Chapter 17 – 2What happens at

equilibrium?

Chemist’s De nition of ”Spontaneous”

• A reaction/process that occurs by itself under a given set of

conditions is said to be spontaneous.

→ spontaneous ≠ instantaneous

• A nonspontaneous process requires some external action

to be continuously applied in order for it to occur.

fi

4Fe(s) + 3O2(g) → 2Fe2O3(s)

How to predict spontaneity?

goal is to nd that chemical

potential energy analog!

fi

mechanical sytem

chemical system

Chemical Energy ➜ Potential Energy

chemical hand warmers:

Fe ( s ) + O 2 ( g ) ⎯⎯

→ Fe 2 O3 (s ) + heat

ΔH < 0
What Makes a Process Spontaneous?
• Hypothesis: Exothermic processes (ΔH < 0) are
spontaneous while endothermic processes (ΔH > 0)

are nonspontaneous.

→ Often true, but there are exceptions:

H2O(s)

ΔH° = +6.01

kJ/mol

Enthalpy

isHnot

criterion

2O(l) a reliable

forH2spontaneous

change!

O

NaCl(s)

Na+(aq) + Cl—(aq) ΔH° = 3.9 kJ/mol

These endothermic reactions occur spontaneously!

The sign of ΔH does not predict the direction of a

spontaneous change!

Another Look at Spontaneity

• An ideal gas expands into a vacuum at constant T:

ΔU = ΔH = 0

→ Expansion is NOT driven by a change in energy

→ Particles are spreading out…

→ Energy is spreading out over more energy levels…

“Spreading out” of energy

Energy levels:

V

2V

• The allowed energy levels for translational motion are

spaced more closely together in a larger volume.

→ Energy of the gas spreads out over more energy levels

• New hypothesis: Spreading out of energy

drives spontaneous processes.

Entropy, S

• Entropy (S) — a thermodynamic property describing

the distribution of a system’s energy over the available

energy levels.

→ The greater the number of con gurations among

the energy levels in a particular system➜ the

greater the entropy of the system.

→ Entropy is a state function

→ unique value for a system with T, P and de ned

composition

→ value is independent of the path taken

fi

fi

ΔS = Sfinal – Sinitial

ΔS > 0

More levels

more ways to

distribute the energy

System A

System B

microstate: the exact energy distribution among the

molecules at any one instant

Microstates Explained…

Spontaneous expansion goes in the direction of

increasing number of microstates ➜ increasing

entropy

Smore microstates > Sfewer microstates

Microstates Explained…

• System A has 2 allowable energy levels while System B has 4

• with only 8 J of energy, one molecule cannot have all 8 J because that

would leave the other molecule with 0 J (motionless)

• violates Heisenberg’s uncertainty principle, there are only certain

“allowed” energy levels because they are composed of atoms that are

governed by the laws of quantum mechanics.

Entropy, a Mathematical De nition

S = k ln W

# of microstates

Boltzmann constant

1.38 x 10–23 J/(K•mol)

Ludwig Eduard Boltzmann

(1844 – 1906)

• microstate — the particular way in which the energy of

a state is distributed within the system

• W = # of microstates = # of energetically equivalent

ways of arranging the energy of a system

fi

• Entropy increases with increasing W

Macro- and Microstates

• Three (of ve) possible ways of

arranging 4 gas particles in two

connected asks — Macrostates

(P,V,T)

• Potential energy of each is

identical (ideal gas assumption)

fl

fi

• State B is more probable

because it has more microstates

→ greater entropy

Microstates Explained…

• Compare the number of microstates (energetically

equivalent arrangements) available for each state:

State C

W=1

State A

W=1

State B, W = 6

Melting of Ice

When ice melts, the arrangement of water molecules changes from

an orderly one to a more disorderly one.

Evaporating Water

When water evaporates, the arrangement of water molecules

becomes still more disorderly.

Dissolving Salt

When salt dissolves in water, the arrangement of the

molecules and ions becomes more disorderly.

Processes in Which Entropy Increases

change in the freedom of motion of particles,

dispersal of their energy of motion,

key factor for predicting the direction of spontaneity!

Sgas > Sliquid > Ssolid

www.youtube.com/watch?v=NQhjAtCKghE

Recall: Molecular Motion

• Kinetic energy of molecules: a result of their motion

Three types:

Processes in Which Entropy Increases

Sgas > Sliquid > Ssolid

Processes in Which Entropy Increases

NaCl(s)

H2O

Na+(aq) + Cl—(aq) ΔH° = 3.9 kJ/mol

S is a key factor in the formation of solutions!

✔︎

Practice Problem

Which of the following processes would NOT result in

an increase in entropy?

× 1.

× 2.

× 3.

× 4.

Melting of an ice cube

Sublimation of a moth ball

Evaporation of a puddle of gasoline

A glass of cool lemonade warming in the sun

5. Condensation of water vapor on a cold windshield

Practice Problem

Which substance in each pair has the higher entropy?

Rationalize…

1.

2.

3.

4.

5.

6.

1 mol of SO2(g) OR 1 mol of SO3(g)

1 mol of CO2(s) OR 1 mol of CO2(g)

3 mol of O2(g) OR 2 mol of O3(g)

1 mol of KBr(s) OR 1 mol of KBr(aq)

Seawater at 2℃ or at 23℃

1 mol of CF4(g) OR 1 mol of CCl4(g)

# of particles

Entropy

es w/

ing T

low temperature

high temperature

more available energy levels

at higher temperatures

Kinetic energy

⬆︎

⬆︎

Maxwell-Boltzmann Distribution of Kinetic Energies

Practice Problem

Which of the following processes would you expect to

result in a greater positive change in entropy?

1. H2O (l, 25 °C) → H2O (l, 75 °C)

2. CO2 (s, 101 kPa) → CO2 (g, 2kPa)

3. CH3OH (l, 101 kPa) → CH3OH (g, 2 kPa)

4. Fe(s) → Fe(l)

5. None of the above processes involve a positive

entropy change.

Chapter 17 – 3

From last lecture…

enthalpy (∆H) is the heat evolved at constant pressure

distinction between endothermic (∆H > 0) versus

exothermic (∆H < 0) reactions
sign of ∆H is not a reliable criteria for predicting
spontaneity
entropy: S = k ln W
microstates ➜ places of energy ➜ energy levels
spontaneous processes go in the direction of
increasing # of microstates or towards higher entropy
Sgas > Sliquid > Ssolid

Review: Entropy, S

S = k ln W

Where k is the Boltzmann constant, and W is the # of

possible arrangements of a particular state.

A microstate is a specific microscopic configuration

describing how particles of a system are distributed among

the available energy levels.

Entropy is related to the way in which the energy of a system

is distributed among the available microscopic energy levels.

R

−23 J

k=

= 1.38 × 10

NA

K

Microstates

microstate: exact energy distribution among the molecules at any

one instant

W: number of possible microstates that can result in a given

macrostate

In these systems, there are only certain “allowed” energy levels because they

are composed of atoms that are governed by the laws of quantum mechanics.

Ways to disperse energy

• Kinetic energy of molecules: a result of their motion

Three types:

Translational motion:

movement from one point in

space to another

A system where a given amount of energy can be dispersed in

Rotational transitions:

many different ways has more entropy

a system whose

rotation ofthan

the molecule

energy can only be dispersed

in centre

a few of

ways.

around its

gravity

Vibrational transitions:

interatomic vibrations

Criteria for Spontaneity

ΔSuniverse = ΔSsystem + ΔSsurroundings > 0

• Second Law of Thermodynamics: All spontaneous

processes produce an increase in the entropy of the

universe.

• Recall the rst law of thermodynamics: Energy cannot

be created or destroyed, only converted from one

form to another.

fi

The total energy of the universe is constant, but it

tends to spread out over more and more energy levels.

How do we quantify

entropy of the system?

How do we measure S of the system?

• The absolute entropy of a substance can be measured!

→ Done so in reference to a pure perfect crystal of a

substance at 0 K for which S = 0 (3rd law of

thermodynamics)

• Standard Molar Entropy (S˚) – absolute entropy of 1

mole of a pure substance in its standard state at 25 ˚C and

at 1 atm.

• Since entropy is a state function, use S˚ to nd ΔSsys

fi

ΔSsys = [ΣnpS°(products) – ΣnrS°(reactants)]

Zero Entropy

3rd Law of Thermodynamics and Standard Molar Entropies (Sº)

S = k ln W

ΔSsys = Sfinal – Sinitial

ΔSsys = Sfinal

The entropy of a perfect crystal at absolute zero

(0 K) is zero.

0

How do we measure S of the system?

• The absolute entropy of a substance can be

measured!

→ Done so in reference to a pure perfect crystal of a

substance at 0 K for which S = 0 (3rd law of

thermodynamics)

• Standard Molar Entropy (S˚) – absolute entropy of 1

mole of a pure substance in its standard state at 25 ˚C

and at 1 atm.

• Since entropy is a state function, use S˚ to nd ΔSsys

fi

ΔSsys = [ΣnpS°(products) – ΣnrS°(reactants)]

Comparing Absolute Entropies

Table 17.1 Standard Molar Entropy Values (S°) for Selected

Substances at 298 K

Substance S°(J K−1 mol−1) Substance S°(J K−1 mol−1) Substance S°(J K−1

mol−1)

Gases

Liquids

Solids

Blank

H2(g)

130.7

H2O(l)

70.0

MgO(s)

27.0

Ar(g)

154.8

CH3OH(l)

126.8

Fe(s)

27.3

CH4(g)

186.3

Br2(l)

152.2

Li(s)

29.1

H2O(g)

188.8

C6H6(l)

173.4

Cu(s)

33.2

N2(g)

191.6

Na(s)

51.3

NH3(g)

192.8

K(s)

64.7

F2(g)

202.8

NaCl(s)

72.1

O2(g)

205.2

CaCO3(s)

91.7

C2H4(g)

219.3

FeCl3(s)

142.3

Cl2(g)

223.1

S = J / K · mol

measure of Energy dispersal per unit temperature

Calculating the Standard Entropy Change (∆rS°)

for a Reaction

ΔrSº= ΣnpSºproducts − ΣnrSºreactants

change in entropy for a process in which all reactants

and products are in their standard states

use Table 17.1 (Appendix II) for calculations

Practice Problem

Using the data in Table 17.1, calculate the standard

entropy changes for the following reaction at 25 ˚C:

N2(g) + 3H2(g)

!

2NH3(g)

Comparing Absolute Entropies

N2(g) + 3H2(g)

2NH3(g)

!

