Chemical Reactions
• The following table summarizes the common Driving forces that
tend to make reactions go in the direction of the arrow:
Reactants  Products
Formation of a solid
Formation of water
Formation of gas
Transfer of electrons
• When two or more chemicals are brought together, if any of the
above can occur, a chemical reaction is likely to take place!
CHEMISTRY
- science of properties and transformations of matter
CHEMICAL REACTIONS - central to CHEMSITRY
CLASSIFICATION of REACTIONS
Precipitation
Oxidation-reduction
Acid-base
Gas forming
Precipitation Reactions
Solubility Rules
• There are simple guidelines that allow one to predict whether or not a
reaction will produce an insoluble compound: these are called the
Solubility Rules
Ionic Equations:
Net Ionic
• The ions that appear on both sides of the equation in identical
forms. These ions are not involved directly in the reaction and
are called spectator ions.
Pb2+(aq) + 2NO3-(aq) + 2K+(aq) + 2I-(aq)  PbI2 (s) + 2K+(aq) + 2NO3-(aq)
spectator ions
• In the net ionic equation, the spectator ions are omitted
leaving only the ions directly involved in the reaction.
Pb2+(aq) + 2I-(aq)  PbI2 (s)
Ionic Equations:
Summary
• Summary of the 3 different chemical equations:
– Molecular equation
Pb(NO3)2(aq) + 2KI(aq)  PbI2 (s) +2KNO3(aq)
– Complete ionic equation
Pb2+(aq) + 2NO3-(aq) + 2K+(aq) + 2I-(aq)  PbI2 (s) + 2K+(aq) + 2NO3-(aq)
– Net ionic equation
Pb2+(aq) + 2I-(aq)  PbI2 (s)
Acid-Base Reactions:
Neutralization
• Consider the reaction between Mg(OH)2 [milk of magnesia] and
HCl (stomach acid)
2HCl(aq)  Mg(OH )2 (aq)  MgCl2 (aq)  2H 2O(l )
• The net reaction is:


H (aq)  OH (aq)  H 2O(l )
The acid has been neutralized
‫یونهای دارای عدد اکسایش و کاهش متفاوت‬
Half-Reaction Method Of
Balancing Redox Equations
•
•
•
•
•
•
•
•
•
•
•
•
•
Assign oxidation numbers to each atom and identify the species
undergoing oxidation and reduction. Write two skeleton halfreactions.
oxidation # increase => oxidation; oxidation # decrease =>
reduction
(1) Balance the numbers of atoms in each half-reaction.
(a) Balance all atoms except O and H.
(b) Balance O atoms by adding H2O molecules.
(c) Balance H atoms by adding H+ ions.
(2) Balance electric charge by adding electrons to more positive
side.
(3) Combine two half-reactions to obtain balanced equation.
(a) Multiply each half-reaction by factors that make the number
of
electrons in two half-reactions equal.
(b) Combine the adjusted half-equations into an overall
equation.
Simplify the balanced equation if necessary.
Verify that the equation is balanced.
•
•
•
•
Balance the following redox equation in acidic solution:
Fe2+ (aq) + Cr2O72- (aq)  Fe3+ (aq) + Cr3+ (aq)
Fe2+ (aq)  Fe3+ (aq)+e
Cr2O72- (aq) +e  Cr3+ (aq)
‫تعریف سینتیک شیمیایی‬
‫• ترمودینامیک شیمیایی علمی است که پیرامون خود بخود‬
‫بودن یا خود بخود نبودن یک تغییر فیزیکی یا شیمیایی از‬
‫نظر انرژی مطالعه می نماید‬
‫• سینتیک شیمیایی مطالعه سرعت یا تندی واکنش های‬
‫شیمیایی است‪ .‬عوامل کمی در تعیین سرعت یک واکنش‬
‫دخالت دارند‪ .‬مطالعه این عوامل به ما کمک می کند تا سر‬
‫نخهایی پیرامون نحوه تبدیل مواد واکنش دهنده به فراورده‬
‫ها در واکنش شیمیایی بدست آوریم‪ .‬مسیر تبدیل مواد اولیه‬
‫به محصول مکانیسم واکنش نام دارد‪.‬‬
‫تعریف سرعت واکنش‬
Reaction rate: changes in a concentration
of a product or a reactant per unit time.
