kinetics and equilibrium chm 130 - 2012.ppt and equilibrium chm 130.pdf · collision theory of...

Chemical Kinetics(Reaction Rates)

andEquilibrium

David A. KatzDepartment of ChemistryPima Community College

Tucson, AZ USA

Chemical Kinetics

• The study of the rates of chemical reactions and how they occur.

• The conditions that affect the rate or speed at which reactions occur.

• The reaction mechanism (how the reaction occurs at the atomic-molecular level).

• Reaction rate = change in concentration of a reactant or product with time.

• Rate can be expressed as initial rateaverage rateinstantaneous rate

Reaction Rates

ΔconcentrationRate=Δtime

Determining a Reaction Rate

Factors Affecting Reaction Rates

• Physical State of the Reactants In order to react, molecules must come

in contact with each other. The more homogeneous the mixture of

reactants, the faster the molecules can react.• Gaseous reactions are preferred• Solutions are most convenient


• Concentration of Reactants As the concentration of reactants

increases, so does the likelihood that reactant molecules will collide.

Burning iron in air

Burning iron in oxygen

Burning an iron nail

Lycopodium powder burning:a) In an evaporating

dishb) In air


• Temperature At higher temperatures, reactant

molecules have more kinetic energy, move faster, and collide more often and with greater energy.


• Catalysis Catalysts speed up reactions by

changing the mechanism of the reaction.

Catalysts are not consumed during the course of the reaction.

Negative catalysts (inhibitors) can be used to slow down the rate of a reaction.

Collision Theory of Reaction Rates

• In a chemical reaction, bonds are broken and new bonds are formed.

• Molecules can only react if they collide with each other.


The possible collisions for three sets of molecules that would result in a possible reaction


For a collision to be effective (i.e., result in a reaction):Molecules must collide with the correct

orientation (or alignment)There must be enough energy to overcome

repulsions so molecules do not reboundBonds must break and new bonds formed.

NO + NO3 2 NO2


“Ineffective” means that this has a lower probability of reacting.

Activation Energy• There is a minimum amount of energy required

for reaction: the activation energy, Ea.• Just as a ball cannot get over a hill if it does not

roll up the hill with enough energy, a reaction cannot occur unless the molecules possess sufficient energy to get over the activation energy barrier.

Reaction Energy DiagramsThis diagram relates the energy for the rearrangement of methyl isonitrile. To visualize the energy changes throughout a process we use a reaction energy or a reaction coordinate diagram

Reaction Energy DiagramsThe species present at the transition state is called the activated complex.

This is the energy of the reactants, Ereact

This is the energy of the products, Eprod

This is the net energy for the reaction, ∆E

The height of the “hill” is called the activation energy, Ea

Reaction energy diagram for CH3Br + OH- → CH3OH + Br -

Reaction energy diagram for CH3Br + OH- → CH3OH + Br -

The transition state or activated complex has definite bond lengths and bond angles

Entropy on the Molecular Scale• Ludwig Boltzmann (1844-1906)

described the concept of entropy on the molecular level.

• Temperature is a measure of the average kinetic energy of the molecules in a sample.

An animation of the Maxwell-Boltzman distribution for molecular speeds in a gas can be found at http://www.chm.davidson.edu/chemistryapplets/KineticMolecularTheory/Maxwell.html

Entropy on the Molecular Scale• Molecules exhibit several types of motion:Vibrational: Periodic motion of atoms within a

molecule. (This is the lowest energy state)• Vibrations occur at absolute zero. Called zero-point energy.

Rotational: Rotation of the molecule on about an axis or rotation about bonds.

Translational: Movement of the entire molecule from one place to another. (This is the highest energy state)

Entropy on the Molecular Scale• Boltzmann envisioned the motions of a sample of

molecules at a particular instant in time.This would be akin to taking a snapshot of all the

molecules.• He referred to this sampling as a microstate of the

thermodynamic system.

Maxwell–Boltzmann Distributions

Temperature is defined as a measure of the average kinetic energy of the molecules in a sample.

• At any temperature there is a wide distribution of kinetic energies.

Maxwell–Boltzmann Distributions

As the temperature increases, the curve flattens and broadens.Thus, at higher temperatures, a larger population of molecules has higher energy.

Maxwell–Boltzmann Distributions• If the dotted line represents the activation energy,

as the temperature increases, so does the fraction of molecules that can overcome the activation energy barrier.

As a result, the reaction rate increases.

