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Rates of Reaction
Chapter 13
Materials are referenced to Chemistry
7th ED. by Chang, 2001
Chemical Kinetics
 | The study of rates of chemical reactions and the mechanisms (series of
steps) by which they occur. |
Rate of Reaction
Measured by the decrease in concentration of a reactant or the increase in
concentration of a product during a given uint of time.
Kinetics
"Kinetics is the study of the rate at which a
chemical reaction occurs."
Units for rate
| 1. Nature of reactants |
| 2. concentration of reactants |
| 3. temperature |
| 4. presence of a catalyst |
Let’s Consider an Example:
| Nitrogen Dioxide decomposes into nitric oxide and oxygen. |
Let’s Consider an Example
(this graphic did not print see fig 13.4 and section 13.1
for a similar discussion and graphic.)
| Nitrogen Dioxide decomposes into nitric oxide and oxygen. |
Concentrate on the Rate of NO2 Loss with Time
| Nitrogen Dioxide decomposes into nitric oxide and oxygen. |
Rates of Formation
| Instantaneous rate |
| The rate during a brief portion of a reaction (a tangent to the curve) |
| Average rate |
| The average of the rate over the entire reaction |
The Rate Law Expression
| The rate law expression for a reaction in which A, B,... are reactants
has the general form |
| Rate = k[A]x[B]y |
Initial Rate Method
| The initial concentrations can be used to deduce
the rate law experimentally |
In such an experiment, we often keep some initial concetrations
the same and vary others by simple factors, such as 2 or 3. This makes it easier to assess
the effect of each change on the rate.
Rate Laws
| Consider a reaction:2N2O5 (soln) ---> 4NO2 (soln) + O2 (g) |
| O2 escapes, so the reverse reaction is negligible. |
| Differential rate law: rate of the reaction depends on concentrations |
Rate Laws
| Rate = k [N2O5]n |
| k is the rate constant, n is the the order of the reactant. |
| n does NOT necessarily correspond to a coefficient in the balanced
chemical equation! |
| The values for n and k must be determined experimentally. |
The specific rate constant (k)
| Is determined ONLY EXPERIMENTALLY |
| applies to a specific reaction |
| depends on the overall order of the reaction |
| does not change with concentration changes |
| does not change with time |
| It IS temperature and catalyst dependent |
Determining the Form of the Rate Law
| What is n for: 2N2O5 (soln) ---> 4NO2 (soln) + O2 (g) |
| Let’s consider some data: [N [N2O5] i
Rate (mol/ L s) i |
0.90 M
5.4 x10-4
0.45 M
2.7 x 10-4
| Rate 2 =
2.7 x 10-4
= k[Nk[N2O5]n |
| Rate 1
5.4
x10-4
k[N2O5]n |
Looking at the Ratio of the Rates
| Rate 2 = 2.7 x 10-4 = k[Nk[N2O5]n Rate 1 5.4 x10-4 k[N2O5]n |
| 0.5 = (0.5)n |
| Our n value is 1, first order reaction in N2O5 |
| Rate = k[N2O5] |
| |
| Rate Laws
|
Nitric oxide, NO, reacts with hydrogen to give nitrous oxide, N 2O, and water.
2NO(g) + H 2(g) N2O(g) + H2O(g)
Rate Laws
In a series of experiments, the following initial rates of disappearance
of NO were obtained:
Initial concentrations Initial rate of [NO] [H 2] reaction of NO
Rate Laws
Find the rate law and the value of the rate constant for the reaction of
NO.
Doubling [NO] quadruples the rate, so the reaction is second order in NO.
Doubling [H 2] doubles the rate, so the
reaction is first order in H2.
