UNIT 3: Energy Changes and Rates of Reaction Chapter 6: Rates of Reaction.

UNIT 3: Energy Changes and Rates of Reaction

Chapter 6: Rates of Reaction


UNIT 3

An important part of studying chemical reactions is to monitor the speed at which they occur. Chemists look at how quickly, or slowly, reactions take place and how these rates of reaction are affected by different factors.

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The light produced by a firefly depends on the speed of a particular chemical reaction that occurs in its abdomen.


UNIT 3 Section 6.1

6.1 Chemical Reaction Rates

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Chemical kinetics is the study of the rate at which chemical reactions occur.


The term reaction rate, or rate of reaction refers to:• the speed that a chemical reaction occurs at, or• the change in amount of reactants consumed or

products formed over a specific time interval

UNIT 3 Section 6.1

Determining Reaction Rates

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The reaction rate is often given in terms of the change in concentration of a reactant or product per unit of time.

The change in concentration of reactant A was monitored over time.


UNIT 3 Section 6.1

Determining Reaction Rates

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The change in concentration of reactant or product over time is often graphed.

For the reaction A → B, over time, the concentration of A decreases, and the concentration of B increases.


UNIT 3 Section 6.1

Average and Instantaneous Reaction Rates

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Average rate of reaction:

• change in [reactant] or [product] over a given time period (slope between two points)

Instantaneous rate of reaction:

• the rate of a reaction at a particular point in time (slope of the tangent line)

Average rate of reaction and instantaneous rate of reaction can be determined from a graph of concentration vs. time.


UNIT 3 Section 6.1

Expressing Reaction Rates in Terms of Reactants or Products

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A known change in concentration of one reactant or product and coefficients of a chemical equation allows determination of changes in concentration of other reactants or products.


Express the rate of formation of ammonia relative to hydrazine, for the reaction on the previous slide.

UNIT 3 Section 6.1

Answer on the next slide

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The mole ratio of ammonia to hydrazine is 4:3

Section 6.1UNIT 3

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UNIT 3 Section 6.1

Methods for Measuring Rates of Reaction

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UNIT 3 Section 6.1

Calculating Reaction Rates from Experimental Data

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The following data were collected in order to calculate the rate of a reaction.

UNIT 3 Section 6.1

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UNIT 3 Section 6.1Chapter 6: Rates of Reaction


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Section 6.1 Review

UNIT 3 Section 6.1

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UNIT 3 Section 6.2

6.2 & 6.3: Collision Theory and Factors Affecting Rates of Reaction

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According to collision theory, a chemical reaction occurs when the reacting particles collide with one another.

Only a fraction of collisions between particles result in a chemical reaction because certain criteria must be met.


UNIT 3 Section 6.2

Effective Collision Criteria 1:The Correct Orientation of Reactants

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For a chemical reaction to occur, reactant molecules must collide with the correct orientation relative to each other (collision geometry).

Five of many possible ways that NO(g) can collide with NO3(g) are shown. Only one has the correct collision geometry for reaction to occur.


UNIT 3 Section 6.2

Effective Collision Criteria 2:Sufficient Activation Energy

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The shaded part of the Maxwell-Boltzmann distribution curve represents the fraction of particles that have enough collision energy for a reaction (ie the energy is ≥ Ea).


For a chemical reaction, reactant molecules must also collide with sufficient energy.

Activation energy, Ea, is the minimum amount of collision energy required to initiate a chemical reaction.

Collision energy depends on the kinetic energy of the colliding particles.

UNIT 3 Section 6.2

Representing the Progress of a Chemical Reaction

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From left to right on a potential energy curve for a reaction:• potential energy increases as reactants become closer• when collision energy is ≥ maximum potential energy,

reactants will transform to a transition state• products then form (or reactants re-form if ineffective)

Exothermic Endothermic

UNIT 3 Section 6.2

Activation Energy and Enthalpy

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The Ea for a reaction cannot be predicted from ∆H.

• ∆H is determined only by the difference in potential energy between reactants and products.

• Ea is determined by analyzing rates of reaction at differing temperatures.

• Reactions with low Ea occur quickly. Reactions with high Ea occur slowly.

Potential energy diagram for the combustion of octane.


UNIT 3 Section 6.2

Activation Energy for Reversible Reactions

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Potential energy diagrams can represent both forward and reverse reactions.

