principles of chemistry ii chem 1212 chapter 13

PRINCIPLES OF CHEMISTRY II

CHEM 1212

CHAPTER 13

DR. AUGUSTINE OFORI AGYEMANAssistant professor of chemistryDepartment of natural sciences

Clayton state university

CHAPTER 13

CHEMICAL KINETICS

- Chemical reactions occur when reactant species strike each other and interact to form products

Reaction kinetics is studied to

- improve production of materials- increase quality and quantity of products

- increase energy efficiency- minimize pollution

etc

RATES OF REACTIONS

RATES OF REACTIONS

Rate = change per unit time

Rate of reaction = change in concentration per unit time

t

cRate

For a chemical reaction

Reactant → Product

- Rate at which reactants are consumed or products are formed in a given period of time is given as

RATES OF REACTIONS

Δt

]Δ[reactant

Δt

Δ[product]Rate

Units: M/s

Square brackets represent molar concentrations [reactant] = reactant concentration[product] = product concentration

Rate of appearance of product = rate of disappearance of reactant

- Reactant concentration decreases during reaction∆[reactant] is negative

- Product concentration increases during reaction∆[product] is positive

- Rate is always positive

- Rate can be measured by following the concentrations of reactants or products

RATES OF REACTIONS

- Rate of reaction is generally not constant

- Rate of reaction changes over the course of reaction

- Concentration of reactants or products are measured at regulartime intervals

- A graph of concentration vs time may be plotted

- Instantaneous rate at a given time is the slope of the tangent to the curve at that time

- Average rate is measured rate over a time interval

INSTANTANEOUS AND AVERAGE RATES

INSTANTANEOUS AND AVERAGE RATES

∆x

∆y

x

yslopeRateousInstantane

- Rate depends on stoichiometry of the reaction

- Rate is the ratio of rate of change of a substance to its coefficient

Consider the reaction

2HBr(g) → H2(g) + Br2(g)

2 mol HBr : 1 mol of each product

REACTION STOICHIOMETRY

Δt

Δ[HBr]x

2

1

Δt

]Δ[H

Δt

]Δ[BrreactionofRate 22

For the decomposition of HBr

2HBr(g) → H2(g) + Br2(g)

If HBr concentration is decreasing at a rate of 0.52 M/sWhat is the rate of the reaction?

What is the rate of appearance of H2 and Br2?

REACTION STOICHIOMETRY

M/s0.27M/s)0.52(x2

1

Δt

Δ[HBr]x

2

1reactionofRate

M/s0.27HBrmol2

BrorHmol1xHBr)M/s0.52(

Δt

]Δ[H

Δt

]Δ[BrreactionofRate 2222

Factors Affecting Rate of Chemical Reaction

- Concentration of reactants

- Reaction temperature

- Physical nature of reactants

- Catalysts

RATES OF REACTIONS

Concentration of Reactants

- An increase in the concentration of reactants causes an increase in the rate of reaction

- Collisions are more frequent in a given time for higher concentrations

RATES OF REACTIONS

Reaction Temperature

- An increase in temperature of a system increases the average kinetic energy of the reacting molecules

- An increase in kinetic energy results in an increase in collisions in a given time

- The rate of a chemical reaction normally doubles for every 10 oC raise in temperature

RATES OF REACTIONS

Physical State of Reactants: solid, liquid, or gas

solid-state reactants

liquid-state reactants

gaseous-state reactants

< <

Increasing rateof reaction

RATES OF REACTIONS

Physical State of Reactants: solid, liquid, or gas

- Most frequent collisions occur in the gaseous state (the most freedom of movement of particles)

Solid-State Particle Size- Smaller particles have larger surface area and

higher reaction rates

- Extremely small particles may result in very fast reaction rates and may lead to explosion

RATES OF REACTIONS

Catalysts

-Catalysts increase the rate of a reaction without being used up

RATES OF REACTIONS

- Rate of reaction is strongly influenced by concentrationsof reacting species

- Rate is proportional to the product of the concentrations of the reactants each raised to some power

aA + bB → cC + dD

Rate = k[A]x[B]y

x and y are usually positive integersk = rate constant

RATE LAW

aA + bB → cC + dD

Rate = k[A]x[B]y

- x and y are not necessarily coefficients of A and B- x and y are the orders of the reaction

