Chapter 16 Kinetics Dinosaurs generated enough heat to sustain its biochemical reactions at high rates. Reaction rate = f(temperature)

Chapter 16Kinetics

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.

Describes the change in concentration as a function of time.

Qualitatively…… Quantitatively……

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

Section 16.1: Factors that influence reaction rate

Each specific reaction has its own characteristic reaction rate.


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

The reaction rate changes throughout the course of a reaction.


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

Crushed coral versus whole coral.

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

solid + aqueous(interface only)

aqueous + aqueous

The more finely divided a solid or liquidreactant, the greater its surface area perunit volume More contact with other

reactants Faster reaction.


(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).

(4) Presence of a catalyst – A catalyst speeds up the reaction rate without being consumed (chemically reacting) in the reaction.

Example: Biological catalysts

Example: refrigeration to preserve food

Section 16.2: Expressing reaction rate quantitatively

Rate – a change in some variable per unit of time

Analogy with speed

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

Reactant concentrations decrease, while product concentrations increase.

Reaction: A B

Reaction: A B


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):


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 istangent to the curve att = 0)


Rates for reactants and products

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

General equation:

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

General equation:


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

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

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?

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?

Section 16.3: Rate laws

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

Rate law constant

CH3COOCH2CH3 (ester) + H2O CH3COOH + CH3CH2OH

Example: Specific reaction, specific temperature

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

Examples: If the rate doubles when [A] doubles [A]1 and m = 1If the rate quadruples when [B] doubles [B]2 and n = 2If the rate does not change when [A] doubles [A]0 and m = 0


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:

(1) 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)

Transparent Reaction Cell

NO2

Light Source Light Detector

Lightin Lightout

Transmittance = Lightout

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)

Tra

nsm

itta

nce

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)

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.


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 doesnot depend on the concentration of that reactant.

Section 16.3: Rate lawsDetermining 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?





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*





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*

Section 16.3: Rate lawsDetermining the Rate Constant

Simply, solve for k.

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

What is k for this reaction?

Section 16.4: Integrated rate laws

Rate = k[O2][NO]2

This equation says what the rate will be when [O2] is X and [NO] is Y, but doesnot tell use how long it will take for X moles of [NO] to be used up (for example).

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

Integrated rate laws – consider the time factor and are derived from equations we have already seen using calculus

For reaction: A BRate = k[A]

Magic

Section 16.4: Integrated rate laws

Example: At 1000 ºC, cyclobutane (C4H8) decomposes in a first-order reaction, withThe 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 a1.0 L container, how long will it take for the concentration of HI to reach 0.00900 mol/L?

Section 16.1 to 16.4: Solving Rate Problems – A Summary

Q: What is the average,instantaneous, or initial

reaction rate?

Use concentration versustime data in either a Table

or a Graph to calculate each.

Q: If a reactant/product is increasing/decreasing at some

rate X, what is the rate of increase/decrease of some

other reactant/product?

*Stoichiometry-based*

(a, b, c, d coefficients)

Questions related to rate lawsaA + bB + … cC + dD + …Rate law: Rate = k[A]m[B]n…

*Use experimental data**NOT stoichiometry*

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?

Reaction order (m, n) = ∆rate ∆concentration

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

If given a rate, can solve for k.

If the rate law is not known, but youhave [reactant] and initial rate data:

Use data to find reaction orders for each reactant, then solve for k.

No rate law, no initialreaction rates. Have[A] vs. time data.

Trial-and-errorgraphical plotting

Q: • What is the [A] after x time?• How long will it take for [A] toreach X mol/L?

Integrated rate laws

time

Section 16.4: Integrated rate laws (continued)

Trial-and-error graphical plotting

For zero-order reactions: For first-order reactions: For second-order reactions:

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 (Ocean Acidification!!!)

Kinetics of Coral Dissolution

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

0 1 2 3 4 5 6 7

Time (Days)

Ma

ss

los

s (

g)

Temp1 Temp2


y = -0.9x + 0.6931

R2 = 1

y = -0.4x + 0.6931

R2 = 1

-6

-5

-4

-3

-2

-1

0

1

2

0 1 2 3 4 5 6 7 8

Time (Days)

ln (

Mas

s L

oss

)

Temp1 Temp2 Linear (Temp2) Linear (Temp1)


0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

0 1 2 3 4 5 6 7Time (Days)

Ma

ss

los

s (

g)

Temp1 Temp2

Does coral dissolution follow zero-order kinetics?

Does coral dissolution follow first-order kinetics?


