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Learning Zone Presents:

                                    Chemical Kinetics

                       Factors affect reaction rates

 
Many factors influence rates of chemical reactions, and these are summarized below.
 
                                                    Nature of Reactants

Acid-base reactions, formation of salts, and exchange of ions are fast reactions.
 
Reactions in which large molecules are formed or break apart are usually slow.
 
Reactions breaking strong covalent bonds are also slow.

                                               Temperature
Usually, the higher the temperature, the faster the reaction.

Increasing the temperature increase the number of effective collisions during a reaction. 

Concentration effect

In general, the more reactants present at the beginning of a reaction, the faster the reaction proceeds.

This is because the number of effective collisions possible are highest when the concentration of the reactants are highest. 

Heterogeneous reactions: reactants are present in more than one phase
For heterogeneous reactions, the rates are affected by surface areas.

http://wps.prenhall.com/wps/media/objects/167/172009/ArrheniusActivity.html

                                                            Surface area

The smaller or more broken up the reacting particles will allow more contact between them. 

This in turn increases the number of effective collisions between reacting particles. 

                                                               Catalysts

Substances used to facilitate reactions that lower the activation energy of the reaction.

http://wps.prenhall.com/wps/media/objects/167/172009/Catalysis.html
                                                    

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                            Potential Energy Diagrams 

The activation energy is the minimum energy required to start a chemical reaction by providing colliding molecules with enough energy for effective collisions to occur.

 

   

 

        

A catalyst provides an alternate reaction pathway, which has a lower activation energy than an uncatalyzed reaction.

 

           

          http://wps.prenhall.com/wps/media/objects/167/172009/Golfer.html

            http://wps.prenhall.com/wps/media/objects/167/172009/EnergyProfile.html

                            http://wps.prenhall.com/wps/media/objects/167/172009/BimolecularReaction.html

      go to pracice problems

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                              Average Reaction Rates 

 

Reaction rates have the general form of (change of concentration / change of time).

 

There are two types of reaction rates.

 
By rate of reaction we mean the change in concentration of a reactant (or a product) in a given period of time.
 
This might, for example, be over a short period of time anywhere during the reaction, or it might be at an instant of time (or corresponding concentration of a reactant).
 

                         Reaction Rate Stoichiometry

Formal definition of reaction rate

Consider a typical chemical reaction:

aA + bB → pP + qQ

The lowercase letters (a, b, p, and q) represent stoichiometric coefficients, while the capital letters represent the reactants (A and B) and the products (P and Q).

The definition the reaction rate v (also r or R) for a chemical reaction occurring in a closed system under constant-volume conditions, without a build-up of reaction intermediates, is defined as:

v = - \frac{1}{a} \frac{d[A]}{dt} = - \frac{1}{b} \frac{d[B]}{dt} = \frac{1}{p} \frac{d[P]}{dt} = \frac{1}{q} \frac{d[Q]}{dt}

or 

v = -1/a Δ[A]/ Δt = -1/b Δ[B]/ Δt = 1/p Δ[P]/ Δt = 1/q Δ[Q]/ Δt

(NOTE:Rate of a reaction is always positive. '-' sign is present in the reactant involving terms because the reactant concentration is decreasing.)

The IUPAC, International Union of Pure and Applied Chemistry, recommends that the unit of time should always be in seconds.

The rate is measured by the rate of increase of concentration of a product P multiplied by a constant factor (the reciprocal of its stoichiometric number) and for a reactant A by minus the reciprocal of the stoichiometric number. (see the balance equation)

In the illustration, d = delta or change.

Reaction rate usually has the units of mol/dm3 s−1. (Molarity/sec)

It is important to bear in mind that the previous definition is only valid for a single reaction, in a closed system of constant volume.

Example:

How is the rate of disappearance of ozone related to the rae of appearance of oxygen in the following equation:

2O3(g) → 3O2(g)

If the rate of appeance of O2, Δ[O2]/Δt, is 6.0 x 10-5 M/s at a particular

 instant, what is the value of the rate of disappearance of O3, -Δ[O3]/Δt at

 the same time?

solution: 

Rate = -1/2Δ[O3]/Δt = 1/3Δ[O2]/Δt 

solve for the fraction:  -Δ[O3]/Δt = 2/3Δ[O2]/Δt = 2/3 (6.0 x 10-5 M/s)

                                                   =  4.0 x 10-5 M/s

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                                          Instantaneous Rate

Average verses instantaneous Reaction rate
  

The average rate   of reaction, as the name suggests, is an average rate, obtained by taking the change in concentration over a time period, for example: -0.3 M / 15 minutes.

