Unit 8Chemical Kinetics & Thermodynamics

Chemical Kinetics• Chemical kinetics is the study of the

factors that affect the speed of a reaction and the mechanism by which a reaction proceeds.• Rate is how much

a quantity changesin a given period of time.

• Collision Theory says that in order for reactions to occur between substances, particles must collide.• For a collision to be

effective, particles must:

1. Collide with sufficient energy.

2. Have a favorable orientation.

Collision Theory

Effect of Orientation on a Collision

Smaller pieces = larger surface area

• The rate of a chemical reaction depends on the number of collisions between particles.

• Rate-influencing factors are:• Nature of Reactants• Surface Area• Temperature• Concentration• Presence of Catalysts

Rate-Influencing Factors

Visual Concept

Effect of Concentration• The rate of a reaction often depends on the

concentration of one or more of the reactants.• Greater concentration of reactant particles =

more collisions = increase in reaction rate.

Rate Law Expressions• Rate law equations are based on the Molarity

concentrations of the reactant(s). (Indicate Molarity concentrations with brackets.)• The Rate (R) involves the product of the

concentrations of the individual reactants.• Experiments have proven that coefficients of

reactants have an exponential effect on R.

Ex: For the equation A + 2 B → C

R = k [A][B]2 (k = rate law constant)

Rate Law ExpressionsSample Problem

For the following single-step chemical reaction:

2SO2 + O2 → 2 SO3

1. Write the rate law expression.

R = k [SO2]2 [O2]

2. What happens to the Rate if the [SO2] is cut by ½?

Rate is 4x slower

3. What happens to the Rate if the [O2] is tripled?

Rate is 3x greater

Reaction Mechanism• Chemical reactions often occur as a

series of consecutive steps (this is called a reaction mechanism.)• Products in an early step that are reactants in a

later step are called intermediates.

Example: Overall reaction:

H2(g) + 2 ICl(g) 2 HCl(g) + I2(g) Mechanism:

1) H2(g) + ICl(g) HCl(g) + HI(g)

2) HI(g) + ICl(g) HCl(g) + I2(g)

intermediate

Rate-Determining Step• One of the steps will always occur at a

slower rate than the others.• The rate-determining step is

the slowest step (bottleneck)in a reaction mechanism.• ONLY the rate-determining step

is used for rate law expressions.

Rate-Determining StepSample Problem

Equation: 2NO2 + F2 → 2 NO2F

Step 1: NO2 + F2 → NO2F + F “slow”

Step 2: F + NO2 → NO2F“fast”

1. Write the rate law expression for this reaction.

R = k [NO2] [F2]

2. Given the following data, calculate the [NO2] used:

R = 0.00905 M/s, k = 1.44 x 10-3 1/M●s, [F2] = 1.11 M

[NO2] = 5.66 M

[NO2] =R

k [F2]=

0.00905 M/s(1.44 x 10-3 1/M●s) (1.11 M)

Rate-determining step

Getting a Reaction Started• Energy is needed to overcome the repulsion

between molecules and get a reaction started.• Just as a ball cannot get

over a hill without added energy, a reactioncannot occur unless the molecules have enough energy to get over this initial energy barrier.

Activation Energy• Activation energy (Ea) is the minimum amount

of energy required to get a reaction started.

Exothermic Reactions• In an exothermic reaction,

energy is released. Therefore, the energy of the products must be less than the energy of the reactants.• The great majority of

chemical reactions innature are exothermic.

Endothermic Reactions• In an endothermic reaction, energy is

absorbed. Therefore, the energy of the products must be greater than the energy of the reactants.

Catalysts• Catalysts are substances which affect the rate

of a reaction without being consumed.• Catalysts increase the rate of a reaction by

changing the reaction mechanism and decreasing the activation energy.

Enzymes• Due to the complexity of

organic molecules, most biological reactions needa catalyst to proceed at a reasonable rate.• Protein molecules that

catalyze biological reactionsare called enzymes.• Enzymes pull the reactants onto

an active site that orients them for the reaction.

Rate Order• Experiments have shown that there are 3 ways

that a reactant’s concentration can affect the overall rate of the reaction:• Rate Order 0 ([X]0) - The concentration of a

reactant has no effect on reaction rate.• Rate Order 1 ([X]1) - The concentration of a

reactant has a direct effect on reaction rate.• Rate Order 2 ([X]2) - The concentration of a

reactant has an exponential effect on reaction rate.

• The order of a reaction can be determined only by experiment.