Rate = k[NO] 2[H2]
Rate Laws
H 2(g) + 2ICl(g) ---> I2(g) + 2HCl(g)
| Rate of reaction = |
| - ( rate of decrease in [H2]
) = |
| -1/2 (rate of decrease in [ICl] = |
| (rate of increase in [I2] ) = |
| 1/2 (rate of increase in [HCl] ) |
| Note: reactant rates have a (-) and rates are based on mole/L per time |
A + 2B ---> AB 2
| Exp. Initial
Initial
Initial Rate of |
| [A]
[B] formation of AB2 |
| 1 1.0 x 10-2 M
1.0 x 10-2 M 1.5 x 10-4 Ms-1 |
| 2 1.0 x 10-2 M
2.0 x 10-2 M 1.5 x 10-4 Ms-1 |
| 3 2.0 x 10-2 M
2.0 x 10-2 M 6.0 x 10-4
Ms-1 |
RATE = k[A]x[B]y
| As [B] changed, the rate did not. Therefore it can be concluded that the
rate is independent of [B]. y = 0 and [B]0 |
| rate ratio = ([A] ratio)x |
| log rate ratio = x log [A] ratio |
rate law continued
| To solve for k, select the data from any one of the trials and replace
the variables in the rate law equation |
| Rate = k[A]2 |
| 1.5 x 10-4 Ms-1 = k[1.0 x 10-2 M]2 |
| k = (1.5 x 10-4 Ms-1)/ [1.0 x 10-2
M]2 |
| k = 1.5 M-1 s-1 |
The complete rate law expression
| RATE = 1.5 M-1
s-1 [A]2 |
2A +B 2 + C ---> A2B +BC
| Exp. Initial
Initial
Initial
Initial Rate of |
|
[A]
[B]
[C]
formation of AB2 |
| 1 0.20 M
0.20 M
0.20M
2.4 x 10-6 Mmin-1 |
| 2 0.40 M
0.30 M
0.20M
9.6 x 10-6 Mmin-1 |
| 3 0.20 M
0.30 M
0.20M
2.4 x 10-6 Mmin-1 |
| 4 0.20 M
0.40 M
0.60M
7.2 x 10-6 Mmin-1 |
RATE = k[A]x[B]y[C]z
| note that as [B] changes the rate does not, therefore the reaction rate
is independent of [B], so y = 0 |
| RATE = k[A]x[C]z |
| for [C], |
| log (7.2x 10-6Mmin-1)/(2.4 x 10-6Mmin-1) = |
| z log (0.60M / 0.20M) |
Rate continued
| log (3) = z log(3) |
| z = log (3) / log (3) |
| z = 1 |
| for [A], |
| log(9.6x 10-6Mmin-1/ 2.4 x 10-6Mmin-1) = |
| x log (0.40M / 0.20M |
rate continued
| x = log(4) / log (2) |
| x = 2 |
| to solve for k, use one of
the trials and put the numbers in the rate law expression |
| 2.4 x 10-6Mmin-1 = k(0.20M)2(0.20M)1 |
| 2.4 x 10-6Mmin-1/ (0.20M)2(0.20M)1 = k |
| k = 3.0 x 10-4 M-2min-1 |
Completed Rate Law
| RATE = 3.0 x 10-4 M-2min-1 [A]2[C] |
3A + 2B ---> 2C + D
| Exp.
Initial
Initial
Initial Rate of |
|
[A]
[B]
formation of D |
| 1
1.0 x 10-2 M 1.0 x 10-2 M
6.0 x 10-3
Ms-1 |
| 2
2.0 x 10-2 M 3.0 x 10-2 M
1.44x 10-1
Ms-1 |
| 3
1.0 x 10-2 M 2.0 x 10-2 M
1.2 x 10-2
Ms-1 |
RATE = k[A]x[B]y
| for y , |
| log (1.2 x 10-2
Ms-1)/(6.0 x 10-3 Ms-1) = |
| y log (2.0 x 10-2 M / 1.0 x 10-2 M ) |
| y = log (2) / log (2) = 1 |
| Since [B] does not stay contant at any time, it must be included when solving for the x
order of [A] |
solving for x
| Rate2 / Rate1 = |
| ([A]2 / [A]1)x ([B]2 /[B]1)y = |
| log (1.44 x 10-1 Ms-1)/(6.0 x 10-3
Ms-1) = |
| log [(2.0 x 10-2 M / 1.0 x 10-2 M)x((3.0 x 10-2 M / 1.0
x 10-2 M )y] |
| log [(2.0 x 10-2 M / 1.0 x 10-2 M)x(3)] |
| bring the two halves together |
rate continued
| log (24) = log(2)x log(3) |
| log(24)/log(3) = log(2)x |
| 8.0 = log(2)x |
| log(8) = x log(2) |
| log(8) / log(2) = x |
| x = 3 |
RATE = k[A]3[B]
| using the first trial data |
| 6.0 x 10-3 Ms-1 = k[1.0 x 10-2 M]3[1.0
x 10-2 M] |
| (6.0 x 10-3 Ms-1) / (1.0 x 10-8) = k |
| k = 6.0 x 10-5Ms-1 |
Integrated Rate Laws
"Integrating the rate law allows us to determine the concentration
at any given time."
First Order
| A ==> product
rate = -D[A]/Dt or
rate = k[A] |
First Order Plots
| ln( [A] / [A]0) = -kt
or ln [A] = -kt + ln [A]0 |
| |
| Second Order
|
| Using the same integrating procedure we can show that:rate = k[A]2 |
| Becomes |
| 1/ [A]t = kt +
1/[A]o |
Half Life
| "The amount of time for half of the reactants to be consumed." |
Half life Second Order
[A]t - [A]0 = [A]0
then t1/2 = 1/k[A]0
A Problem
The recombination of iodine atoms to form molecular iodine in the
gas phase follows second-order kinetics with a rate constant of 7.0
x 109 M-1 s-1 at 23 oC. If the initial concentration of I was 0.086 M, calculate the concentration
after 2.0 minutes.
The Solution
A Question of Half-life
| If a first order reaction proceeds to 75% completion in 32 minutes, what is the
half-life of the reaction? t1/2
= ln2/k = 0.693/k |
| But, what is k?
Use Integrated Rate Law |
| t1/2 resolved |
| Since t1/2 = ln2/k = 0.693/k |
| and k = 7.2 x 10-4 s-1
then |
| t1/2 = 0.693/ 7.2 x 10-4 |
| t1/2 = 9.6 x 102 s
= 16 minutes |
Change of Concentration with Time
Sulfuryl chloride, SO 2Cl2, decomposes when heated.