[ ]
Reaction rate = ——
t
[]
concentration
change
[ ]
t
Define reaction rate and explain
Average reaction rate
Instantaneous reaction rate
(2 tangents shown)
Initial reaction rate
t
‫سرعت نسبی واکنش‬
For a chemical reaction, there are many ways to express the reaction
rate. The relationships among expressions depend on the equation.
Note the expression and reasons for their relations for the reaction
2 NO + O2 (g) = 2 NO2 (g)
[O2]
1 [NO]
1 [NO2]
Reaction rate = – ——— = – — ———— = — ———
t
a
t
b t
Make sure you can write expressions for any reaction and figure out
the relationships. For example, give the reaction rate expressions for
2 N2O5 = 4 NO2 + O2
How can the rate expression be unique and universal?
Calculating reaction rate
The concentrations of N2O5 are 1.24e-2 and 0.93e-2 M at 600 and 1200 s after the
reactants are mixed at the appropriate temperature. Evaluate the reaction rates for
2 N2O5 = 4 NO2 + O2
Solution:
(0.93 – 1.24)e-2
– 0.31e-2 M
Decomposition rate of N2O5 = – ———————— = – ——————
1200 – 600
600 s
= 5.2e-6 M s-1.
Note however,
rate of formation of NO2 = 1.02e-5 M s-1.
rate of formation of O2 = 2.6e-6 M s-1.
Be able to do this type problems
The reaction rates are
expressed in 3 forms
‫سرعت کلی واکنش یا قانون سرعت‬
Dependence of reaction rate on the concentrations of reactants is
called the rate law, which is unique for each reaction.
For a general reaction, a A + b B + c C  products
the rate law has the general form
order wrt A, B, and C, determined experimentally
reaction rate = k [A]X [B]Y [C]Z
the rate constant
For example, the rate law is
rate = k [Br-] [BrO3-] [H+]
for
5 Br- + BrO3- + 6 H+  3Br2 + 3 H2O
Use differentials
to express rates
The reaction is 1st order wrt all three reactants, total order 3.
‫مرتبه واکنش‬
‫‪ -1‬مجموع توانهای عبارتهای غلظت در معادله‬
‫قانون سرعت مرتبه واکنش نامیده می شود‬
‫‪ -2‬مرتبه واکنش به دو صورت مرتبه کلی یا مرتبه‬
‫گردد‬
‫می‬
‫بیان‬
‫جزئی‬
‫‪ -3‬مرتبه کلی یا جزئی یک واکنش را تنها از‬
‫طریق آزمایش می توان بدست آورد‬
‫نمودار تغییرات سرعت برای واکنش با مرتبه های مختلف‬
2nd order, rate = k [A]2
rate
First order, rate = k [A]
k = rate, 0th order
[A] = ___?
[A]
The variation of reaction rates as functions of concentration for
various order is interesting.
Mathematical analysis is an important scientific tool, worth noticing.
‫مثال‬
‫در واکنش زیر مرتبه واکنش نسبت به تک تک واکنشگرها را بدست آورده سپس‬
‫مقدار ثابت سرعت واکنش را محاسبه نمایید‪ .‬در صورتی که غلظت تمام واکنشگرها‬
‫یک موالر باشد سرعت واکنش چقدر است؟‬
‫‪H2O2 + 3 I- + 2 H+  I3- + 2 H2O‬‬
‫‪Initial rate M s-1‬‬
‫‪1.15e-6 R = k [H2O2]x [I-]y [H+]z‬‬
‫‪2.30e-6 R = k [H2O2]1 [I-]y [H+]z‬‬
‫‪2.30e-6 R = k [H2O2]1 [I-]1 [H+]z‬‬
‫‪1.15e-6 R = k [H2O2]1 [I-]1 [H+]0‬‬
‫]‪[H+‬‬
‫‪0.0050‬‬
‫‪0.0050‬‬
‫‪0.0050‬‬
‫‪0.0100‬‬
‫]‪[I-‬‬
‫‪0.010‬‬
‫‪0.010‬‬
‫‪0.020‬‬
‫‪0.010‬‬
‫]‪[H2O2‬‬
‫‪0.010‬‬
‫‪0.020‬‬
‫‪0.010‬‬
‫‪0.010‬‬
‫‪Exprmt‬‬
‫‪1‬‬
‫‪2‬‬
‫‪3‬‬
‫‪4‬‬
‫)‪R = 1.15e-6 = k (0.010)(0.010‬‬
‫‪k =0.0115 M-1 s-1‬‬
‫‪R = k [H2O2] [I-]=0.0115(1)(1)=0.0115‬‬
‫‪Solution next‬‬
Differential Rate Law determination - continue
From the following reaction rates observed in 4 experiments, derive the
rate law for the reaction
A + B + C  products
where reaction rates are measured as soon as the reactants are mixed.
