Organic Chemistry Compounds Questions

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
Copyright © 2020 Pearson Canada Inc.
[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
Copyright © 2020 Pearson Canada Inc.
[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…