This is an approximation of the reaction rate in the interval; it does not necessarily mean that the reaction has that specific rate throughout the time interval or even at any instant during that time.

The instantaneous rate  of reaction, on the other hand, depicts a value that is more accurate and absolute.

The instantaneous rate of reaction is defined to be the change of concentration in an infinitely small time interval, expressed as the limit or derivative.

The instantaneous rate can be obtained from the experimental data by first graphing a graph of concentration of the system as function of time, then finding the slope of the tangent line at a specific point in the graph that corresponds to a particular time of interest (ie: t = 200 may corresponds to 200 seconds after the start of the reaction).

Alternatively, one can try to measure the change in concentration two times in as small time interval in between as possible to get an average rate that is closer to the instantaneous rate. (not advised)

Example: 

sample reaction rate graph1.JPG

For the reaction rate  graph of a certain hypothetical unimolecular reaction shown above, the average rate of reaction between, say t = 0~1000 sec (blue line), is

                                                      (Δ[conc.] / Δt)

                           (0M - 0.75M)/(1000s - 0s) = -7.5 x 10-4 M/s = -0.75 mM/s

But the instantaneous rate of reaction at t=400 sec, or the slope of the tangent (red line) at t=400, is

                                                    

                                (0M - 0.18M)/(600s - 200s) = -4.5 x 10-4 M/s

Note that the instantaneous rate at time=400 is not the same as the average rate of the interval, it is slower.

Also note that the red tangent line is shallower campared to the blue average line, indicating a slower rate.

A shallower line (closer to the horizontal) indicates a slower rate and a sharper slope (closer to the vertical) slope indicates a faster rate.

                                           http://www.t2i2edu.com/WebMovie/Ch13Th3.htm

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Initial  rate of Reaction

The initial rate of reaction is the rate of reaction when the reagents are first brought together.

 

Like the instantaneous  rate mentioned above, the initial rate can be obtained either experimentally or graphically.

 

To find it experimentally, one will bring the reagents together and quickly measure the reaction rate as quickly as possible. 

 

If this is not possible to do, one will have to find the initial rate graphically.

 

Simply find the slope of the line tangent to the reaction curve when t=0.

 

                                                  Example

For the graph shown above, the intial reaction rate (green line) will be about  -40 x 10-4 M/s. = 0.4M-0.8M/100 sec - 0 sec)

 

Again, this makes sense graphically - the initial rate has a steeper slope

 

(compared to the average), which corresponds to a faster rate.


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                         Determining Reaction Order

Initial Rates and the Method of Isolation

A fundamental challenge in Chemical Kinetics is the determination of the reaction order (or, in general, the Rate Law) from experimental information.

We know that the rate law closely related to the reaction mechanism, and the knowledge of the mechanism of a given reaction allows us to control that reaction.

But how does one actually determine a rate law?

The first step is to control the conditions under which a reaction occurs, and then determine the rate of that reaction.

If we measure the reaction rate just after we define the reaction conditions, then this is a measurement of the initial rate of the reaction.

Consider a reaction A + B -> Products. An experimenter prepares a reaction mixture of 1.00 M of each of the reactants and measures the rate of reaction initially (at very early times, i.e. before the reactant concentrations have had a chance to change much due to the progress of the reaction) to be 1.25 x 10-2 M/s.

What is the rate constant for the reaction?

  
Note:  The reaction stoichiometry does NOT determine the reaction order except in the special case of an ELEMENTARY reaction. (An elementary reaction is one in which the reaction takes place with a mechanism implied by the reaction equation. Most reactions are NOT elementary)

  

If the above reaction was elementary, then the rate law would be:

Rate = k[A][B]

and the reaction would be first order in A, first order in B and second order overall. The rate constant can then be determined from initial rate information easily:

k = (1.25 x 10-2 M/s)/ (1.00 M)2

k = 1.25 x 10-2 l/mol.s

Remember: The units of the rate constant depend on the overall order of the reaction.

More information about the reaction rate law is needed than a single measurement of an initial rate! In order to determine the rate law experimentally, we must use more than one measurement of rate versus concentration!

