Energy profile for a reaction
“activated complex”
Rate-determining
Quantity
Ea
Energy
Chem 107: More Kinetics
Arrhenius Theory
reactants
∆Erxn
products
Thermodynamic Quantity
“Reaction Coordinate”
Ea, The Activation Energy
The reverse direction...
Energy
“activated complex”
“products”
■  Energy
Rate-determining
for reverse reaction
Ea(reverse)
of activation for forward reaction:
Ea = Etransition state - Ereactants
■  A
reaction can’t proceed unless reactants
possess enough energy to give Ea.
■  ∆E,
“Reaction Coordinate”
the thermodynamic quantity, tells us
about the net reaction. The activation
energy, Ea , must be available in the
surroundings for the reaction to proceed
at a measurable rate.
Temperature, qualitative effect
Maxwell-Boltzmann Distribution
∆Erxn
“reactants”
Thermodynamic Quantity
Increases in temperature
increase the fraction of
molecules that have the the
activation energy (Ea).
T = 300 K
Number
of
Molecules
T = 900 K
0
5
Ea
10
15
20
25
Kinetic Energy (10-21 J)
⎡ M ⎤
N (u)
= 4π ⎢
⎥
N total
⎣ 2π RT ⎦
3/ 2
u 2 e– Mu
2
2 RT
“kills it off”
at high u
N (u)
N total
30
u
Temp. and Rate Acceleration
Arrhenius Equation
■  Arrhenius
noted that reaction rates could
be understood to depend on Ea and T
with the exponential form:
k = Aexp(-Ea/RT)
■  Or,
in logarithmic form:
lnk = lnA - (Ea/RT)
using base 10 logs: [logk = logA - (Ea/2.303RT)]
Arrhenius Eqn., Alternative Form
Arrhenius Eqn., Graphical Form
A “best fit” to many data is better!
■  Taking
two measured values of the rate
(at two different temperatures) one can
write:
lnk = -(Ea/R)(1/T) + lnA
lnk 1 = lnA - (Ea /RT1)
lnk 2 = lnA - (Ea /RT2)
lnk 2 - lnk 1 = - (Ea /R)[(1/T2)-(1/T1)]
ln(k 2/k 1) = - (Ea/R)[(1/T2)-(1/T1)]
Arrhenius Equation, Example
■  If
a reaction has an activation energy of 50
kJ/mol, then how much should the rate of
the reaction accelerate if the temperature
is raised from 300 K to 310 K?
Arrhenius Equation, Example
■  If
a reaction has an activation energy of 50
kJ/mol, then how much should the rate of
the reaction accelerate if the temperature
is raised from 300 K to 310 K?
ln(k 310/k 300) = - (Ea/R)[(1/T2) - (1/T1)]
- (50,000 J/mol)/(8.314 J/mol K)•
[(1/310K)-(1/300K)] = 0.647
k 310 = e0.647k 300
roughly, rate doubles
=1.9k 300 for every 10 ˚C.
Catalysts - lowering Ea . Mechanism may change.
Pt
Catalytic Converters, 2NO → O2+N2
Pt
surface
Energy
Ea(uncatalyzed)
Ea(catalyzed)
reactants
∆Erxn
products
“Reaction Coordinate”
MnO2
→ 2H2O + O2
2H2O2
Reversible Reactions, Equilibrium
CHEM 107
T. Hughbanks
Reversible Elementary Reaction
2 NO ⇌
N2O2
■  At
equilibrium, forward and backward
rates equal:
forward rate = reverse rate
k f [NO]2 = k r [N2O2]
■  From
k
f
kr
Equilibrium Constants
■  In
general, for ANY reversible chemical
reaction:
αA + βB ⇌ γC + δD
■  We define an equilibrium constant:
K eq =
this:
=
[N 2 O 2 ]
[NO]
2
= constant = K eq
■  Value
[C]γeq [D]δeq
" products"
α
β =
" reactants"
[A]eq [B]eq
of Keq depends on specific
reaction and temperature.
Keq - important features
■  Although
we have introduced the concept
of an equilibrium constant in discussing
reaction rates, the value of the
equilibrium constant depends only on
thermodynamics! The quantitative
relationship is ∆G° = – RT lnKeq (more on
this later).
■  If a catalyst decreases Ea, it will increase
both k f and k r and does not affect Keq.
Equilibrium: Connection to
Thermodynamics
Magnitude of Keq
Magnitude of Keq
■  What
does it mean if Keq is very large?
CH4 + O2 → CO2 + 2H2O
25˚C, Keq is about 10140, so we don’t
usually write the “ ⇌ ” symbol.
■  Recall form of Keq :
“products/reactants”
■  With huge Keq, lots of products, no
reactants. Reaction goes to completion.