- Described as xth order in A and yth order in B

If x = 1 and y = 2The reaction is first order in A and second order in B

Overall order = 1 + 2 = 3

RATE LAW

Example

For the reaction

2NO2(g) + F2(g) → 2NO2F(g)

Rate = k[NO2][F2]

The reaction is first order in NO2 and first order in F2

RATE LAW

The decomposition of nitrosyl chloride was studied:2NOCl(g) ↔ 2NO(g) + Cl2(g)The following data were obtained

INITIAL RATE OF REACTION

[NOCl]0 (molecules/cm3)

3.0 x 1016

2.0 x 1016

1.0 x 1016

4.0 x 1016

Initial Rate (molecules/cm3·s)

5.98 x 104

2.66 x 104

6.64 x 103

1.06 x 105

What is the rate law? Calculate the rate constant

Rate = k[NOCl]2, k = 6.64 x 10-29 cm3/molecules∙s

The reaction below was studied at -10 oC2NO(g) + Cl2(g) → 2NOCl(g)The following data were obtained

INITIAL RATE OF REACTION

[NO]0 (mol/L)

0.100.100.20

Initial Rate (mol/L)

0.180.361.45

What is the rate law? Calculate the rate constant

Rate = k[NO]2[Cl2], k = 1.8 x 102 L2/mol2

[Cl2]0 (mol/L)

0.100.200.20

- Rate of reaction decreases with time

- Rate of reaction eventually goes to zero

- Concentrations of reactants decrease

- Concentrations of products increase

CONCENTRATION AND TIME

- Rates are independent of the concentrations of the reactants

R → product

Rate = k[R]0

Rate = k

- Called differential rate law

Unit of k = unit of reaction rate = M/s

ZERO-ORDER RATE LAW

- Graph of concentration vs time

ZERO-ORDER RATE LAWC

once

ntr

atio

n

Time

- Rates are independent of the concentrations of the reactants

R → product

[R]t = [R]0 - kt

- Called integrated rate law

Unit of k = unit of reaction rate = M/s

ExamplesMetabolism of ethyl alcohol in the body

Biochemical reactions involving enzymes

ZERO-ORDER RATE LAW

- Graph of concentration vs time is a straight line

ZERO-ORDER RATE LAWC

once

ntr

atio

n

Time

Slope = −kIntercept = [R]0

- A large value of k implies a fast reaction

- The half-life (t1/2) is also used to describe the speed of a reaction

- Half-life is the time needed for the concentration of a reactant to decrease to half its original value

- A short half-life indicates a fast reaction

HALF - LIFE

ZERO-ORDER RATE LAW

At t = 0Initial concentration = [R]0

At half-life t = t1/2

[R]t = ½[R]0

- Substitute in zero-order equation and simplify

HALF - LIFE

ZERO-ORDER RATE LAW

HALF - LIFE

- Using the zero-order rate equation

[R]t = [R]0 – kt

- Simplifying gives

ZERO-ORDER RATE LAW

2k

[R]t 0

1/2

- Half-life for zero-order depends on concentration

ZERO-ORDER RATE LAW

The reaction A → B + C

is known to be zero order in A and to have a rate constant of 5.0 x 10-2 mol/L·s at 25 oC. An experiment was run at 25 oC

where [A]0 = 1.0 x 10-3 M.

a) What is the integrated rate law for this reaction?b) Calculate the half-life for the reaction.

c) Calculate the concentration of B after 5.0 x 10-3 s has elapsed.

a) [A] = [A]0 - ktb) 1.0 x 10-2 sc) 2.5 x 10-4 M

- Rate is proportional to the concentration of the reactant

R → product

FIRST-ORDER RATE LAW

k[R]Δt

Δ[R]Rate

- Called the differential form of the rate law

- Relates differences in concentration and time

Unit of k = s-1

- A graph of concentration vs time describes an exponential decay

FIRST-ORDER RATE LAWC

once

ntr

atio

n

Time

- Rate is proportional to the concentration of the reactant

R → product

FIRST-ORDER RATE LAW

- Called the integrated form of the rate law (describes an exponential decay)

- Relates instantaneous concentrations

[R]t = concentration of R at any time[R]0 = initial concentration at t = 0

e = base of natural logarithms ≈ 2.718

kt0t e[R][R]

FIRST-ORDER RATE LAW

From the first-order rate equation

ktln[R]ln[R] 0t

kt0t e[R][R]

Take natural logarithm on both sides and simplify

kt[R]