  

The Method of Initial Rates

Consider a reaction A + B -> Products (this time we know the reaction is NOT elementary). An experimenter prepares several reaction mixtures and determines the initial reaction rates under these different conditions. The data obtained for several experimental runs in tabulated below

Run #

Initial [A] ([A]0)

Initial [B] ([B]0)

Initial Rate (v0)

1

1.00 M

1.00 M

1.25 x 10-2 M/s

2

1.00 M

2.00 M

2.5 x 10-2 M/s

3

2.00 M

2.00 M

2.5 x 10-2 M/s

What is the rate constant for this reaction?

ANS: Before we do anything, we must determine the order of the reaction with respect to every reactant, i.e. we must determine the rate law for the reaction.

In the above example, the order can be easily determined by inspection.

Method:

a.  Find a pair of experimental runs that the concentration of only one reactant changes.

This is called the method of isolation and a good experimental design always has one such pair of experimetal runs.

b.  We can see that in runs 1 and 2, only the initial concentration of B has been varied. In fact, [B]0 has doubled from run 1 to run 2 and the reaction rate has also doubled.

Therefore the reaction must be First Order in B.

c.  Examination of runs 2 and 3 show that in these experimenal runs the concentration dependence of A has been isolated.

In this case the doubling of the initial concentration of A has No EFFECT on the reaction rate so the reaction must be Zeroth Order in A.

d.  Thus the rate law for the reaction is

rate = k[B]1[A]0 = k[B]

e.  Now, the rate constant can be determined from any of the experimental runs.

     Let's use run # 3

rate = 2.5 x 10-2 M/s= k (2.0 M)

k = 1.25 x 10-2 s-1

Note that the initial rate of any expermental run (any prepared experimental condition) can now be determined (calculated) since we have the rate law and the rate constant.

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A More Complicated Example

Consider a reaction A + B + C-> Products (this, too, is not an elementary reaction). Several initial conditions of this reaction are investigated and the following data are obtained:

Run #

[A]0

[B]0

[C]0

v0

1

0.151 M

0.213 M

0.398 M

0.480 M/s

2

0.251 M

0.105 M

0.325 M

0.356 M/s

3

0.151 M

0.213 M

0.525 M

1.102 M/s

4

0.151 M

0.250 M

0.480 M

0.988 M/s

What is the initial rate of the reaction when all the reactants are at 0.100 M concentrations?

 
 ANS: First we must determine the rate law for the reaction, then the rate constant and then evaluate the rate law under the desired conditions.

Method:

a.  Isolate the effect of the concentration of one of the reactants.

In runs number 1 and 3, the only change of initial concentrations is that of reactant C, a change from 0.398 M to 0.525 M with a change in rate from 0.480 M/s to 1.102 M/s.

These numbers are not so easy as to be able to guess the order by inspection, so lets think mathematically.

b.  If we define the order of the reaction in reactantys A, B, and C as a, b, and c, we can write down the rate law (with unknown orders)

rate = k [A]a [B]b [C]c

c.  The ratio of the initial rates of runs 1 and 3 is then:

Note that we have chosen this ratio so that many terms on the right hand side cancel, i.e. k and the concentrations of species A and B. The ratio reduces to:

Note that none of the numbers in this equation have any units anymore because we have divided them out in a dimensionless ratio.

 

d.  Now, how to solve for c?

Take the natural log (i.e. ln) of both sides of the equation

or

So the reaction is Third Order in reagent C!

To continue, we pick another pair of runs that have a change in the initial concetration of C and just one other reactant. Let us choose the runs 1 and 4.

The ratio of the initial rates of these runs is:

Note that everthing cancels, is known, or is the order b. So:

We can determine b as before by taking the ln of both sides of this equation:

and b = 1.00. The reaction is First Order in B!

Now that we know the order of Reactants A and B, we can use another pair of experimental runs to determine the remaining unknown order.

We must use a pair of runs where the initial concentration of A changes, so we pick runs 1 and 2. As before, the ratio of initial rates is

The rate constant cancels as before and we can solve for a:


and, as usual, take the ln of both sides:

which results in a = 2.00 and the reaction is Second Order in A.

The rate constant for the reaction may now be evaluated from any of the experimental runs:

or

k = 1.57 x 103    l5 / mol5.s

These units are appropriate for a reaction that is sixth order overall.

Use another experimental run to check your answer.

If the rate law is correct, every experimetal run will give the same value of the rate constant.

