CHEMICAL KINETICS
Four factors affect reaction rates

Concentration of reactants

Temperature of reaction

Presence or absence of a catalyst

Surface area of liquid or solid reactants
Reaction Rates

Reaction speed defined as the change that occurs per unit time o It is determined by measuring the disappearance of a reactant or the appearance of a product o The speed of a reaction is the reaction rate

For reaction A

B : Avg rate = change in the number of moles of B / change in time o 
B= moles of B at final time- moles of B initial o At t= 0, there is 1 mole of A and no B o At t= 10 min, there is .74 mol A and .26 mol B o At t=20 min, there is.54 mol A and .46 mol B o Avg rate=

B/

T= Mol B at 10min- Mol B at 0 min/
10min –0 min= .026mol/min o Or Avg rate= -

Mol A /
 t ; note the negative sign. It means reactants are disappearing
Rates in Terms of Concentration

The most useful measure is molarity

Since volume is constant, molarity and moles are directly proportional

Units for average rate are mol/L sec or M/sec

The average rate of a reaction decreases over time. Why?

The rate at any specific point in time is the instantaneous rate

If we plot the concentration of a reactant vs time, the rate at any instant is the slope of the straight tangent to the curve at that point
Stoichiometry

For the reaction :
C
4
H
9
Cl + H
2
O

C
4
H
9
OH + HCl
Rate= -

[C
4
H
9
]/
 t =

[C
4
H
9
OH] /
 t

What if stoichiometry is not one to one?

2HI

H
2
+ I
2
Rate = -

[HI]/ 2
 t =

[H
2
]/
 t =

[I
2
]/
 t
Generalized : For the reaction aA + bB

cC + d D
Rate = -

[A]/a
 t = -

[B]/ b
 t=

[C]/c
 t =

[D]/d
 t
Dependence of rate on concentration

In general, rate increases with increased concentration of
Reactants and decreases with decreasing concentration

We often examine the effect of concentration by measuring the reaction rate at the beginning of a reaction

Consider A + B

C + D
We find that if we hold [B] constant and double [A], the rate doubles. Therefore , the rate is directly proportional to [A].
If we hold [A] constant and double [B], then the rate doubles. Therefore, the rate is directly proportional to [B]

The overall concentration dependence is given by the rate law. For this example the rate law is:
Rate = k [A] [B] where k is the rate constant
Reaction Order

For any reaction with rate law: o Rate= k[A] m [B] n
, the exponents are called reaction orders. The overall reaction order is the sum of the exponents. o This reaction is first order in[A] and first order in [B]
And second order overall

Reaction orders must be determined experimentally o They do not necessarily correspond to stoichiometric coefficients o They are typically 0,1 or 2

o However , they can be fractional or negative
Units of rate constants

Units of the rate constant depend on the overall reaction order

For a second order reaction:
Units of rate = (Units of rate constant) (units of concentration) (units of concentration)

Therefore : Units of rate constant= Units of rate/ (Units of concentration) 2 = M/sec/M 2 = M -1 S -1
Using Initial Rates to determine rate laws

To determine the rate law, we observe the effect of changing the initial concentrations o If the reaction is zero order, changing the concentration has no effect on rate o If the reaction is first order, doubling the concentration will double the rate( change is directly proportional) What if concentration is halved? o If the reaction is second order, doubling the concentration will result in a 4X increase (2 2 ).
Tripling the concentration will cause a 9X increase o A reaction is nth order if doubling the concentration causes a 2 n increase in rate o The reaction rate, NOT THE RATE CONSTANT, depends on concentration.
First order Reaction
 ln[A] t
= -kt + ln[A]
0 is the equation for a first order reaction. Note the similarity to y = mx + b. In this case , a plot of ln[A] t
vs t yields a straight line with a slope –k and an intercept of ln[A]
0

Half-life is the time required for the initial concentration of a reactant to decrease to one half its original value

T
1/2
= .693/k Note that the half life of a first order reaction is independent of the initial concentration of the reactant
Second order reactions

1/ [A] t
= kt + 1/[A]
0
A plot of 1/[A] t
versus t is a straight line with a slope k and an intercept of 1/[A]
0

For a second order reaction a plot of ln[A] t is not a straight line

T
1/2
= 1/k[A]
0
Half life is dependent on initial concentration


Note that a second order process can have the form : Rate= k[A] [B] . The reaction is second order overall but first order in reactant concentration
Temperature and Rate

As temperature increases, the rate increases

Since the rate law does not have temperature in it, the rate constant must be temperature dependent
The Collision Model

Rates of reaction are effected by concentration and temperature

The collision model, which is based on kinetic molecular theory, explains this o In order for molecules to react they must collide o The greater the number of collisions the faster the rate o If you have more molecules the probability of collision increases and the frequency of collision o Higher temperature ( higher KE ) gives the molecules higher energy and increases the number of collisions(moving faster) o However , not all collisions lead to reaction. Only a small fraction of collisions lead to reaction
Activation Energy

