Lecture Notes 1 Kinetics PDF

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3/1/2018

Chemical Kinetics: Rates of Reactions Chemical kinetics is concerned with how fast things react, and with the factors that influence how fast they react Reading: Chapter 13 (all sections) Topics: What is a reaction rate Stoichiometry and reaction rate Rate laws and reaction order Zero, first, and second order reactions Integrated rate laws Pseudo first order reactions Half life Elementary reactions Effect of temperature on the reaction rate Rate laws for elementary reactions Reaction mechanisms Catalysts and enzymes

Reactions can be divided into two classes: Homogeneous (where the reactants are in the same phase) Give examples: H2 + O2 Acid + base (2 solutions) Rection between 2 gases or 2 solutions

Heterogeneous (where the reactants are in different phases) Give examples: Na + Cl (g) Combustion Gas reacting with surface of liquid, liquid reacting with surface of a solid e.g. rxns in catalytic converter; H + H -> H2: which occurs in interstellar space

Surface area can strongly influence the rates of heterogeneous reactions e.g. aluminum as a finely divided powder is a a rocket propellant

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The Reaction Rate The reaction rate is a measure of how fast a reaction occurs! Example: [Fe(H2O)6]3+(aq) + SCN−(aq) → [Fe(SCN)(H2O)5]2+(aq) + H2O(l) If we followed the amount of reactant and products present as a function of time we would see:

Product [Fe(SCN)(H2O)5]2+ (blood-red color) Reactant [Fe(H2O)6]3+ (colorless)

[Fe(H2O)6]3+(aq) + SCN−(aq) → [Fe(SCN)(H2O)5]2+(aq) + H2O(l)

∆ = final - initial ∆[R] ∆t Note the minus sign Reactionrate   

Note units: . 

∆ 0.0117  0.0415 /   8.95  10 ∆  555  222

This is the average rate over the time range ∆t.

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If we make the time range ∆t smaller (i.e., ∆t → 0) the rate becomes an instantaneous rate (the rate at a specific time rather than an average over a range of times)

The instantaneous rate is equal to the slope at a particular time

The rate decreases with time……..why? There is less reactant involved — reactant concentration depends on amount of reagents

We can define the reaction rate in terms of the reactant or product concentration: ∆󰇟󰇠 ∆󰇟󰇠     ∆ ∆ Note the different signs so that the reaction rate is always positive

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A Word About Units For Concentrations are usually expressed in (or Molarity, M) and the concentrations are usually represented by [square brackets] i.e., [H2SO4] For Concentrations can be expressed in moles/L but it is more common to express them in partial pressures (which are usually given in Atmospheres, ) and represented by Psubscript i.e.   0.2 For ideal gases: Ptotal = PA + PB +PC + ….. From the ideal gas law (PV = nRT):

(Dalton’s Law of Partial Pressures)

    and so for a given T:  ∝   

Stoichiometry and the Reaction Rate For this reaction: Cu(H2O)62+ + NH3 → Cu(H2O)5NH32+ + H2O One mole of Cu(H2O)62+ reacts with one mole of NH3 to give one mole of Cu(H2O)5NH32+ and one mole of H2O. It doesn’t matter which reactant or product we monitor we get the same reaction rate. For reactions with stoichiometry coefficients that are not equal, the rates of consumption of different reactants and the rates of formation of different products can differ.

w/ coefficients

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Stoichiometry and the Reaction Rate For example, for the reaction CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

Partial Pressure

The O2 partial pressure decreases twice as fast as the CH4 partial pressure, and the H2O partial pressure increases twice as fast as the CO2 partial pressure.

CH4

O2 Time

Stoichiometry and the Reaction Rate In general, for the reaction ∆ = final - initial

aA + bB → cC + dD

we define

rate stoichiometric coefficient



1 '[ A] a 't



1 '[ B] b 't

Reactants = negative



1 ' [ C] c 't



1 '[ D] d 't

products = positive

Including the stoichiometry coefficient means that we always get the same reaction rate regardless of which reactant or product we use to measure it. However, if we change the stoichometry coefficients the rate changes Reaction rates are asociated with a particular chemical reaction

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The Rate Law and Reaction Order In general, for the reaction aA + bB → Products the rate of a reaction can be described by an equation like

  

1∆    ∆ Rate Constant

This equation is called the rate law for the reaction. The values of the powers in the rate law must be determined from experiments (except for elementary reactions discussed later) One way to get information on the powers in the rate law is to mix known amounts of the reagents together and then very quickly determine the reaction rate (this reaction rate is called the initial rate). This experiment is then repeated for a number of different initial concentrations.

