The Reaction Rate decreases as the reaction proceeds.
Reason: the concentration of reactants decreases
Any substance in the reaction can be used to express the Rxn. Rate
Rate of disappearance of
Rate of disappearance of
Rate of formation of
Reaction Rate is always positive
(a) Rate of formation of a product
(b) Rate of decomposition of a reactant
Since: [decrease in concentration of reactant] has a negative value the Reaction Rate will be positive
Reaction Rate in terms of stoichiometry
OR
Reason: 2 moles of HI are formed from reaction of 1 mole of
REACTION RATES
Examples:
For the hypothetical reaction shown below, write a rate expression based on each reactant and product:
Rate based on A =
Rate based on C =
Rate based on
Rate based on D =
Rate =
2. Consider the following reaction:
In the first 10.0 seconds of the reaction, the concentration of dropped from 1.000 M to 0.868 M .
a) Calculate the average rate of this reaction in this time interval.
b) Predict the rate of change in concentration of during this time interval.
Examples (cont'd):
The following data was collected for the reaction shown below:
a) What is the average rate of reaction between 0 and 10 seconds?
Time
0
1.000
10
0.913
20
0.835
30
0.763
40
0.697
50
0.637
b) What is the rate of formation of between 20 and 30 seconds?
4. For the reaction , the initial rate is , at a given condition. What is under the same conditions?
a)
b)
c)
RATE DETERMINATION
Rate determination is done by monitoring the change in concentration of reactant or product over time.
Method:
Withdraw samples from the reaction vessel at various times and analyze some of their physical or chemical properties that give information on the concentrations present.
Properties appropriate and easy to monitor
Color (if product or reactant absorbs light); a spectrophotometer (shown on bottom of page) can be used to quantify the difference in color difference of reactants and products.
Volume, at constant P and T ; useful if gaseous substances are present, and # of mol of gaseous reactants of mol of gaseous products
Pressure at constant V and T ; useful if gaseous substances are present, and # of mol of gaseous reactants # of mol of gaseous products
Density; useful is density of reactants density of products
Properties not appropriate to monitor
Flammability (difficult to measure quantitatively)
The Rate of Reaction depends on the concentration of one or more of the reactants and, for the simple reaction ( products) it can be written as:
The relationship above is called the Rate Law, where is a constant of proportionality called the rate constant and is the reaction order. This relationship describes how the rate of a reaction changes with respect to the concentration of a reactant.
For a more general reaction with general equation of the form:
the Rate Law has the form: Rate
where,
, and
are reaction orders with respect to each reactant or the catalyst,
are are determined experimentally but not always integers
and
is the Rate Constant
with its units depending upon the form of Rate Law
and its value depending on temperature
REACTION ORDER
Reaction Order is the exponent ( n ) of the concentration of that species in the Rate Law, as determined experimentally.
When , the reaction is zero order, when , the reaction is first order and when , the reaction is second order.
Although other orders are possible, including fractional or negative orders, they are not common.
Zero-Order Reaction:
In a zero-order reaction, the rate of the reaction is independent of the reactant concentration.
Zero-order reactions occur under conditions where the actual amount of reactant available for a reaction does not change as the overall amount of reactant changes.
For example, sublimation is usually zero order, since the amount of substance on the surface available to sublime does not change as the substance sublimes.
First-Order Reaction:
In a first-order reaction, the rate of a reaction is directly proportional to the concentration of the reactant.
Therefore, the rate of a first-order reaction slows down as the reaction proceeds because the concentration of the reactant decreases.
Furthermore, the rate of the reaction increases with the same change factor in concentration of the reactant. For example, doubling the concentration of a reactant in a first-order reaction, doubles the rate of reaction.
Second-Order Reaction:
In a second-order reaction, the rate of a reaction is proportional to square of the concentration of the reactant.
Therefore, the rate of a second-order reaction is more sensitive to the concentration of the reactant. Furthermore, since the rate of the reactant increases with the square of the change factor in concentration of the reactant, doubling the concentration of a reactant in a second-order reaction, quadruples the rate of reaction.
REACTION ORDER
Effect of concentration change on reaction rates and orders can be summarized as:
Order
Effect of Doubling Concentration
Rate
Zero
No Effect
First
x 2
Second
x 4
Negative
Half
Diagram A below shows how the concentration of reactant changes with time for the three common reaction orders, with identical rate constant (k) and initial concentrations.