0

50

100

150

200

250

300

0 1 2 3 4 5 6 7 8

Time (Days)1

/ Mas

s L

oss

Temp1 Temp2Does coral dissolution follow second-order kinetics?

If coral dissolution followed first-order kinetics, what would that tell youabout the dependence of the overall reaction rate of CaCO3 dissolution?

(What does the rate depend on?)


Example #2: Trial-and-error graphical plotting


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 independentof the number of other particles present)

t1/2 = 0.693 / k


Radioactive decay – a common application of half-life (t1/2)

Carbon-14t1/2 = 5568 yrs

How old is the Ice Man? Found in1991 in the Alps. (~3,330 yrs old)

Suggested Problems

16.39, 16.41, 16.43

(1) What is the average reaction rate, the initial reaction rate, and the instantaneousreaction rate for the reaction below given the data in the Table below?

0.50

0.70

0.90

1.10

1.30

1.50

1.70

0 2 4 6Time (Minutes)

[O2

] (m

ol/L

)

Time (min)

[O2] (mol/L)

0 1.60

1 1.21

2 0.98

3 0.82

4 0.70

5 0.62

6 0.55

(2) For the same reaction (above), how would the reaction rate change if you doubled theH2 concentration? How would you need to change the O2 concentration if you wanted toquadruple the reaction rate?

(3) If you know that the rate law for the reaction shown below is rate = k[C2H4]2[O3], thenif you doubled the [O3], how would the reaction rate change? How would you need to Change the [C2H4] to quadruple the rxn rate?

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?

(6) At 700 ºC, H2S gas breaks down into to diatomic hydrogen and sulfur gases. Therate = k[H2S]0 If the rate constant is 9.30 x 10-8 mol/L sec and the initial concentrationof H2S gas is 0.18 mol/L, what will be the concentration of this gas 1 minute after thereaction is started?

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

Section 16.4: Half-life (continued)

Half-lifet1/2 = 0.693 / k

16.43 In a first-order decomposition reaction, 50.0% of a compound decomposes in10.5 minutes. (a) What is the rate constant of the reaction? (b) How long does it takefor 75 % of the compound to decompose?

Suggested Problems

16.44, 16.45

Section 16.5: Effect of Temperature on Reaction Rate

Integrated Rate Laws

Rate Law

rate = k[A]m[B]n…

Half-lifet1/2 = 0.693 / k

Q10 – the factor by which the reaction rate is acceleratedby raising the temperature by 10 degrees (2 – 3)

An increase of 10 ºC (or K) causes a doubling ortripling of the reaction rate.

Expressed by relationship to the rate constant, k.

Q10 – the factor by which the reaction rate is acceleratedby raising the temperature by 10 degrees (2 – 3)

Metabolic rates – the rates at which organisms use energy and materials Thefundamental rate of biology (it comes down to biochemistry)

Universal temperature dependence (UTD) of biological processes

2001, Science, Volume 293

“Despite a hundredyears of research,

ecology has little in the way of universallaws of gravity andthermodynamics in

physics or theMendelian laws of

inheritance in biology.Is ecology really devoid

of universal laws?”

2004, New Scientist

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)

Back to Chemistry……Section 16.5: Effect of Temperature on a Chemical Reaction Rate

Using this equation to find Ea from experimental data…

First, the math

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 and2.43 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 thefollowing data from a kinetics experiment.

Section 16.6: Effect of Concentration on Reaction Rate

Collision Theory – Reactant particles (atoms, molecules, and ions) must collide witheach other in order to react. Therefore, the # of collisions per time puts an upper limiton how fast a reaction can take place.

Explains several things:(1) Why reactant concentrations are multiplied together in the rate law

A + B products

Rate Law

rate = k[A]m[B]n…

(2) How temperature affects the rate


In most collisions, the molecules rebound without reacting. Every reaction has anenergy threshold that colliding molecules must exceed in order to react.

This minimum energy threshold is called the activation energy (Ea). Only collisionswith E > Ea will react.

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

(3) the influence molecular structure has on rate

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


Collision Theory – Reactant particles (atoms, molecules, and ions) must collide witheach other in order to react. Therefore, the # of collisions per time puts an upper limiton how fast a reaction can take place.

Molecules must:(1) 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 crucialHow the activated molecules look

Transition state theory – focuses on the high-energy species that form through an effective collision

Suggested Problems

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

Documents

Chapter 16 Kinetics Dinosaurs generated enough heat to sustain its biochemical reactions at high rates. Reaction rate = f(temperature)