Arrhenius Equation

Svante Arrhenius developed a mathematical relationship between kand Ea:

k = A e−Ea/RT

where A is the frequency factor, a number that represents the likelihood that collisions would occur with the proper orientation for reaction.A must be determined experimentally

Arrhenius Equation

If k is determined experimentally at two or more temperatures, T1 and T2, you can select two points from your data or graph, and Ea can be calculated from the equation:

[ ]2 1

2 a

1

[k ] E 1 1ln = - -[k ] R T T

R = gas constant = 8.31 J/mol KT = temperature in K

Reaction Rates

Rates of reactions can be determined by monitoring the change in concentration of either reactants or products as a function of time.

Reaction Rates

In this reaction, the concentration of butyl chloride, C4H9Cl, was measured at various times.

C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)

Reaction Rates

The average rate of the reaction over each interval is the change in concentration divided by the change in time:

Average rate =[C4H9Cl]

t


Reaction Rates

• Note that the average rate decreases as the reaction proceeds.

• This is because as the reaction goes forward, concentration of reactant decreases and there are fewer collisions between reactant molecules.


Reaction Rates

• A plot of concentration vs. time for this reaction yields a curve like this.

• The slope of a line tangent to the curve at any point is the instantaneous rate at that time.


Reaction Rates

• All reactions slow down over time.

• Therefore, the rate of a reaction is generally stated as the instantaneous rate near the beginning.


Reaction Rates and Stoichiometry

• In this reaction, the mole ratio of C4H9Cl to C4H9OH is 1:1.

• Thus, the rate of disappearance of C4H9Cl is the same as the rate of appearance of C4H9OH.


Rate = -[C4H9Cl]t = [C4H9OH]

t


• What if the ratio is not 1:1?

2 HI(g) H2(g) + I2(g)

•Then:

Rate = − 12[HI]t

= [I2]t

The ½ represents the stoichiometery of the reaction


• To generalize, for the reaction

aA + bB cC + dD

Rate = − 1a

[A]t = − 1

b[B]t = 1

c[C]t

1d

[D]t=

One can gain information about the rate of a reaction by seeing how the rate changes with changes in concentration

Concentration and Rate

NH4+(aq) + NO2

−(aq) N2(g) + 2 H2O(l)

For the reaction

Data is collected for several experiments:

Note that the concentrations are always changed, one at a time, in a systematic manner.

Concentration and Rate

• Comparing Experiments 1 and 2, when [NH4+] doubles, the

initial rate doubles. Note: [NO2−] is constant.

• Also, comparing Experiments 2 and 3, when [NH4+] doubles,

the initial rate doubles.

NH4+(aq) + NO2

−(aq) N2(g) + 2 H2O(l)

Rate Laws• A rate law equation shows the relationship

between the reaction rate and the concentrations of reactants.

Rate = k [NH4+] [NO2

−]• The exponents of the concentration terms tell the

order of the reaction with respect to each reactant.

• For a zero order reaction, the rate is independent of concentration of reactants

• For a first order reaction, the rate depends on the concentration of one reactant

• For a second order reaction, the rate depends on the concentration of two molecules of reactant(s)

Rate Laws


−]• This reaction is

First-order in [NH4+]

First-order in [NO2−]

The overall reaction order can be found by adding the exponents on the reactants in the rate law.

• m + n = 1 + 1 = 2• This reaction is second-order overall.

NH4+(aq) + NO2

−(aq) N2(g) + 2 H2O(l)

For the reaction

Rate Laws


−]• To calculate the value of k, substitute data from any

experiment• Using Exp. 1:

5.4 x 10-7 M/s = k [0.0100] [0.200]• Solve for k

k = 2.7 x 10-4 M-1s-1

• All the k values should be approximately the same at the same temperature.

NH4+(aq) + NO2

−(aq) N2(g) + 2 H2O(l)

For the reaction

First-Order Processes

If a reaction is first-order, a plot of ln [A] vs. t will yield a straight line, and the slope of the line will be -k.


Consider the process in which methyl isonitrile is converted to acetonitrile.

CH3NC CH3CN


This data was collected for this reaction at 198.9°C.

CH3NC CH3CN


• When ln P is plotted as a function of time, a straight line results.

• Therefore,The process is first-order.k is the negative slope: 5.1 x 10-5 s−1.

Second-Order Processes

If a process is second-order in A, a plot of 1/[A] vs. t will yield a straight line, and the slope of that line is k.