Diagram B shows how the rate of reaction changes as a function of reactant concentration.
In determining the rate law for a reaction, first the order of reaction with respect to each reactant and catalyst must be found.
Commonly a series of experiments are done in which the initial concentrations of reactants are varied (usually doubled).
The results of these experiments will yield the respective orders of reaction.
Example:
Initial Conc.
Initial Rate of Disappearance
of
Experiment 1
Experiment 2
Rate m = ? (must be determined from experimental data)
Experimental: - Doubling [ ] doubled the rate of reaction
It follows:
The Reaction is: - First order in
First order overall
Rate
Rate
Examples:
Use the rate data below to
A) determine the reaction orders with respect to [ ], [ ] and [ ] and
B) find the rate constant k for the reaction shown below:
Initial Concentrations (M)
Initial Rate (M/s)
[ ]
[ ]
[ ]
Experiment 1
0.010
0.010
0.00050
Experiment 2
0.020
0.010
0.00050
Experiment 3
0.010
0.020
0.00050
Experiment 4
0.010
0.010
0.00100
Solution
A) Compare two rate experiments in which all concentrations of reactants but one are held constant:
[ ] : Compare Experiment 1 and Experiment 2
When is doubled, the Rate is doubled
It follows: 1st order in [ ]
[I-]: Compare Experiment 1 and Experiment 3
When [ ] is doubled, the Rate is doubled
It follows: 1st order in [ ]
[ ]: Compare Experiment 1 and Experiment 4
When [ ] is doubled, the Rate does not change
It follows: order in
Summary: 1st order in
1st order in
0 order in
commonly written :
NOTE: Reaction Orders are not related to the coefficients of the overall equation
Examples (cont'd):
B) Find the Rate Constant, k
From Experiment 1: (any other experiment could be used)
Use the rate data below to
A) determine the rate law for the reaction , and
B) calculate the rate constant k , and
C) calculate the rate of reaction for experiment 4
Experiment
Number
Initial Rate
1
0.100
0.100
2
0.100
0.200
3
0.200
0.100
4
0.050
0.100
Examples (cont'd):
The reaction below was experimentally determined to be order with respect to and order with respect to NO. Which mixture in the diagram below has the fastest initial rate?
2. The graph below shows a plot of the rate of reaction versus the concentration of the reactant A for the reaction products.
a. What is the order of the reaction with respect to A?
b. Write the rate law for the reaction and determine the value of k .
INTEGRATED RATE LAW (1 ORDER REACTION)
The rate laws discussed earlier examine the relationship between concentration of a reactant and the rate of a reaction. But often, the relationship between the concentration of a reactant and time is of interest.
The integrated rate law for a reaction is a relationship between the concentrations of the reactants and time. For example, consider a simple first order decomposition reaction shown below:
Since Rate , the following can be written:
Differential Rate Law
Calculus can be used to integrate the differential rate law and obtain the order integrated rate law:
Integrated Rate Law
The integrated rate law can be used to calculate the concentration of reactant at any time during the reaction.
Examples:
Sulfuryl chloride, , decomposes when heated:
In an experiment, the initial concentration of was 0.0248 M . If the reaction is first order with the Rate Constant of , what is the concentration of after 2.5 hours?
Taking antilog of both sides,
Examples (cont'd):
Cyclobutane, decomposes, when heated to give ethylene:
The reaction is first order. In an experiment, the initial . After heating at for 455 seconds, . Based on this information, determine the rate constant for this reaction at this temperature.
Cyclopropane rearranges to form propene in the gas phase:
The reaction is first order and has a rate constant of at 720 K . If the initial concentration of cyclopropane is 0.0445 M , what will the cyclopropane concentration be after 235.0 min ?
HALF-LIFE ( ) OF A FIRST ORDER REACTION
Half-life is the time it takes for the reactant concentration to decrease to one-half of its initial value.
NOTES:
Half-Life does not depend on the initial concentration,
Half-Life for a First-Order Reaction
Example:
Dinitrogen pentoxide, decomposes when heated in carbon tetrachloride solvent:
The rate law is first order in with a rate constant of . What is the half-life for this reaction?