• follow left to right for the forward reaction

• follow right to left for the reverse reaction


UNIT 3 Section 6.2

Analyzing Reactions Using Potential Energy Diagrams

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The BrCH3 molecule and OH- collide with the correct orientation and sufficient energy and an activated complex forms. When chemical bonds reform, potential energy decreases and kinetic energy increases as the particles move apart.


Describe the relative values of Ea(fwd) and Ea(rev) for an exothermic reaction

UNIT 3

Answer on the next slide

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LEARNING CHECK

Ea(rev) is greater than Ea(fwd)

Section 6.2UNIT 3

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LEARNING CHECK

UNIT 3 Section 6.2

Factors Affecting Reaction Rate

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1. Nature of reactants

• reactions of ions tend to be faster than those of molecules

2. Concentration

• a greater number of effective collisions are more likely with a higher concentration of reactant particles

3. Temperature

• with an increase in temperature, there are more particles with sufficient energy needed for a reaction (energy is ≥ Ea)


UNIT 3 Section 6.2

Factors Affecting Reaction Rate

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4. Pressure

• for gaseous reactants, the number of collisions in a certain time interval increases with increased pressure

5. Surface area

• a greater exposed surface area of solid reactant means a greater chance of effective collisions

6. Presence of a catalyst

• a catalyst is a substance that increases a reaction rate without being consumed by the reaction

UNIT 3 Section 6.2

A Catalyst Influences the Reaction Rate

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A catalyst lowers the Ea of a reaction.

• this increases the fraction of reactants that have enough kinetic energy to overcome the activation energy barrier

• a catalyzed reaction has the same reactants, products, and enthalpy change as the uncatalyzed reaction

A catalyst decreases both Ea(fwd) and Ea(rev).


UNIT 3 Section 6.2

Catalysts in Industry

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A metal catalyst is used for industrial-scale production of ammonia from nitrogen and hydrogen.

Hydrogen and nitrogen molecules break apart when in contact with the catalyst. These highly reactive species then recombine to form ammonia.


A catalyst (V2O5) is used for industrial-scale production of sulfuric acid from sulfur.

UNIT 3 Section 6.2

Catalysts in Industry

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The Ostwald process uses a platinum-rhodium catalyst for the industrial production of nitric acid.


Many industries use biological catalysts, called enzymes, which are most often proteins.

For example: the use of enzymes decreases the amount of bleach (an environmental hazard) needed to whiten fibres used in paper production.

Section 6.2 Review

UNIT 3 Section 6.2

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UNIT 3 Section 6.3

6.5: Rate Law

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Initial rate is found by determining the slope of a line tangent to the curve at time zero.


Initial rate is the rate of a chemical reaction at time zero.

• products of the reaction are not present, so the reverse reaction cannot occur

• it is a more accurate method for studying the relationship between concentration of reactant and reaction rate

UNIT 3 Section 6.3

The Rate Law

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The rate law shows the relationship between reaction rates and concentration of reactants for the overall reaction.


rate = k[A]m[B]n

m: order of the reaction for reactant An: order of the reaction for reactant Bk: rate constantm + n: order of the overall reaction

UNIT 3 Section 6.3

Graphing Reaction Rate in Terms of Concentration

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To study the effects of concentration on reaction rate:

• different starting concentrations of reactant are used

• initial rates are calculated using the slopes of the tangent lines from concentration vs time curves

• initial rates are plotted against starting concentration


Initial rates are determined (A) and these are plotted against concentration (B).

UNIT 3 Section 6.3

First-order Reactions

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The initial rate vs starting concentration graph on the previous slide is a straight line.


• the equation of the line can be expressed as:

rate = k[A]

• This represents a first-order reaction

For reactions with more than one reactant (e.g. A and B):• if experiments for each reactant produce straight lines,

the rate is “first order with respect to reactant A and first order with respect to reactant B.”

Example1:

The empirically determined rate law equation for the reaction

between nitrogen dioxide and fluorine gas :

2NO2 + F2 → 2NO2F

r = K[NO2]1[F2]1

The order of reaction with respect to NO2, F2 is 1 respectively.

the overall order of reaction is (1 + 1) = 2 ( The reaction is second order overall)

Example 2:

The discomposition of dinitrogen pentoxide is first order with respect to N2O5. If the initial rate of consumption is 2.1 x 10-4 mol/(L*s)

when the initial concentration of N2O5 is 0.40 mol/L. Predict what the rate would be if another experiment were performed in which the initial concentration of N2O5 were 0.80 mol/L