Determining the Order of a Reaction

[A] (M)

Rate (M/s)

0.10 0.015

0.20 0.030

0.40 0.060

x2 x2

[A] (M)

Rate (M/s)

0.10 0.015

0.20 0.015

0.40 0.015

x2 x2

[A] (M)

Rate (M/s)

0.10 0.015

0.20 0.060

0.40 0.240

x1 x4

When [A] doubles, the rate doubles

When [A] doubles, rate doesn’t change

When [A] doubles, the rate quadruples

Rate = k[A]1

First Order

Rate = k[A]0 = k

Zero Order

Rate = k[A]2

Second Order

Rate Order of a ReactionSample Problem

From the data above, determine the following:1. Rate order of each reactant.

second order in [NO2] and zero order in [CO]

2. Overall rate order of the reaction.2 + 0 = 2 second order reaction

3. The rate law for the reaction.Rate = k [NO2]2

[NO2] (M) [CO] (M) Rate (M/s)

0.10 0.10 0.0021

0.20 0.10 0.0082

0.20 0.20 0.0083

0.40 0.10 0.033

x2 x4

constant

constant

x2 x1

Finding the Rate ConstantSample Problem

Using the data above, and the rate law we just found, (R = k [NO2]2), determine the following:

1. Value of k.

2. Reaction rate when the concentration of [NO2] is 0.50 M and [CO] is 1.00 M.

[NO2] (M) [CO] (M) Rate (M/s)

0.10 0.10 0.0021

0.20 0.10 0.0082

0.20 0.20 0.0083

0.40 0.10 0.033

k =R

[NO2]2=

0.0021 M/s(0.10 M)2

= 0.21 M-1s-1

R = k [NO2]2 = (0.21 M-1s-1) (0.50 M)2 = 0.053 M/s

Use data from any one line of chart to find k

Thermodynamics• Thermodynamics is the study of energy

transfer in reactions.

Energy• Energy is the ability to do work.• Some forms of energy are:• Mechanical (kinetic and potential)• Electrical• Heat or thermal• Light or radiant• Nuclear• Chemical

• Energy is commonly measured in joules.• 1 joule is the amount of energy needed to move a

1 kg mass at a speed of 1 m/s. (1 J = 1 )2

2

s

mkg

First Law of Thermodynamics• The First Law of Thermodynamics is the

Law of Conservation of Energy which states that energy cannot be created or destroyed.• However, energy can

be transferred betweenobjects, or between forms. (Ex: heat → light → sound)

Energy Exchange• Conservation of Energy requires that the total

energy change in a system and its surroundings must be zero.• Energy is exchanged between system

and surroundings through either heat exchange or work being done.• Heat is the exchange of thermal energy between

a system and its surroundings.

Enthalpy• the Enthalpy change (DH) of a

reaction is the heat exchangedin a reaction under conditionsof constant pressure.

• DH for a reaction is equal to the difference between the DHof formation of the products andthe reactants.

DHrxn = S nDHf(products) - S nDHf(reactants)

• S means sum• n is the coefficient• Elements in their standard state have a DHf of zero.

Calculating Enthalpy of ReactionSample Problem

Calculate the Enthalpy Change in the Reaction: 2C2H2(g) + 5O2(g) ® 4CO2(g) + 2H2O(l)

Reactant/Product

DHf

(kJ/mol)

C2H2(g) +227.4

CO2(g) -393.5

H2O(l) -285.8

DHrxn = S nDHf(products) - S nDHf(reactants)

DHrxn = [(4•DHCO2 + 2•DHH2O) – (2•DHC2H2 + 5•DHO2)]

DHrxn = [(4•(-393.5) + 2•(-285.8)) – (2•(+227.4) + 5•(0))]

DHrxn = -2145.6 kJ – 454.8 kJ

DHrxn = -2600. kJ

Stoichiometry Involving ∆H• The amount of heat generated or absorbed

during a reaction depends on the amount of the substances reacting.

Example:C3H3(g) + 5O2(g) → 3CO2(g) + 4H2O(g) ∆H = -2044 kJ

Means 2044 kJ of heat is given off when 1 mole of C3H3 reacts, or when 3 moles of CO2 is produced.

• These relationships can be used as ratios in stoichiometric conversions.