Arrhenius: molecules must possess a minimum amount of energy to react. Why? o To form products , bonds in the reactants must be broken o Breaking bonds requires energy o Molecules moving too slowly, KE too low, do not react

Activation energy, E a is the minimum energy required to start a reaction o E a varies with the reaction

Consider the rearrangement of methyl nitrile to acetonitrile o Energy is required to stretch the bond between the methyl group and the nitrile group and to allow the nitrile group to rotate o The carbon carbon bond forms o The energy associated with the molecule drops

o The energy level between the starting molecule and the highest energy state found along the reaction pathway is the activation energy.
 The chemical species at the top of the barrier is the activated complex or the transition state

The energy change for the reaction is the difference in energy between the products and the reactants o 
E rxn
has no effect on reaction rate

The activation energy is the difference in energy between the reactant and the activated complex. o Rate depends on activation energy. The lower E a
, the faster the reaction

How does this relate to temperature o At any particular temperature, molecules present have an average kinetic energy associated with the population o In the same population , some molecules have more energy than average , some have less.

The fraction of molecules with energy equal to or greater than E a is given by f = e -E a
/RT
 As we increase temperature, f becomes larger
Orientation Factor

Orientation has a major effect on whether or not a reaction will occur

Consider the reaction between NOCl and Cl
Cl + NOCl

NO + Cl
2

If the Cl collides with the Cl the reaction proceeds
 If the Cl collides with the O, no reaction occurs
Arrhenius Equation

Arrhenius discovered that most reaction rate data obeys an equation based on 3 factors o Number of collisions per unit time o Fraction of collisions with the correct orientation o Fraction of molecules that have energy greater than or equal to E a
 lnk = -E a
/RT + ln A where k is the rate constant, E a
is the activation energy, R is the gas constant(8.314J/Kmol) T is the temperature in K


A is the frequency factor. It is related to the frequency of collision and the probability that the orientation will be correct

Both A and E a
are specific to a reaction

Rearranging the equation we get lnk = -E a
/RT + lnA. Once again , if you graph the lnk vs 1/T you get a straight line with a slope –E a
/R
Reaction Mechanisms

Balanced chemical equations provide information about substances present at the beginning and end of a reaction

The reaction mechanism is the process by which the reaction occurs
Elementary Steps

Any process that occurs in a single step

The number of molecules present in an elementary step gives the molecularity o Unimolecular-one molecule o Bimolecular- two molecules o Termolecular- three molecules

A multistep mechanism consists of a series of elementary steps o Elementary step must give a balanced reaction o Some multistep reactions include intermediates

Species that appear in an elementary step but not as a product or reactant

Intermediates are formed in one elementary step and consumed in another
 They are not found in the balanced equation for the reaction
Rate Laws for Multistep Reactions

Rate laws of the elementary steps determine the overall rate law

Rate law of the elementary step is determined by molecularity o Uni- processes are first order o Bi- processes are second order o Ter- processes are third order

Most reactions have more than one elementary step o Often one step is much slower

o This step limits the reaction rate and governs the rate law
Consider: NO
2
+ CO

NO + CO
2

The experimentally derived rate law is : rate = k[NO
2
] 2

A proposed mechanism is o Step 1 : NO
2
+ NO
2

NO
3
+ NO slow
 o Step 2 : NO
3
+ CO

NO
2
+ CO
2
fast
Note NO
3
is an intermediate

The rate law for the slow step is rate = k[NO
2
] 2

This theoretical rate law is in agreement with the experimental

This supports(does not prove) a proposed mechanism
Mechanisms with an initial fast step

Consider the reaction 2NO + Br
2

2NOBr

The experimentally determined rate is : Rate = k[NO] 2
[Br
2
]

Proposed Mechanism o Step 1 : NO + Br
2

NOBr o Step 2 : NOBr
2
2
fast
+ NO

2NOBr slow o The theoretical rate law for the slow step is : rate = k
2
[NOBr
2
] [NO]

Problem: this rate law depends on the concentration of an intermediate o Intermediates are usually unstable and have low/unknown concentrations o We can express [NOBr
2
] in terms of NOBr and Br
2 by assuming an equilibrium in Step 1 o In a dynamic equilibrium the e forward rate equals the reverse rate
 Therefore we get k
1
[NO] [Br
2
] = k
-1
[NOBr
2
]

Rearranging : [NOBr
2
] = k
1
/k
-1
[NO] [Br
2
]

Substituting in the rate law : rate = k
2
k
1
/k
-
1
[NO] [Br
2
] [NO] = k[NO] 2 [Br
2
]
 This is consistent with the experimental rate law
Catalysis

A catalyst is a substance that changes the rate of a reaction without undergoing a permanent change


There are 2 types of catalysts : homogeneous and heterogeneous
Homogeneous Catalysts

A catalyst that is present in the same phase as the reactants

How do catalysts increase rates? o By lowering the activation energy o Increasing the number of collisions o Catalysts provide a completely different mechanism for reaction
Heterogeneous Catalysts

A catalyst that exists in a different phase from the reactants.
Ie. Solid catalyst and gas reactants: the catalytic convertor on a car