2N2O5 → 2NO2 +O2

Solution:

First: write the rate law equation for the system:

r = K[N2O5]1 (first order reaction, so exponent is 1)

r = 0.8 mol/L = x

0.4 mol/L 2.1 x 10-4mol/(L*s)

Since it is first order, any change in [N2O5] will have the same effect

on the rate. The [N2O5] is doubled from 0.4 to 0.8 mol/L, so the rate

of consumption will double from 2x (2.1 x 10-4) to 4.2 x 10-4

mol/(L*s)

UNIT 3 Section 6.3

Second-order Reactions

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For chlorine dioxide in this reaction:

• the initial rate vs concentration curve is parabolic

• the reaction is proportional to the square of [ClO2]

• it is a second order reaction with respect to this reactant


rate = k[A]2

Example 3:

The dimerization reaction of 1,3-butadiene (C4H6) is a second

order with respect to C4H6. If the initial rate of reaction were

32 mmol/(L*min) C4H6 at a given initial concentration of C4H6,.

What would be the initial rate of reaction if the initial

concentration of C4H6 were doubled

2C4H6 → C8H12

Solution:

The rate equation is r = K[C4H6]2 (let x be the initial concentration)

x = 32

2x2

If the initial concentration is doubled (multiplied by 2), the initial

rate will be multiplied by 22, or 4. The new rate is 4 x 32 mmol

C4H6/(L*min) or 0.13 mol /(L*min) C4H6

DETERMINATION OF EXPONENT OF THE RATE LAW

The exponents of the Rate Law must be determined

experimentally. To find the exponent of the rate law, we must

study how changes in concentration affect the rate of reaction.

Example: -

[A]+ B → products

 Rate = k[A]m[B]n

DETERMINATION OF EXPONENT OF THE RATE LAW

The exponents of the Rate

Law must be determined

experimentally by study

how changes in

Concentration affect the

Rate of reaction.

Example: -

 A + B → products

 Rate = k[A]m[B]n

Example: Experimental DataInitial Concentration Initial rate of formation

of products

[A] [B]

1. 0.10 0.10 0.20

2. 0.20 0.10 0.40

3. 0.30 0.10 0.60

4. 0.30 0.20 2.40

5. 0.30 0.30 5.40 To determine the value of m, we use the results from Exp.# 1 and 2 or 3, in which only [A] changes: rate 2 = 0.4 mol/L.s =k(0.2 mol/L)m

rate 1 0.2 mol/L.s k(0.1mol/L)m

2.0 = (2.0)m

Meaning, if [A] is doubled the rate doubles (#1 & 2).  Also, if [A] is tripled (#1 &3), the rate is tripled .Therefore, the exponent of A must be 1 ie [A]1

To determine n, we use the results from Exp. #3 and 4 or 5, in which only [B] changes: This time it is the concentration of B that affects the rate

rate 4 = 2.4 mol/L.s = k(0.20 mol/L)n rate 3 0.6 mol/L.s K(0.10 mol/L)n

n

4 = 2

When [B] is doubled the rate increases by factor of 4 which equals 22

[A] [B] Rates

1) 0.10 0.10 0.20

2) 0.20 0.10 0.40

3) 0.30 0.10 0.60

4) 0.30 0.20 2.40

5) 0.30 0.30 5.40

rate 5 = 5.4 mol/L.s = k(0.30 mol/L)n

rate 3 0.60 mol/L.s k(0.10 mol/L)n

9 = 3

Calculating Constant, K, from the Rate Law:

Using the data from set # 1, K can be calculated from the

r= k [A]1 [B]2

K=Rate

[A][B]

K =0.20 mol L-1s-1

(0.10 mol L-1)(0.10 mol L-1)

K=0.2 mol L-1s-1

0.010 mol2 L-2

K=2.0 x 101 L mol-1 s-1

Let’s check other sample on P.378-379

[A] [B] Rates0.10 0.10 0.200.20 0.10 0.400.30 0.10 0.600.30 0.20 2.400.30 0.30 5.40

Try this Example: - Determine the exponents of the Rate Law when the following data were measured for the reduction of Nitric Oxide with Hydrogen 2NO(g)+2H2(g)→N2(g)+2H2O(g)

Initial Concentrations Initial Rate of Disappearance of NO

(mol L-1s-1)

[NO] [H2]