Second-Order ProcessesThe decomposition of NO2 at 300°C is described by the equation

NO2 (g) NO (g) + 1/2 O2 (g)

and yields data comparable to this:

Time (s) [NO2], M0.0 0.01000

50.0 0.00787100.0 0.00649200.0 0.00481300.0 0.00380

Second-Order Processes• Graphing ln [NO2] vs. t

yields:

Time (s) [NO2], M ln [NO2]0.0 0.01000 −4.610

50.0 0.00787 −4.845100.0 0.00649 −5.038200.0 0.00481 −5.337300.0 0.00380 −5.573

• The plot is not a straight line, so the process is notfirst-order in [A].

Second-Order Processes• Graphing ln

1/[NO2] vs. t, however, gives this plot.

Time (s) [NO2], M 1/[NO2]0.0 0.01000 100

50.0 0.00787 127100.0 0.00649 154200.0 0.00481 208300.0 0.00380 263

• Because this is a straight line, the process is second-order in [A].

Summary of the Rate Law Graphs

Half-Life

• Half-life is defined as the time required for one-half of a reactant to react.

• Because [A] at t1/2 is one-half of the original [A],

[A]t = 0.5 [A]0.

Half-life

Half-Life: First Order ReactionFor a first-order process, the equation

for half-life is

= t1/20.693

k

Where k is the rate constant for the reaction0.693 = ln 2

Temperature and Rate

• Generally, as temperature increases, so does the reaction rate.

• This is because k is temperature dependent.

Catalysts• Catalysts

increase the rate of a reaction by decreasing the activation energy of the reaction.

• Catalysts change the mechanism by which the process occurs.

A Homogeneous CatalystBromide ion catalyzes the reduction of hydrogen peroxide in acid solution

2 H2O2 + 2 Br - + 2 H+ → H2O2 + 2 H2O + Br2 → 2 H2O + O2 + 2 Br - + 2 H+

A Heterogeneous Catalyst

H2C=CH2 (g) + H2 (g) → CH3-CH3 (g)

The hydrogenation of ethene (ethylene):

CATALYSISCatalytic converters in auto exhaust systems

The catalytic converter contains a ceramic surface that is coated with a thin later of platinum, palladium, and/or rhodium.

Older catalytic converters used ceramic beads.

Newer catalytic converters use a honeycomb structure.

Reduction of exhaust gases, mixed with air, takes place on the metal surface:

2 CO + O2 → 2 CO2

2 NO → N2 + O2

EnzymesEnzymes are catalysts in biological systems.

In one type of reaction, the substrates fit into the active site of the enzyme much like a key fits into a lock.

In a second type of reaction the substrates fit into a more generalize active site of the enzyme

Chemical Equilibrium

The Concept of Equilibrium

Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate.

Kinetics applies to the speed of a reaction, the concentration of product appearing (or of reactant disappearing) per unit time

Equilibrium applies to the extent of a reaction, the concentration of product that has appeared given unlimited time, or when no further macroscopic change occurs.

The Cobalt(II) Chloride EquilibriumCo(H2O)6

2+(aq) + 4 Cl(aq) CoCl4 (aq) + 6 H2O (l)

The equilibrium systemN2O4 (g) 2 NO2 (g)

Initially, N2O4 (colorless) is added to the reaction container

When equilibrium is reached, the numbers of N2O4 and NO2molecules is constant and the color remains unchanged

As the system moves toward an equilibrium condition, the color changes as more NO2 (red-brown) is formed

The Concept of Equilibrium

• As a system approaches equilibrium, both the forward and reverse reactions are occurring.

• At equilibrium, the forward and reverse reactions are proceeding at the same rate.

A System at Equilibrium

Once equilibrium is achieved, the amount of each reactant and product remains constant.

A system at equilibrium is dynamic on the molecular level, that is, both the forward and the reverse reactions are still taking place at the same rate.

rate forward = rate backward

No net change is observed because changes in one direction are balanced by changes in the other.

Depicting Equilibrium

To show an equilibrium system, we write its equation with a double arrow

N2O4 (g) 2 NO2 (g)

Chemical EquilibriumIron(III) and thiocyanate

+

Fe(H2O)63+

(aq) Fe(SCN)(H2O)52+

(aq)+ SCN-(aq) + H2O (l)

The Equilibrium

Constant

The Equilibrium Constant

• Forward reaction:N2O4 (g) 2 NO2 (g)

• Rate law:Rate = kf [N2O4]

• Reverse reaction:2 NO2 (g) N2O4 (g)

• Rate law:Rate = kr [NO2]2

The Equilibrium Constant• At equilibrium

Ratef = Rater

• Substitute the rate equations

kf [N2O4] = kr [NO2]2

• Rewriting this, it becomes

kfkr

[NO2]2[N2O4]

=


The ratio of the rate constants is a constant at that temperature, so the constants are combined into a single constant, and the expression becomes

= Keqkfkr

[NO2]2[N2O4]

=

The Equilibrium Constant• To generalize this expression, consider

the reaction

• The equilibrium expression for this reaction would be

• This equation is known as the Law of Mass Action

Keq = [C]c[D]d[A]a[B]b

aA + bB cC + dD

The Equilibrium Constant• For any reaction of the form:

• The equilibrium expression is

The values of a, b, c, and d are those of the coefficients in the balanced chemical equation.