[R]ln

0

t

or

FIRST-ORDER RATE LAWln

[Con

cen

trat

ion

]

Time

Slope = −kIntercept = ln[R]0

A graph of ln[R]t vs time is a straight line

At t = 0Initial concentration = [R]0

At half-life t = t1/2

[R]t = ½[R]0

Substitute in first-order equation and simplify

HALF - LIFE

kt[R]

[R]ln

0

t kt[R]

[R]ln

0

t

FIRST-ORDER RATE LAW

HALF - LIFE

kt[R]

[R]ln

0

t kt[R]

[R]ln

0

t

kt0t e[R][R]

k

.6930t1/2

From the first-order rate equation

Substitute and simplify

FIRST-ORDER RATE LAW

- Half-life of a first-order reaction is independent of the concentration of the reactant

- Depends on only the rate constant (k)

- Constant half-life from concentration vs time plot indicates first-order reaction

ExampleRadioactive decay processes

HALF - LIFE

kt[R]

[R]ln

0

t kt[R]

[R]ln

0

t

FIRST-ORDER RATE LAW

The radioactive isotope 32P decays by first-order kinetics and has a half-life of 14.3 days. How long does it take for 95% of

a sample of 32P to decay?

k = 0.0485 1/dayt = 61.8 days

FIRST-ORDER RATE LAW

kt[R]

[R]ln

0

t kt[R]

[R]ln

0

t

A first-order reaction is 75.0% complete in 320 second.a) What are the first and second half-lives for this reaction?

b) How long does it take for 90% completion?

a) 160 s for both first and second half-livesb) 532 s

FIRST-ORDER RATE LAW

kt[R]

[R]ln

0

t kt[R]

[R]ln

0

t

Calculate the half-life of a first order reaction if the concentration of the reactant is 0.0451 M at 30.5 seconds after

the reaction starts and is 0.0321 M at 45.0 seconds after the reaction starts. How many seconds after the start of the

reaction does it take for the reactant concentration to decrease to 0.0100 M?

a) 29.5 sb) 94.9 s

kt[R]

[R]ln

0

t kt[R]

[R]ln

0

t

FIRST-ORDER RATE LAW

- Rate is proportional to the concentration of the reactant raised to the second power

R → product

SECOND-ORDER RATE LAW

- Called the differential form of the rate law

- Relates differences in concentration and time

Unit of k = M-1s-1 or L/mol·s

2k[R]Δt

Δ[R]Rate

- Graph of concentration vs time

SECOND-ORDER RATE LAWC

once

ntr

atio

n

Time

- Rate is proportional to the concentration of the reactant raised to the second power

R → product

SECOND-ORDER RATE LAW

- Called the integrated form of the rate law

Unit of k = M-1s-1 or L/mol·s

kt[R]

1

[R]

1

0t

- A graph of 1/concentration vs time is a straight line

SECOND-ORDER RATE LAW1/

[Con

cen

trat

ion

]

Time

Slope = kIntercept = 1/[R]0

HALF-LIFE

Half-life depends on starting concentration

01/2 k[R]

1t

SECOND-ORDER RATE LAW

For the reaction A → products

successive half-lives are observed to be 10.0, 20.0, and 40.0 min for an experiment in which [A]0 = 0.10 M. Calculate the

concentration of A at a) 30.0 minb) 70.0 minc) 80.0 min

a) 0.025 Mb) 0.013 Mc) 0.011 M

kt[R]

[R]ln

0

t kt[R]

[R]ln

0

t

SECOND-ORDER RATE LAW

SECOND-ORDER RATE LAW

Consider the following initial rate data for the decomposition of compound AB to give A and B

[AB]0, mol/L: 0.200 0.400 0.600Initial rate, mol/L·s: 3.20 x 10-3 1.28 x 10-2 2.88 x 10-2

Determine the half-life for the decomposition reaction initially having 1.00 M AB present

Rate = k[AB]2

k = 0.0800 L/mol·st1/2 = 12.5 s

RATE AND TEMPERATURE

- Almost all reactions go faster at higher temperatures

- The rate of most reactions increase at increasing temperature

- The order of the reaction usually does not change with temperature

RATE AND TEMPERATURE

Example

For the reactionNO(g) + O3(g) → NO2(g) + O2(g)

The rate constant increases with increasing temperature

k (L

/mol

·s)

T (K)

