Kinetics Chemical kinetics is the study of the time dependence of chemical reactions.

Kinetics

Chemical kinetics is the study of the time dependence of chemical reactions.

Objectives Understand rates of reaction and the

conditions affecting rates Derive the rate equation, rate constant and

reaction order from experimental data Use integrated rate laws Understand the collision theory of reaction

rates and the role of activation energy Relate reaction mechanisms and rate laws

Kinetics

Reaction rate The change in concentration of a

reactant or product per unit time

timeinchange

ionconcentratinchangereactionofRate

t

ARate

][

Rate of Reaction Rate of reaction may be described based on

either the increase in concentration of a product or the decrease in concentration of a reactant per unit time Rates decrease with time Minus sign is required because the concentration

of A decreases with time, and the rate is always expressed as a positive quantity.

Instantaneous rate can be determined by finding the slope of a line tangent to a point representing a particular time.

Rate of Reaction

Decomposition of N2O5

2N2O5 4 NO2 + O2

Rate can be expressed as

t

ONRate

][ 52

Rate of Reaction

Rate could also be expressed in terms of the formation of NO2 or the rate of formation of O2. Rates expressed in these terms will have positive signs.

Formation of NO2 is 2x rate of decomposition of N2O5

Formation of O2 is one half of the rate of decomposition of N2O5

Rate of Reaction

Example Assume the disappearance of N2O5

between 40 min and 55 min is

min15

/12.0

min40min55

)/22.1()/10.1(][ 52 LmolLmolLmol

t

ON

min0080.0 52

L

consumedONmol

Rate of Reaction

Rate in terms of the appearance of NO2

consumedONmol

formedNOmol

L

consumedONmol

t

NORate

52

2522

2

4

min

0080.0][

min016.0 2

L

formedNOmol

Rate of Reaction

Rate in terms of the rate at which O2 is formed

consumedONmol

formedOmol

L

consumedONmol

t

ORate

52

2522

2

1

min

0080.0][

min0040.0 2

L

formedOmol

Rate of Reaction This procedure gives the average rate Graphing concentration vs. time does not

give a straight line because the rate of the reaction changes during the course of the reaction.

The instantaneous rate is given at a single point by drawing a line tangent to the concentration-time curve at a particular time and obtaining the slope of this line.

Rate of Reaction

Rate of Reaction

For this reaction, if the rate is measured at a range of instantaneous concentration during the experiment, a Rate vs. Concentration graph can be plotted. The resulting graph is given below:

Rate of Reaction

Rate of Reaction This graph shows that the rate of reaction is

directly proportional to the concentration of [N2O5]:

Rate = constant x [N2O5]

Rate = k[N2O5]

This expression is called a rate equation, where k is called the rate constant

Rate of Reaction

Summarizing the rate expressions

To equate rates of disappearance or appearance, divide [reagent]/t by the stoichiometric coefficient in the balanced chemical equation

t

O

t

NO

t

ON

][][

4

1][

2

1 2252

Rate of Reaction

PROBLEMGive the relative rates for the disappearance of reactants and formation of products for the following reaction:

4 PH3(g) P4(g) + 6 H2(g)

Rate of Reaction

PROBLEMData collected on the concentration of dye as a function of time are given in the graph below. What is the average rate of change of the dye concentration over the first 2 minutes? What is the average rate of change during the fifth minute (from t = 4 to t = 5)? Estimate the instantaneous rate at 4 minutes.