Table 17.1 Standard Molar Entropy Values (S°) for Selected

Substances at 298 K

Substance

S°(J K−1 mol−1)

Gases

Substance

S°(J K−1 mol−1)

Liquids

Substance

S°(J K−1 mol−1)

Solids

Blank

H2(g)

130.7

H2O(l)

70.0

MgO(s)

27.0

Ar(g)

154.8

CH3OH(l)

126.8

Fe(s)

27.3

CH4(g)

186.3

Br2(l)

152.2

Li(s)

29.1

H2O(g)

188.8

C6H6(l)

173.4

Cu(s)

33.2

N2(g)

191.6

Na(s)

51.3

NH3(g)

192.8

K(s)

64.7

F2(g)

202.8

NaCl(s)

72.1

O2(g)

205.2

CaCO3(s)

91.7

C2H4(g)

219.3

FeCl3(s)

142.3

Cl2(g)

223.1

Make sure you understood…

• Entropy (S) – a thermodynamic property (state function)

describing the distribution of a system’s energy over the

available energy levels.

• Second Law of Thermodynamics: All spontaneous

processes produce an increase in the entropy of the

universe.

• The absolute entropy of a substance can be measured in

reference to a pure perfect crystal of a substance at 0 K for

which S = 0 (3rd law of thermodynamics)

ΔSsys = [ΣnpS°(products) – ΣnrS°(reactants)]

Key equations so far…

S = k ln W

Smore microstates > Sfewer microstates

ΔSuniverse = ΔSsystem + ΔSsurroundings > 0

ΔSsys = [ΣnpS°(products) – ΣnrS°(reactants)]

Sgas > Sliquid > Ssolid

How about factors affecting entropy?

S increases w/ increasing T, # of moles, # of atoms,

complexity of the atom, molar mass

Factors affecting S

1. State of the substance

Sº (J mol–1 K–1)

H2O (l)

70.0

H2O (g)

188.8

Factors affecting S

2. Molar mass

3. Allotropes

Comparing Absolute Entropies

diamond,

So=2.4 J mol-1 K-1

graphite,

So=5.7 J mol-1K-1

Factors affecting S

4. Molecular complexity

Molar Mass (g mol–1)

Sº (J K–1 mol–1)

Ar (g)

39.948

154.8

NO (g)

30.006

210.8

Relative standard entropies

Molar Mass (g mol–1)

Sº (J K–1 mol–1)

CO(g)

28.01

197.7

C2H4(g)

28.05

219.3

Sº (J K–1 mol–1)

NO(g)

210.8

NO2(g)

240.1

NO3(g)

304.4

Sº (J mol–1 K–1)

KClO3(s)

143.1

KClO3(aq)

265.7

Special case: highly charged ions

Sº (J mol–1 K–1)

AlCl3 (s)

110.7

AlCl3 (aq)

–152.2

before dissolution

after dissolution

# of particles

Entropy

es w/

ing T

low temperature

high temperature

more available energy levels

at higher temperatures

Kinetic energy

⬆︎

⬆︎

Maxwell-Boltzmann Distribution of Kinetic Energies

es w/

ing T

# of particles

Entropy

molecular speed (m/s)

⬆︎

⬆︎

https://ibchem.com/IB/ibnotes/full/sta_htm/Maxwell_Boltzmann.htm

The Haber-Bosch Process

➔ 2NH3(g)

N2(g) + 3H2(g) ➔

How do we make it work?

Quantifying Entropy of the Surroundings

• Recall: Entropy increases with increasing temperature

→ Exothermic process, Ssurr increases

→ Endothermic process, Ssurr decreases

→ Magnitude of entropy change is inversely proportional to

temperature

https://chem.libretexts.org

Quantifying Entropy of the Surroundings

∆Ssurr ∝ –qsys

∆Ssurr ∝

1

T

can now calculate

∆S

surr

∆S ∝

T

from

∆H

!

sys

At constant P: q = ∆H

–qsys

surr

sys

∆Ssurr =

sys

–∆Hsys

T

(constant P and T)

Practice Problem

Using data from Appendix II of your text, predict

whether or not the reaction below will proceed

spontaneously at 25 °C:

N2(g) + 3H2(g)

!

2NH3(g)

ΔSuniverse = ΔSsys + ΔSsurr

ΔS°sys = – 198.1 J/mol K

ΔS°surr = – ΔH°sys / T

(previous example)

ΔH°sys = [ΣnpΔH°(products) – ΣnrΔH°(reactants)]

Temperature-dependence of entropy

ΔSuniverse = ΔSsys + ΔSsurr

H2O (l)

H2O (s)

ΔSuniverse > 0 ➜ reaction to be spontaneous

It’s like a balancing act!

ΔSuniverse = ΔSsys + ΔSsurr

we want the entropy of the universe to be positive if

we want the reaction to be spontaneous

Practice Problem

Consider the following reaction at constant pressure.

Use the information here to determine the value of

ΔSsurr at 298 K. Predict whether or not this reaction

will be spontaneous at this temperature.

N2(g) + 2O2(g) → 2NO2(g) ΔrH = +66.4 kJ

A) ΔSsurr = +223 J K-1 mol-1, reaction is spontaneous

B) ΔSsurr = -223J K-1 mol-1, reaction is not spontaneous

C) ΔSsurr = -66.4 J K-1 mol-1, reaction is spontaneous

D) ΔSsurr = +66.4 kJ K-1 mol-1, reaction is not spontaneous

E) ΔSsurr = -66.4 J K-1 mol-1, reaction is not spontaneous

Chapter 13-2

From last lecture

aA + bB → cC + dD

1 Δ[A]

1 Δ[B]

1 Δ[C]

1 Δ[D]

rate of reaction = −

=−

=+

=+

a Δt

b Δt

c Δt

d Δt

Rate Law: Effect of Concentration on Reaction Rates

rate of reaction = k[A]m[B]n ….

A ⎯⎯

→ products

rate = k[A]

zero-order

rate = k

first-order

rate = k[A]1

second-order

rate = k[A]2

n

[13.6]

Determined

Reaction Order

using method

of initial rates

Getting our units right…

rate law:

rate of reaction = k × [HgCl2] × [C2O42−]2

M

min−1

M

M2

−1

M

min

−2 min−1

unit of k =

=

M

M × M2

The Integrated Rate Law:

dependence of concentration on Time

First-Order Reactions

A ⎯⎯

→ products

differential rate law

first-order

integrated rate law

= k[A]

rate rate

= k[A]

n

Δ[A]

−

= k[A]

Δt

ln[A]t = – kt + ln[A]0

[A]t

ln

= – kt

[A]0

[1

Integrated Rate Law: First-Order

First-Order Reactions

ln[A]t = – kt + ln[A]0

y

= mx + b

H2O2(aq) → H2O(l) + ½ O2(g)

rate of reaction = k[H2O2]

[k] = s−1

Test for a First-Order Reaction

H2O2(aq) → H2O(l) + ½ O2(g)

ln [H2O2]

ln[A]t = – kt + ln[A]0

plot of ln[A]t vs Time

yields a straight line

with slope = —k

Time, s

Integrated Rate Law: Second-Order

Second-Order Reactions

A ⎯⎯

→ products

differential rate law

second-order

integrated rate law

raterate

= k=[A]

k[A]

n 2

[1

Δ[A]

2

−

= k[A]

Δt

1

1

= kt +

[A]t

[A]0

[

Integrated Rate Law: Second-Order

Second-Order Reactions

1

1

= kt +

[A]t

[A]0

y

= mx + b

plot of 1/[A]t vs Time

yields a straight line

with slope = k

Integrated Rate Law: Zero-Order

Zero-Order Reactions

A ⎯⎯

→ products

differential rate law

zero-order

integrated rate law

raterate==k[A]

k[A] = k

n0

Δ[A]

−

=k

Δt

[A]t = −kt + [A]0

[1

Integrated Rate Law: Second-Order

Zero-Order Reactions

[A]t = −kt + [A]0

y

= mx + b

plot of [A]t vs Time

yields a straight line

with slope = —k

In the diagram to the right is

a plot of the concentrations of

all reactants and products as a

fu nct io n of t im e fo r a

particular reaction.

ln [X]

Practice Problem

t

Which of the following statements is correct?

1. The reaction is second order.

2. The blue curve represents the time dependence on a particular

product.

3. The rate constant for the reaction could have units of s-1.

4. The product represented by the green line is produced almost

twice as fast as the product represented by the red line.

5. Cannot tell with the information given.

Practice Problem

Reaction A has a rate constant which

is equal to 16.2 L mol-1 s-1. A plot

relating the concentrations of the

reactants and products with time is

plotted for reaction B to the right.

ln [X]

In this question, two reactions are considered.

t

1. Reaction A is first order and Reaction B is second order.

2. Reaction A is second order and Reaction B is first order.

3. Both reactions are first order.

4. Both reactions are second order.

Practice Problem

Use a value of k = 7.30×10-4 s-1 for the first-order decomposition

of H2O2(aq) to determine the percent H2O2 that has decomposed

in the first 500.0 s after the reaction begins.

Half-Life, Lifetime and Decay Time

Half-life

t½ is the time taken for one-half of a reactant to be consumed

Lifetime

𝝉 is the time taken for the reaction to decrease to 1/e (about

1/2.7);

a measure of the average life expectancy of a chemical entity

for first-order reactions

half life t = t½ and n = 2

lifetime t = 𝝉 and n = e

ln(2) 0.693

t1/2 =

=

k

k

ln(e) 1

τ=

=

k

k

Half-Life, t½

• the time taken for one-half of a reactant to be consumed.

First-Order Reactions

ln

[A]t

ln

½[A]0

[A]0

[A]0

= −kt

= −k × t½

−ln 2 = −kt½

ln 2

0.693

t½ =

=

k

k

first order reactions have a

constant half-life, t1/2 !

Half-Life: First-Order Reactions

First-Order Reactions

Examples of First-Order Reactions

Half-Life, Lifetime and Decay Time

First-Order Reactions

general expression for decay time

⎛ 1n [A] 0 ⎞

1n(n)

1n ⎜

⎟ = −kt or t =

⎜ [A] ⎟

k

0 ⎠

⎝

half life t = t ½ and n = 2

t½ is the time taken for one-half of a

[13.18]

1n(2) 0.693

t1/ 2 =

=

k

k

reactant to be consumed

τ and

lifetime t =

n=e

𝝉 is the time taken for the

reaction to decrease to 1/e

1n(e) 1

τ=

=

k

k

[13

Lifetimes of CFCs

*measure of the average life expectancy of a chemical entity

Practice Problem

The first-order decay of radon has a half-life of

3.823 days. How many grams of radon remain after

7.22 days if the sample initially weighs 250.0 grams?