And now for the finale. The rate of the reaction when all the reactants are 0.100 M is:

  

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                                 Time Dependent Concentrations

Suppose you encounter a reaction with some stoichiometry,

say:                                A --> 2B + C

and you are asked to determine the order of the reaction from data

obtained for [A] as a function of time.

time[A]
0.000 min1.000 M
10.00 mins0.8000 M
20.00 mins0.6667 M
40.00 mins0.5000 M

How do you solve such a problem?:

This, unfortunately is not an initial rates problem.

We have not been provided with data spaced close enough to get an instantaneous rate under known reactant concentrations.

So we must directly test the integrated rate laws for validity.

The way we have been shown to test for a particular integrated rate law is to make a plot of the data that should show a straight line.  

Suppose you can't make a physical plot, but want to solve the problem anyway?

If the reaction above were zeroth order, the concentration versus time would be a straight line.

This means the change in concentration (Δ[A]) divided by the elapsed time (Δt) would be a constant, for any pair of data points.

Let's see if this is true for our data 

 (0.800 - 1.000) / 10 = -0.020
(0.6667 - 0.800) / 10 = -0.01333    
 (0.5000 - 0.6667) / 20 = -0.00834

These numbers are NOT constant, so the reaction is NOT zeroth order. 

 

 

If the reaction is first order, a plot of the natural log of the concentrations versus time will be a straight line.

Thus, the change in the natural log of the concentration (Δln([A])) divided by the elapsed time (Δt) will be constant. 

   (ln(0.800) - ln(1.000)) / 10 = -0.0223     
(ln(0.6667) - ln(0.800)) / 10 = -0.0182
(ln(0.5000) - ln(1.000)) / 40 = -0.0173 

Note: doen't matter which points...

These numbers are NOT constant, so the reaction is NOT first order.

If the reaction is second order, a plot of the inverse concentration vesus time will be linear.

Thus, the change in inverse concentration (Δ(1/[A]))divided by the elapsed time (Δt) will be constant.

   (1/0.800 - 1/1.000) / 10 = 0.0250 

                                      (1/0.6667 - 1/0.8000) / 10 = 0.0250

                                         (1/0.5000 - 1/1.000) / 40 = 0.0250

The reaction is second order with a rate constant of  0.0250 M-1min-1

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                     Reaction Mechanisms

 
 Reaction Mechanisms

Reactions may occur in one step (elementary mechanism) or more than one step (multistep mechanism).

ELEMENTARY MECHANISM

Consider the hydrolysis of a tertiary bromomethane,

for example, (CH3)3C-Br.

R-Br + OH-    ®    R-OH + Br-

Changing the concentration of the OH- ions is found to have no effect on the rate of the reaction. The experimentally determined rate equation is:

Rate = k[RBr]   (or   Rate = k[RBr][OH-]0   since [OH-]0 = 1)

The molecularity of the process above is called bimolecular because two reacting particles are required to react.

another example:

 The reaction between nitrogen monoxide gas and bromine gas to produce NOBr takes place without an intermediate.  Propose a mechanism.

 Elementary process:  Just write the balanced equation.

2NO + Br2    →   2NOBr    

rate = k[NO]2[Br2]  termolecular

elementary termolecular processes are not normal.      

The link below demonstrate a unimolecular mechanism where only one reactant is required to produce a product.

http://wps.prenhall.com/wps/media/objects/167/172009/EnergyProfile.html

 

 

 

 

                                       MULTISTEP MECHANISM

So why is it that in some reactions changing the concentration of a particular reactant has no affect on the rate?

Where this is the case, there must be at least two steps involved in the reaction.

The mechanism of a reaction is a proposal about the series of steps involved in going from reactants to products.
It is a suggestion based on the scientific evidence available for the reaction.

Part of the experimental evidence is the rate equation.

Where there are two or more steps, it is the slowest step that determines the overall rate.

This is called the rate-determining step (RDS).

example:

    slow:  H2  +   I2  →   H2I2   (bimolecular)

                                   fast:       H2I2   →   2HI (unimolecular)                                

                                         overall:  H2  +   I2  →  2HI

 

 

the rate law depends on the slowest step:

                                           rate = k[H2][I2]  

                                           1o in H2

                                                   1o in I2

                                                   2o overall

H2I2   is called a transition state intermediate and is the species found at the top of the hump in a PE diagram.

The numbers before the formula in the balanced chemical equation for the rate-determining step respectively correspond to the powers to which the reactant concentrations are raised in the rate equation. 

