Chapter 14

transcript

Chapter 14

Chemical Kinetics

Reaction Rate• The rate of a chemical reaction is measured as

the decrease in the concentration of a reactant or the increase in the concentration of a product in a unit of time.

∆[ ]

∆timeRate =

∆[ ]

∆timeRate =

What units would What units would we use for rate?we use for rate?

2H2H22OO22(aq) → 2H(aq) → 2H22O(l) + OO(l) + O22(g)(g)

∆[ ]

∆timeRate =

∆[ ]

∆timeRate =

How could the rate be expressed for How could the rate be expressed for this reaction in terms of Hthis reaction in terms of H22OO22??

What is the rate of the reaction from 0s to 2.16 x 10What is the rate of the reaction from 0s to 2.16 x 1044s?s?

What is the average rate of appearance of OWhat is the average rate of appearance of O22 from 0s to 2.16 x 10 from 0s to 2.16 x 1044s?s?

1.16 x 101.16 x 10-5-5 mol O mol O22 L L-1 -1 ss-1-1

General Rate of Reactiona A + b B → c C + d D

Rate of reaction = rate of disappearance of reactants

We can use the coefficients in the equation to compare the reaction rates for all the substances in the reaction.

Rate of reaction = rate of appearance (formation) of productsor

15-1 The Rate of a Chemical Reaction

• Rate is change of concentration with time.

2 Fe3+(aq) + Sn2+(aq) → 2 Fe2+(aq) + Sn4+(aq)

t = 38.5 s [Fe2+] = 0.0010 M

Rate of formation of Fe2+= = = 2.6 x 10-5 M s-1Δ[Fe2+]

0.0010 M

38.5 s

∆t = 38.5 s ∆[Fe2+] = (0.0010 – 0) M

Rates of Chemical Reaction2 Fe3+(aq) + Sn2+(aq) → 2 Fe2+(aq) + Sn4+(aq)

Rate of formation of Fe2+ = 2.6 x 10-5 mol L-1 s-1

What is the rate of formation of Sn4+?

What is the rate of disappearance of Fe3+?

1.3 x 101.3 x 10-5 -5 mol Snmol Sn4+4+ L L-1-1 s s-1-1

2.6 x 102.6 x 10-5-5 mol Fe mol Fe3+3+ L L-1-1 s s-1-1

What does the slope of the line represent?

What is the concentration at 100s for the reaction: 2H2O2(aq) → 2H2O(l) + O2(g)?

Given:

Δ[H2O2] = (1.7 x 10-3 M s-1) (∆t)

Rate = 1.7 x 10-3 M s-1

Δ[H2O2]

∆[H2O2] = (1.7 x 10-3 M s-1)(100 s) = 0.17M

= 2.15 M

= 2.32 M - 0.17 M [H2O2]100 s

[H2O2]i = 2.32 M

What does it mean when the rate of a reaction reaches zero?

• For a normal reaction it means that one or more of the reactants are used up and the reaction has stopped.

• For a reversible reaction it means that the reaction has reached equilibrium.

Factors Affecting Reaction Rates

1.1. The nature of the The nature of the reacting substances.reacting substances.

Factors Affecting Reaction Rates2.2. The state of subdivision of the reacting The state of subdivision of the reacting

substances (surface area).substances (surface area).

Lycopodium Powder

Factors Affecting Reaction Rates3. The temperature of the 3. The temperature of the

reacting substances.reacting substances.

4. The concentration of the reacting substances. 4. The concentration of the reacting substances. (Except in zero order reactions)(Except in zero order reactions)

Air (21% oxygen)Air (21% oxygen) 100% oxygen 100% oxygen

5.5. The presence of a catalyst.The presence of a catalyst.

CatalystsCatalysts speed up reactions speed up reactions but are left unchanged by the but are left unchanged by the reaction.reaction.

The Rate Law

a A + b B …. → g G + h H ….

Rate = k [A]m[B]n ….

Rate constant = k (k is constant for a particular reaction at a specific temperature)

Order of A = m Order of B = n

Overall order of reaction = m + n + ….

Temperature and Rate• Generally, as temperature

increases, so does the reaction rate.

• This is because k is temperature dependent.

• Therefore the temperature dependence of reaction rates is contained in the temperature dependence of the rate constant.