A first order reaction has a half-life of 26.4 seconds. How long does it take for the concentration of the reactant to fall to of its initial value?
The radioactive decay of U-238 is first order with a half-life of years. How long would it take for of the U-238 atoms in a sample to U-238 to decay?
GRAPHING FIRST ORDER REACTIONS
The order of a reaction can be determined by graphing the experimental data. For a first order reaction, the integrated rate law can be written as:
First order Rate Law:
This equation may be rewritten:
(This is the equation of a straight line, )
A plot of versus time should give a straight line for a first order reaction with a slope of and a y-intercept of .
For example, for the decomposition of (shown below), the concentration vs. time data can be graphed to show that the reaction is first order.
Use the graph provided on the previous page to determine the concentration of at 1900 s . (Round to 3 significant figures)
Shown below is a graph of concentration vs. time for decomposition of . Determine the order of the reaction and the value of rate constant at this temperature.
INTEGRATED RATE LAW ( ORDER REACTION)
Products
Differential Rate Law
By using calculus:
Integrated Rate Law
Example:
For the reaction
At 330 K :
HALF-LIFE ( ) OF A ORDER REACTION
Half-life is the time it takes for the reactant concentration to decrease to one-half of its initial value.
NOTES:
Half-Life depends on the initial concentration,
Half-Life increases as reaction progresses
Example:
For a particular order reaction .
a) How long does it take to the concentration to decrease from 0.0030 M to 0.0015 ?
For concentration to drop to half of its original value, half-life
b) How long does it take for the concentration to be halved again?
Twice as long
GRAPHING ORDER REACTIONS
The second order integrated rate law is in the form of a an equation for a straight line.
A plot of versus time (s) should give a straight line for a second order reaction with a slope of and y-intercept of .
Summary:
To determine the order of a reaction,
Collect time vs. concentration data for the reaction.
Assuming order, plot vs. time.
If a linear graph is obtained, then it is order.
If no linear graph, then plot vs. time.
If a linear graph is obtained, then it is order.
Example:
Given the experimental data listed on the right, determine if the reaction shown below is or order, and determine the rate constant ( k ) for the reaction shown below:
Strategy:
We must first see if the data fits the Rate Law for Order Reactions. If it does, the reaction is of the order
If it does not, we must see if the data fits the Rate Law for Order If it does, the reaction is of the order
First Order Rate Law:
Plot as a function of t
Second order Rate Law
Time (s)
[ ] (M)
0
0.01000
50
0.00887
100
0.00797
150
0.00723
200
0.00662
250
0.00611
300
0.00567
350
0.00528
400
0.00495
450
0.00466
500
0.00440
550
0.00416
600
0.00395
650
0.00376
700
0.00359
750
0.00343
800
0.00329
850
0.00316
900
0.00303
950
0.00292
1000
0.00282
SUMMARY OF KINETICS EQUATIONS
Rate
Units
of
Integrated
Rate Law
Straight-Line
Plot
Half-Life
Expression
0
Rate
Time
Examples:
The data shown below was collected for the reaction:
Initial Rate (M/s)
0.050
0.050
0.014
0.100
0.050
0.029
0.100
0.100
0.041
0.200
0.200
0.115
Write the rate law and calculate the rate constant for this reaction.
Examples (cont'd):
The reaction shown below is found to be first order with respect to A, second order with respect to B and zero order with respect to C . Answer the following questions based on this information.
a) Write a rate law for this reaction.
b) By what factor does the reaction rate change if the concentrations of all the three reactants double?
c) By what factor does the reaction rate change if the concentration of A is doubled and concentration of B is halved?
3. In a kinetic study of the reaction shown below, the following data were obtained for the initial rates of disappearance of A and B
Exp.
Initial Rate
1
0.010
0.010
2
0.030
0.010
3
0.020
0.020
Write the rate law and calculate the rate constant for this reaction.
In the presence of excess thiocyanate ion (SCN-) the following reaction is 1st order with respect to with a rate constant of .
a) What is the Half-Life in seconds?
b) How many seconds would be required for the initial concentration of to decrease to each of the following values? ( and )
c) How many seconds are required for of reaction to be completed?
5. The graph below was prepared for the reaction:
a) Determine the order of the reaction and the value of the rate constant.
b) Predict the concentration of AB at 25 seconds.
c) Determine the rate of the reaction when the concentration of AB is 0.10 M ?