                                                See below:

                                        rate = k[H2][I2

 

Acetylene gas (C2H2) reacts with hydrogen chloride gas to produce

dichloroethane gas. 

This reaction involves the intermediate C2H3Cl. 

Propose a 2 step mechanism consistent with this intermediate product.

Slow:  C2H2  +   HCl  →   C2H3Cl 

Fast:  C2H3Cl   +   HCl   →   C2H4Cl2

                                        

                                 overall:  C2H2  +  2HCl  →  C2H4Cl2

                         1o in C2H2

                                1o in HCl

                                2o overall

The numbers before the formula in the balanced chemical equation for the rate-determining step respectively correspond to the powers to which the reactant concentrations are raised in the rate equation. 

 

 

 

 

                                                See below:

                                        rate = k[C2H2][HCl] 

A zero order reactant does not appear in the chemical equation for the rate-determining step.

The term molecularity refers to the number of chemical species in the rate-determining step. Read more about Mechanisms.

Reaction mechanisms

Most chemical reactions occur by mechanisms that involve more than one step.

As a result, the rate law can not be directly deduced from the stoichiometry of a balanced chemical equation.

Only elementary reactions (those which occur in one step) can give the rate law.

The rate law of an elementary process does follow from the coefficients of its balanced equation.

Among the steps in a multi-step reaction, there is always one that acts like a bottleneck.

There is usually one step that is slowest compared to the rest.

This slowest step limits the overall reaction rate and is called the rate-determining step.

The rate-determining step determines the rate law for the overall reaction.

Example:

The reaction,

NO2(g) + CO(g) --> NO(g) + CO2(g)

is found to be second order with respect to [NO2] and zero order with respect to [CO]:

rate = k[NO2]2

The following mechanism has been proposed:

Step 1: NO2(g) + NO2(g) --> NO3(g) + NO(g) (slow)

Step 2: NO3(g) + CO(g) --> NO2(g) + CO2(g) (fast)

Overall: NO2(g) + CO(g) --> NO(g) + CO2(g)

Since Step1 is the slowest step, it determines the rate of the overall reaction.

 Thus, the above mechanism agrees with the observed rate law.

Here is another example:

2NO(g) + O2(g) --> 2 NO2(g)

with the following observed rate law:

rate = kobs [NO]2[O2]

Of course, one possible mechanism is a one-step reaction between two molecules of nitrogen oxide and a molecule of oxygen. 

This is a termolecular step, which is very unlikely since three way collissions are highly improbable. 

An alternative mechanism is provided by the following:

Both steps are bimolecular, which are more likely to happen than a three-way collision. 

The slow step determines the overall reaction rate:

which is sort of useless since it contains a term that does not correspond to either any of the reactants or the product.

A rate law should only consist of concentrations of reactants and/or products, no intermediates.

Since the intermediate, dinitrogen dioxide, reacts slowly with oxygen, the reverse reaction of Step 1 is possible. 

What we are allowed to do then is to assume that Step 1 will reach a dynamic equilibrium, which will lead to a constant concentration of dinitrogen dioxide, that is, the forward and reverse rates of Step 1 are equal:

which allows us to express the concentration of the intermediate dinitrogen dioxide in terms of the starting material:

Therefore allowing us to express:

as:

which then relates the observed rate constant to the rate constants of the elementary steps as follows:

 
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                                  PE Diagrams

TEST YOUR UNDERSTANDING

1.  Given the balanced equation representing a reaction at 101.3 kPa and      298 K: 

N2(g) + 3H2(g) ---> 2NH3(g) + 91.8 kJ

Which statement is true about this reaction?

(1) It is exothermic and ΔH equals –91.8 kJ.   

(2) It is exothermic and ΔH equals +91.8 kJ.

(3) It is endothermic and ΔH equals –91.8 kJ.   

(4) It is endothermic and ΔH equals +91.8 kJ.

2.  Base your answers to questions 5 through 7 on the reaction represented by the balanced equation below.

2H2(g) + O2(g) ---> 2H2O(l) + 571.6 kJ

3. Identify the information in this equation that indicates the reaction is exothermic.

4. Draw a potential energy diagram for the reaction represented by this equation.

5. Explain why the entropy of the system decreases as the reaction proceeds.

6.  In a chemical reaction, the difference between the potential energy of the products and the potential energy of the reactants is defined as the

(1) activation energy                                     (2) ionization energy

(3) heat of reaction                                       (4) heat of vaporization

 

7.  Given the balanced equation:

 

   

 

8.  Given the potential energy diagram for a chemical reaction:

Which statement correctly describes the energy changes that occur in the forward reaction?

