equation, and applying your understanding of collision theory, you would assume that this reaction proceeds when NO2 and CO collideThe activation energy is exceededThe atoms rearrange to form NO and CO2
β’ Thus, we would expect the rate of the reaction to depend on both [NO2] and [CO]
π΅ πΆπ (π )+πͺπΆ (π )βπ΅πΆ (π )+πͺπΆπ(π)
Reaction Mechanisms
However, according to experimental data, the reaction is 2nd order with respect to NO2 and zero order with respect to CO. The rate law is:
πππ‘π=π ΒΏ
β’ This means that the rate of the reaction does not depend on [CO] at all. How can this be? What does this tell us about the reaction mechanism?
β’ This particular reaction must proceed through multiple steps, and the rate-determining step must only depend on NO2
π΅ πΆπ (π )+πͺπΆ (π )βπ΅πΆ (π )+πͺπΆπ(π)
Reaction Mechanisms
(1 ) ππ2 (π )+ππ2 (π)βππ3 (π )+ππ (π)(π πππ€)β’ For this particular reaction, there are two steps:
(3 )ππ2 (π)+πΆπ (π)βππ (π )+πΆπ2 (π )
β’ This step-by-step description of the molecular pathway is the reaction mechanism. From reaction (1), we can physically understand why the reaction is 2nd order with respect to [NO2].
β’ The 1st step is substantially slower than the 2nd. Therefore, step (1) acts as the βreaction bottleneckβ, and the overall reaction speed can only be as fast as its slowest step.
β’ Thus, step (1) is the rate determining step.
π΅ πΆπ (π )+πͺπΆ (π )βπ΅πΆ (π )+πͺπΆπ(π)
Each step in a reaction mechanism is called an elementary reaction.
An elementary reaction is any reaction that proceeds in a single step. Mechanisms consist of sums of elementary reactions.
Going back to rate laws, each elementary reaction must have a corresponding rate constant
*** For elementary reactions only, you can assume that the rate law depends directly on the number of species present. Thus, the rate law of an elementary reaction follows the stoichiometry.
(1 ) ππ2 (π )+ππ2 (π) ππ3 (π)+ππ (π )(π πππ€)
k1
k2
πππ‘π1=π1 [ππ2 ] [ππ2 ]=π1ΒΏ rate of overall reaction
intermediate (transition step)
Catalysis
The rates of reactions can be increased by using catalysts. Catalysts work by providing a different reaction path (mechanism)
between reactants and products with a lower activation energy The catalyst itself is not consumed in the reaction.
Example: Oxidation of Sulfur Dioxide Pt-Catalyzed
2SO2(g) + O2(g) 2SO3(g)
(1) O2(g) ---> 2O ΞHorxn (1) = 498.4 kJ/mol (slow step)
(2) SO2(s) + O(g) ---> SO3(g) ΞHorxn (2) = -696.2 kJ/mol
Pt
2SO2(g) + O2(g) 2SO3(g)
(1) 2SO2(g) ---> 2S(s) + 2O2(g) ΞHo(1) = 593.6 kJ/mol (slow step)
(2) 2S(s) + 3O2(g) ---> 2SO3(g) ΞHo (2) = -791.4 kJ/mol
uncatalyzed
catalyzed
βπ» ππ₯ππ =β197.8
ππ½πππ
βπ» ππ₯ππ =β197.8
ππ½πππ
Example: Oxidation of Sulfur Dioxide Pt-Catalyzed