Expt
[A]o
[B]o
[C]o
rate
1
0.100
0.100
0.100
0.100
2
0.200
0.100
0.100
0.800
3
0.200
0.300
0.100
7.200
4
0.100
0.100
0.400
0.400
This example illustrates the strategy to determine, and a
reliable method to solve rate-law experimentally.
Solution next
Differential Rate Law determination - continue
From the following reaction rates, derive the rate law for the reaction
A + B + C  products
where reaction rates are measured as soon as the reactants are mixed.
Expt
[A]o
[B]o
[C]o
rate
1
0.100
0.100
0.100
0.100
2
0.200
0.100
0.100
0.800
3
0.200
0.300
0.100
7.200
4
0.100
0.100
0.400
0.400
order
3 from expt 1 & 2
2 expt 1, 2 & 3
1 expt 1 & 4
Assume rate = k [A]x[B]y[C]z
Therefore 8 = 2x
log 8 = x log 2
x = log 8 / log 2
=3
0.800 k 0.2x 0.1y 0.1z
----- = ---------------------0.100 k 0.1x 0.1y 0.1z
‫نمایش مشتقی و انتگرالی قانون سرعت‬
‫نمایش مشتقی‬
‫نمایش انتگرالی‬
R= – d[A] / dt = k
[A] = [A]o – k t
d[A]
R= – —— = k [A]
dt
[A] = [A]o e – k t or ln [A] = ln [A]o – k t
d[A]
R= – —— = k [A]2
dt
1
1
—— – —— = k t
[A]
[A]o
Describe, derive and apply the integrated rate laws
Learn the strategy to determine rate-law
[A] conc at t
[A]o conc at t = 0
‫نمودار تغییرات غلظت در سرعت مرتبه اول‬
Describe the features of plot of [A] vs. t and ln[A] vs. t for 1st order
reactions. Apply the technique to evaluate k or [A] at various times.
[A]
ln[A]
ln [A] = ln [A]o – k t
[A] = [A]o e – k t
t½
t
t
‫نمودار تغییرات غلظت در سرعت مرتبه دوم‬
t=
[B]
1
5
10
15
30
35
[B] = 1.67 1.0 0.67 0.50 0.29 0.25
1
—
[B]
[B]o
1
1
[B] = ——————
——= k t + ——
[B]o k t + 1
[B]
[B]o
t
What kind of plot is linear for 1st and 2nd reactions?
t
‫نیمه عمر‬
The time required for half of A to decompose is called half life t1/2.
Since
[A] = [A]o e – k t
or
When
t = t1/2, [A] = ½ [A]o
Thus
ln ½ [A]o = ln [A]o – k t1/2
ln [A] = ln [A]o – k t
– ln 2 = – k t1/2
k t1/2 = ln 2 = 0.693  relationship between k and t1/2
Radioactive decay usually follow 1st order kinetics, and half life of an
isotope is used to indicate its stability.
Evaluate t½ from k or k from t½
1st order reaction calculation
N2O5 decomposes according to 1st order kinetics, and 10% of it
decomposed in 30 s. Estimate k, t½ and percent decomposed in 500 s.
If the rate-law is known, what are the key parameters?
Solution next
1st order reaction calculation
N2O5 decomposes according to 1st order kinetics, and 10% of it decomposed
in 30 s. Estimate k, t½ and percent decomposed in 500 s.
Solution: Assume [A]o = [N2O5]o = 1.0, then [A] = 0.9 at t = 30 s or
0.9 = 1.0 e – k t apply [A]o = [A] e– k t
ln 0.9 = ln 1.0 – k 30 s
– 0.1054 = 0 – k * 30
k = 0.00351 s – 1
t½ = 0.693 / k = 197 s apply k t ½ = ln 2
[A] = 1.0 e – 0.00351*500 = 0.173
Percent decomposed: 1.0 – 0.173 = 0.827 or 82.7 %
After 2 t½ (2*197=394 s), [A] = (½)2 =¼, 75% decomposed.
After 3 t½ (3*197=591 s), [A] = (½)3 =1/8, 87.5% decomposed.