    (1) The activation energy is 10. kJ and the reaction is endothermic.

    (2) The activation energy is 10. kJ and the reaction is exothermic.

    (3) The activation energy is 50. kJ and the reaction is endothermic.

    (4) The activation energy is 50. kJ and the reaction is exothermic.

 

9. Which expression represents the ΔH for a chemical reaction in terms of the potential energy, PE, of its products and reactants?

 (1) PE of products + PE of reactants            (2) PE of products – PE of reactants

 (3) PE of products × PE of reactants            (4) PE of products ÷ PE of reactants

 

10.  Which balanced equation represents an endothermic reaction?

 

 

11. Which statement best describes how a catalyst increases the rate of a reaction?

    (1) The catalyst provides an alternate reaction pathway with a higher activation energy.

    (2) The catalyst provides an alternate reaction pathway with a lower activation energy.

    (3) The catalyst provides the same reaction pathway with a higher activation energy.

    (4) The catalyst provides the same reaction pathway with a lower activation energy.

 

12.  Given the balanced equation representing a reaction:

 

 

 

 

Which statement is true about energy in this reaction?

(1) The reaction is exothermic because it releases heat.

(2) The reaction is exothermic because it absorbs heat.

(3) The reaction is endothermic because it releases heat.

(4) The reaction is endothermic because it absorbs heat.

 

13.  Given the potential energy diagram for a reaction:

Which interval on this diagram represents the difference between the potential energy of the products and the potential energy of the reactants?   

(1) 1                          (2) 2                             (3) 3                              (4) 4


 

 

14.   The letter B represents which chemical formula or formulas in the equation?

15.   If 682.2 kilojoules are absorbed, how many moles of C2H2(g) are produced?

16.  Describe how the potential energy diagram will change if a catalyst is added.

 

Base your answer on the information below.

Given the reaction at equilibrium:

 

2NO2(g) + 7H2(g) <--> 2NH3(g) + 4H2O(g) + 1127 kJ

 

17.  Complete a potential energy diagram for the forward reaction. Be sure your drawing shows the activation energy and the potential energy of the products.


18.   Which statement best explains the role of a catalyst in a chemical reaction?
        (1) A catalyst is added as an additional reactant and is consumed but not regenerated.
        (2) A catalyst limits the amount of reactants used.

        (3) A catalyst changes the kinds of products produced.

        (4) A catalyst provides an alternate reaction path-way that requires less activation energy.

 

Base your answers to questions 19 through 21 on the information and potential energy diagram below. Chemical cold packs are often used to reduce swelling after an athletic injury. The diagram represents the potential energy changes when a cold pack is activated.

19.  Which lettered interval on the diagram represents the potential energy of the products? 


20.  Which lettered interval on the diagram represents the heat of reaction?


21 A catalyst increases the rate of a chemical reaction by 
(1) lowering the activation energy of the reaction

(2) lowering the potential energy of the products 

(3) raising the temperature of the reactants

(4) raising the concentration of the reactants


Given the reaction:

S(s) + O2(g) à SO2(g) + energy 

22.  Which diagram best represents the potential energy changes for this reaction?

23. Which statement correctly describes an endothermic chemical reaction?

(1) The products have higher potential energy than the reactants, and the ΔH is negative.

(2) The products have higher potential energy than the reactants, and the ΔH is positive. 

(3) The products have lower potential energy than the reactants, and the ΔH is negative.

(4) The products have lower potential energy than the reactants, and the ΔH is positive.

 

24.  A catalyst is added to a system at equilibrium. If the temperature remains constant, the activation energy of the forward reaction

(1) decreases

 

(2). increases

 

(3) remains the same

 

25. The potential energy diagram below represents a reaction.

Which arrow represents the activation energy of the forward reaction?

(1) A                                      (2) B                                   (3) C                          (4) D  

 

Base your answers on the information and diagram below, which represent the changes in potential energy that occur during the given reaction.

 

26.  Given the reaction: A + B --> C

 

 a)  Does the diagram illustrate an exothermic or an endothermic reaction? State one reason, in terms of energy, to support your answer.

 

b) On the diagram provided in your answer booklet, draw a dashed line to indicate a potential energy curve for the reaction if a catalyst is added.