Temperature dependence of “k”

• After finding the trials to compare:• A reactant is zero order if the change in concentration of that

reactant produces no effect on the rate.• A reaction is first order if doubling the concentration of that

reactant causes the rate to double.• A reactant is nth order if doubling the concentration of that

reactant causes an 2n increase in rate.• Note that the rate constant does not depend on concentration.

Concentration and Rate SummaryConcentration and Rate Summary

Use the data provided to write the rate law and indicate the order of the reaction with respect to HgCl2 and C2O4

2- and also the overall order of the reaction.

First determine the order of HgCl2

Next determine the order of C2O42-

Now write the rate law and determine the order of the reaction.

Calculate the rate constant “k” and its units.

Initial rate of disappearance HgCl2

mol L-1 min-1

What is the average rate of disappearance of C2O42- in trial 1?

mol L-1 min-1

Use the data provided to write the rate law and indicate the order of the reaction with respect to NO2 and CO (support

your answers). Also give the overall order of the reaction.

Calculate the rate constant “k” and its units.

What is the average rate of disappearance of CO in trial 2?

How do we make these charts?

mol L-1 min-1

•Rates can be measured experimentally using a variety techniques:•moniter pH changes•Titrations •Change in volume or mass (gas production)

•Basically we can use any method to follow a reaction that produces a measurable change.

How do we make these charts?

mol L-1 min-1

One important method involves the spectroscopic determination of concentration through Beer’s Law.

Using Beer’s Law to Determine [ ] vs. time.

• For each trial, the reactants are mixed and the reaction mixture is transferred into a test tube or cuvette.

• Without any delay, the reaction vessel is placed into a spectrophotometer. The absorbance data is then collected at the wavelength of maximum absorbance as a function of time.

• This absorbance data is then converted to concentration data using Beer’s Law: A = ɛ l c

Fe(s)+CuSO4(aq)→Fe2SO4(aq)

+Cu(s) • The solution gradually gets paler as the

concentration of copper sulfate decreases and the concentration of iron sulfate increases.

Concentration of copper sulfate solution

1M 0.8M 0.6M 0.4M 0.2M

0s 30s 90s 200s 500s

Using Beer’s Law to Determine [ ] vs. time.

• A graph of concentration vs. time can be prepared and then used to experimentally determine the rate.

•What does this tangent allow us to measure?

Half Life of a First Order Reaction

• Half-life is the time required to convert one half of a reactant to product.

• For first-order reactions, half-life is often used as a representation for the rate constant.

• This is because the half-life of a first-order reaction and the rate constant are inversely proportional, and the half-life is independent of concentration.

Radioactivity

• Radioactive decay is the spontaneous breakdown of unstable atoms into more stable atoms with the simultaneous emission of particles and rays.

• Radioactive decay occurs at a constant rate that is a first order process.

Radioactivity and Half - Life• The half-life of carbon-14 is 5730 years.

• How old is a bone that has about 12.5% of the carbon-14 that a living organism would have in it?

Carbon Dating

Big Question

• How can we experimentally determine the order of a reaction?

Make “3” Graphs

• In order to determine order of reactant, A. We must collect data consisting of concentration versus time.

• One common way to determine concentration vs. time data is through the use of a spectrophotometer.

Make “3” Graphs

• We then use the data to make three graphs.– [A] versus t – ln [A] versus t– 1 / [A] versus t

• By examining these graphs we can determine the order of the reaction with respect to a particular reactant and determine the rate constant.

[A] versus t (linear for a zero order reaction)

k must be a positive number.

ln [A] versus t (linear for a 1st order reaction)

1 / [A] versus t (linear for a 2nd order reaction)

Collision Model

•Key Idea: Molecules must collide to react.

•However, only a small fraction of collisions produces a reaction. Why?

Two Factors

- Collisions must have enough energy to produce the reaction (must equal or exceed the activation energy).

- Orientation of reactants must allow formation of new bonds.

2HI → H2 + 2I

Concentration and Collision Theory

• Why does an increase in concentration cause an increase in reaction rate?

Concentration and Collision Theory

• Why does an increase in concentration cause an increase in reaction rate?

• Increasing the concentration increases the number of collisions and therefore there are more collisions leading to product.