The Rate Law and Reaction Order

Initial Concentration mol/L

Rate at t=0

Experiment

[CH3COOCH3]

[OH–]

Initial Rate mol/L.s

1 2 3

0.040 0.040 0.080

0.040 0.080 0.080

0.00022 0.00045 0.00090

We can write: Rate = k [CH3COOCH3]m[OH–]n What are n and m? Rate is proportional to [A]^3

n = 1 b/c doubled; rate = proportional to [OH-] m=1

There may be other terms in the rate law that are not in the overall chemical equation for the reaction

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The Rate Law and Reaction Order So, we can write Rate = k [CH3COOCH3]1 [OH–]1 = k [CH3COOCH3] [OH–] The exponents give the order of the reaction: If the exponent is 1 then the reaction is said to be first order with respect to the reagent, if the exponent is 2 the reaction is second order, etc. The overall reaction order = sum of the exponents

=2

The reaction above is first order with respect to CH3COOCH3 and first order with respect to OH– and second order overall (at least according to the experiments that were performed). The rate law and reaction orders must be determined experimentally – they cannot be determined from the balanced chemical equations (except for elementary reactions discussed later). Sometimes catalysts or even products can appear in the rate law of a reaction. 2nd order - goes up by 4

The Rate Law and Reaction Order For the reaction A + B → Products, the initial rate is measured for several different initial concentrations of A and B: Experiment 1 2 3 4

[A], M 0.100 0.100 0.200 0.400

[B], M 0.100 0.200 0.300 0.200

Initial Rate, M/s 8.0x10–5 8.0x10–5 11.3x10−5 16.0x10–5

What is the rate law? What is the effect of changing the [A]? what is the exponent: [A]^? Exponent = 1/2

B - doesn’t change; exponent = 0 Rate law = k[A]^1/2[B]^0 = k[A]^1/2 1. exponents are not always integers 2. exponent can be 0 (the rate is independent of one of the reagent concentrations 3. You cannot get rate laws from the chemical equation for a reaction (unless it’s an elementary reaction)

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The Rate Law and Reaction Order Some problems are not easy to solve by inspection. Here is an example: The following information was obtained for the reaction A + B → Products Experiment

Initial [A], M

Initial [B], M

Initial Rate, M/s

1

24

10

64.55

2

24

17

243.2

3

89

10

33.52

Here is an equation to solve these hard to solve cases:

       

The Rate Law and Reaction Order Determine Order for A: Experiments 1 and 3

1 64.55  33.52 3    24  0.50 1   3 89 

Determine Order for B: Experiments 1 and 2

 

64.55 243.2  2.50 10  17



So Rate = k [A]−0.5 [B]2.50

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Determining the Rate Constant from the Rate Law Experiment 1 2 3

[A]0, M 0.150 0.150 0.300

[B]0, M 0.100 0.200 0.100

Initial Rate, M/s 4.0x10–5 16.0x10–5 8.0x10–5

When we know the rate law, a value for the rate constant can be obtained from the initial rates: Step 1: Determine the rate law Step 2: Use the rate law to determine k

Units for Rate Constants Units of k depend on the order of the reaction For zero order:

  󰇟󰇠    

For first order:

  󰇟󰇠  

 





 For second order:   󰇟󰇠  





  

In general:

  󰇟󰇠  

󰇟󰇠



    󰇟󰇠  

            

Note that x is the overall reaction order: For   󰇟󰇠 󰇟󰇠 󰇟󰇠 … ,

        …

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Zero Order Reactions Rate = k[A]0 Rate = k Changing concentration has no effect on the rate

Zero order kinetics can occur in a gas/surface reaction where the surface becomes saturated with the adsorbed reagent. Then changing the pressure of the reagent has no effect on the coverage and no effect on the rate.