Ammonia reacts with oxygen as follows:4NH3(g) + 5O2(g)→ 4NO(g) + 6 H2O(g) ∆H = -906 kJ

Calculate the heat (in kJ) associated with the complete reaction of 155 g of NH3.

mol NH3

mol NH3

Stoichiometry Involving ∆HSample Problem

155 g NH3g NH3

-2070 kJ NH3=17.0

1

4

kJ-906

Hess’s Law• Hess’s Law – the overall enthalpy change in a

reaction is equal to the sum of enthalpy changes for the individual steps in the process.• Possible steps in Hess’s Law:

1. Reverse equation/change sign on ΔH.

2. Multiply or Divide coefficients/multiply or divide ΔH.

Hess’s LawSample Problem

Calculate the enthalpy of formation for CH4:

C(s) + 2H2(g) → CH4(g) ∆Hf = ?

The component reactions are:C(s) + O2(g) → CO2(g) ∆Hc = -393.5 kJ

H2(g) + ½O2(g) → H2O(l) ∆Hc = -285.8 kJ

CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) ∆Hc = -890.8 kJ

CO2(g) + 2H2O(l) → CH4(g) + 2O2(g) ∆Hc = +890.8 kJ

- 2 moles of H2 are used to make CH4,

so multiply the 2nd equation by 2 (including ∆H.)- CH4 is on the products side, not the reactants side, so

reverse the 3rd reaction and change the sign on ∆H.- Cancel unwanted terms and add the ∆H’s.

2 2 -571.6 kJ

-74.3 kJ

-74.3 kJ

Sign of ∆H• In an exothermic reaction, the energy of

the products is less than the energy of the reactants, so ∆H is negative.• In an endothermic reaction, the energy of

the products is greater than the energy of the reactants, so ∆H is positive.

Exothermic∆H < 0

Endothermic∆H > 0

Spontaneous Processes• A spontaneous process is one that occurs

without ongoing outside intervention.• A nonspontaneous process is not impossible,

but it requires energy input.• if a process is spontaneous in one direction,

it must be nonspontaneous in the other.

Enthalpy and Reaction Tendency

• Most spontaneous reactions are exothermic (∆H is negative).• Products have less energy than reactants.• Less energy = greater stability.

• But not all spontaneous reactions are exothermic.• Ex: The melting of ice

is endothermic andspontaneous (at >0oC).

• Enthalpy must not be thesole criteria for spontaneity.

Entropy• Entropy (S) is a measure of the

degree of disorder in a system.• Changes that increase entropy:• Physical state (phase) changes

(solid < liquid < gas)• A larger number of moles

of products than reactants• Increase in temperature• Solids dissociating into

ions upon dissolving• Expansion (greater

volume) of gases

Second Law of Thermodynamics

• The Second Law of Thermodynamics states that for any spontaneous process, the entropy of the universe increases (DSuniv is positive).• The entropy of the

system could decreaseas long as the entropyof the surroundings increases by a greateramount.(DSuniv = DSsys + DSsurr)

Temperature Dependence of DS • The increase in DSsurr often comes from the

heat released in an exothermic reaction.• the amount the entropy

of the surroundings changes depends on the temperature it is at originally.• the higher the original temperature, the less

effect addition or removal of heat has.

T

HS system

gssurroundin

Gibbs Free Energy• The total amount of energy available in the system

to do work on the surroundings is called the Gibbs Free Energy (DG).• DG can be calculated using the following equation

(T is the Kelvin temperature):

• Systems tend toward lower Gibbs free energy (lower chemical potential.)• A negative DG = a spontaneous reaction.• A positive DG = a nonspontaneous reaction.

DG = DH – TDS

Gibbs Free EnergySample Problem

Given: CCl4(g) C(s, graphite) + 2 Cl2(g)

DH = +95.7 kJ and DS = +142.2 J/K at 25°C.

1. Calculate DG and determine if it is spontaneous:

DG = DH – TDS

DG = 95.7 x 103 J – (298K) (142.2 J/K)

DG = +5.33 x 104 J

2. Determine at what temperature (if any) the reaction becomes spontaneous:

DG = DH – TDS < 0

95.7 x 103 J – T (142.2 J/K) < 0

95.7 x 103 J < T (142.2 J/K)

Nonspontaneous

T > 673 K

Spontaneous at:

Effect on Spontaneity• When DH and DS have opposite signs,

spontaneity does not depend on the temperature, but when they have the same sign it does.

DG = DH – TDS

Third Law of Thermodynamics• The Third Law of Thermodynamics states

that for a perfect crystal at absolute zero, absolute entropy (S) = 0.• So every substance that is not a

perfect crystal at absolute zero has some energy from entropy.• Perfect crystals never occur in

practice; imperfections just get"frozen in" at low temperatures.• Scientists have achieved

temperatures extremely close to absolute zero.