Unit 9: Kinetics and Equilibrium

Chapter 13

Pre-AP Chemistry I

Edmond North High School

Collision Theory• The following three statements summarize the collision

theory. – 1. Particles must collide in order to react. – 2. The particles must collide with the correct orientation.– 3. The particles must collide with enough energy to form an

activated complex, which is an intermediate particle made up of the joined reactants.

• An effective collision is one that results in a reaction; new products are formed.

Activated Complex• The Activated Complex is an

intermediate particle formed when reactants collide and stick together.– Old bonds are breaking while

new bonds are forming.• The minimum amount of

energy colliding particles must have in order to form an activated complex is called the activation energy. – Particles that collide with less

than the activation energy cannot form an activated complex.

Exothermic Reactions Review• An exothermic reaction releases heat,

and an endothermic reaction absorbs heat.

Endothermic Reactions Review• The endothermic reaction absorbs heat

because the products are at a higher energy level than the reactants.

Kinetics• Kinetics is the study

of reaction rates and the factors that effect them.

• Why is it important?– When we understand

reaction rates we can control chemical reactions and use them for specific purposes.

Kinetics• As a reaction occurs:

– concentration of reactants decreases– concentration of products increases

• Kinetics studies the rate at which a chemical process occurs.• Besides information about the speed at which reactions occur,

kinetics also sheds light on the reaction mechanism (exactly how the reaction occurs).

Reaction Rate• Reaction Rate is the number of atoms,

ions or molecules that react in a given time to form products– We measure either the rate of disappearance

of the reactants or the rate of appearance of one of the products

– In simpler terms:• The average rate is the change in a given quantity

during a specific period of time.

Measuring Reaction Rates• Change in Electrical Conductivity• Change in Color• Change in Pressure• Change in Volume

Reaction Rate Graph• For the reaction H2 + I2 → 2HI,

– What is happening to the concentration of the reactants?

– What is happening to the concentration of the products?

Reaction Rates & Stoichiometry• We use coefficients to give us a ratio of

rates between species in a reaction.– In the reaction 3H2 + N2 2NH3, since N2 has a

coefficient of 1 and NH3 has a coefficient of 2, the rate N2 is used = 2(rate NH3 is created)

– Likewise, since H2 has a coefficient of 3 and N2 has a coefficient of one, it is used 3 times as fast as N2.

Factors Affecting Reaction Rates• The reaction rate for almost any chemical

reaction can be modified by varying the conditions of the reaction. – An important factor that affects the rate of a

chemical reaction is the reactive nature of the reactants. As you know, some substances react more readily than others. • The more reactive a substance is, the faster the

reaction rate.

Factors That Affect 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.

• Concentration of Reactants– As the concentration of reactants increases, so does the

likelihood that reactant molecules will collide.

Factors that Effect Reaction Rate• Particle Size (surface

area)– For solids, breaking

up big pieces increases surface area, increasing rate by having more places for the molecules to interact

– Ionic compounds have more surface area than covalent

Factors that Effect Reaction Rate• Pressure – Gases

only!– As pressure

increases the concentration increases, so you will have more collisions

Factors That Affect Reaction Rates• Temperature

– At higher temperatures, reactant molecules have more kinetic energy, move faster, and collide more often and with greater energy.

• Presence of a Catalyst– Catalysts speed up reactions by changing the

mechanism of the reaction.• Catalysts are not consumed during the course of

the reaction.

Catalyst

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.

– One way a catalyst can speed up a reaction is by holding the reactants together and helping bonds to break.

How Do Catalysts Work?• Instead of increasing KE, the catalyst lowers the

Ea

– It creates a new pathway with a lower activation energy

• It is never consumed in the reaction so is not considered a reactant– Enzymes are biological catalysts.

• Homogenous Catalysts are in the same phase as the reactants.

• Heterogeneous Catalysts are in a different phase as the reactants.

Factors that Effect Reaction Rate• Inhibitors

– Compounds added to a reaction that slow it down• Examples: Lead in diesel, preservatives in food.

