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Chapter 16 Kinetics. Dinosaurs generated enough heat to sustain its biochemical reactions at high rates . Reaction rate = f(temperature). What is chemical kinetics?. Deals with the speed (rate) of a chemical reaction and its reaction mechanism .

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Chapter 16 Kinetics

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Chapter 16 kinetics l.jpg

Chapter 16Kinetics

Dinosaurs generated enough heat to sustain its biochemical reactions at high rates.

Reaction rate = f(temperature)


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What is chemical kinetics?

Deals with the speed (rate) of a chemical reaction and its reaction mechanism.

Describes the change in concentration as a function of time.

Quantitatively……

Qualitatively……


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Section 16.1: Factors that influence reaction rate

Each specific reaction has its own characteristic reaction rate.

We can control 4 factors that affect the rate of a given reaction:

• Concentration of reactants

• Physical state of reactants

• Temperature of reaction

• Presence of a catalyst


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Section 16.1: Factors that influence reaction rate

(1) Concentration – Molecules must collide in order to react.

The reaction rate changes throughout the course of a reaction.


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Section 16.1: Factors that influence reaction rate

Crushed coral versus whole coral.

(2) Physical state of reactants – Molecules must collide in order to react.

Reactants in same physical state  random thermal motion brings them into contact

Reactants in different physical states  contact between reactants occurs only at the

interface between the phases

Example: reactants – orange + blue

The more finely divided a solid or liquid

reactant, the greater its surface area per

unit volume  More contact with other

reactants  Faster reaction.

solid + aqueous

(interface only)

aqueous + aqueous


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Section 16.1: Factors that influence reaction rate

(4) Presence of a catalyst – A catalyst speeds up the reaction rate without being

consumed (chemically reacting) in the reaction.

Example: Biological catalysts

(3) Temperature of reaction – Molecules must collide with enough energy to react.

Two aspects to this:

(1) At higher temperatures, more collisions occur at a given time.

(2) At higher temperatures, the energy of collisions is higher (K.E. of

molecules is higher).

Example: refrigeration to preserve food


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Section 16.2: Expressing reaction rate quantitatively

Rate – a change in some variable per unit of time

Analogy with speed

Reaction: A  B

reaction rate – the changes in concentrations of reactants or products per unit time

Reactant concentrations decrease, while product concentrations increase.

Reaction: A  B


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Section 16.2: Expressing reaction rate quantitatively

In most reactions, not only the concentration changes, but the reaction rate also changes.

Therefore, we can define three reaction rates:

Average reaction rate – how fast the concentration changes over the entire time period

Instantaneous reaction rate – the reaction rate at an instant in time

Initial reaction rate – the instantaneous rate at the moment the reactants are mixed

Example: Reaction involved in the decrease of photochemical smog (ethylene + ozone)

Reaction rate (rate of decrease of reactants):


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Section 16.2: Expressing reaction rate quantitatively

Average reaction rate – how fast the concentration changes over the entire time period

Instantaneous reaction rate – the reaction rate at an instant in time

Initial reaction rate – the

instantaneous rate the

moment the reactants are

mixed (slope of line that is

tangent to the curve at

t = 0)


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Section 16.2: Expressing reaction rate quantitatively

Rates for reactants and products

General formula: (a, b, c, d  coefficients)

General equation:


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2 N2O5 (g)  4 NO2 (g) + O2 (g)

(3) When [N2O5] is decreasing at 0.95 mol/L sec, at what rate is [NO2] increasing?

Section 16.2: Expressing reaction rate quantitatively

General formula: (a, b, c, d  coefficients)

General equation:

  • Express the rate in terms of changes in [H2], [O2], and [H2O] with time.

  • When [O2] is decreasing at 0.23 mol/L sec, at what rate is [H2O] increasing?

2 N2O5 (g)  2 NO2 (g) + N2O3(g) + 3 O2 (g)

(4) When [O2] is increasing at 0.54 mol/L sec, at what rate is [N2O5] decreasing?


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Section 16.3: Rate laws

Example: Specific reaction, specific temperature

Rate law constant

CH3COOCH2CH3 (ester) + H2O 

CH3COOH + CH3CH2OH

Experimentally determined not determined from reaction stoichiometry

General reaction: aA + bB + …  cC + dD + …

Rate law: rate = k[A]m[B]n…

where [A] and [B] are concentrations of reactants A and B

k is the rate law constant – specific for a given reaction at a given temperature

m and n are the reaction orders – defines how the rate is affected by reactant

concentration

Reaction orders – m and n = rate change / concentration change

Examples:

If the rate doubles when [A] doubles  [A]1 and m = 1

If the rate quadruples when [B] doubles  [B]2 and n = 2

If the rate does not change when [A] doubles  [A]0 and m = 0


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Section 16.3: Rate laws

All terms in the rate law (rate = k[A]m[B]n….) must be determined experimentally.