COLLISION THEORY

- Explains the rate of reactions in terms of molecular-scale collisions

- The basic assumption is that molecules must collide to react

- The collision frequency (Z) is the number of collisions per second

- Z depends on the concentrations of the reacting species

COLLISION THEORY

- The collision frequency (Z) between two molecules is proportional to the product of their concentrations

- For two reacting molecules XY and AB

Z α [XY][AB]

Z = Z0[XY][AB]

Z0 = is the proportionality constant

Z0 depends on sizes and speed of reacting species

COLLISION THEORY

- Collision frequency increases with increasing temperature as molecules move faster

- However, the increase in collision frequency cannot accountfor the temperature dependence of reaction rate

- Not every collision results in a chemical reaction

ACTIVATION ENERGY (Ea)

- Not all collisions result in the formation of products (by Svante Arrhenius)

- Molecules must collide with enough energy to rearrange the bonds

- Molecules bounce off if the total energy of colliding species is not enough

- Activation energy (Ea) is the minimum collision energy required for a reaction to occur

THE ACTIVATED COMPLEX

- Is a transition state

- The least stable or highest energy transition state

- Very unstable and concentration is extremely small

- The energy needed to from the activated complex from the reactants is the activation energy

- Reactions with high activation energies are generally slower than reactions with low activation energies

Reaction coordinate

pote

nti

al e

nerg

y

ENERGY LEVEL DIAGRAM

Products

Activated complex

ReactantsEa

EFFECT OF TEMPERATURE

- The effect of temperature on reaction rate is influenced by the magnitude of the activation energy

- The number of molecules with high enough kinetic energies to initiate a reaction is directly related to temperature

- The fraction of collisions (fr) with energy in excess of Ea

/RTEr

aef

R is the gas constant = 8.314 J/mol·KT is the temperature in Kelvin

EFFECT OF TEMPERATURE

- fr is between 0 and 1

- fr gets closer to 1 as T increases

- Ea does not change with T

- Number of collisions exceeding Ea increases exponentially with T

Energy

Fra

ctio

n

High Temperature Gas

Low Temperature Gas

ENERGY DISTRIBUTION IN GAS MOLECULES

Ea

EFFECT OF TEMPERATURE

Rate of reaction = (collision frequency) x (fraction exceeding Ea)

Rate = Z x fr

Z = Z0[XY][AB]

Experimental rate = k[XY][AB] /RTE

oa[XY][AB]eZratePredicted

/RTEo

aeZk

k is the rate constant

STERIC FACTOR

- The expression predicts faster rates than experimentally observed

- Not all collisions with energies greater than Ea result in a reaction

- The correct orientation of reactants is an important factor

- The steric factor (p) expresses the need for the correct orientation

Rate = (steric factor) x (collision frequency) x (fraction exceeding Ea)

/RTEo

a[XY][AB]eZRate p

STERIC FACTOR

A = pZo

A is known as the pre-exponential term

/RTEa[XY][AB]eRate A

/RTEaek A

The Arrhenius equation

- A includes the steric factor and cannot be predicted by theory- A can only be determined by experiment

THE ARRHENIUS EQUATION/RTEaek A

Take natural log of both sides

/RTEAlnkln a

For an Arrhenius plot

- That is a graph of ln k versus 1/T

Slope = -Ea/R

Intercept = ln A

THE ARRHENIUS EQUATION

/RTEaek A

Consider rate constants k1 and k2 at temperatures T1 and T2

21

a

2

1

T

1

T

1

R

E

k

kln

THE ARRHENIUS EQUATION

The activation energy for the decomposition of HI(g) to H2(g) and I2(g) is 186 kJ/mol. The rate constant at 555 K is

3.52 x 10-7 L/mol·s. What is the rate constant at 645 K?

9.60 x 10-5 L/mol·s

THE ARRHENIUS EQUATION

A first order reaction has rate constant of 4.6 x 10-2 s-1 and 8.1 x 10-2 s-1 at 0 oC and 20 oC, respectively. What is the value of the activation energy?

Ea = 19 kJ/mol

THE ARRHENIUS EQUATION

A certain reaction has an activation energy of 54.0 kJ/mol.As the temperature is increased from 22 oC to a higher temperature,

the rate constant increases by a factor of 7.00. Calculate the higher temperature.