TEMPERATURE AND RATE
The rate of reaction depends on temperature, and the rate law for a reaction can be expressed as:
The temperature dependence of the rate is contained in the rate constant ( k ), which is a constant only when the temperature remains constant.
An increase in temperature generally results in an increase in k , which results in a faster rate. This relationship is expressed in the Arrhenius equation, which relates the rate constant (k) to the temperature in Kelvin.
In this equation,
gas constant
Activation energy (energy barrier)
frequency factor (the number of times reactants approach the activation barrier per unit time)
To better understand these factors, we examine the reaction in which (methyl isonitrile) rearranges to form (acetonitrile):
TEMPERATURE AND RATE
Activation Energy:
Activation energy ( ) is defined as the energy barrier that must be overcome for reactants to transform into products.
To get from reactants to products, the molecule must go through a high energy intermediate state called the activated complex or transition state.
For transformation of to , even though the overall reaction is energetically downhill (exothermic), it must first go uphill (endothermic) to reach the activated complex, because energy is required to initially weaken the bond to allow rotation.
Frequency Factor:
For this reaction, the frequency factor represents each time the NC part of the molecule begins to rotate (wags) and approaches the activation barrier.
Note that approaching the activation barrier is not equivalent to overcoming the barrier. Most of the approaches do not have enough energy to overcome the barrier.
TEMPERATURE AND RATE
Exponential Factor:
The exponential factor is a number between 0 and 1 that represents the fraction of molecules that possess enough energy to overcome the activation barrier. The exponential factor is the fraction of approaches that are actually successful and result in the product.
The exponential factor depends on both the temperature ( T ) and the activation energy ( ) of the reaction:
A low activation energy and a high temperature make the negative exponent small, so the exponential factor approaches 1 . A high activation energy and a low temperature make the exponent a very large negative number, so the exponential factor becomes small.
The effect of increase in temperature on the energy of the particles is shown in the diagram below. As the temperature increases, the fraction of molecules with enough energy to surmount the activation energy increases.
Summarizing Temperature and Reaction Rates:
The frequency factor is the number of times the reactants approach the activation barrier per unit time.
The exponential factor is the fraction of approaches that are successful in surmounting the activation barrier and forming products.
The exponential factor increases with increasing temperature but decreases with increasing activation energy.
ARRHENIUS PLOTS: DETERMINATION OF A &
The frequency factor (A) and the activation energy (Ea) are important quantities in understanding the kinetics of any reaction. These factors can be measured in the laboratory by manipulating and rearranging the Arrhenius equation as shown below:
Taking natural of both sides of the equation above,
Rearranging the last equation above we can obtain:
The equation above is in the form of a straight line. A plot of vs (in Kelvin) results a straight line with slope equal to . Such plot is called the Arrhenius plot and is commonly used in the analysis of kinetic data.
Examples:
The kinetic data, shown below, was collected for the decomposition of ozone. Determine the frequency factor and the activation energy (in ) for this reaction.
Activation energy can be calculated from the slope:
The frequency factor can be determined from the y-intercept:
Use the results obtained in problem above to determine the rate constant at 298 K for the decomposition of ozone.
Examples (cont'd):
Reaction A and reaction B have identical frequency factors. However, reaction B has a higher activation energy than reaction A . Which reaction has a greater rate constant at room temperature?
Use the Arrhenius plot shown below to determine the frequency factor (A) and activation energy ( , in ) for the first order decomposition of :
COLLISION THEORY
Earlier we discussed the frequency factor as representing the number of approaches to the activation barrier per unit time. Let's now apply this to a reaction involving two gas-phase reactants:
Based on collision theory, a chemical reaction occurs after a sufficiently energetic collision between two reactant molecules.
Furthermore, collision theory separates the frequency factor ( A ) into two distinct parts, as shown in the equations below:
Orientation factor Collision frequency
The collision frequency (z) is the number of collisions that occur per unit time, which can be calculated from the pressure of the gases and the reaction temperature. The orientation factor (p) represents the fraction of the collisions with orientation that allows the reaction to occur.