Note that this is equilibrium, not kinetics.

Equilibrium is a State Function, that is, the value of Kc depends on the concentrations of the reactants and the products.

Keq = [C]c[D]d[A]a[B]b

aA + bB cC + dD


Because pressure is proportional to concentration for gases in a closed system, the equilibrium expression for 4 gases, A, B, C, and D, can also be written

Kp = (PC)c (PD)d

(PA)a (PB)b

Equilibrium Can Be Reached from Either Direction

Examining the data in the table, above, the equilibrium ratio of [NO2]2 to [N2O4] remains constant at this temperature no matter what the initial concentrations of NO2 and N2O4 are.

Note: There must be sufficient quantities of compounds to reach an equilibrium condition


This is the data from the last two trials from the table on the previous slide.


It does not matter whether we start with N2 and H2or whether we start with NH3. We will have the same proportions of all three substances at equilibrium at the specified temperature.

What Does the Value of KMean?

• If K >> 1, the reaction is product-favored; product predominates at equilibrium.

What Does the Value of KMean?

• If K >> 1, the reaction is product-favored; product predominates at equilibrium.

• If K << 1, the reaction is reactant-favored; reactant predominates at equilibrium.

Heterogeneous Equilibria

The Concentrations of Solids and Liquids Are Considered to

be Constant

Both can be obtained by dividing the density of the substance by its molar mass—and both of these are constants at constant temperature.

The Concentrations of Solids and Liquids Are Considered to be

ConstantTherefore, the concentrations of solids and liquids do not appear in the equilibrium expression

Kc = [Pb2+] [Cl−]2

PbCl2 (s) Pb2+(aq) + 2 Cl−(aq)

As long as some CaCO3 or CaO remain in the system, the amount of CO2 above the solid will remain the same.

CaCO3 (s) CO2 (g) + CaO(s)

What Are the Equilibrium Expressions for These Equilibria?

SnO2 (s) + 2 CO (g) Sn (s) + 2 CO2 (g)

CaCO3 (s) CaO (s) + CO2 (g)

Zn (s) + Cu2+ (aq) Cu (s) + Zn2+

(aq)

Equilibrium Calculations

Equilibrium Calculations

A closed system initially containing1.000 x 10−3 M H2 and 2.000 x 10−3 M I2At 448C is allowed to reach equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 x 10−3 M. Calculate Kcat 448C for the reaction taking place, which is

H2 (g) + I2 (g) 2 HI (g)

What Do We Know?(This is our initial data)

[H2], M [I2], M [HI], M

Initially 1.000 x 10-3 2.000 x 10-3 0

Change

At equilibrium

1.87 x 10-3

Note: We assume there is no HI present initially

Determine changes:[HI] Increases by 1.87 x 10-3 M

[H2], M [I2], M [HI], M

Initially 1.000 x 10-3 2.000 x 10-3 0

Change +1.87 x 10-3

At equilibrium

1.87 x 10-3

Stoichiometry tells us [H2] and [I2]decrease by half as much

[H2], M [I2], M [HI], MInitially 1.000 x 10-3 2.000 x 10-3 0

Change -9.35 x 10-4 -9.35 x 10-4 +1.87 x 10-3

At equilibrium

1.87 x 10-3

H2 (g) + I2 (g) 2 HI (g)

We can now calculate the equilibrium concentrations of all

three compounds…

[H2], M [I2], M [HI], M

Initially 1.000 x 10-3 2.000 x 10-3 0

Change -9.35 x 10-4 -9.35 x 10-4 +1.87 x 10-3

At equilibrium

6.5 x 10-5 1.065 x 10-3 1.87 x 10-3

…and substitute the concentrations into the

equilibrium constant expression

Kc = [HI]2[H2] [I2]

= 51

=(1.87 x 10-3)2

(6.5 x 10-5)(1.065 x 10-3)

Note that the equilibrium value has no units

Le Châtelier’s Principle

Le Châtelier’s PrincipleHenri Louis Le Châtelier (1850-1936) In 1884, Le Chatelier stated:

Any system in stable chemical equilibrium, subjected to the influence of an external cause which tends to change either its temperature or its condensation (pressure, concentration, number of molecules in unit volume), either as a whole or in some of its parts, can only undergo such internal modifications as would, if produced alone, bring about a change of temperature or of condensation of opposite sign to that resulting from the external cause.