Reaction Rate Laws• Rate Laws relate the reaction rate and

reactant concentration• We write the rate laws for elementary

reactions from the equation.• They include only the gaseous or aqueous

reactants!– Solids & liquids are not included because they

have a constant concentration that does not increase or decrease

Rate Law• A Rate Law is an expression that shows how the rate

depends on concentration of reactants

• Example:– Rate laws use reactants only– k is proportionality constant (called the rate constant)– n is the order (exponent) of the reactant determined

experimentally• n is generally the coefficient of the chemical in the balanced

equation HOWEVER experimental data overrides the coefficient.– Reaction is xth order in A– Reaction is yth order in B– Reaction is (x +y)th order overall

Rate = k [A]x[B]yaA + bB cC + dD

Order of Reaction• Order of Reaction

– For the reaction with the rate = k[A]2[B]• Rate is 2nd order with respect to A• Rate is 1st order with respect to B• Rate is 3rd order overall (add exponents)

• How does concentration affect reaction rate?– If [A] doubles what happens to the rate?– If [B] triples, what happens to the rate?

Initial Rate Method• Reactions are reversible.

– As products accumulate they can begin to turn back into reactants.

• Early on the rate will depend on only the amount of reactants present.– We want to measure the reactants as soon as

they are mixed.– This is called the Initial Rate Method.

Method of Initial Rates• Used to find the form of the rate law

1) Choose one reactant to start

2) Find two experiments where the concentration of that reactant changes but all other reactants stay the same

3) Write the rate laws for both experiments

4) Divide the two rate laws

5) Use log rules to solve for the order

6) Follow the same technique for other reactants

Determining Experimental Rate LawRun # Initial [A]

([A]0) Initial [B] ([B]0)

Initial Rate (v0)

1 1.00 M 1.00 M 1.25 x 10-2 M/s

2 1.00 M 2.00 M 2.5 x 10-2 M/s

3 2.00 M 2.00 M 2.5 x 10-2 M/s

• What is the order with respect to A?• What is the order with respect to B?• What is the overall order of the reaction?• What is the value for k?

Determining Experimental Rate Law

[NO(g)](mol dm-3)

[Cl2(g)](mol dm-3)

Initial Rate (mol dm-3 s-1) 

0.250  0.250  1.43 x 10-6

0.250  0.500  2.86 x 10-6

0.500  0.500  1.14 x 10-5

• What is the order with respect to Cl2?

• What is the order with respect to NO?• What is the overall order of the reaction?• What is the value for k?

Remember!• Two key points

1) The concentration of the products do not appear in the rate law because this is an initial rate.

2) The order must be determined experimentally, can’t be obtained from the equation

Temperature and Rate• Generally, as

temperature increases, so does the reaction rate.

• This is because k is temperature dependent.

Reaction Mechanisms• In a complex reaction, a certain series of

steps must occur with correct fit and in an exact sequence in order to yield the reaction products. This series of steps is called a reaction mechanism.– When we write a chemical equation we show

the overall reaction (reactants → products)• We call the compounds that

form/decompose during the steps but are not in the overall reaction Intermediates.

Reaction Mechanisms• Elementary Reactions, are the simplest one-step

reactions in a reaction mechanism.• Example:

– HBr + O2 → HOOBr (slow)

– HOOBr + HBr → 2HOBr (fast)– 2HOBr + 2HBr → 2H2O + 2Br2 (fast)

• The Rate-determining step is the slowest step in the chemical reaction.

• Which is the rate-determining step for the reaction above?

Multistep Mechanisms• There is an activation energy for each

elementary step.• In a multistep process, one of the steps

will be slower than all others.– Slowest step (rate determining) must have

the highest activation energy.• The overall reaction cannot occur faster

than this slowest, rate-determining step.

Intermediates2NO2 + F2 2NO2F

• Rate = k[NO2][F2]

• The proposed mechanism is– NO2 + F2 NO2F + F (slow)

– F + NO2 NO2F (fast)

• F is called an intermediate. It is formed then consumed in the reaction.

Reaction Intermediates• Example:

– 2 Steps: NO2(g) + NO2 (g) NO3 (g) + NO(g)– NO3(g) + CO(g) NO2 (g)

+CO2 (g)

• Overall: NO2(g) + CO(g) NO(g) + CO2(g)

Slow Initial Step

• A proposed mechanism for this reaction is• Step 1: NO2 + NO2 NO3 + NO (slow)

• Step 2: NO3 + CO NO2 + CO2 (fast)– The NO3 intermediate is consumed in the second

step.

• As CO is not involved in the slow, rate-determining step, it does not appear in the rate law: Rate = k [NO2]2

– CO is necessary for this reaction to occur, but the rate of the reaction does not depend on its concentration.