Rate of Reaction

Rate Laws

Reversibility of Reactions For the reaction 2N2O5 4 NO2 + O2

the reverse reaction may also take place The reverse reaction effects the rate of

change in concentration [N2O5] depends on the difference in the

rates of the forward, kf, and the reverse, kb, reactions

Reaction Conditions and Rate

Molecular collisions required for reaction to take place. Atoms and molecules are mobile (mixture of gases or using solutions of reactants)

Under these conditions several factors affect the rate of reaction

Reaction Conditions and Rate

Concentration Temperature Catalyst Surface area (if reactant is a solid)

Effect of Concentration

Rate Equation Relationship between reactant

concentration and reaction rate is expressed by the rate equation or rate law. For our example, N2O5

Rate of Reaction = k[N2O5]

Effect of Concentration

k = proportionality constant = rate constant at a given temperature

Rate equation indicates reaction rate is proportional to the concentration of the reactant

[N2O5] doubles as the rate doubles

Effect of Concentration

In general,aA + bB xX

Rate equation formRate = k[A]m[B]n

m and n are not necessarily the stoichiometric coeffcients (a and b). The exponents must be determined by experiment. m and n may be positive, negative, fractions or zero

Order of Reaction

Order of a reaction is with respect to a particular reactant is the exponent of its concentration in the rate expression

Total reaction order is the sum of the exponents

Order of Reaction

Example Decomposition of H2O2 in the

presence of iodide ion (clock reaction) Reaction rate = k[H2O2][I-] Reaction is first order with respect to

H2O2 and I-; second order overall.

Order of Reaction

So, what does it mean? Rate doubles if either H2O2 or I- is

doubled and rate increases by a factor of 4 if both concentrations are doubled.

Order of Reaction

Let’s consider the following rate law

Rate = k[NO]2[Cl2]

Reaction rate is second order in NO Reaction rate is first order in Cl2 Overall third order

Order of Reaction

Let’s look at some data

Exp [NO] M [Cl2] M Rate M∙s-1

1 0.250 0.250 1.43 x 10-6

2 0.500 0.250 5.72 x 10-6

3 0.250 0.500 2.86 x 10-6

4 0.500 0.500 11.4 x 10-6

Order of Reaction

Exp 1 & 2 – [Cl2] held constant; [NO] is 2x; reaction rate increases by factor of 4

Exp 1, 3, 4 – Compare 1 & 3; [NO] held constant, [Cl2] 2x; rate doubles compare 1& 4; [NO] and [Cl2] 2x; rate 8x original value

1

4

/1043.1

/1072.5

1

26

6

sLmolx

sLmolx

ExpRate

ExpRate

Rate Constant k

Reaction rates ([A]/t) have units of mol/L∙time when [ ] are given as moles/L

Rate constants must have units consistent with the units for the other terms in the rate equation

Rate Constant k

1st order reactions: units of k are time-

1

2nd order reactions: units of k are L/mol∙time

zero order reactions: units of k are mol/L∙time

Determining a Rate Equation

“Method of Initial Rates” Initial rate (instantaneous reaction

rate at start of reaction (rate at t = 0) Mix reactants then determine

[products]/t or -[reactants]/t after 1% or 2% of limiting reactant has been used up.

Determining a Rate Equation

ExampleCH3CO2CH3(aq) + OH-(aq) CH3CO2

-(aq) + CH3OH(aq)

Initial Concentrations Initial Rxn Rate

Exp [CH3CO2CH3 ] M

[OH-] M M∙s-1

1 0.050 0.050 0.00034

no change x 2 x 2

2 0.050 0.10 0.00069

x 2 no change x 2

4 0.10 0.10 0.00137

Determining a Rate Equation

When initial [reactant] is 2x while other [reactant] held constant, initial rxn rate 2x

Reaction is directly proportional to the concentrations of both reactants

Reaction is first order in each reactantRate = k[CH3CO2CH3][OH-]

Determining a Rate Equation

If rate equation is known, the value of k can be found by substituting values for the rate and concentration into rate equation

Determining a Rate Equation

Using our example Substitute data from one of the

experiments into the rate equation

L

mol

L

molk

sL

molRate

050.0050.000034.0

smolLLmolLmol

sLmolk

/14.0

/050.0/050.0

/00034.0

Integrated Rate Laws

What are they? Equations that describe

concentration-time curvesWhy use them? Calculate a concentration at any

given time Find length of time needed for a given

amount of reactant to react

First Order Reactions

If reaction ‘A products’ is first order, the reaction rate is directly proportional to [A]1

A little calculus…

][][

Akt

A

First Order Reactions

Integrated rate equation

Where [A]0 and [A]t are concentrations of reactants at time t = 0 and at a later time, t

This ratio is the fraction of reactant that remains after a given time has elapsed.

ktA

A t 0][

][ln

First Order Reactions

Negative sign (-) means ratio is less than 1 because [A]t is always less than [A]0.