27. Given the reaction: 2 H2(g) + O2(g) --> 2 H2O(l) + 571.6 kJ What is the approximate ΔH for the formation of 1 mole of H2O(l)?

(1) -285.8 kJ                                (2) +285.8 kJ                    (3) -571.6 kJ               (4) +571.6 kJ

 

28.  According to Table I, which potential energy diagram best represents the reaction that forms H2O(l) from its elements?

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                           Average Rate Practice

a.  Consider the hypothetical aqeous reaction A(aq)  = >   B(aq).  A flask is charged with 0.065 mole of A in a total volume of 100.0 mL.  The following data were collected. 

 time (s)

 0

 10

 20

 30

 40

 moles of A

 0.065

 0.051

 0.042

 0.036

 0.031

1.  Calculate the number of moles of B at each time period.

2.  Calculate the average rate of disappearance of A for each time period.

3.  During time period 10 - 20 minutes what is the average rate of appearance of B in M/sec?

b.  A 1 liter flask is charged with 0.100 mol of A and allowed to react to form B according to the following reaction.

A(g)     = >   B(g) 

The following data were collected. 

 time (s)

 0

 40

 80

 120

 160

 moles of A

 0.100

 0.067

 0.045

 0.030

 0.020

1.  Calculate the number of moles of B at each time period.

2.  Calculate the average rate of disappearance of A for each time period.

3.  During time period 10 - 20 minutes what is the average rate of appearance of B in M/sec?

c.  The rate of disappearance of HCl was measured for the following reaction.

CH3OH(g) +  HCl(g)     = >   CH3Cl(g) + H2O(g)

The following data were collected. 

 time (s)

 [HCl]

 0.0

 1.85

 54.0

1.58

 107.0

1.36 

 215

 1.02

 430.0

 0.580

1.  Calculate the average rate of reaction for each time period.

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                           Instantaneous Rate Practice

sample reaction rate graph1.JPG

1.  Using the graph above, calculate the reaction rate at the following instants during the reaction:

a.  200 seconds                      b.  100 seconds                 c.  600 seconds

2.  Using the graph above, calculate the average rate between the following time intervals:

a.  0 to 100 seconds               b. 0 to 400 seconds           c.   0 to 800 seconds

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                  Reaction Order Practice Problems

2 A + 2 B = > C + D

The following data about the reaction above were obtained from three experiments:

 

Experiment

 

[A]

 

[B]

Initial Rate of Formation of C (mole.liter-1min-1)

1

0.60

0.15

6.3 x 10-3

2

0.20

0.60

2.8 x 10-3

3

0.20

0.15

7.0 x 10-4

 

(a)     What is the rate equation for the reaction?

(b)     What is the numerical value of the rate constant k? What are its dimensions (units)?

(c)     Propose a reaction mechanism for this reaction.

A(aq) + 2 B(aq) = >3 C(aq) + D(aq)

For the reaction above, carried out in solution of 30°C, the following kinetic data were obtained:

 

 

Experiment

Initial Conc. of Reactants

(mole.liter-1)

Initial Rate of Reaction

(mole.liter-1.hr-1)

 

Ao

Bo

 

1

0.240

0.480

8.00

2

0.240

0.120

2.00

3

0.360

0.240

9.00

4

0.120

0.120

0.500

5

0.240

0.0600

1.00

6

0.0140

1.35

?

(a)     Write the rate-law expression for this reaction.

1.  Rate = k[A][B]

2.  trial 2 and 4 holds the concentration of B at 0.120 M, at this time concentration A = 0.24 to 0.12 which gives a factor of 2, the rate during this period goes from 2.0 M/hr (5.6 x 10-4 M/sec) to 0.5 M/hr (1.39 x 10-4 M/sec)

3.  dividing the larger rate value by the smaller rate value give a factor of 4.

4.  2x = 4; x = 2; at this point the rate law is:  Rate = k[A]2[B]

5.  trial 1 and 2  holds the concentration of A at 0.240 M, at this time concentration B = 0.48 to 0.12 which gives a factor of 4, the rate during this period goes from 8.0 M/hr (2.2 x 10-3 M/sec) to 2 M/hr (5.6 x 10-4 M/sec)

6.  dividing the larger rate value by the smaller rate value give a factor of 1.

7.  4x = 4; x = 1; at this point the rate law is:  Rate = k[A]2[B]

(b)     Calculate the value of the specific rate constant k at 30°C and specify its units.