Reaction Rate

Zero order reactions are rare

First Order Reactions In a first order reaction, the reaction rate is proportional to the concentration or partial pressure of the reagent Rate = k[A] or kPA This means that a constant fraction of the molecules react in a given time period Let’s say that fraction is 10% and that we start off with 1000 molecules

Zero order

Reagent Pressure

Time

Nt

0.1 x Nt

0 1 2 3 4 5 6 7 8 10 11 12 13

1,000 900 810 729 656 590 531 478 430 387 348 313 282

100 90 81 73 66 59 53 48 43 39 35 31

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First Order Reactions Time

Nt

0.1 x Nt

0 1 2 3 4 5 6 7 8 10 11 12 13

1,000 900 810 729 656 590 531 478 430 387 348 313 282

100 90 81 73 66 59 53 48 43 39 35 31

Mathematically, this is an exponential decay, and the products will show an exponential growth

Integrated Rate Law for First Order Reactions 󰇟󰇠   

      󰇟󰇠 

󰇟󰇠   󰇟󰇠 󰇟󰇠



 kt

󰇟󰇠

Constant Fraction Reacts

t





 󰇟󰇠     󰇟󰇠 

     



 

󰇟󰇠   



 

󰇟󰇠  󰇟󰇠  

These expressions are called the integrated rate law

Use log form to calculate k or t, use exponential form to calculate [A]t or [A]0

Many other processes show first order kinetics. What are some examples?

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Second Order Reactions For first order kinetics all molecules have the same probability of reacting in a particular time window. In a second order reaction the rate is given by Rate = k[A][B] or k[A]2 The reaction rate depends on the concentration of A and B or on the concentration of A squared. What does this tell you about the reaction?

Integrated Rate Law for Second Order Reactions 1 1     󰇟󰇠

For A → Products   󰇟󰇠 

󰇟󰇠   󰇟󰇠 󰇟󰇠



󰇟󰇠



t

   

 󰇟󰇠      󰇟󰇠 

1 1      

1 1    󰇟󰇠 

This expression is the integrated rate law for a second order reaction

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Integrated Rate Law for Second Order Reactions For A → Products

  󰇟󰇠  1 1 Integrated rate law:    󰇟󰇠     

For aA + bB → Products

Rate



1 d[ A] a dt

Integrated rate law:



1 d[ B] b dt

kt

k[ A][ B]

1 [ A] t [ B] 0 ln b[ A] 0  a[ B] 0 [ A] 0[ B] t won’t have to use this equation

Pseudo First Order Kinetics For the reaction: A + B → Products If one of the reagents is present in excess, i.e., [B]>>[A] then things get much simpler because [B] is effectively constant. The rate expression simplifies to      

 󰆒    󰇟󰇠

The reaction will now follow first order kinetics and k1' is called the pseudo first order rate constant (with units of . For reactions with [B]>>[A]: 1) Determine k1' using first order kinetics with [A] 2) Then if [B] is known determine k2

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Integrated Rate Laws Summary Zeroth Order Reactions rate = k[A]o = k Integrated rate law:

󰇟󰇠  



 

First Order Reactions rate = k[A] Integrated rate law:

󰇟󰇠  󰇟󰇠  

Second Order Reactions rate = k [A]2 Integrated rate law:

1 1    󰇟󰇠 󰇟󰇠

Determining Reaction Order From Concentration Versus Time Data The following data was obtained for the gas phase decomposition of nitrogen dioxide at 300oC: Time, s 0.0 50.0 100.0 200.0 300.0

[NO2], M 0.01000 0.00787 0.00649 0.00481 0.00381

The balanced equation for the reaction is: NO2(g) → NO(g) + ½O2(g) What is the order of the reaction, and what is the rate constant? use graphs to solve

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Strategy: 1) Make plot of ln[NO2] versus time. If linear, reaction is first order, else 2) Make plot of 1/[NO2] versus time. If linear, reaction is second order, else 3) Make plot of [NO2] versus time. If linear, reaction is zeroth order. Zero = rare

Plot of ln[NO2] against time is not linear, thus the reaction is first order

Plot of 1/[NO2] against time is linear, thus the reaction is Second order

Now determine rate constant For a second order reaction:

1 [ A]t

kt 

1 [ A]o

How do we determine k? From the slope Slope = k

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The Half-life

Concentration or Number

The half-life t½ is the time for the concentration of a reactant to fall by a factor of two. The same concept is used for radioactivity. Here the half-life is the time it takes for the radiation to decrease by a factor of two. For a first order reaction the half-life is

t½

t½

t½

For a second order reaction the half-life depends on the concentration.