SO 2Cl2(g) SO2(g) + Cl2(g)
In an experiment, the initial concentration of sulfuryl chloride was 0.0248 mol/L.
If the
rate constant is 2.2 x 10 -5/s,
what is the concentration of SO2Cl2 after 4.5
hrs? The
reaction is first order.
Change of Concentration with Time
Let [SO 2Cl2] = 0.0248 M
and [SO2Cl2]t = the concentration after 4.5 hrs.
Substituting these
and k = 2.2 x 10 -5/s into the
first-order rate equation gives: (This calculation was a graphics and did not print. I
will add at a later date)
Change of Concentration with Time
Taking the antilns of both sides gives:
Collision Theory of Reactions
| If molecules in a solution are going to react, they must collide with one
another. |
| Factors affecting rate: |
| Rate of collision |
| Orientation of collision |
| Energy of collision |
Activation Energy
| The minimum energy of collision required for two molecules to react is called the
activation energy, Ea . |
Activation Barriers
| k = A exp -Ea/RT |
Transition States:
A Temporary Step on the Reaction Path
| The collisional encounter generates new species on the reaction path. |
Activated complex
| An activated complex (transition state) is an unstable grouping of atoms that can break
up to form products. |
Temperature Dependence
| The Arrhenius Equation
k = A exp -Ea/RT |
| Where: |
| k is the rate constant |
| A is the frequency factor |
| Ea is the activation energy |
| R and T are as defined previously |
Reaction Mechanisms
"A sequential series of simple reactions which combine to form a
larger, balanced chemical equation."
Reaction Mechanisms
A singular molecular event, such as a collision of molecules,
resulting in a reaction is called a step or elementary reaction .
| The set of elementary reactions whose overall effect is given by the net
chemical equation is called the reaction mechanism. |
Reaction Mechanisms
A reaction intermediate is a species produced during a reaction
that does not appear in the net equation because it is used up in a subsequent step in the
mechanism.
Example 1: Writing an Overall Equation
Carbon tetrachloride, CCl4, is obtained by chlorinating a methane or an incompletely chlorinated methane
such as chloroform, CHCl3.
The mechanism for the gas-phase chlorination of CHCl3 is
Ex. 1 continued
| Cl2
---> 2Cl (elementary reaction) |
| Cl + CHCl3 ---> HCl + CCl3 ( elem. reac.) |
| Cl + CCl3 ---> CCl4 (elem. reac.) |
| Obtain the net, or overall, chemical equation from this mechanism. |
Ex. 1 cont.
| Cl2 -----> 2Cl |
| Cl + CHCl3 ---> HCl + CCl3 |
| Cl + CCl3 ---> CCl4 |
| Cl2 + CHCl3 ---> HCl + CCl4 |
| (this is the overall equation) |
Molecularity
| The molecularity is the number of molecules on the reactant side of an
elementary reaction. |
Molecularity of Intermediates
| Unimolecular Reaction |
| One molecule reacts alone to form product |
| Bimolecular Reaction |
| Two molecules combine in the transition state |
| Termolecular Reaction |
| Three or more molecules combine in the transition state |
Example 2: Determining the Molecularity
of an Elem. Reac.
| What is the molecularity of each step in the mechanism described in
this example? |
| Cl2 ---> 2Cl |
| Cl + CHCl3 ---> HCl + CCl3 |
| Cl + CCl3 ---> CCl |
Example 2 solution
The molecularity of any elementary reaction
equals the number of reactant molecules. Thus, the forward part of the first step is
unimolecular; the reverse of the first step, the second step and the third step are each
bimolecular.
Rate-Determining Step
| The slowest step in the reaction mechanism is the rate-determining step. |
| Since this step controls the speed of the reaction, it is also used to
derive the Rate Law for that reaction. |
Let’s Consider
| 2NO2 (g) + F2 (g) ----> 2NO2F (g) |
| What is the mechanism? |
| The rate law is: RATE = k[NO2][F2] |
| What is the intermediate? |
| Experimentally, we can see evidence for NO2F being
formed! |
What If...?
| Experimentally, we can see evidence for NO2F being formed. We might speculate two elementary reactions occur. |
What If...?
| Experimentally, we can see evidence for NO2F being formed. We might speculate two elementary reactions occur. |
A Reaction Mechanism MUST!
| Sum to the balanced chemical equation |
| Must agree with the experimentally determined rate law |
What If...?
| Experimentally, we can see evidence for NO2F being formed. We might speculate two elementary reactions occur. |
What If...?
| Experimentally, we can see evidence for NO2F being formed. We might speculate two elementary reactions occur. |
Catalysis
In order to facilitate beneficial reactions that are very slow,
we often use catalysts to decrease the activation barrier for a reaction.
Types of Catalysis
| Homogeneous Catalysis |
| The interaction of reactants and catalysts in the same phase. |
| e.g., CFC’s (gas/gas) |
| Heterogeneous Catalysis |
| The interaction of reactants and catalysts in different phases. |
| e.g., catalytic converters (solid/gas) |
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