The orientation factor can be better understood by considering the reaction shown below:
In order for the above reaction to occur, two molecules of sufficient energy and proper orientation must collide. Therefore, as shown below, not all collisions of sufficient energy will lead to products, since they do not possess proper orientation.
COLLISION THEORY
Some reactions have orientation factors that are much smaller than one. This indicates that the orientation requirements for these reactions are very stringent, and molecules must be aligned in a very specific way for the reaction to occur. One such example is shown below:
Reactions between atoms usually have orientation factors of approximately 1 because atoms are spherical and thus any orientation will lead to the formation of product.
Examples:
Which reaction below would you expect to have the smallest orientation factor? Explain.
a)
b)
c)
POTENTIAL - ENERGY DIAGRAMS FOR REACTIONS
A potential energy diagram shows how the energy of the reactants, activated complex and products are related for an endothermic and for an exothermic reaction.
Endothermic Reactions
(activated complex)
Exothermic Reactions
(activated complex)
Examples:
The diagram below shows the energy of a reaction as the reaction progresses. Label each blank box in the diagram.
Sketch a potential energy diagram for the decomposition of nitrous oxide:
The activation energy for the forward reaction is 251 kJ ; the is +167 kJ . What is the activation energy for the reverse reaction? Label the diagram appropriately.
REACTION MECHANISMS
Reaction Mechanisms show the steps involved in the change from Reactants to Products
Mechanisms consist of a set of "Elementary Reactions" whose overall effect is the Net Chemical Equation.
Elementary Reactions:
Elementary reactions are single molecular events that result in a reaction and are caused by molecular collisions
Below 500 K , this reaction takes place in 2 steps:
Elementary Reaction 1:
Elementary Reaction 2:
Adding the steps (Elementary Reactions) yields the Overall Equation:
Examples:
The decomposition of Ozone ( ) is believed to occur in 2 steps:
Elementary Reaction 1:
Elementary Reaction 2:
Identify any Reaction Intermediate:
What is the Overall equation?
Examples (cont'd):
Sodium hydrogen carbonate ( ), also called sodium bicarbonate can been synthesized through a sequence of 3 elementary steps:
Identify the Reaction Intermediates:
Write the Overall Equation:
MOLECULARITY
Molecularity is the number of molecules on the reactant side of an Elementary Reaction.
Unimolecular Reactions:
Unimolecular reactions are Elementary Reactions that involve one reactant molecule.
These are commonly the decomposition reactions of unstable species.
Bimolecular Reactions:
Bimolecular reactions are Elementary Reactions that involve two reactant molecules.
These are very common reactions.
Termolecular Reactions:
Termolecular reactions are Elementary Reactions that involve three reactant molecules.
These are less common because the chance of three molecules coming together with the right orientation is unlikely.
Examples:
What is the molecularity of each of the following elementary reactions?
Rate Equation for an Elementary Reaction
I. For an Overall Reaction, the Rate Law cannot be predicted by the Overall Equation
Reasons:
The majority of reactions consist of several elementary steps.
The Rate Law is the combined result of the elementary steps
The rate of all the elementary reactions must be known in order to predict the rate law for the overall equation.
II. For an Elementary Reaction, the Rate Law can be written directly from the Elementary Equations
The rate of an elementary reaction is proportional to the product of the concentrations of each reactant molecule.
Unimolecular Elementary Reactions
Bimolecular Elementary Reactions
Reason:
The frequency of collisions is proportional to the number of A molecules and the number of B molecules ( )
Termolecular Elementary Reactions
Examples:
Write Rate Equations for each of the elementary reactions shown below:
Rate =
2)
Rate =
3)
Rate =
NOTE:
For elementary reactions , the coefficients of the balanced chemical elementary reaction are the exponents to which the concentrations of the reactants are raised.
THE RATE LAW AND THE MECHANISM
The Suggested Mechanism for a particular reaction:
cannot be observed directly,
is a rationalized explanation based on experimental data,
is accepted provisionally, and may be replaced by another suggested mechanism based on further experimentation.
A Suggested Mechanism
Is considered correct if it agrees with the experimentally determined Rate Law
Is considered incorrect if it does not agree with the experimentally determined Rate Law
Example 1:
Overall Equation:
Experimental Rate Law:
Rate
NOTE: order with respect to order with respect to order Overall
Predicted Possible Mechanisms:
One Single Elementary Reaction
(Termolecular)
Predicted Rate Law: Rate:
NOTE: order with respect to order with respect to order overall
The predicted Rate Law does not agree with the experimental Rate Law
This mechanism must be incorrect.