1) 0.10 0.10 1.37 X 10-3

2) 0.10 0.20 2.75 X 10-3

3) 0.20 0.10 5.47 X 10-3

1. What is the rate law for this reaction?

2. What is the unit of the constant?

Solution:1. Find the order of the reaction2. Solve for K value and unit3. Then write the rate law Equation

Solution:The general form of the rate law for this reaction:Rate = K[NO]m[H2]n

To determine m, use result where only NO changes: Rate 3 = 5.47x10-3 mol/L*s= 0.20 mol/L Rate 1 1.37 x 10-3mol/L 0.10 mol/L 4 = [2]n, ie [2]2 , Thus m = 2

To determine n, use result where only [H2] changes: Rate 2 = 2.75x10-3 mol/L*s= 0.20 mol/L Rate 1 1.37 x 10-3mol/L 0.10 mol/L 2.0 = [2]n, ie [2]1 , Thus m = 1

Therefore, r = k[NO]2[H2]1

Using trial # 1, find k0.00137 mol/L.s = k x 0.100 mol/L x (0.100 mol/L)2

k = 0.00137 mol/L.s = 1.37 L2/mol2.s 0.100 mol/L x (0.100 mol/L)2

r = 1.37 L2/(mol2.s)[NO]2[H2]

Note: Monitor the Unit of the rate constant, K, closely as it varies with different order of reactions

UNIT 3 Section 6.3

6.6: Reaction Mechanisms

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A reaction mechanism is the series of elementary steps that occur as reactants are converted to products.


For example, oxygen and nitrogen are not formed directly from the decomposition of nitrogen dioxide:

It occurs in two elementary steps:

A balanced equation does not tell us how the reactants become product, it only tells us the reactant, the product, and the stoichiometry, but gives no information about the reaction mechanism.

O is an intermediate, it is neither a reactant nor a product, but formed and consumed during the reaction sequence.

Each of the two reactions is called an elementary step, a reaction whose rate law can be written from its molecules.

A reaction with one molecule is called a unimolecular step, two or three molecules are called bimolecular and termolecular steps respectively

A unimolecular step is always first order,bimolecular, second order Termolecular steps are uncommon

Elementary step Molecularity Rate law

A → Product Unimolecular Rate = [A]A + A → Products Bimolecular Rate = [A]2

A + B → Products Bimolecular Rate = [A][B]

a reaction mechanism must satisfy two requirements:

1.The sum of the elementary steps must give the overall balanced equation for the reaction2. The mechanism must agree with the experimentally determined rate law.

UNIT 3 Section 6.3

The Rate-determining Stepthe slowest of the elementary steps in a complex

reaction is called the rate-determining step.

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This reaction occurs in three elementary steps:


Step 2 is the rate-determining step:• it is the slowest elementary step• the overall rate of the reaction is dependent on this step• the Ea for this step is higher than Ea for each of the other steps

Potential Energy graph of Reaction Mechanism

A

B

C

D

E

UNIT 3 Section 6.3

A Proposed Reaction Mechanism

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• Experiments show that this reaction is zero order with respect to OH– (i.e. its rate does not depend on [OH–])

• This can be explained by a two-step mechanism

Step 2 is very fast and depends on completion of Step 1, not on the concentration of OH–.


Potential Energy graph of Reaction Mechanism

Example:The balanced equation for a reaction of the gases nitrogen dioxide and fluorine is

2NO2(g) + F2(g) → 2NO2F(g)

Experimentally determined rate law is

Rate = k[NO2][F2]

The suggested mechanism for this reaction is:

NO2 + F2 → NO2F + F slow

F + NO2 → NO2F fast

Is this an acceptable mechanism?

Does it satisfy the requirements for reaction mechanism?

1. The sum of the elementary steps must give the overall balanced equation for the reaction

2. The mechanism must agree with the experimentally determined rate law.

3. The reactant (s) of the slowest reaction gives the rate law equation.

Solution:

First:The sum of the steps should give the balanced equationNO2 + F2 → NO2F + F

F + NO2 → NO2F ___________________

NO2 + F2 + F + NO2 →NO2F + F + NO2F

Overall reaction : 2NO2 + F2 → 2NO2F

Second: Mechanism must agree with the experimentally determine rate. Since the first step is the slowest (the rate determining step), it must determine the reaction rate

Rate = k[NO2][F2]

UNIT 3: Energy Changes and Rates of Reaction Chapter 6: Rates of Reaction.

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Transcript of UNIT 3: Energy Changes and Rates of Reaction Chapter 6: Rates of Reaction.