A) 4.21 g

B) 183 g

C) 54.8 g

D) 76.3 g

E) 67.5 g

Second-order Half-Life, t½

half-life for second-order reactions

After time t1/2

integrated

rate

law

integrated

rate law

the [A]

t

becomes the new [A]0, so

t1/2 is not constant

for decayfortodecay

1/n toof

1/2[A]

of [A]

for second order reactions!

0 0

th

1

1

= kt +

[A]t

[A]0

1

1

=

kt

+

( 1n )[A]0

[A]0

kt1/2 =

1

1

–

1 [A]

0 [A]0

2

for decay to 1/n of [A]0

n–1

kt =

[A]0

kt1/2

t1/2

2

1

–

=

[A]0

[A]0

=

1

k[A]0

Practice Problem

The half-life for the second-order decomposition of

HI is 15.4 s when the initial concentration of HI is

0.67 mol L-1. What is the rate constant for this

reaction??

A) 1.0 × 10 -2 L mol-1 s-1

B) 4.5 × 10 -2 L mol-1 s-1

C) 9.7 × 10 -2 L mol-1 s-1

D) 2.2 × 10 -2 L mol-1 s-1

E) 3.8 × 10 -2 L mol-1 s-1

Practice Problem

A certain second-order reaction (A ➜ products) has a rate

constant of 1.40×10−3 M—1 s—1 at 27℃ and an initial half-life

of 212 s. What is the concentration of the reactant after one

half-life?

(A)

(B)

(C)

(D)

(E)

0.594 M

0.297 M

3.37 M

1.68 M

7.57 ×10−4 M

Zero-order Half-Life, t½

half-life for zero-order reactions

integrated rate law

for decay to 1/2 of [A]0

[A]t = –kt + [A]0

1 [A]

0

2

kt1/2

t1/2

= –kt1/2 + [A]0

=

=

[A]0 –

[A]0

2k

1

[A]0

2

Zero-order Reaction

integrated rate law

[A]t = –kt + [A]0

for decay to 1/nth of [A]0

1 [A]

0

n

= –kt + [A]0

1

n

kt = [A]0 –

t =

1

n [A]0

(n – 1) [A]0

nk

The half-life (t½), when the concentration is half its original

concentration (n=2) is:

t1/2 =

[A]0

2k

Reactions Involving Gases

ln

[A]t

[A]0

PV = nRT

= −kt

ln

M = n/V

n/V = P/RT

ln

[A] = P/RT

Pt/RT

P0/RT

Pt

P0

= −kt

= −kt

C4H9OOC4H9 (g) → 2 CH3COCH3 (g) + CH3CH3(g)

DTBP

di-tert-butyl peroxide

acetone

ethane

Reactions Involving Gases

C4H9OOC4H9 (g) 2 CH3COCH3 (g) + CH3CH3(g)

DTBP

di-tert-butyl peroxide

acetone

ethane

The first-order reaction above, with a half-life of 8.0×101 min, is

started with pure DTBP at 147℃ and 800.0 mmHg in a flask of

constant volume.

(a) Determine the value of the rate constant, k

(b) At what time will the partial pressure of DTBP be 50.0 mmHg

0

1

2

Chapter 13-4

From last lecture

Rate constant and T

k = Ae–Ea/RT

Arrhenius

Equation

ln k =

ln

Collision Model

effective collision

k = pze–Ea/RT

–Ea 1 + lnA

R T

k2

k1

=

–Ea 1

R

T2

–

1

T1

Transition State Theory

Reaction Mechanisms

A series of individual chemical steps by which an overall

chemical reaction occurs.

H 2 (g ) + 2 ICl (g ) ⎯⎯

→ 2 HCl (g ) + I 2 ( g )

Step 1

Step 2

H 2 (g ) + ICl (g ) ⎯⎯

→

HI (g ) + ICl (g ) ⎯⎯

→

HCl (g ) + HI ( g )

HCl (g ) + I 2 ( g )

Overall Rxn H 2 (g ) + 2 IC l (g ) ⎯⎯

→ 2 HCl (g ) + I 2 ( g )

Reaction Mechanisms

Step-by-step description of a reaction.

Each step is called an elementary process.

• Any molecular event that significantly alters a

molecules energy or geometry or produces a new

molecule.

Plausible reaction mechanisms must:

(1) be consistent with the stoichiometry for the overall

reaction

(2) account for the experimentally determined rate law

Rate Laws for Elementary Steps

an elementary process is a specific collisional event or

molecular process

(1 of 2)

molecularity refers to the number of molecular entities

involved in an elementary process

Rate Laws for Elementary Steps

A ¾¾

® products

unimol ec ular

A + A ¾¾

® products

bimo lecular

A + B ¾¾

® products

bimo lecular

Rate Laws for Elementary Steps

Table 13.3 Rate Laws for Elementary Steps

Elementary Step

Molecularity

Rate Law

A → products

1

Rate = k[A]

A + A → products

2

Rate = k[A]2

A + B → products

2

Rate = k[A][B]

A + A + A → products

3 (rare)

Rate = k[A]3

A + A + B → products

3 (rare)

Rate = k[A]2[B]

A + B + C → products

3 (rare)

Rate = k[A][B][C]

Characteristics of Elementary Processes

Unimolecular or bimolecular.

Exponents for concentration terms in the rate law for an

elementary process are the same as the stoichiometric factors in

the balanced equation for the process.

Elementary processes are reversible.

Intermediates are produced in one elementary process and

consumed in another.

One elementary step is usually slower than all the others and is

known as the rate-determining step.

Multistep Reactions and

the Rate-Determining Step

Reaction profile for a hypothetical two-step reaction

step 1

A

step 2

B

overall

A

k1

k−1

k2

k−2

B

C

C

A mechanism with a Slow Step

Followed by a Fast Step

H2(g) + 2 ICl(g) → I2(g) + 2 HCl(g)

rate (observed) = k[H2][ICl]

Postulate a plausible mechanism:

(1) slow

H2(g) + ICl(g) → HI(g) + HCl(g)

rate (1) = k[H2][ICl]

(2) fast

HI(g) + ICl(g) → I2(g) + HCl(g)

rate (2) = k[HI][ICl]

H2(g) + 2 ICl(g) → I2(g) + 2 HCl(g)

Reaction profile for a two-step mechanism

Example 1: A mechanism with a Slow

Step Followed by a Fast Step

H2(g) + 2 ICl(g) → I2(g) + 2 HCl(g)

Postulate a plausible mechanism:

(1) slow

H2(g) + ICl(g) → HI(g) + HCl(g)

rate (1) = k[H2][ICl]

(2) fast

HI(g) + ICl(g) → I2(g) + HCl(g)

rate (2) = k[HI][ICl]

H2(g) + 2 ICl(g) → I2(g) + 2 HCl(g)

rate (observed) = k[H2][ICl]

✔

Example 2: A mechanism with a Fast

Reversible First Step Followed by a Slow Step

2NO(g) + O2(g) → 2 NO2(g)

rateobs = kobs[NO]2[O2]

Postulate a plausible mechanism:

fast

slow

2NO(g)

k1

⇌ N2O2(g)

k−1

N2O2(g) + O2(g)

2NO2(g)

2NO(g) + O2(g) → 2 NO2(g)

2NO(g) + O2(g) → 2 NO2(g)

✔

rateobs = kobs[NO]2[O2]

Postulate a plausible mechanism:

fast

slow

2NO(g)

k1

k−1

N2O2(g) + O2(g)

N2O2(g)

2NO2(g)

2NO(g) + O2(g) → 2 NO2(g)

rate of reaction = k2 [N 2O 2 ][O 2 ]

rate of forward reaction = rate of reverse reaction

k2 k1

=

= k2 K1

k−1

k1[NO] = k−1[N 2O 2 ]

kobs

k1 [N 2O 2 ]

2

[N

O

]

=

K

[NO]

K1 =

=

➜

2 2

1

2

k−1 [NO]

rateobs = kobs[NO]2[O2]

2

rate of reaction = k2 [N 2O 2 ][O 2 ] = k2 K1[NO] [O 2 ]

2

✔

Practice Problem

What is the rate law for the following mechanism?

CH3COOC2H5 + H2O ➜ CH3COOC2H6+ + OH – (slow)

CH3COOC2H6+ ➜ CH3COOH + C2H5+ (fast)

C2H5+ + OH – ➜ C2H5OH (fast)

(A) rate = k[CH3COOC2H5][H2O]2

(B) rate = k[C2H5OH]

(C) rate = k[CH3COOH]

(D) rate = k[CH3COOC2H5]

(E) rate = k[CH3COOC2H5][H2O]

Practice Problem

Given the following proposed mechanism, predict the rate

law for the overall reaction.

A2 + 2B ➜ 2AB (overall reaction)

Mechanism

A2 ⇌ 2A

(fast)

A + B ➜ AB

(A) rate = k[A][B]

(B) rate = k[A2][B]

(C) rate = k[A2][B]1/2

(D) rate = k[A2]

(E) rate = k[A2]1/2[B]

(slow)

The Steady State Approximation

The Steady State Approximation

rate-determining step

2NO(g)

find a form for [intermediates]

N2O2(g)

rate law deduced

k1

k−1

N2O2(g) + O2(g)

Problem: in complex multistep reaction mechanisms,

more than one step may control the reaction

N2O2(g)

2NO(g)

k2

2NO2(g)

rateobs = kobs[NO]2[O2]

rate of reaction = k2 [N 2O 2 ][O 2 ]

Δ[N 2O 2 ]/Δt = rate of formation of N 2O 2 − rate of disappearance of N 2O 2 = 0

rate of formation of N 2O 2 = rate of disappearance of N 2O 2

2

rate of formation of N 2O 2 = k1[NO]

rate of disappearance of N 2O 2 = k−1[N 2O 2 ] + k2 [N 2O 2 ][O 2 ]

2NO(g)

N2O2(g)

k1

k−1

N2O2(g) + O2(g)

N2O2(g)

rateobs = kobs[NO]2[O2]

2NO(g)

k2

2NO2(g)

rate of formation of N 2O 2 = rate of disappearance of N 2O 2

k1[NO]2 = k−1[N 2O 2 ] + k2 [N 2O 2 ][O 2 ]

2

k1[NO]

[N 2O 2 ] =

k−1 + k2 [O 2 ]

2

k2 k1[NO] [O 2 ]

rate = k2 [N 2O 2 ][O 2 ] =

k−1 + k2 [O 2 ]

compare to last example:

compare to last example:

2

[N 2O 2 ]=K1[NO]

rate law based on

steady-state analysis

rate = k2 [N 2O 2 ][O 2 ] = k2 K1[NO]2 [O 2 ]

Kinetic consequences of assumptions

2

k2 k1[NO] [O 2 ]

rate = k2 [N 2O 2 ][O 2 ] =

k−1 + k2 [O 2 ]

k−1[N 2O 2 ] > k2 [N 2O 2 ][O 2 ]

k−1 > k2 [O 2 ]

2NO(g)

N2O2(g)

k1

k−1

N2O2(g) + O2(g)

✔ rate

obs

k−1 + k2 [O 2 ] ≈ k−1

2

k1[NO]

[N 2O 2 ] =

k−1

2

k2 k1[NO] [O 2 ]

2

rate =

= kobs [NO] [O 2 ]

k−1

✔

N2O2(g)

2NO(g)

k2

2NO2(g)

= kobs[NO]2[O2]

A schematic of the potential energy

diagram for the reaction of iodine with

hydrogen is shown to the right. Based

on this diagram, which of the following

statements is incorrect.