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

Fast Reactions• 2 IBr I2+ Br2

– Mechanism• IBr I + Br (fast)• IBr + Br I + Br2 (slow)

• I + I I2 (fast)

• Rate = k[IBr][Br] but [Br]= [IBr]• Rate = k[IBr][IBr] = k[IBr]2

Indicators of Completed Reactions• So how do you know

when a reaction has gone to completion?– A gas is produced and

escapes (open container)

– A precipitate forms from two aqueous solutions.

– A covalent product is formed (usually water)

What is Equilibrium?• When a reaction results in

almost complete conversion of reactants to products the reaction goes to completion.

• Most reactions, however, do not go to completion. They appear to stop. – The reason is that these

reactions are reversible.

• A reversible reaction is one that can occur in both the forward and the reverse directions.

Chemical Equilibrium• Chemical equilibrium occurs when

two opposite reactions occurring at the same time and rate

A + B ↔ AB• Forward & Reverse reactions don’t

stop at equilibrium, just looks that way because concentration remains constant–Rateforward reaction = Ratereverse reaction

Equilibrium Example

• You have a bridge between 2 cities. The number of cars going in either direction on the bridge are equal (at equilibrium)

• The populations of the cities on either side of the bridge do not have to be equal!

Equilibrium Position• The equilibrium position of a reaction is

determined by:– Initial concentrations– Energy of reactants and products– Degree of organization of reactants and

products• GOAL of reactions at equilibrium is to:

– Lower energy

Progression of Equilibrium• As a reaction progresses:

– [A] decreases to a constant,– [B] increases from zero to a constant.– When [A] and [B] are constant, equilibrium is

achieved.

• In a system at equilibrium, both the forward and reverse reactions are being carried out; as a result, we write its equation with a double arrow.

A B

A System at Equilibrium

• Once equilibrium is achieved, the amount of each reactant and product remains constant.– Rates are equal; concentrations are not.– The concentrations do not change at equilibrium.

Equilibrium Positions • If equilibrium lies “to the left”:

– There are more reactants and less products.• If equilibrium lies “to the right”:

– There are less reactants and more products.• If reactants are mixed and concentrations

do not change:– The reaction could already be at equilibrium.– Reaction rates are so slow that change is too

difficult to detect.

Equilibrium Expressions• Equilibrium expressions relate

concentrations of reactants to those of the products–These can be written from balanced

equations–They look similar to rate laws! BE

CAREFUL!• Uses both reactants & products• Include only the gaseous or aqueous

phases as with rate laws

Deriving an Equilibrium Expression• 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

• Therefore, at equilibrium: Ratef = Rater

– kf [N2O4] = kr [NO2]2

• Rewriting this, it becomesKeq =

kfkr

[NO2]2

[N2O4]=

General Equilibrium Expressions• A (s) + 2B (g) ↔ 2C (g) + 3D (g)

Keq = [C]2[D]3

[B]2

• Keq = equilibrium constant (capital K)– Numerical value of the ratio of product

concentrations to reactant concentrations• [ ] = concentration in M (mol/L)• Order = coefficient becomes exponent

Solids and Liquids Are 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.– Therefore, the concentrations of solids and

liquids do not appear in the equilibrium expression.

Kc = [Pb2+] [Cl−]2

PbCl2 (s) Pb2+ (aq) + 2 Cl−(aq)

Equilibrium Constant, K• If we know the value of K, we can predict:

– The tendency of a reaction to occur.– If a set of concentrations could be at

equilibrium.– The equilibrium position, given initial

concentrations.• K will always have the same value at a

certain temperature.– No matter what amounts are added, the ratio

at equilibrium will always be same

Equilibrium Constant (Keq)

• Tells you whether the products or reactants are favored– Keq > 1

• Products are favored (forward rxn); lots of product is made

– Keq < 1• Reactants are favored (reverse

rxn); not much product made

• The size of K and time needed to reach equilibrium are NOT related– This is determined by reaction

rate (Ea)

The Units for K• Are determined by the various powers and

units of concentrations.• They depend on the reaction.• At any temperature.

– Temperature affects rate.– Equilibrium position is a set of concentrations

at equilibrium.

Solving Problems Using the Equilibrium Constant (RICE Tables)1) Write equation and set up the RICE table.

2)  Begin filling in the RICE table with the initial concentration of each species given in the problem.

3) Fill in line C of the RICE table, the change in concentrations of all species.

4) Fill in line E the RICE table, the concentration of species when the system is at equilibrium.

5)  Write the equilibrium expression and plug in the equilibrium concentrations as they are represented in the “E” line in the ICE table.

Solving Problems Using the Equilibrium Constant (RICE Tables)6) Make any assumptions if convenient.