From trial one:  8 = k[0.24]2[0.48]

K = 286 (1/hr x M2)

Or

From trial one:  2.2 x 10-3 = k[0.24]2[0.48]

K = 0.018 (1/sec x M2)

(c)     Calculate the value of the initial rate of this reaction at 30°C for the initial concentrations shown in experiment 6.

                                                                 Rate = 286 [0.014]2[1.35] =  0.0756 M/hr

                                                       Rate = 0.018 [0.014]2[1.35] = 4.9 x 10-6 M/sec

Note:  to convert rate per hour to rate per seconds, (1 M/hr) x (1 hr/3600 sec)

or

(2.77 x 10-4 M/sec)

(d)    Assume that the reaction goes to completion. Under the conditions specified for experiment 2, what would be the final molar concentration of C?

                                                              from the balanced equation:

A(aq) + 2 B(aq) = >3 C(aq) + D(aq)

Molar ratio

1  :  2   :  3  :  1

Initial amounts of reactants:  0.240 M   0.120 M

Initial amounts of products:  0 and 0

Upon completion  amounts of reactants:  0 M   0  M

Initial amounts of products: will follow stoichiometry where C = 3/1 [A] and C = 3/2 [B],

 therefore [C] = 3/2[B] or 3/2[0.12] 0.18 M

notice that according to the molar ratio and the amounts given in trial 2, [B] is limiting and [A] is excess.

 

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                         Reaction Rate Stoichiometry

1.  For each of the following gas-phase reactions, write the rate expression in terms of the appearance of each product or disappearance of each reactant.

a.  H2O2(g)    ==>   H2(g)   +   O2(g)

 

 

b.  2N2O(g)   ==> 2N2 (g)   +   O2(g)

 

 

c.  N2(g)    +   3H2(g)  ==>   2NH3(g)

 

2.  For each of the following gas-phase reactions, write the rate expression in terms of the appearance of each product or disappearance of each reactant.

a.  2HBr (g)    ==>   H2 (g)   +   Br2 (g)

 

 

b.  2SO2 (g)  +  O2(g)  ==> 2SO3(g)

 

 

c.  C4H8(g)    ==>   2C2H4 (g)

 

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                                                    Practice Problems

Practice Problems

Chemical Kinetics

1.  For the reaction given below, what is the instantaneous rate for each of the reactants and products?

                              3 A   +   2 B  ==>  4 C

2.  Given the following experimental data, find the rate law and the rate constant for the reaction:

 

                              NO (g)   +   NO2  (g)   +  O2 (g) ==>  N2O5 (g)

Run      [NO]o , M        [NO2[o , M       [O2]o , M         Initial Rate, Ms-1 

   1          0.10 M          0.10 M               0.10 M             2.1 x 10-2

   2          0.20 M          0.10 M               0.10 M             4.2 x 10-2

   3          0.20 M          0.30 M              0.20 M             1.26 x 10-1

   4          0.10 M          0.10 M               0.20 M             2.1 x 10-2

3.  The half-life of a radioisotope is found to be 4.55 minutes. If the decay follows first order kinetics, what percentage of isotope will remain after 2.00 hours?

4.  The mechanism of a reaction is shown below.   

 a)  What is the overall reaction?

b)  Which compounds are intermediates? 

c) Predict the rate law based on this mechanism.

d)  What is the overall order of the reaction?

            HOOH  +  I¯   ==> HOI  +  OH¯             (slow)

            HOI  +  I¯  ==> I2   +   OH¯                   (fast)

            2OH¯   +   2H3O+  ==> 4 H2O               (fast)

5.  For the reaction A + B ž C, the rate constant at 215 oC is 5.0 x 10-3 and the rate constant at 452o C is        1.2 x 10-1.

            a)  What is the activation energy in kJ/mol?

            b)  What is the rate constant at 100o C.

Answers:

1.   

2.  Rate =  k[NO] [NO2]          k  =  2.1 M-1s-1

3.  k = 0.152 min-1       At  =  1.15 x 10-6 %  (not much!!)

4.  a)   Overall reaction:          HOOH  +  2 I¯   +  2 H3O+    ==>  I2   +  4 H2O

     b)   Intermediates:  OH¯  and HOI

     c)   Predicted mechanism:   Rate =  k [HOOH][I¯ ]

     d)  Overall order:  2nd order

5.  a) 39.4 kJ/mol

     b) 2.50 x 10-4 s-1