Energy

Practice Problem

+

2

+

2

1. The mechanism for the reaction

could be

I 2 ⇌ 2I

reaction coordinate

H 2 + 2I → 2HI

2. Since iodine atoms are intermediate species, their

concentration is undetectable during the course of the

reaction.

3. The second step is the rate-determining step.

4. The overall reaction is exothermic.

Practice Problem

The exothermic iodide catalyzed decomposition of peroxide

was determined to occur via two separate steps in which the

first step is the rate determining step. Which of the

following three potential energy diagrams best summarizes

these findings.

+

2.

R

Energy

Energy

R

reaction coordinate

3.

R

P

reaction coordinate

Energy

1.

P

reaction coordinate

The

Steady

State

Approximation

The Steady State Approximation (1 of 6)

NO 2 (g ) + CO (g ) ¾¾

® NO(g ) + CO 2 ( g )

Step 1

Step 2

k1

k

à

à

NO 2 ( g ) + NO 2 ( g ) á k àÜà

k

1

-1

−1

k

NO3 ( g ) + CO( g ) ¾¾

®

2

Overall Rxn N O 2 ( g ) + CO( g ) ¾¾

®

NO3 ( g ) + NO( g )

NO 2 ( g ) + CO 2 ( g )

NO( g ) + CO 2 ( g )

D[CO 2 ]

ra t e =

= k2 [NO3 ][CO]

Dt

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[13.2 8]

13 – 49

The Steady State Approximation

The Steady State Approximation (3 of 6)

k1

Step 1

¾¾

® NO3 ( g ) + NO( g )

NO 2 (g ) + NO 2 (g ) ¬¾

¾

k

Step 2

NO3 (g ) + CO ( g ) ¾¾

® NO 2 ( g ) + CO 2 ( g )

-1

k2

Rate of production of NO3

D[NO3 ]produced

Dt

= k1 [NO 2 ]2

Rate of consumption of NO3

D[NO3 ]consumed

= k-1[NO3 ][NO] + k2 [NO 3 ][CO]

Dt

Copyright © 2020 Pearson Canada Inc.

13 –

The

Steady

State

Approximation

The Steady State Approximation (4 of 6)

k1

Step 1

¾¾

® NO3 ( g ) + NO( g )

NO 2 (g ) + NO 2 (g ) ¬¾

¾

k

Step 2

NO3 (g ) + CO ( g ) ¾¾

® NO 2 ( g ) + CO 2 ( g )

-1

k2

Steady state approximation

D[NO3 ]produced

Dt

D[NO3 ]consumed

=

Dt

k1[NO 2 ]2 = k-1[NO3 ][NO] + k2 [NO 3 ][CO]

k1[NO 2 ]

[N O3 ] =

k-1[NO] + k2 [CO]

2

Copyright © 2020 Pearson Canada Inc.

13

The

Steady

State

Approximation

The Steady State Approximation (5 of 6)

k1

Step 1

¾¾

® NO3 ( g ) + NO( g )

NO 2 (g ) + NO 2 (g ) ¬¾

¾

k

Step 2

k

NO3 (g ) + CO ( g ) ¾¾

® NO 2 ( g ) + CO 2 ( g )

Steady state

Overall Rate

-1

2

k1[NO 2 ]

[NO3 ]=

k-1[NO]+ k2 [CO]

2

D[CO 2 ]

rate =

= k2 [NO3 ][CO]

Dt

D[CO 2 ] k1k2 [NO 2 ]2 [CO]

=

Dt

k-1[NO]+ k2 [CO]

Copyright © 2020 Pearson Canada Inc.

13

The

Steady

Approximation

The

Steady

State State

Approximation

(6 of 6)

k1

Step 1

¾¾

® NO3 ( g ) + NO( g )

NO 2 (g ) + NO 2 (g ) ¬¾

¾

k

Step 2

k

NO3 (g ) + CO ( g ) ¾¾

® NO 2 ( g ) + CO 2 ( g )

-1

2

Δ [CO 2 ] k1k2 [NO 2 ]2 [CO]

=

Δt

k-1[NO] + k2 [CO]

Overall Rate

If k2 >> k1

If [CO] is small

Δ [CO 2 ] k1k2 [NO 2 ] [C O]

=

Δt

k2 [CO]

2

Δ [CO 2 ]

= k1[NO 2 ]2

Δt

k-1[NO] + k2 [CO] » k-1[ NO]

D[CO 2 ] k1k2 [NO 2 ]2 [CO]

=

Dt

k-1[NO]

Copyright © 2020 Pearson Canada Inc.

1

Relationship between Equilibrium

Constant and Rate Constants

aA+b

k1

B ….

k−1

g G + h H ….

a

b

g

h

rate of forward reaction = k1[ A] [B]

rate of reverse reaction = k−1[G] [H]

k1[ A] [B] …= k−1[G] [H] …

a

k1 [G] [H] …

=

a

b

k−1 [ A] [B] …

g

h

b

g

h

k1

=K

k−1

thermodynamic

equilibrium

constant

Chapter 13-2

From last lecture

aA + bB → cC + dD

1 Δ[A]

1 Δ[B]

1 Δ[C]

1 Δ[D]

rate of reaction = −

=−

=+

=+

a Δt

b Δt

c Δt

d Δt

Rate Law: Effect of Concentration on Reaction Rates

rate of reaction = k[A]m[B]n ….

A ⎯⎯

→ products

rate = k[A]

zero-order

rate = k

first-order

rate = k[A]1

second-order

rate = k[A]2

n

[13.6]

Determined

Reaction Order

using method

of initial rates

Getting our units right…

rate law:

rate of reaction = k × [HgCl2] × [C2O42−]2

M min−1

M

M2

−1

M

min

−2 min−1

unit of k =

=

M

M × M2

The Integrated Rate Law:

dependence of concentration on Time

First-Order Reactions

A ⎯⎯

→ products

differential rate law

first-order

integrated rate law

= k[A]

rate rate

= k[A]

n

Δ[A]

−

= k[A]

Δt

ln[A]t = – kt + ln[A]0

[A]t

ln

= – kt

[A]0

[1

Integrated Rate Law: First-Order

First-Order Reactions

ln[A]t = – kt + ln[A]0

y

= mx + b

H2O2(aq) → H2O(l) + ½ O2(g)

rate of reaction = k[H2O2]

[k] = s−1

Test for a First-Order Reaction

H2O2(aq) → H2O(l) + ½ O2(g)

ln [H2O2]

ln[A]t = – kt + ln[A]0

plot of ln[A]t vs Time

yields a straight line

with slope = —k

Time, s

Integrated Rate Law: Second-Order

Second-Order Reactions

A ⎯⎯

→ products

raterate

= k=[A]

k[A]

differential rate law

Δ[A]

2

−

= k[A]

Δt

second-order

integrated rate law

n 2

1

1

= kt +

[A]t

[A]0

[1

[

Integrated Rate Law: Second-Order

Second-Order Reactions

1

1

= kt +

[A]t

[A]0

y

= mx + b

plot of 1/[A]t vs Time

yields a straight line

with slope = k

Integrated Rate Law: Zero-Order

Zero-Order Reactions

A ⎯⎯

→ products

differential rate law

zero-order

integrated rate law

raterate==k[A]

k[A] = k

n0

Δ[A]

−

=k

Δt

[A]t = −kt + [A]0

[1

Integrated Rate Law: Second-Order

Zero-Order Reactions

[A]t = −kt + [A]0

y

= mx + b

plot of [A]t vs Time

yields a straight line

with slope = —k

In the diagram to the right is

a plot of the concentrations of

all reactants and products as a

fu nct io n of t im e fo r a

particular reaction.

ln [X]

Practice Problem

t

Which of the following statements is correct?

1. The reaction is second order.

2. The blue curve represents the time dependence on a particular

product.

3. The rate constant for the reaction could have units of s-1.

4. The product represented by the green line is produced almost

twice as fast as the product represented by the red line.

5. Cannot tell with the information given.

Practice Problem

Reaction A has a rate constant which

is equal to 16.2 L mol-1 s-1. A plot

relating the concentrations of the

reactants and products with time is

plotted for reaction B to the right.

ln [X]

In this question, two reactions are considered.

t

1. Reaction A is first order and Reaction B is second order.

2. Reaction A is second order and Reaction B is first order.

3. Both reactions are first order.

4. Both reactions are second order.

Practice Problem

Use a value of k = 7.30×10-4 s-1 for the first-order decomposition

of H2O2(aq) to determine the percent H2O2 that has decomposed

in the first 500.0 s after the reaction begins.

Half-Life, Lifetime and Decay Time

Half-life

t½ is the time taken for one-half of a reactant to be consumed

Lifetime

𝝉 is the time taken for the reaction to decrease to 1/e (about

1/2.7);

a measure of the average life expectancy of a chemical entity

for first-order reactions

half life t = t½ and n = 2

ln(2) 0.693

t1/2 =

=

k

k

lifetime t = 𝝉 and n = e

ln(e) 1

τ=

=

k

k

Half-Life, t½

• the time taken for one-half of a reactant to be consumed.

First-Order Reactions

ln

[A]t

ln

½[A]0

[A]0

[A]0

= −kt

= −k × t½

−ln 2 = −kt½

ln 2

0.693

t½ =

=

k

k

first order reactions have a

constant half-life, t1/2 !