(∆G°≪ 0)
■  At
Equilibrium Calculations
■  If
Keq is neither huge nor tiny, both
products and reactants will be present
at equilibrium. (∆G is neither very
positive nor very negative.)
■  We can use value of Keq to find
equilibrium amounts from initial
amounts.
■  Use stoichiometry to relate changes in
concentration of various substances.
■  What
does it mean if Keq is very small?
N2 + O2 ⇌ 2NO
■  At 25˚C, Keq is 4.6 × 10–31
■  Recall form of Keq:
products/reactants
■  With tiny Keq, lots of reactants, no products.
No detectable reaction. (∆G° ≫ 0)
Equilibrium Calculation
■  Consider
the reaction CO + H2O ⇌ CO2 + H2
— called the “water-gas shift reaction”
■  Place 1.00 mol each of CO, H2O in a
50.0 L vessel at 1000 ˚C.
■  If Keq = 0.58, what will the equilibrium
concentrations of all 4 substances be?
■  Some hints on next slide.
Equilibrium Calculation: Strategy
■  Start
by writing expression for Keq
■  Think about what will happen. Which
concentrations will increase? Decrease?
■  Use stoichiometry to write changes in
concentrations in terms of one variable.
■  Set up equilibrium constant expression
and solve for the variable introduced
above, then calculate final
concentrations.
CO + H2O ⇌ CO2 + H2
Use stoichiometry to write equilibrium
concentrations in terms of one
variable.
CO + H2O ⇌ CO2 + H2
initial: 0.02 M 0.02 M
0
0
change: – x
–x
+x
+x
final: 0.02 – x 0.02 – x
x
x
Direction of Reaction
■  In
that example, we knew which way
reaction had to go because only
reactants were present at start.
both reactants & products are
present. How can we tell which way
reaction would go to reach equilibrium?
CO + H2O ⇌ CO2 + H2
■  Write
equilibrium constant expression in
terms of equilibrium concentrations:
K eq =
[CO 2 ][H 2 ]
[CO][H 2O]
Think about the chemistry. Starting with
only CO & H2O, reaction must go to the
right to form some CO2 & H2.
CO + H2O ⇌ CO2 + H2
■  Find
x by plugging into expression for Keq
and solving for x.
K eq =
[CO 2 ][H 2 ]
(x)( x)
(0.02 - x)(0.02 - x)
x2
0.58 =
(0.02 - x) 2
[CO][H 2 O]
■  Solve
=
and find x = 0.0086
use x to find concentrations.
■  Finally,
Reaction Quotient (Q)
■  Define
a “reaction quotient,” Q. Looks
just like Keq, except that concentrations
need not be equilibrium ones.
αA + βB ⇌ γ C + δ D
■  Suppose
K eq =
[C]γeq [D]δeq
[A]αeq [B]βeq
Q =
[C]γ [D]δ
[A]α [B]β
Reaction Quotient
Q and K to see which way
reaction should go.
■  Q < K : need more products, so
reaction goes from left to right
■  Q > K : need more reactants, so
reaction goes from right to left
■  Q = K : reaction is already at equilibrium
Example Problem (Haber Process)
N2(g) + 3H2(g) ⇌ 2NH3(g)
■  Compare
Haber Process - answer, 1st part
■  Compare
the value of Q with Keq
[NH 3 ]2
(0.01)2
=
= 23.15
[N2 ][H2 ]2
(0.02)(0.06)3
This larger than 0.5 (= Keq ). Rxn. shifts left.
Q=
■  [NH3]
will decrease from its initial value,
[H2] and [N2] will increase.
■  1.00
mol N2, 3.00 mol H2, and 0.500
mol NH3 are placed in a 50.0 L vessel at
400 ˚C. At this T, Keq = 0.500.
■  Will more NH3 form, or will NH3
dissociate to form more N2 and H2?
■  Set up an equation to find the
equilibrium concentrations.
Haber Process - answer, 2nd part
N2(g) + 3H2(g) ⇌ 2NH3(g)
define x as the amount that [NH3] decreases:
initial:
0.02M
0.06M
0.01M
change: +(1/2)x
+(3/2)x
–x
Keq =
[NH3 ]2
(0.01- x)2
=
[N2 ][H2 ]2 (0.02 + x/2)(0.06 + 3x/2)3
0.500 =
(0.01- x)2
⇒ x ≈ .0079
(0.02 + x/2)(0.06 + 3x/2)3
(Solution found graphically)
Le Châtelier's Principle
■  “When
a change is imposed on a
system at equilibrium, the system will
react in the direction that reduces the
amount of change.”
■  “Changes”
include adding or removing
material, or changes in pressure or
temperature.