[A] and [B] measured and rate, rate constant (k), reaction orders (m, n)

are deduced from these measurements.

  • Measuring rates – many methods:

  • Conductometric methods – used when a nonionic reactant forms ionic products

  • Example: (CH3)3C–Br (l) + H2O (l)  (CH3)3C–OH (l) + H+ (aq) + Br- (aq)

(2) Manometric methods – used when a reaction involves a change in the number of

moles of a gaseous reactant or product  reaction rate determined by the change

in pressure over time

Example: Zn (s) + 2 CH3COOH (aq)  Zn2+ (aq) + 2 CH3COO- (aq) + H2 (g)


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Section 16.3: Rate laws

Transmittance

Lightout

Transmittance =

Lightin

(3) Spectrometric methods – used when one of the reactants of products absorbs (or

emits) certain wavelengths of light

Example: NO (g, colorless) + 2 O3 (g, colorless)  O2(g, colorless) + NO2 (g, brown)

Lightin

Lightout

NO2

Light Source

Transparent Reaction Cell

Light Detector


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Section 16.3: Rate laws

(4) Direct chemical methods – used for reactions that can be easily slowed or stopped

Example: Measure respiration rate by killing bacteria with HgCl2 to stop respiration.

Respiration: O2 + CH2O  CO2 + energy

BOD bottle

(Biological Oxygen Demand)

• Measure O2 concentration at t=0 and t=24 hrs.

• Change in O2 concentration =

O2 (t=0) – O2 (t=24 hrs)

• Respiration rate = change in O2 concentration

time (24 hours)


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Section 16.3: Rate laws

Determining Reaction Order

First, some terminology………individual reaction order vs. overall reaction order

Example: 2 NO (g) + 2 H2 (g)  N2 (g) + 2 H2O (g) Rate = k[NO]2[H2]

Individual: Reaction is second order with respect to NO and first order with respect to H2.

Overall: Reaction is third order overall (sum of individual reaction orders).

Rate = k[NO][O3]

Rate = k[(CH3)3CBr][H2O]0

Rate = k[CHCl3][Cl2]1/2

*A zero order reaction order means that the reaction does

not depend on the concentration of that reactant.


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Section 16.3: Rate laws

Determining Reaction Order

We have been determining reaction orders from a known rate law (i.e. Rate = k[NO][O3])

When the rate law is not known, use data from a series of experiments

with different reactant concentrations to determine initial reaction rates.

Change one reactant concentration, while keeping the other constant.

Reaction is what order with respect to O2? With respect to NO? Overall?


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Section 16.3: Rate laws

Determining Reaction Order

We have been determining reaction orders from a known rate law (i.e. Rate = k[NO][O3])

When the rate law is not known, use data from a series of experiments

with different reactant concentrations to determine initial reaction rates.

Change one reactant concentration, while keeping the other constant.

Reaction is what order with respect to O2?

Double O2, double reaction rate: 1st order w.r.t O2

*reaction order = rate change / concentration change*


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Section 16.3: Rate laws

Determining Reaction Order

We have been determining reaction orders from a known rate law (i.e. Rate = k[NO][O3])

When the rate law is not known, use data from a series of experiments

with different reactant concentrations to determine initial reaction rates.

Change one reactant concentration, while keeping the other constant.

Reaction is what order with respect to NO?

Double NO, quadruple reaction rate: 2nd order w.r.t NO

*reaction order = rate change / concentration change*


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Section 16.3: Rate laws

Determining the Rate Constant

Simply, solve for k.

Rate law: rate = k[O2][NO]2

What is k for this reaction?


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Section 16.4: Integrated rate laws

Rate = k[A]

Magic

So far, we have not considered the time factor in the rate law equations.

Rate = k[O2][NO]2

This equation says what the rate will be when [O2] is X and [NO] is Y, but does

not tell use how long it will take for X moles of [NO] to be used up (for example).

Integrated rate laws – consider the time factor and are derived from equations

we have already seen using calculus

For reaction: A  B


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Section 16.4: Integrated rate laws

Example: At 1000 ºC, cyclobutane (C4H8) decomposes in a first-order reaction, with

The very high rate constant of 87 s-1, to two molecules of ethylene (C2H4).