Half-Life: First-Order Reactions

First-Order Reactions

Examples of First-Order Reactions

Half-Life, Lifetime and Decay Time

First-Order Reactions

general expression for decay time

⎛ 1n [A] 0 ⎞

1n(n)

1n ⎜

⎟ = −kt or t =

⎜ [A] ⎟

k

0 ⎠

⎝

half life t = t ½ and n = 2

t½ is the time taken for one-half of a

[13.18]

1n(2) 0.693

t1/ 2 =

=

k

k

reactant to be consumed

τ and

lifetime t =

n=e

𝝉 is the time taken for the

reaction to decrease to 1/e

1n(e) 1

τ=

=

k

k

[13

Lifetimes of CFCs

*measure of the average life expectancy of a chemical entity

Practice Problem

The first-order decay of radon has a half-life of

3.823 days. How many grams of radon remain after

7.22 days if the sample initially weighs 250.0 grams?

A) 4.21 g

B) 183 g

C) 54.8 g

D) 76.3 g

E) 67.5 g

Second-order Half-Life, t½

half-life for second-order reactions

After time t1/2

integrated

rate

law

integrated

rate law

the [A]

t

becomes the new [A]0, so

t1/2 is not constant

for decayfortodecay

1/n toof

1/2[A]

of [A]

for second order reactions!

0 0

th

1

1

= kt +

[A]t

[A]0

1

1

=

kt

+

( 1n )[A]0

[A]0

kt1/2 =

1

1

–

1 [A]

0 [A]0

2

for decay to 1/n of [A]0

n–1

kt =

[A]0

2

1

kt1/2 =

–

[A]0

[A]0

t1/2

=

1

k[A]0

Practice Problem

The half-life for the second-order decomposition of

HI is 15.4 s when the initial concentration of HI is

0.67 mol L-1. What is the rate constant for this

reaction??

A) 1.0 × 10 -2 L mol-1 s-1

B) 4.5 × 10 -2 L mol-1 s-1

C) 9.7 × 10 -2 L mol-1 s-1

D) 2.2 × 10 -2 L mol-1 s-1

E) 3.8 × 10 -2 L mol-1 s-1

Practice Problem

A certain second-order reaction (A ➜ products) has a rate

constant of 1.40×10−3 M—1 s—1 at 27℃ and an initial half-life

of 212 s. What is the concentration of the reactant after one

half-life?

(A) 0.594 M

(B) 0.297 M

(C) 3.37 M

(D) 1.68 M

(E) 7.57 ×10−4 M

Zero-order Half-Life, t½

half-life for zero-order reactions

integrated rate law

for decay to 1/2 of [A]0

[A]t = –kt + [A]0

1 [A] = –kt

+

[A]

1/2

0

0

2

kt1/2

t1/2

=

=

[A]0 –

[A]0

2k

1

[A]0

2

Zero-order Reaction

integrated rate law

[A]t = –kt + [A]0

for decay to 1/nth of [A]0

1 [A] = –kt

+ [A]0

1

0

n

n

1

[A]

–

kt =

0

n [A]0

t =

(n – 1) [A]0

nk

The half-life (t½), when the concentration is half its original

concentration (n=2) is:

t1/2 =

[A]0

2k

Reactions Involving Gases

ln

[A]t

[A]0

PV = nRT

= −kt

ln

M = n/V

n/V = P/RT

ln

[A] = P/RT

Pt/RT

P0/RT

Pt

P0

= −kt

= −kt

C4H9OOC4H9 (g) → 2 CH3COCH3 (g) + CH3CH3(g)

DTBP

di-tert-butyl peroxide

acetone

ethane

Reactions Involving Gases

C4H9OOC4H9 (g) 2 CH3COCH3 (g) + CH3CH3(g)

DTBP

di-tert-butyl peroxide

acetone

ethane

The first-order reaction above, with a half-life of 8.0×101 min, is

started with pure DTBP at 147℃ and 800.0 mmHg in a flask of

constant volume.

(a) Determine the value of the rate constant, k

(b) At what time will the partial pressure of DTBP be 50.0 mmHg

0

1

2

Chapter 13-3

From last lecture

Integrated Rate Law: dependence of Concentration on Time

zero

first

[A]t = −kt + [A]0

ln[A]t = – kt + ln[A]0

second

1

1

= kt +

[13.12]

[A]t

[A]0

Half-Life, Lifetime and Decay Time

Half-life

t½ is the time taken for one-half of a reactant to be consumed

Lifetime

𝝉 is the time taken for the reaction to decrease to 1/e (about

1/2.7);

a measure of the average life expectancy of a chemical entity

for first-order reactions

half life t = t½ and n = 2

ln(2) 0.693

t1/2 =

=

k

k

lifetime t = 𝝉 and n = e

ln(e) 1

τ=

=

k

k

Please clarify…

What is the difference between rate and

the rate constant (k)?

Rate

how a concentration of a product or reactant changes with time

—∆[A]/∆t

from a tangent to a concentration-time curve

calculation from a rate law

Rate constant, k

relates the rate of a reaction to reactant concentrations

can be used to calculate rates of reaction

fixed at a given T regardless of reactant concentrations

Please clarify…

Order of reaction and overall order of

reaction?

rate of reaction = k[A]m[B]n ….

overall order of reaction = m + n + ….

zero

sum of m + n + …. = 0

first

sum of m + n + …. = 1

second

sum of m + n + …. = 2

Reaction Kinetics: Summary

1. Calculate the rate of a reaction from a known

rate law using:

Rate of reaction = k [A]m[B]n …

2. Determine the instantaneous rate of the reaction

by:

• Find the slope of the tangent line of [A] vs t

• Evaluate −Δ[A]/Δt, with a short Δt interval.

Reaction Kinetics: Summary

3. Determine the order of reaction by:

• Using the method of initial rates.

• Find the graph that yields a straight line.

• Test for the half-life to find first order reactions.

• Substitute data into integrated rate laws to find

the rate law that gives a consistent value of k.

Reaction Kinetics: Summary

4. Find the rate constant k by:

• Determine k from the slope of a straight line

graph.

• Substitute concentration-time data into the

appropriate integrated rate law.

• Measure the half life of first-order reactions.

5. Find reactant concentrations or times for certain

conditions using the integrated rate law after

determining k.

Effect of Temperature on Reaction Rate

Arrhenius Equation

rate = k[A]n

Activation energy

–E

/RT

a

k = Ae

Frequency factor

Exponential factor

Effect of Temperature on Reaction Rate

Arrhenius Equation

k = Ae

– Ea / RT

rate constant

the temperature dependence of the reaction rate is

contained in the rate constant

frequency factor

the frequency factor represents the number of

approaches to the activation barrier per unit time

fraction of molecules with enough kinetic energy to

exponential factor

go over the activation energy barrier

activation energy

energy barrier that needs to be overcome for

reactants to be transformed into products

The Activation Energy Barrier

2 H2 (g) + O2 (g) ⇌ 2 H2O (g)

The Activation Energy Barrier

An analogy for a reaction profile and activation energy

Thermal Energy Distribution

exponential factor = e–Ea/RT

* diff values of Ea

the higher the

activation energy,

the slower the

reaction rate at a

given T

Arrhenius Plots

Experimental Measurements of the Frequency Factor and the

Activation Energy

k = Ae–Ea/RT

ln k = ln(Ae–Ea/RT)

ln k = ln(A)–Ea/RT

− Ea 1

ln k =

+ ln A

R T

y = mx+b

Determining Ea from Arrhenius Plots

− Ea 1

ln k =

+ ln A

R T

− Ea

4

= −1.2 × 10 K

R

Ea = 1.0 × 10 kJ mol

2

Ea = 100 kJ mol

−1

−1

Arrhenius Plots

Ea can be calculated if you know the rate at two temperatures:

–Ea 1

ln k2 =

+ ln A

R T2

ln k2 – ln k1 =

–Ea 1

ln k1 =

+ ln A

R T1

–Ea 1

R

1

–E

a

+ ln A –

–ln A

T2

R T1

k 2 − Ea ⎛ 1 1 ⎞

ln =

−

⎜

⎟

k1

R ⎝ T2 T1 ⎠

Practice Problem

A particular first-order reaction has a rate constant of

1.35 × 102 s-1 at 25.0 °C. What is the magnitude of k at

75.0 °C if Ea = 85.6 kJ/mol?

A) 3.47 × 104 s-1

B) 1.92 × 104 s-1

C) 670 s-1

D) 3.85 × 106 s-1

E) 1.36 × 102 s-1

Practice Problem

A first order reaction is found to have an activation energy

of 105.2 kJ mol-1. If the rate constant for this reaction is

4.60 × 10 -6 s-1 at 275 K, and the initial reagent

concentration is 2.50 mol L-1 , what is the remaining

concentration after 500 seconds at a higher temperature of

325 K?

(A)

(B)

(C)

(D)

(E)

1.19 mol L-1

5.27 mol L-1

0.16 mol L-1

5.4 × 10 -4 mol L-1

none of the above

Some facts:

In gases the collision density is of the order of ~1032

collisions per second.

If each collision produced a reaction, the rate would be

about 106 M s−1.

Actual rates are on the order of 10−4 M s−1.

Only a fraction of collisions lead to chemical reaction.

Kinetic-Molecular theory can be used to calculate the

collision frequency.

Molecular Collisions and

Chemical Reactions

H• + •H

H2

N≡N−O + N=O

N≡N + O−N=O

effective collision

ineffective collision

The Collision Model

The Collision Model: A Closer Look at the Frequency Factor

k = Ae–Ea/RT

k = pze–Ea/RT

Orientation

Factor

Collision

Frequency

The Collision Model

k = Ae–Ea/RT = pze–Ea/RT

steric factor decreases as the complexity of the reactant

molecules increases (fewer collisions will occur with the

proper orientation to produce chemical reactions)

Transition State Theory

reactants

activated complex

products

activated complex:

hypothetical transitory

species that exists in highenergy transition state

between reactants and

products

A reaction profile for the reaction

N2O(g) + NO(g)

N2(g) + NO2(g)

The Activated Complex (TS)

to weaken the N-CH3 bond and

for NC group to rotate

the higher the activation energy,

the slower the reaction rate at a

given T

k = Ae–Ea/RT

Collision Model and Arrhenius Eqn

From collision theory:

(2)

(3)

k = zpe–Ea/RT = Ae–Ea/RT

(1)

Arrhenius equation was determined experimentally before

the development of collision theory. It is consistent with

important aspects of collision theory such as:

(1) the frequency factor of molecular collisions

(2) the fraction of collisions energetic enough to

produce a reaction

(3) the need for favorable orientations during collisions

Practice Problem

Which of the following statements is correct?

A) A zero order reaction depends on the concentration of

reactants.

B) A reaction rate cannot be calculated from the collision

frequency alone.

C) The activated complex is a chemical species that can be

isolated and analysed.

D) The number of collisions has no effect on the rate constant.

E) The orientation of a collision does not affect the rate constant.