In 1888, he restated this as:Every change of one of the factors of an equilibrium occasions a rearrangement of the system in such a direction that the factor in question experiences a change in a sense opposite to the original change.


Our modern statement is: If a system at equilibrium is disturbed by a change in temperature, pressure, or the concentration of one of the components, the system will shift its equilibrium position so as to counteract the effect of the disturbance.


Effect of a change in T:change in K therefore change in P or concentrations at

equilibriumChange in concentration (Add or take away reactant or

product):K does not changeReaction adjusts to new equilibrium “position”

Use a catalyst:reaction comes more quickly to equilibrium. K not changed.

The Effect of Changes in TemperatureCo(H2O)6

2+(aq)+ 4 Cl- (aq) + HEAT CoCl42-

(aq) + 6 H2O (l)

SOLUTION:

Predicting the Effect of a Change in Concentrationon the Equilibrium Position

PROBLEM: To improve air quality and obtain a useful product, sulfur is often removed from coal and natural gas by treating the fuel contaminant hydrogen sulfide with O2;

2H2S(g) + O2(g) 2S(s) + 2H2O(g)

What happens to(a) [H2O] if O2 is added? (b) [H2S] if O2 is added?

(c) [O2] if H2S is removed? (d) [H2S] if sulfur is added?

Write an expression for Q and compare it to K when the system is disturbed to see in which direction the reaction will progress.

Q = [H2O]2[H2S]2[O2]

(a) When O2 is added, the denominator increases, so Q decreases. The reaction must progress to the right to come back to K. Therefore [H2O] increases.

Predicting the Effect of a Change in Concentration on the Equilibrium Position

(b) When O2 is added, the denominator increases and Q decreases. The reaction must progress to the right to come back to K. Therefore [H2S] decreases.

Q = [H2O]2

[H2S]2[O2]

(c) When H2S is removed, the denominator decreases and Q increases. The reaction must progress to the left to come back to K. Therefore [O2] increases.

(d) Sulfur is not part of the Q (K) expression because it is a solid. Therefore, as long as some sulfur is present the reaction is unaffected. [H2S] is unchanged.

+

lower P(higher V)

less moles per liter of gas shift to the left

higher P(lower V)

More moles per liter of gas shift to the right

The effect of pressure (volume) on an equilibrium system

Catalysts increase the rate of both the forward and reverse reactions.

Equilibrium is achieved faster, but the equilibrium composition remains unaltered.

The Effect of Various Changes on an Equilibrium System

The Haber-Bosch ProcessThe Haber-Bosch process is the transformation of atmospheric nitrogen and hydrogen into ammonia (NH3) for the production of ammonia-based fertilizers.Initially developed by Fritz Haber (1868-1934) in 1905 by passing a mixture of N2 and H2 over an iron catalyst at 1000C. Later, Haber modified the process by increasing the pressure to 150-200 atm. over a catalyst at 500C.In 1908, BASF acquired the process and assigned Carl Bosch (1874-1940) the task of scaling the process up to industrial quantities. Bosch’s modifications of the Haber process provided ammonium sulphate for use as a fertilizer for the soil.

N2 (g) + 3 H2 (g) 2 NH3 (g) + 92 kJ.

In 1914, Germany’s supplies of sodium and potassium nitrates for making explosives were blocked off by the Allied forces. Using the Haber-Bosch process, they were able to produce explosives, prolonging World War I.

Fritz Haber

Carl Bosch

The Haber-Bosch Process

If H2 is added to the system, N2 will be consumed and the two reagents will form more NH3.

K = 3.5 x 108 at 298 K

The Haber-Bosch Process

This apparatus helps push the equilibrium to the right by removing the ammonia (NH3) from the system as a liquid.

T (K) Kc

200.300.400.500.600.700.800.

7.17 x 1015

2.69 x 108

3.94 x 104

1.72 x 102

4.53 x 100

2.96 x 10-1

3.96 x 10-2

Effect of Temperature on Kc for Ammonia Synthesis

Percent yield of ammonia vs. temperature (C) at five different operating pressures

kinetics and equilibrium chm 130 - 2012.ppt and equilibrium chm 130.pdf · collision theory of...

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