If the initial cyclobutane concentration is 2.00 M, what is the concentration after 0.010 s?

What fraction of cyclobutane has decomposed in this time?

At 25 ºC, hydrogen iodide breaks down very slowly to hydrogen and iodine: rate = k[HI]2

The rate constant at 25 ºC is 2.4 x 10-21 L/mol sec. If 0.0100 mol of HI(g) is placed in a

1.0 L container, how long will it take for the concentration of HI to reach 0.00900 mol/L?


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Section 16.1 to 16.4: Solving Rate Problems – A Summary

Q: What is the average,

instantaneous, or initial

reaction rate?

Q: If a reactant/product is

increasing/decreasing at some

rate X, what is the rate of

increase/decrease of some

other reactant/product?

Questions related to rate laws

aA + bB + …  cC + dD + …

Rate law: Rate = k[A]m[B]n…

Use concentration versus

time data in either a Table

or a Graph to calculate each.

*Use experimental data*

*NOT stoichiometry*

*Stoichiometry-based*

Q:

• If the [A] doubles what will happen to the reaction rate?

• To quadruple the reaction rate, how would you need to change [A]?

• What is reaction order w.r.t. [A]?

• What is overall reaction order if the rate law is ‘rate = k[A]2[B]’?

• What is k for this reaction?

(a, b, c, d coefficients)

time

If the rate law (i.e. rate = k[A]2[B]) is known:

If given a rate, can solve for k.

Reaction order (m, n) = ∆rate

∆concentration

Q:

• What is the [A] after x time?

• How long will it take for [A] to

reach X mol/L?

No rate law, no initial

reaction rates. Have

[A] vs. time data.

If the rate law is not known, but you

have [reactant] and initial rate data:

Use data to find reaction orders for

each reactant, then solve for k.

Trial-and-error

graphical plotting

Integrated rate laws


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Section 16.4: Integrated rate laws (continued)

What if you have concentration and time data, but do NOT

have the rate law (rate = k[A]m[B]n…) or the initial rate data?

Trial-and-error graphical plotting

For zero-order reactions:

For first-order reactions:

For second-order reactions:


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Section 16.4: Integrated rate laws (continued)

Trial-and-error graphical plotting

(Ocean Acidification!!!)


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Does coral dissolution

follow zero-order kinetics?

Does coral dissolution

follow first-order kinetics?


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Does coral dissolution

follow second-order kinetics?

If coral dissolution followed first-order kinetics, what would that tell you

about the dependence of the overall reaction rate of CaCO3 dissolution?

(What does the rate depend on?)


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Section 16.4: Integrated rate laws (continued)

Example #2: Trial-and-error graphical plotting


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Section 16.4: Integrated rate laws (continued)

Half-life (t1/2)

It is the time required for a reactant concentration to reach half of its initial value.

Applies to first-order reactions only  independent of the starting concentration

(In other words, if independent of starting concentration  the t1/2 is independent

of the number of other particles present)

t1/2 = 0.693 / k


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Section 16.4: Integrated rate laws (continued)

Radioactive decay – a common

application of half-life (t1/2)

Carbon-14

t1/2 = 5568 yrs

How old is the Ice Man? Found in

1991 in the Alps. (~3,330 yrs old)


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Suggested Problems

16.39, 16.41, 16.43

  • What is the average reaction rate, the initial reaction rate, and the instantaneous

  • reaction rate for the reaction below given the data in the Table below?

(2) For the same reaction (above), how would the reaction rate change if you doubled the

H2 concentration? How would you need to change the O2 concentration if you wanted to

quadruple the reaction rate?

(3) If you know that the rate law for the reaction shown below is rate = k[C2H4]2[O3], then

if you doubled the [O3], how would the reaction rate change? How would you need to

Change the [C2H4] to quadruple the rxn rate?


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Suggested Problems (Continued)

(4) At 50 ºC, H2Cl2 breaks down into to H2 and Cl2 gases: rate = k[H2Cl2]2 The rate

constant 1.8 x 10-2 L/mol sec. If 0.40 mol of H2Cl2 is placed in a 2.5 L container, how

long will it take for the concentration of H2Cl2 to reach 0.25 mol/L?