Chapter 13-4

From last lecture

Rate constant and T

k = Ae–Ea/RT

Arrhenius

Equation

ln k =

ln

Collision Model

effective collision

k = pze–Ea/RT

–Ea 1 + lnA

R T

k2

k1

=

–Ea 1

R

T2

–

1

T1

Transition State Theory

Reaction Mechanisms

A series of individual chemical steps by which an overall

chemical reaction occurs.

H 2 (g ) + 2 ICl (g ) ⎯⎯

→ 2 HCl (g ) + I 2 ( g )

Step 1

Step 2

H 2 (g ) + ICl (g ) ⎯⎯

→

HI (g ) + ICl (g ) ⎯⎯

→

HCl (g ) + HI ( g )

HCl (g ) + I 2 ( g )

Overall Rxn H 2 (g ) + 2 IC l (g ) ⎯⎯

→ 2 HCl (g ) + I 2 ( g )

Reaction Mechanisms

Step-by-step description of a reaction.

Each step is called an elementary process.

• Any molecular event that significantly alters a

molecules energy or geometry or produces a new

molecule.

Plausible reaction mechanisms must:

(1) be consistent with the stoichiometry for the overall

reaction

(2) account for the experimentally determined rate law

Rate Laws for Elementary Steps

an elementary process is a specific collisional event or

molecular process

(1 of 2)

molecularity refers to the number of molecular entities

involved in an elementary process

Rate Laws for Elementary Steps

A ¾¾

® products

unimol ec ular

A + A ¾¾

® products

bimo lecular

A + B ¾¾

® products

bimo lecular

Rate Laws for Elementary Steps

Table 13.3 Rate Laws for Elementary Steps

Elementary Step

Molecularity

Rate Law

A → products

1

Rate = k[A]

A + A → products

2

Rate = k[A]2

A + B → products

2

Rate = k[A][B]

A + A + A → products

3 (rare)

Rate = k[A]3

A + A + B → products

3 (rare)

Rate = k[A]2[B]

A + B + C → products

3 (rare)

Rate = k[A][B][C]

Characteristics of Elementary Processes

Unimolecular or bimolecular.

Exponents for concentration terms in the rate law for an

elementary process are the same as the stoichiometric factors in

the balanced equation for the process.

Elementary processes are reversible.

Intermediates are produced in one elementary process and

consumed in another.

One elementary step is usually slower than all the others and is

known as the rate-determining step.

Multistep Reactions and

the Rate-Determining Step

Reaction profile for a hypothetical two-step reaction

step 1

A

step 2

B

overall

A

k1

k−1

k2

k−2

B

C

C

A mechanism with a Slow Step

Followed by a Fast Step

H2(g) + 2 ICl(g) → I2(g) + 2 HCl(g)

rate (observed) = k[H2][ICl]

Postulate a plausible mechanism:

(1) slow

H2(g) + ICl(g) → HI(g) + HCl(g)

rate (1) = k[H2][ICl]

(2) fast

HI(g) + ICl(g) → I2(g) + HCl(g)

rate (2) = k[HI][ICl]

H2(g) + 2 ICl(g) → I2(g) + 2 HCl(g)

Reaction profile for a two-step mechanism

Example 1: A mechanism with a Slow

Step Followed by a Fast Step

H2(g) + 2 ICl(g) → I2(g) + 2 HCl(g)

Postulate a plausible mechanism:

(1) slow

H2(g) + ICl(g) → HI(g) + HCl(g)

rate (1) = k[H2][ICl]

(2) fast

HI(g) + ICl(g) → I2(g) + HCl(g)

rate (2) = k[HI][ICl]

H2(g) + 2 ICl(g) → I2(g) + 2 HCl(g)

rate (observed) = k[H2][ICl]

✔

Example 2: A mechanism with a Fast

Reversible First Step Followed by a Slow Step

2NO(g) + O2(g) → 2 NO2(g)

rateobs = kobs[NO]2[O2]

Postulate a plausible mechanism:

fast

slow

2NO(g)

k1

⇌ N2O2(g)

k−1

N2O2(g) + O2(g)

2NO2(g)

2NO(g) + O2(g) → 2 NO2(g)

2NO(g) + O2(g) → 2 NO2(g)

✔

rateobs = kobs[NO]2[O2]

Postulate a plausible mechanism:

fast

slow

2NO(g)

k1

k−1

N2O2(g) + O2(g)

N2O2(g)

2NO2(g)

2NO(g) + O2(g) → 2 NO2(g)

rate of reaction = k2 [N 2O 2 ][O 2 ]

rate of forward reaction = rate of reverse reaction

k1[NO] = k−1[N 2O 2 ]

2

k1 [N 2O 2 ]

2

[N

O

]

=

K

[NO]

K1 =

=

➜

2 2

1

2

k−1 [NO]

rate of reaction = k2 [N 2O 2 ][O 2 ] = k2 K1[NO] [O 2 ]

2

k2 k1

kobs =

= k2 K1

k−1

rateobs = kobs[NO]2[O2]

✔

Practice Problem

What is the rate law for the following mechanism?

CH3COOC2H5 + H2O ➜ CH3COOC2H6+ + OH – (slow)

CH3COOC2H6+ ➜ CH3COOH + C2H5+ (fast)

C2H5+ + OH – ➜ C2H5OH (fast)

(A) rate = k[CH3COOC2H5][H2O]2

(B) rate = k[C2H5OH]

(C) rate = k[CH3COOH]

(D) rate = k[CH3COOC2H5]

(E) rate = k[CH3COOC2H5][H2O]

Practice Problem

Given the following proposed mechanism, predict the rate

law for the overall reaction.

A2 + 2B ➜ 2AB (overall reaction)

Mechanism

A2 ⇌ 2A

(fast)

A + B ➜ AB

(A) rate = k[A][B]

(B) rate = k[A2][B]

(C) rate = k[A2][B]1/2

(D) rate = k[A2]

(E) rate = k[A2]1/2[B]

(slow)

The Steady State Approximation

The Steady State Approximation

rate-determining step

2NO(g)

find a form for [intermediates]

N2O2(g)

rate law deduced

k1

k−1

N2O2(g) + O2(g)

Problem: in complex multistep reaction mechanisms,

more than one step may control the reaction

N2O2(g)

2NO(g)

k2

2NO2(g)

rateobs = kobs[NO]2[O2]

rate of reaction = k2 [N 2O 2 ][O 2 ]

Δ[N 2O 2 ]/Δt = rate of formation of N 2O 2 − rate of disappearance of N 2O 2 = 0

rate of formation of N 2O 2 = rate of disappearance of N 2O 2

2

rate of formation of N 2O 2 = k1[NO]

rate of disappearance of N 2O 2 = k−1[N 2O 2 ] + k2 [N 2O 2 ][O 2 ]

2NO(g)

N2O2(g)

k1

k−1

N2O2(g) + O2(g)

N2O2(g)

rateobs = kobs[NO]2[O2]

2NO(g)

k2

2NO2(g)

rate of formation of N 2O 2 = rate of disappearance of N 2O 2

k1[NO]2 = k−1[N 2O 2 ] + k2 [N 2O 2 ][O 2 ]

2

k1[NO]

[N 2O 2 ] =

k−1 + k2 [O 2 ]

2

k2 k1[NO] [O 2 ]

rate = k2 [N 2O 2 ][O 2 ] =

k−1 + k2 [O 2 ]

compare to last example:

compare to last example:

2

[N 2O 2 ]=K1[NO]

rate law based on

steady-state analysis

rate = k2 [N 2O 2 ][O 2 ] = k2 K1[NO]2 [O 2 ]

Kinetic consequences of assumptions

2

k2 k1[NO] [O 2 ]

rate = k2 [N 2O 2 ][O 2 ] =

k−1 + k2 [O 2 ]

k−1[N 2O 2 ] > k2 [N 2O 2 ][O 2 ]

k−1 > k2 [O 2 ]

2NO(g)

N2O2(g)

k1

k−1

N2O2(g) + O2(g)

✔ rate

N2O2(g)

2NO(g)

k2

2NO2(g)

2

obs = kobs[NO] [O2]

k−1 + k2 [O 2 ] ≈ k−1

2

k1[NO]

[N 2O 2 ] =

k−1

2

k2 k1[NO] [O 2 ]

2

rate =

= kobs [NO] [O 2 ]

k−1

✔

A schematic of the potential energy

diagram for the reaction of iodine with

hydrogen is shown to the right. Based

on this diagram, which of the following

statements is incorrect.

Energy

Practice Problem

+

2

+

2

1. The mechanism for the reaction

could be

I 2 ⇌ 2I

reaction coordinate

H 2 + 2I → 2HI

2. Since iodine atoms are intermediate species, their

concentration is undetectable during the course of the

reaction.

3. The second step is the rate-determining step.

4. The overall reaction is exothermic.

Practice Problem

The exothermic iodide catalyzed decomposition of peroxide

was determined to occur via two separate steps in which the

first step is the rate determining step. Which of the

following three potential energy diagrams best summarizes

these findings.

+

2.

R

Energy

Energy

R

reaction coordinate

3.

R

P

reaction coordinate

Energy

1.

P

reaction coordinate

The

Steady

State

Approximation

The Steady State Approximation (1 of 6)

NO 2 (g ) + CO (g ) ¾¾

® NO(g ) + CO 2 ( g )

Step 1

Step 2

k1

k

à

à

NO 2 ( g ) + NO 2 ( g ) á k àÜà

k

1

-1

−1

k

NO3 ( g ) + CO( g ) ¾¾

®

2

Overall Rxn N O 2 ( g ) + CO( g ) ¾¾

®

NO3 ( g ) + NO( g )

NO 2 ( g ) + CO 2 ( g )

NO( g ) + CO 2 ( g )

D[CO 2 ]

ra t e =

= k2 [NO3 ][CO]

Dt

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[13.2 8]

13 – 49

The Steady State Approximation

The Steady State Approximation (3 of 6)

k1

Step 1

¾¾

® NO3 ( g ) + NO( g )

NO 2 (g ) + NO 2 (g ) ¬¾

¾

k

Step 2

NO3 (g ) + CO ( g ) ¾¾

® NO 2 ( g ) + CO 2 ( g )

-1

k2

Rate of production of NO3

D[NO3 ]produced

Dt

= k1 [NO 2 ]2

Rate of consumption of NO3

D[NO3 ]consumed

= k-1[NO3 ][NO] + k2 [NO 3 ][CO]

Dt

Copyright © 2020 Pearson Canada Inc.