Using Le Châtelier's Principle
CO(g) + 3H2(g) ⇌ CH4(g) + H2O(g)
■  An
equilibrium mixture at 1200 K
contains 0.613 mol CO, 1.839 mol H2,
0.387 mol CH4 and 0.387 mol H2O, all
in a 10.0 L vessel. (What is Keq?)
■  All of the H2O is somehow removed,
and equilibrium is re-established.
■  What will happen to the amount of CH4?
■  Set up eqn. for final amount of CH4.
Using Le Châtelier's Principle - Set up
K eq =
[CH 4 ][H 2 O]
[CO][H 2 ]3
Using Le Châtelier's Principle - Set up
(.0387 + x)x
(.1839 - 3x) 3 (.0613 - x)
K eq = 3.93 =
CO(g) + 3H2(g) ⇌ CH4(g) + H2O(g)
init. .0613
.1839
.0387 .0387 M
chg. .0613
.1839
.0387
0.0 M
■  Equilibrium is re-established by production
of more H2O (let the final [H2O] = x):
fin. .0613-x .1839-3x .0387+x x M
Changes in Temperature
■  The
equilibrium “constant” is not constant
with temperature.
Le Châtelier’s Principle would suggest:
■  Qualitatively, if a reaction is endothermic
then the equilibrium “constant” increases
with temperature
■  If a reaction is exothermic then the
equilibrium “constant” decreases with
temperature
Equilibria Involving Solutions
■  Many
equilibrium reactions occur in
solutions. Weak acids & bases, etc.
Changes in Pressure
Consider the equilibrium (significant at 700 ˚C):
CaO(s) + CO2(g) ⇌ CaCO3(s)
Le Châtelier’s Principle would suggest:
If the pressure is suddenly increased, say by
suddenly compressing the container, more
CO2 would react with CaO to produce
more CaCO3.
Equilibria Involving Solutions
NH3(aq) + H2O
K eq =
■  Example:
solution of ammonia
NH3(aq) + H2O ⇌ NH4+(aq) + OH–(aq)
■  Write
an equilibrium constant for this.
■  But
⇌
NH4+(aq) + OH–(aq)
[NH +4 ][OH – ]
[NH 3 ][H 2 O]
“[H2O]” is constant, so we can
incorporate it into the constant itself.
Acids - Definition of Ka
Equilibria Involving Solutions
NH3(aq) + H2O ⇌ NH4+(aq) + OH–(aq)
[NH +4 ][OH – ]
K eq =
= Kb
[NH 3 ]
■  By
convention, tabulated K’s are written like
this, without including water term.
■  NH3 produces OH– in water and is therefore
a base. Keq for a base is called Kb.
■  For NH3, Kb = 1.8 × 10-5 . What % of NH3 is
converted to NH4+ in a 1.0 M solution?
HA(aq) + H2O ⇌ H3O+(aq) + A–(aq)
K eq =
[H 3 O + ][A – ]
[HA]
= Ka
■  Once
again, the water term is omitted in
tabulating the equilibrium constants of acids.
■  An acid, say HA, produces H3O+ in water.
Keq for an acid is called Ka.
■  For CH3COOH (acetic acid), Ka = 1.8 × 10-5 .
What is [H3O+] in a 0.1 M solution?
Equilibria Involving Solutions
Equilibria Involving Solutions
Water self-dissociates, even in the absence
of added acids or bases:
In all aqueous solutions, equilibrium between
H3O+ & OH– is continually re-established:
H2O + H2O ⇌ H3O+(aq) + OH–(aq)
H2O + H2O ⇌ H3O+(aq) + OH–(aq)
K eq = [H 3O ][OH ] = 1.0 × 10
+
–
−14
= Kw
■  Again,
“[H2O]” is constant, so Kw does
not include it.
■  In pure water, what are [H3O+] & [OH–]?
K eq = [H 3O + ][OH – ] = 1.0 × 10 −14 = K w
■  This
■  In
is true even with added acids or bases.
a 1.0 M NH3 solution, what is [H3O+]?
pH Scale
pH and pOH
H3O+ concentration is conveniently measured
using the logarithmic pH scale:
pH = -log[H3O+]
pH = -log [H3O+]
pOH = -log [OH-]
■  In
pure H2O,
[H3O+] = 10–7, pH = -log 10–7 = 7
■  pOH is defined similarly: pOH = -log[OH-]
■  What are the pH and pOH of a 1.0 M NH3
solution?
aqueous solutions, [H3O+][OH-] = 10–14
■  This is true even with added acids or bases.
-log {[H3O+][OH-]} = -log {10–14}
-log [H3O+] - log [OH-] =14
pH + pOH = 14
■  In