(5) At 50 ºC, C4H8 decomposes. The rate law is rate = k[C4H8]. If the concentration of

cyclobutane is 0.80 mol/L after 20 minutes, what the the rate constant (k) at this

temperature? The initial concentration of cyclobutane is 1.38 mol/L.

(6) At 700 ºC, H2S gas breaks down into to diatomic hydrogen and sulfur gases. The

rate = k[H2S]0 If the rate constant is 9.30 x 10-8 mol/L sec and the initial concentration

of H2S gas is 0.18 mol/L, what will be the concentration of this gas 1 minute after the

reaction is started?


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Section 16.4: Half-life (continued)

Half-life

t1/2 = 0.693 / k

16.43 In a first-order decomposition reaction, 50.0% of a compound decomposes in

10.5 minutes. (a) What is the rate constant of the reaction? (b) How long does it take

for 75 % of the compound to decompose?

Suggested Problems

16.44, 16.45


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Section 16.5: Effect of Temperature on Reaction Rate

Rate Law

rate = k[A]m[B]n…

An increase of 10 ºC (or K) causes a doubling or

tripling of the reaction rate.

Q10 – the factor by which the reaction rate is accelerated

by raising the temperature by 10 degrees (2 – 3)

Expressed by relationship to

the rate constant, k.

Integrated Rate Laws

Half-life

t1/2 = 0.693 / k


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Q10 – the factor by which the reaction rate is accelerated

by raising the temperature by 10 degrees (2 – 3)

Metabolic rates – the rates at which organisms use energy and materials  The

fundamental rate of biology (it comes down to biochemistry)

Universal temperature dependence (UTD) of biological processes

“Despite a hundred

years of research,

ecology has little in

the way of universal

laws of gravity and

thermodynamics in

physics or the

Mendelian laws of

inheritance in biology.

Is ecology really devoid

of universal laws?”

2004, New Scientist

2001, Science, Volume 293


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Back to Chemistry……

Section 16.5: Effect of Temperature on a Chemical Reaction Rate

Arrhenius equation (Svante Arrhenius)

where A is the frequency factor (Next  Section 16.6)

T is the absolute temperature (*Must be in units of Kelvins)

R is the Universal Gas Constant (8.31447 J/mol K)

Ea is the activation energy (the minimum energy the molecules must

have to react)

Using this equation to find Ea from experimental data…

First, the math


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Section 16.5: Effect of Temperature on Reaction Rate

The decomposition of HBr has rate constants of 1.37 x 10-11 L/mol s at 350 K and

2.43 x 10-9 L/mol s at 450 K. Find Ea.

Practice problem for you: Find Ea for the reaction of an ester with water using the

following data from a kinetics experiment.


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Section 16.6: Effect of Concentration on Reaction Rate

Rate Law

rate = k[A]m[B]n…

  • Collision Theory – Reactant particles (atoms, molecules, and ions) must collide with

  • each other in order to react. Therefore, the # of collisions per time puts an upper limit

  • on how fast a reaction can take place.

  • Explains several things:

  • Why reactant concentrations are multiplied together in the rate law

A + B  products


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Section 16.6: Effect of Concentration on Reaction Rate

(2) How temperature affects the rate

In most collisions, the molecules rebound without reacting.  Every reaction has an

energy threshold that colliding molecules must exceed in order to react.

This minimum energy threshold is called the activation energy (Ea). Only collisions

with E > Ea will react.

Temperature rise increases the number of collisions with enough energy to exceed Ea.


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Section 16.6: Effect of Concentration on Reaction Rate

(3) the influence molecular structure has on rate

In addition to colliding, and colliding with enough energy, molecules must also collide

so that the reacting atoms make contact.

Arrhenius equation (Svante Arrhenius)

The effect of molecular orientation is contained in the term A (frequency factor)

A = pZ

Z  collision frequency

p  orientation probability

(effectively oriented collisions / all possible collisions)


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Section 16.6: Effect of Concentration on Reaction Rate

  • Collision Theory – Reactant particles (atoms, molecules, and ions) must collide with

  • each other in order to react. Therefore, the # of collisions per time puts an upper limit

  • on how fast a reaction can take place.

  • Molecules must:

  • Collide

  • (2) Collide with a certain minimum energy (Ea – activation energy)

  • (3) Collide with the right orientation (the A term in Arrhenius equation)

Does not explain:

Why the activation energy is crucial

How the activated molecules look

Transition state theory – focuses on the high-energy species that

form through an effective collision


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Suggested Problems

16.58 – 16.46, 16.72, 16.74, 16.77, 16.78, 16.82


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