• If K is very small, change in the reactants can be ignored.

7)  Determine the numerical value of x.

8)  Check any assumptions that were made. If the assumption was not valid, use the quadratic equation to determine x.

9)  Use the value of x to determine the equilibrium concentrations of all species. The equations written in line E of the ICE table should be used to do this.

Initial Concentration of I2: 0.50 mol/2.5L = 0.20 MI2 2 I

Initialchangeequi

0.20 0-x +2x0.20-x 2x 10

2

10

2

2

1094.2]20.0[

]2[

1094.2][

][

xx

x

xI

IKeq

With an equilibrium constant that small, whatever x is, it’s near nothing, and 0.20 minus nothing is 0.20 (like a million dollars minus a nickel is still a million dollars).

0.20 – x is the same as 0.20

102

1094.220.0

]2[ xx

x = 3.83 x 10-6 M

More than 3orders of mag.between thesenumbers. The simplification willwork here.

Example

Manipulating K: Reverse Rx• The equilibrium constant of a reaction in

the reverse reaction is the reciprocal of the equilibrium constant of the forward reaction.

10.212

=

Kc = = 0.212 at 100C[NO2]2

[N2O4]N2O4 (g) 2 NO2 (g)

Kc = = 4.72 at 100C

[N2O4][NO2]2N2O4 (g)2 NO2 (g)

Manipulating K: Multiplied Rx• The equilibrium constant of a reaction that

has been multiplied by a number is the equilibrium constant raised to a power that is equal to that number.

Kc = = 0.212 at 100C[NO2]2

[N2O4]N2O4 (g) 2 NO2 (g)

Kc = = (0.212)2 at 100C[NO2]4

[N2O4]22 N2O4 (g) 4 NO2 (g)

Manipulating K: Adding Rx• The equilibrium constant for a net reaction

made up of two or more steps is the product of the equilibrium constants for the individual steps.

The Reaction Quotient (Q)• To calculate the reaction quotient or Q,

one substitutes the initial concentrations on reactants and products into the equilibrium expression.– Q gives the same ratio the equilibrium

expression gives, but for a system that is not at equilibrium.

– It is used to tell if a reaction is at equilibrium or not.

Reaction Quotient: Comparing Q and K

• The relationship between Q and K tells which way the reaction will shift– Q = K: at equilibrium,

no shift– Q > K: too large, forms

reactants, shift to left– Q < K: too small,

forms products, shift to right

Le Chatelier’s Principle• Le Chatelier’s Principle states

that if stress is added to a system at equilibrium, the reaction will speed up in the direction that will relieve the stress.– Once the stress is relieved,

equilibrium is re-established and no further changes are noticed.

• 4 Types of stress– Concentration– Temperature– Pressure– Volume

Stress Factors1. Concentration

– Shifts away from an increase or addition– Shifts toward a decrease or subtraction

Change in Concentration• Adding product makes Q > K• Removing reactant makes Q > K• Adding reactant makes Q < K• Removing product makes Q < K • Determining the effect on Q, will tell you

the direction of shift– System will shift away from the added

component or towards a removed component.

Stress Factors2. Temperature (treat heat as a reactant if

endothermic or product if exothermic)– Shifts away from an addition– Shifts toward a subtraction

Change in Temperature• Temperature affects the rates of both the

forward and reverse reactions.– Doesn’t just change the equilibrium position,

changes the equilibrium constant.

Stress Factors

3. Pressure (only effects gases)– Increase pressure,

shifts to side with lower total moles of gas

– Decrease pressure, shift to side with higher total moles of gas

CO (g) + H2 (g) ↔ CH4 (g) + H20 (g)

Change in Pressure• Adding or removing gaseous reactant or product

is same as changing conc.• Adding inert or uninvolved gas

– Increase the total pressure – Doesn’t effect the equilibrium position

Stress Factors4. Changes in the volume • Suppose the volume of the reaction

vessel for the system is decreased, resulting in an increase in pressure.– The equilibrium will shift to relieve the stress

of increased pressure.

Change in Volume• Decrease V:

– Decreases # gas molecules– Shifts towards the side of the reaction with

less gas molecules• Increase V:

– Increase in # of gas molecules– Shifts towards the side of the reaction with

more gas molecules

Stress Factor PracticeN2 (g) + 3H2 (g) ↔ 2NH3 (g) + heat

• Which way will the reaction shift if you:– Add N2

– Decrease pressure– Remove NH3

– Heat– Add a catalyst

The End.

Be Prepared for Unit 9 Test.