The Half-life For a

For a

For a second order reaction the half life becomes longer as the reaction proceeds. Example: The decomposition of hydrogen peroxide is a first order reaction with a rate constant of 1.767x10–5 s–1. What is the half-life? ___ln2___ 1.767x10^-5

This is how long it takes for the initial concentration to drop by a factor of 2. How long does it take to drop by factor of 4? Double

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A Microscopic View of Reactions: Elementary Reactions

This is called the overall reaction

The reaction Ni(CO)4 + P(CH3)3 → Ni(CO)3P(CH3)3 + CO Occurs through two steps that are called elementary reactions: Ni(CO)4 → Ni(CO)3 + CO Ni(CO)3 + P(CH3)3 → Ni(CO)3P(CH3)3

(unimolecular) (bimolecular)

Definition: An elementary reaction is a one step process whose equation describes which molecules collide, and make or break bonds.

There Are Two Basic Types of Elementary Reaction: Unimolecular:

where an individual molecule spontaneously changes its structure or dissociates: A→B A→C+D

Bimolecular:

(isomerization) (dissociation)

where two molecules collide and associate or rearrange into products: A+B→C (association) A + B → C + D (reaction)

More complex reactions can be built up from sequences of elementary reactions. But some reactions occur in a single elementary step

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An Example of an Elementary Unimolecular Reaction

This reaction is exothermic – the trans form is lower in energy than the cis. So why don’t all the cis molecules just instantaneously convert into trans?

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An Example of an Elementary Bimolecular Reaction

The collision geometry is important here, if the I– approached from this direction it is much more difficult to make the reaction occur. Do bimolecular reactions have energy barriers associated with them?

What factors play a role in determining the reaction rate?

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Effect of Temperature on the Reaction Rate The plot shows rate constants as a function of temperature for the reaction: I–(aq) + CH3Br(aq) → ICH3(aq) + Br–(aq)

Reaction rates usually increase when the temperature is raised. Why?

As the temperature increases the kinetic energy increases and so the collision energy increases, and a larger fraction of the collisions lead to reaction:

What is the relationship between the rate constant and temperature? rate constant increases as temp increases but falls off rapidly at higher temps

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The Arrhenius Equation Ae  E A /RT

k Frequency factor

Activation energy

A gives the number of times the barrier is encountered per sec (A is related to collision frequency x steric factor for a bimolecular reaction)

e  EA/ RT is the fraction of the collisions with sufficient energy to overcome the energy barrier, EA. Taking natural logs

ln k

ln A 

EA 1 R T

Plot of lnk against 1/T is linear with a slope of −EA/R and an intercept of lnA.

For the reaction: I–(aq) + CH3Br(aq) → ICH3(aq) + B–(aq) Intercept = lnA = 23.53 A = e23.53 L.mol–1s–1 = 1.66x1010 L.mol–1s–1 Slope = –EA/R = –9.18x103K EA = –Slope x R = 9.18x103K x 8.314J/K.mol = 76.3 kJ/mol

Gas constant in J/K.mol

The Arrhenius equation can now be used to calculate the rate constant at any temperature Temperature in K An example, what is the rate constant at 50oC? E in J/mol

A k = A x exp(–EA/RT) 10 –1 –1 = (1.66x10 L.mol s ) exp(–76,300J/mol/(8.314J/mol.K x 323.15K)) = (1.66x1010 L.mol–1s–1) x e–28.4 = 7.70x10–3 L.mol–1s–1

The same units as k because exp(–EA/RT) is unit-less

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EA and A From Rate Constants at Two Temperatures In some cases you may be provided with rate constants at just two temperatures and asked to determine the activation energy ⁄  ⁄  So we have      and     

We divide the second equ...