Temperature and Collision Theory

• Why does a temperature increase cause the reaction rate to increase?

Temperature and Collision Theory

• Why does a temperature increase cause the reaction rate to increase?

• At higher temperatures there are more collisions and a greater percentage of the collisions have the energy necessary to create a successful collision.

Activation Energy

• The activation energy is the minimum amount of energy necessary for a reaction to occur.

Temperature and Activation Energy (Ea)

Activation Energy

• The activation energy can also be thought of as the energy necessary to form an activated complex during a collision between reactants.

Transition State Theory• The activated complex is a hypothetical species

lying between reactants and products at a point on the reaction profile called the transition state.

The activated complex is a transition state between reactants and products where old bonds have begun to break and new bonds have started to form. It cannot be isolated.

For two reactions at the same temperature, the reaction with the higher activation energy has the lower rate constant (k) and the slower rate.

2O3 3O2

- A chemical equation like the one above does not tell us how reactants become products - it is simply a summary of the overall reaction.

The reaction: 2O3 3O2

- Is proposed to occur through the two step process given below:

O3 O2 + O

O3 + O 2O2

This two step process is an example of a reaction mechanism

Reaction Mechanisms

• A reaction mechanism is a step-by-step description of a chemical reaction.

• Each step is called an elementary reaction.

Often Used Terms

•Intermediate: formed in one step and used up in a subsequent step and so is never seen as a product.

•Molecularity: the number of species that must collide to produce the reaction indicated by that step.

•Elementary Step: A step within a reaction mechanism whose rate law can be written from its molecularity.

Elementary Steps

• Molecularity: the number of molecules present in an elementary step.– Unimolecular: one molecule in the elementary step.– Bimolecular: two molecules in the elementary step.– Termolecular: three molecules in the elementary step.

Reaction MechanismsReaction Mechanisms

Elementary Steps• It is not common to see termolecular processes

(statistically improbable).- Unimolecular reactions occur because collisions with other

molecules provide the activation energy for the molecule to react.

- Bimolecular reactions involve the collision of two particles with sufficient energy and proper orientation.

- Termolecular reactions involve the simultaneous collision of three particles with sufficient energy and proper orientation.

Rate Laws for Elementary Steps• The rate law of an elementary step is determined by its

molecularity:– Unimolecular processes are first order,

– Bimolecular processes are second order, and

– Termolecular processes are third order.

Rate Laws for Elementary Steps

The Rate Determining Step

Rate-Determining Step

•In a reaction mechanism, the rate determining step is the slowest step. It therefore determines the rate of reaction.

Reaction Mechanisms

• Reaction mechanisms must be consistent with:1.Stoichiometry for the overall reaction.

2.The experimentally determined rate law.

Reaction mechanism must be consistent with the stoichiometry of the overall reaction.

• Is the mechanism below consistent with the overall reaction above?

NO2(g) + NO2(g) NO3(g) + NO(g)

NO3(g) + CO(g) NO2(g) + CO2(g)

NO2(g) + CO(g) NO(g) + CO2(g)

Determining the stoichiometry of a reaction mechanism.

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• The reaction mechanism must also support the rate law.

Rate Laws for Multistep Mechanisms

with an initial fast step.• Consider the reaction:

2NO(g) + Br2(g) 2NOBr(g)

Mechanisms with an Initial Fast Step

2NO(g) + Br2(g) 2NOBr(g)

• The experimentally determined rate law is

Rate = k[NO]2[Br2]

• Consider the following mechanism

NO(g) + Br2(g) NOBr2(g)k1

NOBr2(g) + NO(g) 2NOBr(g)k2

Step 1:

Step 2:

(fast)

(slow)

• The rate law is (based on Step 2):

Rate = k2[NOBr2][NO]

• The rate law should not depend on the concentration of an intermediate (intermediates are usually unstable).

• NOBr2 is an unstable intermediate, so we express the concentration of NOBr2 in terms of NO and Br2 Since there is an equilibrium in step 1 we have

]NO][Br[]NOBr[ 21

Step 1:

Step 2:

(fast)

(slow)

• By definition of equilibrium:

• Therefore, the overall rate law becomes

• Note the final rate law is consistent with the experimentally observed rate law.