13 –

The

Steady

State

Approximation

The Steady State Approximation (4 of 6)

k1

Step 1

¾¾

® NO3 ( g ) + NO( g )

NO 2 (g ) + NO 2 (g ) ¬¾

¾

k

Step 2

NO3 (g ) + CO ( g ) ¾¾

® NO 2 ( g ) + CO 2 ( g )

-1

k2

Steady state approximation

D[NO3 ]produced

Dt

D[NO3 ]consumed

=

Dt

k1[NO 2 ]2 = k-1[NO3 ][NO] + k2 [NO 3 ][CO]

k1[NO 2 ]

[N O3 ] =

k-1[NO] + k2 [CO]

2

Copyright © 2020 Pearson Canada Inc.

13

The

Steady

State

Approximation

The Steady State Approximation (5 of 6)

k1

Step 1

¾¾

® NO3 ( g ) + NO( g )

NO 2 (g ) + NO 2 (g ) ¬¾

¾

k

Step 2

k

NO3 (g ) + CO ( g ) ¾¾

® NO 2 ( g ) + CO 2 ( g )

-1

2

Steady state

k1[NO 2 ]

[NO3 ]=

k-1[NO]+ k2 [CO]

Overall Rate

D[CO 2 ]

rate =

= k2 [NO3 ][CO]

Dt

2

D[CO 2 ] k1k2 [NO 2 ]2 [CO]

=

Dt

k-1[NO]+ k2 [CO]

Copyright © 2020 Pearson Canada Inc.

13

The

Steady

Approximation

The

Steady

State State

Approximation

(6 of 6)

k1

Step 1

¾¾

® NO3 ( g ) + NO( g )

NO 2 (g ) + NO 2 (g ) ¬¾

¾

k

Step 2

k

NO3 (g ) + CO ( g ) ¾¾

® NO 2 ( g ) + CO 2 ( g )

-1

2

Δ [CO 2 ] k1k2 [NO 2 ]2 [CO]

=

Δt

k-1[NO] + k2 [CO]

Overall Rate

If k2 >> k1

If [CO] is small

Δ [CO 2 ] k1k2 [NO 2 ] [C O]

=

Δt

k2 [CO]

2

Δ [CO 2 ]

= k1[NO 2 ]2

Δt

k-1[NO] + k2 [CO] » k-1[ NO]

D[CO 2 ] k1k2 [NO 2 ]2 [CO]

=

Dt

k-1[NO]

Copyright © 2020 Pearson Canada Inc.

1

Relationship between Equilibrium

Constant and Rate Constants

k1

a A + b B ….

k−1

g G + h H ….

a

b

g

h

rate of forward reaction = k1[ A] [B]

rate of reverse reaction = k−1[G] [H]

k1[ A] [B] …= k−1[G] [H] …

a

k1 [G] [H] …

=

a

b

k−1 [ A] [B] …

g

h

b

g

h

k1

thermodynamic

= K equilibrium

constant

k−1

Chapter 17 – 4

What we learned last lecture…

2nd Law of Thermodynamics: entropy of the universe is

increasing

ΔSuniverse = ΔSsystem + ΔSsurroundings > 0

3rd Law of Thermodynamics: entropy of a perfect crystal at 0 K

is zero

absolute entropy and standard molar entropy (S°)

ΔSsys = [ΣnpS°(products) — ΣnrS°(reactants)]

S increases w/ increasing T, # of moles, # of atoms, complexity

of the atom, molar mass

@ constant T and P:

∆Ssurr =

–∆Hsys

T

Temperature-dependence of entropy

ΔSuniverse = ΔSsys + ΔSsurr

H2O (l)

H2O (s)

ΔSuniverse > 0 ➜ reaction to be spontaneous

How about ΔS at equilibrium?

• @ equilibrium, ΔSuniv = 0

ΔSuniverse = ΔSsystem + ΔSsurroundings = 0

ΔSsystem = —ΔSsurroundings

• ΔSuniv for the evaporation of water?

• ΔSuniv for the condensation of water?

• ΔSuniv for the freezing of water?

• ΔSuniv for the melting of ice?

How about ΔS at equilibrium?

Energy changes in phase transitions:

0

ΔSuniv = ΔSsys + ΔSsurr

ΔSsurr = —ΔSsys

ΔrSsys =

ΔrSsys =

ΔrSsys =

@ equilibrium

Recall:

— ΔrHsys

T

— ΔrHvaporization

T

T

where “r” denotes “reversible” processes

— ΔrHfusion

T

ΔSsurr =

— ΔHsys

Reversible Process

key word is “in nitesimal” changes!

Weight of sand exactly matches pressure at each increment.

fi

fi

Thermodynamic de nition: a process is reversible if the changes

produced in the system, surrounding (and universe) can be

completely undone by “reversing” the steps.

Practice Problem

Calculate the ΔSuniv for the evaporation of water?

H2O (l, 373 K) ➜ H2O (g, 373 K)

Gibbs Free Energy

• Cumbersome to determine both ΔSsys and ΔSsurr when

we are only concerned with the system.

ΔSuniv = ΔSsys + ΔSsurr

ΔSuniv = ΔSsys — ΔHsys

∆Ssurr =

–∆Hsys

T

(constant P and T)

T

TΔSuniv = TΔSsys — ΔHsys

—TΔSuniv = ΔHsys — TΔSsys

(now only focusing on the system)

Gibbs Free Energy

—TΔSuniv = ΔHsys — TΔSsys

ΔGsys = ΔHsys — TΔSsys

G, Gibbs free energy

∆G is also called the

“chemical potential”…

ΔGsys energy available to do useful work

Josiah Willard Gibbs

90%

80%

60%

30%

single most useful criterion for predicting the direction of a chemical

reaction and the composition of the system at equilibrium!

New Criterion for Spontaneous Change

chemical systems tend towards lower Gibbs energy (a.k.a chemical potential)

ΔGsys = ΔHsys – TΔSsys

(-)

(+)

ΔGsys < 0 (negative) ➜ process is spontaneous
ΔGsys = 0 (zero) ➜ process is at equilibrium
ΔGsys > 0 (positive) ➜ process is non-spontaneous

New Criterion for Spontaneous Change

(-)

ΔGsys = ΔHsys – TΔSsys

2

(+)

Practice Problem

Which of the following processes would you expect to

be spontaneous at all temperatures?

a)

∆H = 6 kJ/mol

b) 2SO2(g) + O2(g) → 2SO3(g) ∆H = -200 kJ/mol

c) 2NH4NO3(aq) → 2N2(g) + 4H2O(g) + O2(g)

∆H = -236 kJ/mol

a) None are spontaneous at all temperatures

b) All are spontaneous irrespective of temperature

Practice Problem

The following generic reaction is endothermic:

2A2B (g) !

2A2 (g) + B2 (g)

this means the reaction…

a) Will be spontaneous at all temperatures

b) Will be spontaneous only at high temperatures

c) Will be spontaneous only at low temperatures

d) Is not spontaneous at any temperature

e) Cannot be determined from the information given

Practice Problem

Consider a reaction that has a negative ΔrH and a positive

ΔrS. Which of the following statements is TRUE?

A) This reaction will be spontaneous only at high temperatures.

B) This reaction will be spontaneous at all temperatures.

C) This reaction will be nonspontaneous at all temperatures.

D) This reaction will be nonspontaneous only at high temperatures.

E) This reaction will be spontaneous ONLY at absolute zero (0

Kelvin).

Standard Change in Free Energy, ΔG°

• Chemists interested in reactions, not “systems”

• Standard change in free energy of a reaction (ΔG°rxn)

change in free energy for a reaction in which the

reactants and products are present in their standard

states at 1 bar (0.987 atm) often @ 25 °C.

• Three ways to calculate:

(1) From enthalpy and entropy data

(2) From tabulated ΔGf° data

(3) From tabulated ΔG°rxn data

(1) ΔG°rxn from Enthalpy and Entropy Data

• Tabulated values of ∆H°f can be used to nd ∆H°rxn

(Section 6.9, Table 6.5, page 225)

• Tabulated values of S° can be used to nd ∆S°rxn

• Together, allow calculation of ∆G°rxn

ΔG°rxn = ΔH°rxn – TΔS°rxn

fi

fi

• NOTE:

→ Method is only valid for reactions at 25 °C

→ However, ΔH and ΔS are only weakly T dependent

→ Method estimates ∆G° for reactions where T ≠ 25°C

(2) ΔG°rxn from ΔG°f data

• Standard free energy of formation, ΔGf° — The change in free

energy when 1 mole of a substance forms from its constituent

elements in their standard states (25 °C, 1 bar)

→ For pure elements in their standard states, ΔGf° = 0

Table 17.3 Standard Molar Gibbs Energies of Formation (!fG°)

for Selected Substances at 298 K

Substance

!fG° (kJ mol−1)

Substance

!fG° (kJ mol−1)

H2(g)

0

CH4(g)

−50.5

O2(g)

0

H2O(g)

−228.6

N2(g)

0

H2O(l)

−237.1

C(s, graphite)

0

NH3(g)

−16.4

C(s, diamond)

2.900

NO(g)

87.6

CO(g)

−137.2

NO2(g)

51.3

CO2(g)

−394.4

NaCl(s)

384.6

ΔG°rxn = [ ΣnpΔGf°(products) – Σ nrΔGf°(reactants)]

Practice Problem

Using data from appendix II, determine ΔG° at 298.15 K for

2NO(g) + O2(g) !

2NO2(g)

and decide whether the reaction is spontaneous. Estimate ΔG°

at 1000 K and decide whether the reaction will be spontaneous

at that temperature.

(3) ΔG°rxn From Tabulated ΔG°rxn Data

• ΔG°rxn values are known for many reactions

→ Use data from known reactions to create a thermodynamic cycle

(Hess’s Law)

Consider:

A + 2B !

2D

ΔG°rxn = ?

∆G°1

G

∆G°2

∆G°3 = ∆G°1 + ∆G°2

ss’s Law

Hess’s Law

ΔrGº is a state function and can be handled similarly to ΔrHº

Calculating ΔrGº for a Stepwise Reaction from the Changes

in Free Energy for Each of the Steps

1. If a chemical equation is multiplied by some factor,

then ΔrGº is also multiplied by the same factor.

2. If a chemical equation is reversed, then ΔrGº

changes sign.

3. If a chemical equation can be expressed as the sum

of a series of steps, then ΔrGº for the overall

equation is the sum of the free energies of

reactions for each step.