T2 = 324 K or 51 oC

CATALYSIS

Rate of reaction can be increased in two ways

1) Increase the temperature

2) Reduce the activation energy or increase the steric factor(addition of catalyst)

- A catalyst is a substance that increases the rate of reaction but is not consumed in the reaction

- A catalyzed reaction generally has lower activation energy

- Catalysts increase the rate of a reaction without being used up

- Provide alternative reaction pathways with lower activation energies

Uncatalyzed reaction: X + Y → XY

Catalyzed reaction: Step 1 X + C → XCStep 2 XC + Y → XY + C

CATALYSIS

Reaction pathway

pote

nti

al e

nerg

y uncatalyzed activation

energy

catalyzed activation

energy

CATALYSIS

HOMOGENEOUS CATALYSIS

- Present in the same phase as the reactants

ExampleN2(g) + O2(g) → 2NO(g)

The formation of ozone

HETEROGENEOUS CATALYSIS

- Present in a different phase from the reactants

ExamplesUse of solid metal catalysts such as platinum, nickel, palladium, titanium

Use of platinum catalyst for the production of methanol from hydrogen and carbon monoxide

2H2(g) + CO(g) → CH3OH(g)

- Catalysts can determine the nature of products formedPlatinum catalyst produces methanol

Nickel catalyst produces methane and water

ENZYME CATALYSIS

- Enzymes are large molecules that catalyze specific biochemical reactions

- An enzyme is specifically tailored to facilitate a given reaction

- Enzymes increase the rate of reaction by increasing the steric factor rather than decreasing the activation energy

- Enzymes are generally named after the reactions they catalyze(that is their functions)

ExamplesCarboxypeptidase-A, Alcohol dehydrogenase (ADH)

COLLISIONS BETWEEN MOLECULES

- The sequence of steps leading from reactants to products is known as the reaction mechanism

- Some reactions require only one step (a single collision)

- Other reactions require more than one collisions leading to the formation of intermediates

INTERMEDIATES

- Compounds that are produced in one step and consumed in another

- Not observed among the products of the reaction

- Differ from activated complex

- An intermediate is in a shallow minimum in the energy level diagram

- An activated complex occurs at the maximum in the energy level diagram

Reaction coordinate

pote

nti

al e

nerg

yENERGY LEVEL DIAGRAM

Products

Intermediates

Reactants

ELEMENTARY STEP

- Chemical equation that describes an actual molecular-level event

- The overall reaction is the sum of the elementary reactions

ExampleNO2 + NO2 → NO3 + NO step 1NO3 + CO → NO2 + CO2 step 2

NO2 + CO → NO + CO2 overall

NO3 is an intermediate

RATE LAW FOR ELEMENTARY STEP

- Rate law of an elementary step can be written directly from the stoichiometry of that step

Consider an elementary stepiA + jB → products

Rate = k[A]i[B]j

- The rate law for an overall reaction cannot be determined from the stoichiometry

Molecularity- The number of species involved in a single elementary step

Unimolecular Step- Involves the spontaneous decomposition of a single molecule

- First-order rate law describes the kineticsHCl → H + Cl

Bimolecular Step- Involves the collision of two species

- Second-order rate law describes the kineticsNO2 + NO2 → N2O4

RATE LAW FOR ELEMENTARY STEP

Termolecular Step- Involves the collision of three species

- Third-order rate law describes the kinetics- Uncommon

NO2 + NO + O2 → NO3 + NO2

- Collisions involving four or more species are very rare

RATE LAW FOR ELEMENTARY STEP

RATE-LIMITING STEP

- The slowest elementary step in a given reaction

- The rate of a chemical reaction is limited by the rate of the slowest step

- The rate law of the slowest step is consistent with the experimental rate law of the overall reaction

Example2NO ↔ N2O2 fast, reversible

N2O2 + Cl2 → 2NOCl slow step (rate limiting)

COMPLEX REACTION MECHANISMS

- Reactions in which the rate limiting step is not the first step

- Reaction rate may depend on intermediates

- Intermediates are unstable and their concentrations are difficult to measure

- Rate laws are not written in terms of intermediates

- Other complex reactions contain rapid and reversible steps before the rate-limiting step

ENZYME METABOLISM

- Many enzyme catalyzed reactions follow the Michaelis-Menten mechanism

E + S ↔ ES → E + P

- Rate of reaction is zero order in substrate (S)

Substrate- The compound on which the enzyme acts

- Product (P) does not bind to enzyme (E)- First step is fast and reversible

- Second step is irreversible