Logarithm of ratio is negative therefore other side of equation must be negative

First Order Reactions

Equation is useful in three ways: If [A]t/[A]0 is measured after some amount

of time has elapsed then k can be calculated

2. If [A]0 and k are known than [A]t can be calculated (amount remaining after a given amount of time

3. If k is known, then time elapsed until a specific fraction ([A]t/[A]0) remains can be calculated

First Order Reactions

Note k for 1st order reactions is

independent of concentration; k has units of time-1

Therefore any unit for [A]t and [A]0 can be chosen (M, mol, g, atoms, molecules, pressure)

First Order ReactionsPROBLEM

When heated cyclopropane rearranges to propene in a first order process

Rate = k[cyclopropane] k = 5.4 x 10-2 h-1

If the initial concentration of cyclopropane is 0.050 mol/L, how much time (in hours) must elapse for its concentration to drop to 0.010 mol/L

First Order Reactions

SOLUTION

ktnecyclopropa

necyclopropa t 0][

][ln

thx )104.5(]050.0[

]010.0[ln 12

hhxhx

t .30104.5

)61.1(

104.5

)20.0ln(1212

Second Order Reactions

If reaction ‘A products’ is second order, the reaction rate equation is

A little calculus again…

2][][

Akt

A

Second Order Reactions

Integrated rate equation

In this case k is the second-order rate constant (L/mol ∙ time)

ktAA t

0][

1

][

1

Second Order ReactionsPROBLEM

The gas-phase decomposition of HIHI(g) ½ H2(g) + ½ I2(g)

has the rate equation

Where k = 30. L/mol ∙ min at 443C. How much time does it take for the concentration of HI to drop from

0.010 mol/L to 0.0050 mol/L at 443C?

2][][

HIkt

HI

Second Order Reactions

SOLUTION

tmolLLmolLmol

min/.30/010.0

1

/0050.0

1

tmolLmolLxmolLx min/.30/100.1/100.2 22

min3.3t

Zero Order Reactions

If reaction ‘A products’ is zero order, the reaction rate equation is

A little calculus again…

0][][

Akt

A

Zero Order Reactions

Integrated rate equation

Where k has the units mol/L ∙ s

ktAA t ][][ 0

Graphical Methods for Determining Reaction Order and the Rate Constant

First Order

[A]t = - kt + [A]0

y mx b

Graphical Methods for Determining Reaction Order and the Rate Constant

Second Order

1/[A]t = + kt + 1/[A]0

y mx b

Half-Life and First Order Reactions

Half-life t1/2

Time required for the concentration of a reaction to decrease to one-half its initial value; reactant R remaining is ½

Used primarily when dealing with 1st order reactions

Indicates the rate at which a reactant is consumed – is reaction fast or slow?

Longer the half-life – slower the reaction

Half-Life and First Order Reactions

Where:[R]0 = initial concentration

[R]t = concentration after the reaction is half complete

2

1

2

1

00

R

RorRR t

t

Half-Life and First Order Reactions

Let’s evaluate t1/2 for 1st order reaction

Substitute [R]t/[R]0 = ½ and t = t1/2 in integrated rate equation

ktR

R t 0

ln

2/12

1ln kt

Half-Life and First Order Reactions

Rearrange equation (ln 2 = 0.693)

Equations relates half-life to 1st order rate constant

t1/2 is independent of concentration

kt

693.02/1

Half-Life, Zero Order & Second Order Reactions

Zero order reaction

Second order reaction

k

Rt

20

2/1

02/1

1

Rkt

Half-LifePROBLEM Sucrose, C12H22O11, decomposes to fructose

and glucose in acid solution with the rate law

Rate =k[sucrose]k = 0.208 h-1 at 25C

What amount of time is required for 87.5% of the initial concentration of sucrose to decompose?