A + 2B ¾¾

®C

D r H1

2A + 4B ¾¾

® 2C

D r H 2 = 2 ´ D r H1

Relationships

involving

6.8 Relationships

Involving

ΔrH ΔrH

multiple of reaction

sum of reactions

2B¾¾

¾¾

D1r H1

AA++2B

®®CC

Dr H

6.8 Relationships Involving ΔrH

2A

+®

4BA¾¾

D r2H=2 -D

= 2r H

´ 1D r H1

C ¾¾

+®

2B2C

Dr H

multiple of reaction

sum

of reactions

reverse

of reaction

A

+¾¾

2B

+ 2B

¾¾

®

CC

A +A2B

®¾¾

C®

4BA

® 2C

C2A

¾¾

+2D

2B

C+®

¾¾

®¾¾

D1 r H1

DDr H

H

r

1

D2r H

= 2 ´ D r H1

DDrrH

=

2-D r H 1

H

2

reverse

of

sum A

of+reactions

2 Breaction

¾¾

® 2D

D r H 3 = D r H1 + D r H 2

+ 2B

¾¾

®C

D rr H11

A +A2B

¾¾

®

C

Copyright © 2020 Pearson Canada Inc.

CC

¾¾

®®

A +2D

2B

Drr H22 = -D r H1

¾¾

reverse

A + of

2 Breaction

¾¾

® 2D

D r H 3 = D r H1 + D r H 2

A + 2B ¾¾

®C

DH

6-

Practice Problem

Determine the free energy change in the following

reaction at 298 K given the known ∆G°rxn data below:

2 H2O(g) + O2(g) → 2 H2O2(g) ∆G°rxn = ?

Practice Problem

Which of the following processes would you expect to

result in a greater positive change in entropy?

1. H2O (l, 25 °C) → H2O (l, 75 °C)

2. CO2 (s, 101 kPa) → CO2 (g, 2kPa)

3. CH3OH (l, 101 kPa) → CH3OH (g, 2 kPa)

4. Fe(s) → Fe(l)

5. None of the above processes involve a positive

entropy change.

Chapter 18-2

Chapter 18-2

from last lecture: The Voltaic Cell

Zn(s)│Zn2+(aq, 1M) ║ Cu2+(aq, 1M)│Cu(s)

salt bridge

OIL RIG

Anode

(oxidation)

Oxidaton Is Lost

Reduction is Gain

Zn ➜Zn2+ + 2e-

Cathode

(reduction)

Cu2+ + 2e-➜Cu

Zn2+

1.00 M Zn(NO3)2(aq)

Cu2+

1.00 M Cu(NO3)2(aq)

Electrochemical Cell Notation

5 Fe(s) + 2 MnO 4 – ( aq) + 16 H + (aq ) ¾¾

® 5 Fe 2 + (aq ) + 2 Mn 2 + (aq ) + 8 H 2O(l )

Oxidation

Fe(s) ¾¾

® Fe 2 + (aq ) + 2e –

Reduction

MnO 4 ( aq) + 8H ( aq) + 5e ¾¾

® Mn (aq ) + 8 H 2 O(l )

–

+

–

2+

Driving force behind flow of e—?

• Cell potential, Ecell, is the potential

difference between the two

electrodes

*aka the electromotive force (emf)

*measured in volts (V):

energy per unit charge (1V = 1 Joule / 1 Coulomb)

1 Coulomb = 6.24×1018 electrons

Charles-Augustin de Coulomb

James Prescott Joule

Alessandro Giuseppe Antonio Anastasio Volta

Standard cell potential: Eºcell

Standard cell potential, E°cell : a measure of the overall

tendency of the reaction to occur spontaneously

E°cell = E°cathode — E°anode

E°cell is positive for

spontaneous reactions

and negative for

nonspontaneous

reactions.

analogy for electrical current

An analogy for electrode potential

Driving force behind the moving electrons

Quantifying Cell Potential

The standard electrode potential, Eº

the potential difference, or voltage, of a cell formed from

two standard electrodes. Difference is always taken as

Eºcell = Eº cathode − Eº anode

Measures the tendency for a reduction process to occur at

an electrode.

→All ionic species present at a = 1 (approximately 1 M).

→All gases are at 1 bar (approximately 1 atm).

→Where no metallic substance is indicated, the potential is

established on an inert metallic electrode (ex. Pt).

8

Standard Reduction Potential

E°cell = E°cathode — E°anode

From the cell voltage we cannot determine

the values of either — we must know one to

get the other

Enter the standard hydrogen electrode

(SHE)

All potentials are referenced to the SHE

(=0 V)

Standard reduction potential

SHE

Cu2+ + 2e— ➜ Cu

E°(Cu2+/Cu ) = 0.340V

SHE

greater tendency

than SHE to

undergo reduction

Zn2+ + 2e— ➜ Zn

E°(Zn2+/Zn) = —0.763V

2H+(1 M) + 2e- ⇌ H2(g, 1 bar)

E° = 0 V

SHE

greater tendency

than SHE to

undergo oxidation

Standard Electrode Potential, E°

stronger

oxidizing

agent

weaker

reducing

agent

Standard Electrode Potentials, E°

Redox couples/pairs with positive E°cell (more positive than

SHE ➜ greater tendency to undergo reduction (cathode)

Redox couples/pairs with negative E°cell (more negative than

SHE ➜ greater tendency to undergo oxidation (anode)

Standard reduction potentials

more positive E° ➜ electrode in any half-cell

with a greater tendency to undergo

reduction

more negative E° ➜ electrode in any halfcell with a lesser tendency to undergo

reduction (or greater tendency to undergo

oxidation)

Combining Standard Electrode Potentials

Zn(s)│Zn2+(aq, 1M) ║ Cu2+(aq, 1M)│Cu(s)

Zn + Cu2+ ! Zn2+ + Cu

Cu2+ + 2e- !Cu

E°(Cu2+/Cu) = 0.340 V

Zn2+ + 2e- !Zn

E°(Zn2+/Zn) = -0.763 V

E°cell = E°cathode – E°anode

Eºcell = 0.340 — (—0.763) = +1.103 V

Practice Problem

A voltaic cell consists of a Mg electrode in 1.0 M

Mg(NO3)2(aq) and a Ag electrode in 1.0 M AgNO3(aq).

Write the cell diagram, the overall reaction, and

determine E°cell at 25 ˚C.

Practice Problem

Determine E°cell for the reaction:

2 Al(s) + 3 Zn2+(aq) → 2 Al3+(aq) + 3 Zn(s).

The half reactions are:

Al3+(aq) + 3 e- → Al(s) E° = -1.676 V

Zn2+(aq) + 2 e- → Zn(s) E° = -0.763 V

A) 0.913 V

B) -2.439 V

C) 2.439 V

D) -1.063 V

E) -0.913 V

How do we know

if the redox reaction is spontaneous?

Identify the relationships between

Ecell, ΔG, and Keq

Use the Nernst equation to determine the

spontaneous direction of a reaction for

given initial conditions

Relationship between E°cell, ΔG°, and K

Spontaneous

Non-spontaneous

K

>1

0

0)

are nonspontaneous.

→ Often true, but there are exceptions:

H2O(s)

ΔH° = +6.01

kJ/mol

Enthalpy

isHnot

criterion

2O(l) a reliable

forH2spontaneous

change!

O

NaCl(s)

Na+(aq) + Cl—(aq) ΔH° = 3.9 kJ/mol

These endothermic reactions occur spontaneously!

The sign of ΔH does not predict the direction of a

spontaneous change!

Another Look at Spontaneity

• An ideal gas expands into a vacuum at constant T:

ΔU = ΔH = 0

→ Expansion is NOT driven by a change in energy

→ Particles are spreading out…

→ Energy is spreading out over more energy levels…

“Spreading out” of energy

Energy levels:

V

2V

• The allowed energy levels for translational motion are

spaced more closely together in a larger volume.

→ Energy of the gas spreads out over more energy levels

• New hypothesis: Spreading out of energy

drives spontaneous processes.

Entropy, S

• Entropy (S) — a thermodynamic property describing

the distribution of a system’s energy over the available

energy levels.

→ The greater the number of con gurations among

the energy levels in a particular system➜ the

greater the entropy of the system.

→ Entropy is a state function

→ unique value for a system with T, P and de ned

composition

→ value is independent of the path taken

fi

fi

ΔS = Sfinal – Sinitial

ΔS > 0

More levels

more ways to

distribute the energy

System A

System B

microstate: the exact energy distribution among the

molecules at any one instant

Microstates Explained…

Spontaneous expansion goes in the direction of

increasing number of microstates ➜ increasing

entropy

Smore microstates > Sfewer microstates

Microstates Explained…

• System A has 2 allowable energy levels while System B has 4

• with only 8 J of energy, one molecule cannot have all 8 J because that

would leave the other molecule with 0 J (motionless)

• violates Heisenberg’s uncertainty principle, there are only certain

“allowed” energy levels because they are composed of atoms that are

governed by the laws of quantum mechanics.

Entropy, a Mathematical De nition

S = k ln W

# of microstates

Boltzmann constant

1.38 x 10–23 J/(K•mol)

Ludwig Eduard Boltzmann

(1844 – 1906)

• microstate — the particular way in which the energy of

a state is distributed within the system

• W = # of microstates = # of energetically equivalent

ways of arranging the energy of a system

fi

• Entropy increases with increasing W

Macro- and Microstates

• Three (of ve) possible ways of

arranging 4 gas particles in two

connected asks — Macrostates

(P,V,T)

• Potential energy of each is

identical (ideal gas assumption)

fl

fi

• State B is more probable

because it has more microstates

→ greater entropy

Microstates Explained…

• Compare the number of microstates (energetically

equivalent arrangements) available for each state:

State C

W=1

State A

W=1

State B, W = 6

Melting of Ice

When ice melts, the arrangement of water molecules changes from

an orderly one to a more disorderly one.

Evaporating Water

When water evaporates, the arrangement of water molecules

becomes still more disorderly.

Dissolving Salt

When salt dissolves in water, the arrangement of the

molecules and ions becomes more disorderly.

Processes in Which Entropy Increases

change in the freedom of motion of particles,

dispersal of their energy of motion,

key factor for predicting the direction of spontaneity!

Sgas > Sliquid > Ssolid

www.youtube.com/watch?v=NQhjAtCKghE

Recall: Molecular Motion

• Kinetic energy of molecules: a result of their motion

Three types:

Processes in Which Entropy Increases

Sgas > Sliquid > Ssolid

Processes in Which Entropy Increases

NaCl(s)

H2O

Na+(aq) + Cl—(aq) ΔH° = 3.9 kJ/mol

S is a key factor in the formation of solutions!

✔︎

Practice Problem

Which of the following processes would NOT result in

an increase in entropy?

× 1.

× 2.

× 3.

× 4.

Melting of an ice cube

Sublimation of a moth ball

Evaporation of a puddle of gasoline

A g…

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