Two Single Elementary Reactions
NOTE:
The Rate Law is determined by the slow step
THE SLOWEST STEP ⟶ THE RATE DETERMINING STEP
Predicted Rate Law RATE (rate-determining-step)
NOTE: order with respect to order with respect to order overall
The predicted Rate Law agrees with the experimental Rate Law
This mechanism must be correct.
Predicted Rate Law
Experimental Rate Law
Rate
Rate
The two Rate Laws are identical if:
Example 2:
The following mechanism has been proposed for decomposition of ozone to oxygen gas:
Determine the Rate Law based on this mechanism.
CATALYSIS
Earlier we learned that reaction rates can be increased by various factors, such as increasing concentration or increasing temperature. But sometimes these methods are not feasible. Alternately, reaction rates can also be increased by use of a catalyst.
A catalyst is a substance that speeds up a reaction without being consumed. In theory the catalyst may be used over and over again. In practice, however, there is some loss of catalyst through other reactions that occur at the same time (side-reactions).
A catalyst works by providing an alternate mechanism for the reaction-one in which the ratedetermining step has a lower activation energy. For example, consider the noncatalytic destruction of ozone in the upper atmosphere:
In this reaction, ozone reacts with an oxygen atom to produce two oxygen molecules. This reaction is slow due to its high activation energy, and therefore proceeds at a fairly slow speed.
However, the addition of Cl atoms (caused by photodissociation of man-made chlorofluorocarbons) in the upper atmosphere, provide an alternate pathway by which can be destroyed.
Since this alternate pathway has a lower activation energy than the uncatalyzed pathway, it occurs at a much faster rate.
Energy Diagram for Catalyzed and Uncatalyzed Pathways
Although the reaction described above is an undesirable use of a catalyst, many catalysts are used to speed up desirable reactions. For example, the catalytic converter in your car is used to convert harmful exhaust pollutants such as nitrogen monoxide and carbon monoxide into less harmful substances:
The catalytic converter also promotes the complete combustion of any fuel fragments present in the exhaust, preventing potential formation of air pollutants in the atmosphere.
HOMOGENEOUS & HETEROGENEOUS CATALYSIS
Catalysts can be categorized into two types: homogeneous and heterogeneous.
In homogeneous catalysis the catalyst exists in the same phase (state) as the reactants. The catalyzed destruction of ozone by Cl is an example of this type of catalysis.
In heterogeneous catalysis the catalyst exists in a different phase than the reactants. The solid catalysts used in the automobile catalytic converters is an example of this type of catalysis.
Another example of heterogeneous catalysis is the hydrogenation of double bonds in alkenes. For example, the reaction between ethene and hydrogen is a relatively slow reaction at normal temperatures:
However, in presence of a catalyst, the reaction occurs rapidly. The catalysis occurs by the fourstep process shown below.
The large activation energy of the hydrogenation reaction-due primarily to the strength of the hydrogen-hydrogen bond in - is greatly lowered when the reactants adsorb onto the surface of the catalyst.
ENZYME CATALYSIS
Enzymes are the catalysts of biological organisms.
Enzymes are large protein molecules that are highly specific sites in their structure, called active sites. The properties and shape of the active site is just right to bind the reactant molecule, called the substrate.
The general mechanism by which an enzyme (E) binds a substrate (S) and then reacts to form the products is:
Enzyme-Substrate Binding
Breakdown of sucrose (table sugar) into glucose and fructose in the body, a slow reaction at normal body temperatures, is facilitated by the catalysis with the enzyme sucrase.
Answers to In-Chapter Problems:
Page
Example No.
Answer
4
1
Rate
2
a) b)
5
4
b)
5
c
12
2A
Rate
2B
2C
Rate (4)
13
3
c
4
a) First order
b)
15
3
0.0277 M
17
2
79.2 s
3
19
1
2
order
24
1
Rate
25
2
a) Rate
b) 8 c)
3
Rate
26
4
a) b)
c)
5
a) order; b)
c)
31
2
32
3
Reaction A is faster and therefore has a larger rate constant