Half-Life

SOLUTION After 87.5% of sucrose has decomposed,

12.5% remains (fraction remaining = 0.125). This will require 2 half-lives to reach this point.

Time elapsed = 3 x 3.33 h = 9.99 h

hh

t 33.3208.0

693.012/1

Half-LifeYOUR TURN

The following reaction is first order with respect to [NH2NO2]. The rate constant, k, is 9.3 x 10-5 s-1. What is the half-life of this reaction?

NH2NO2(aq) N2O(g) + H2O(l)

The following reaction at 400K is second order with respect to [CF3] and the value of the rate constant, k, is 2.51 x 1010 M-1s-1. If the initial [CF3] = 2.0M, what is the half-life of the reaction?

2 CF3(g) C2F6(g)

Collision Theory of Reaction Rates

Theory states that for any reaction to occur 3 conditions must be met The reacting molecules must collide with

one another The reacting molecules must collide with

sufficient energy to break bonds The molecules must collide in an

orientation that can lead to rearrangement of the atoms

Collision Theory of Reaction Rates

So, molecules must collide with one another.

The rate of their reaction is primarily related to the number of collisions, which is in turn related to their concentration

The number of collisions between the two reactant molecules is directly proportional to the concentration of each reactant, and the rate of the reaction shows a first order dependence on each reactant

Collision Theory of Reaction Rates

Temperature, Reaction Rate, & Activation Energy

How and why does temperature affect reaction rates?

In any sample of gas or liquid some molecules have very low energies, others have high energies, most have intermediate energies.

As temperature increases average energy of the molecules in a sample increases as does the fraction having higher energies

Activation Energy Molecules require some minimum energy to

react; ‘energy barrier’ Energy required to surmount the barrier is

called the activation energy, Ea

Barrier low, energy requirement (Ea) is low, a high proportion of molecules in a sample have sufficient energy to react – reaction is fast

Barrier high, energy requirement (Ea) is high and only a few reactant molecules in a sample have sufficient energy to react – reaction is slow

Activation Energy

Reaction Coordinate Diagram

Consider HI H2 + I2

Reaction can not occur without input of energy

Energy reaches maximum at transition state; sufficient energy has been concentrated in bonds, bonds can break, reaction can move forward.

Activation Energy

Effects of Temperature Increase

Raising the temperature always increases the reaction rate by increasing the fraction of molecules with enough energy to surmount the activation energy barrier

Effects of Molecular Orientation on Reaction Rate

The lower the probability of achieving alignment, the smaller the value of k, the slower the reaction

The Arrhenius Equation Reaction rates depend on energy, frequency of

collision between reactants, temperature, and correct geometry. Arrhenius equation summarizes these factors:

R = gas constant 8.314 x 10-3 kJ/K∙mol A = frequency factor L/mol∙s (related to number of

collision and fraction of collisions w/correct geometry. A is specific to each reaction and temperature dependent

Fraction of molecules having the minimum energy required for reaction. Value is always <1

RTEaAeconstratek /

RTEae /

The Arrhenius Equation

Use to calculate: Value of Ea

Rate constant for a given temperature if Ea and A are known

RT

EAk alnln

The Arrhenius Equation

A little rearrangement…

ATR

Ek a ln

1ln

y mx + b

The Arrhenius Equation

PROBLEM Calculate Ea for the reaction

2N2O (a) 2N2 (g) + O2 (g)

Exp T (K) K (L/mols)

1 1125 11.59

2 1053 1.67

3 1001 0.380

4 838 0.0011

The Arrhenius Equation

Kinetics problems that deal with changing rates, or rate constants (k), and temperature changes require use of the Arrhenius equation

121

212

11lnlnln

TTR

E

k

kkk a

The Arrhenius Equation

PROBLEM What is the activation energy of a

reaction that has a rate constant of 2.50 x 102 kJ/mol at 325K and a rate constant of 5.00 z 102 kJ/mol at 375K?