]NOBr[]NO][Br[ 2121 kk

][BrNO][NO][]NO][Br[Rate 22

Step 1:

Step 2:

(fast)

(slow)

Student Example:Determine the rate law for the reaction and the balanced equation given the mechanism below:

2NO ↔ N2O2 fast

N2O2 + O2 → 2NO2 slow

Assume the rate law is:

Rate = k[H2O2][H3O+][I-]

Which step would be the rate – determining step?

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This diagram shows a two-step mechanism for a reaction with the first step being rate determining.

What is the mechanism for the reaction?

Overall Reaction

Mechanism for Previous Reaction

NO + H2 → NOH2 slow

NO + NOH2 → N2O + H2O fast

Catalysts

• A catalyst is a substance that increases the rate of a chemical reaction by reducing the activation energy, but which is left unchanged by the reaction.

What is the overall reaction?

O3 O2 + O

O3 + O 2O2

What is the overall reaction?

Identify the intermediates.

NO is a catalyst

A homogeneous catalyst is of the same phase as the reacting substances. It lowers the activation energy by forming intermediates which allow the reaction to proceed by a different pathway.

Heterogeneous Catalysts

• A heterogeneous catalyst is of a different phase than the reacting substances.

• It provides a surface on which the transition state is stabilized thus lowering the activation energy and increasing the reaction rate.

Catalytic Converter: A Heterogeneous Catalyst

• In a catalytic converter, the catalyst (in the form of platinum and palladium) is coated onto a ceramic honeycomb that are housed in a muffler-like package attached to the exhaust pipe. The catalyst helps to convert harmful exhaust gases into safer ones.

Catalysis

• Catalysis is the process of using a catalyst to speed up a reaction.

Heterogeneous and Homogeneous Catalysts can Catalysis different

Types of Reactions:• Acid – Base Catalysis

• Enzyme Catalysis

• Surface Catalysis

Acid – Base Catalysis

• A chemical reaction is catalyzed by an acid or a base.

• A reactant either gains or loses a proton (H+) which causes an increase in the rate of the reaction.

• In acid catalysis all species capable of donating protons contribute to reaction rate acceleration with the strongest acids being most effective.

Enzyme Catalysis

• Some enzymes accelerate reactions by binding to the reactants in a way that lowers the activation energy.

• Other enzymes react with a reactant species to form a new intermediate.

• Enzyme catalysis essentially occurs when substances catalyze reactions within a living organism.

Enzyme Catalysis

• There is a coulombic attraction between the substrate and the enzyme.

• After the reaction occurs the products do not exhibit the coulombic attraction with the enzyme due to changes in their structure and are therefore are released by the enzyme.

Substrate Interactions with the Active Sites in Enzyme Catalysis

Surface Catalysis

• In surface catalysis, either a new reaction intermediate is formed, or the probability of a successful collision is modified.

• Most catalysts fall into this category.

Reaction Rate Lab

Reaction Rate Lab – Part ATRIAL 0.010M KI

Reaction Mixture I

0.001M Na2S2O3

Reaction Mixture I

H2O Reaction Mixture I

0.040M KBrO3

Reaction Mixture II

0.10M HCl

Reaction Mixture II

1 10 ml 10 ml 10 ml 10 ml 10 ml 2 20 ml 10 ml 0 10 ml 10 ml 3 10 ml 10 ml 0 20 ml 10 ml

4 10 ml 10 ml 0 10 ml 20 ml 5 8 ml 10 ml 12 ml 5 ml 15 ml

•Use different containers for Reaction Mixtures I and II.

•Don’t forget the starch.

Reaction Rate Lab – Part BTRIAL 0.010M KI

Reaction Mixture I

0.001M Na2S2O3

Reaction Mixture I

0.040M KBrO3

Reaction Mixture II

0.10M HCl

Reaction Mixture II

•In part A you will perform five different trials with various concentrations.

Reaction Rate Lab – Part CTRIAL 0.010M KI

Reaction Mixture I

0.001M Na2S2O3

Reaction Mixture I

0.040M KBrO3

Reaction Mixture II

0.10M HCl

Reaction Mixture II

•In part B you will perform trial 1 using a catalyst.

We will not be performing part C of the lab.

Chapter 14

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