One - step, irreversible, unimolecular reactions One - step, irreversible, bimolecular reactions One - step, reversible reactions

Catalysis Catalyst increases

the rate of a reaction without being consumed by it.

Catalyst lowers the activation energy required thus speeding up the reaction

Heterogeneous & Homogeneous Catalysis Heterogeneous – catalyst is in a different phase from

the reaction mixture; generally a solid that increases the rate of a gas phase or liquid phase reaction

Au

N2O(g) N2(g) + ½ O2(g)

N2O is chemically absorbed on the surface of the solid catalyst.

A bond is formed between the N2O molecule and Au (covalent bonds) thus weakening the bond between nitrogen and oxygen making it easier for N2O to break apart.

Heterogeneous & Homogeneous Catalysis

N N – O(g) + Au(s) N N --- O --- Au(s) N N(g) + O2(g) + Au(s)

Heterogeneous catalyst generally used in industrial processes: Preparation of

ammonia sulfuric acid methanol

Pt catalyst used to reduce automobile emissions

Heterogeneous & Homogeneous Catalysis

Homogeneous – catalyst that is present in the same phase as the reactants.

Speeds up reaction by forming a reactive intermediate that decomposes to give products thus providing an alternative path of lower activation energy for the reaction

Heterogeneous & Homogeneous Catalysis

Clock reaction (decomposition of hydrogen peroxide)

Uncatalyzed2H2O2 (aq) 2H2O + O2 (g)

CatalyzedStep 1:H2O2 (aq) + I- (aq) H2O + IO- (aq)

Step 2:H2O2 (aq) + IO- (aq) H2O + O2 (g) + I- (aq)

2H2O2 (aq) 2H2O + O2(g)

Reaction Mechanisms

Sequence of bond-making and bond-breaking steps during a reaction

A ‘guess’ to help better understand the chemistry

Nano level (atoms and molecules) Use rate equation to understand

mechanisms

Reaction Mechanisms

Most reactions occur in a sequence of steps

Br2(g) + 2 NO(g) 2 BrNO(g) Written as a single step, this reaction

would require the 3 reactants to collide simultaneously with the right velocity and orientation to react.

Low probability of this happening

Reaction Mechanisms

Using a sequence of steps involving only one or two molecules increases the probability of a collision (production of an intermediate)

Step 1 Br2(g) + NO(g) Br2NO(g)

Step 2 Br2NO(g) + NO(g) 2 BrNO(g)

Br2(g) + 2 NO(g) 2 BrNO(g)

Reaction Mechanisms

Each step of mechanism is referred to as an elementary step Describes a single molecular event Each step has its own Ea and k Steps must add up to give overall

balance chemical equation

Molecularity

Classification of elementary steps based on # of reactant molecules colliding Unimolecular – one molecule in an

elementary step O2(g) O2(g) + O(g) Bimolecular – two molecules (same or

different)  O2(g) + O(g) 2 O2(g)

Molecularity

Termolecular – three molecules (same or different)

O(g) + O2(g) + N2(g) O3(g) + energetic N2(g)

Molecularity

Elementary Step Molecularity Rate Equation

A product Unimolecular Rate = k[A]

A + A product Bimolecular Rate = k[A]2

A + B product Bimolecular Rate = k[A][B]

2 A + B product Termolecular Rate = k[A]2[B]

Molecularity

The rate equation of an elementary step is defined by the reaction stoichiometry. The rate equation of an elementary step is given by the product of the rate constant and the concentrations of the reactants in that step. 

The molecularity of an elementary step and its order are the same. This is not true for the overall reaction.

Rate Equations & Mechanisms

Products of a reaction can never be produced at a rate faster than the rate of the slowest step

Slowest step = Rate-determining step

Rate Equations & Mechanisms

For the slowest step Rate = k1[A][B] Note: the rate law must be written with respect

to the reactants only

MXBAk

elEaSlow

1

arg,

YAMk

smallEaFast

2

,

YXBA 2