Apply integrated rate law to solve problems
Typical Problem wrt
Reaction
st
1
Order
The decomposition of A is first order, and [A] is monitored. The following
data are recorded:
t / min 0
[A]/[M] 0.100
2
0.0905
4
0.0819
8
0.0670
Calculate k (What is the rate constant? k = 0.0499)
Calculate the half life (What is the half life? Half life = 13.89)
Calculate [A] when t = 5 min. (What is the concentration when t = 5 min?)
Calculate t when [A] = 0.0100. (Estimate the time required for 90% of A
to decompose.)
Work out all the answers
‫مقایسه واکنش با مرتبه های مختلف‬
‫مرتبه دو‬
‫مرتبه یک‬
‫مرتبه صفر‬
‫قانون سرعت‬
‫واحد ثابت سرعت‬
‫معادله خط راست‬
‫شیب و عرض از مبدا‬
‫نیمه عمر‬
The Arrhenius Equation
The temperature dependence of the rate constant k is best
described by the Arrhenius equation:
or
k = A e – Ea / R T
ln k = ln A – Ea / R T
If k1 and k2 are the rate constants at T1 and T2 respectively, then
k1
Ea 1
1
ln —— = – — — – —
k2
R T1 T2
1903 Nobel Prize citation” …in recognition of the extraordinary services he has
rendered to the advancement of chemistry by his electrolytic theory of dissociation”
How does temperature affect reaction rates?
Derive and apply these relationship to solve problems,
and recall the Clausius-Clapeyron equation.
Energy in chemical reactions
Potential
energy
R +A 
H
exothermic
RA-PD
activated
complex
Ea Ea for reverse reaction
 P+D
Endothermic rxn
Progress of reaction
Explain the various terms and energy changes in a chemical reaction
‫نظریه های موجود برای تفسیر پدیده های‬
‫سینتیکی‬
‫‪ -1‬نظریه برخورد‬
‫‪Collision theory‬‬
‫‪ -2‬نظریه حالت گذار‬
‫‪Transition state theory‬‬
A---B---C
The height of the barrier is called the
activation energy, Ea.
The configuration of atoms at the
Ea
maximum in the P.E. profile is called
Potential Energy
the transition state.
A + BC
Reactants
AB + C
Products
Reaction Progress
A2(g) +
B2(g)
2AB(g)
‫برخورد موثر‬
‫فرضیه تشکیل کمپلکس فعال‬
‫انواع واکنشها از نظر مکانیسمی‬
‫‪ -1‬واکنشهای بنیادی واکنشهایی هستند که تنها از یک مرحله تشکیل شده اند و‬
‫در آنها مرتبه واکنش برابر مجموع ضرایب استوکیومتری واکنش است‪.‬‬
‫]‪O3  O2 + O rate = k [O3‬‬
‫‪rate = k [NO2]2‬‬
‫‪NO2 + NO2  NO3 + NO‬‬
‫‪ -2‬واکنشهای غیر بنیادی واکنشهایی که در آنها مراحل مختلفی‬
‫برای تبدیل مواد اولیه به محصول و جود دارد را گویند در این گونه واکنشها مرتبه‬
‫واکنش با ضرایب استوکیومتری برابر نمی باشد تقریبا همه واکنشها از این نوع‬
‫می باشند‪.‬‬
‫واکنشهای بنیادی در یک واکنش واقعی‪ ،‬فرضی بوده و به کمک مکانیسم‪ ،‬تشخیص‬
‫حد واسطها و قانون سرعت واکنش می توان به صحت آنها پی برد‬
‫‪NO2 + CO  NO + CO2‬‬
‫)‪(an elementary reaction‬‬
‫‪Step 1 NO2 + NO2  NO3 + NO‬‬
‫مولکوالریته‬
‫مرتبه کل واکنش در یک واکنش بنیادی را مولکوالریته گویند‬
‫واکنشهایی بنیادی که که در آن مرتبه واکنش یک باشد را تک مولکولی‬
‫واکنشهایی بنیادی که که در آن مرتبه واکنش دو باشد را دو مولکولی‬
‫و واکنشهایی بنیادی که که در آن مرتبه واکنش سه باشد را سه مولکولی گویند‬
Molecularity of elementary reactions Example
Some elementary reactions for the
reaction between CH4 and Cl2 are
Cl2  2 Cl
2 Cl  Cl2
2Cl + CH4  Cl2 + CH4*
Cl + CH4  HCl + CH3
CH3 + Cl  CH3Cl
CH3 + CH3  CH3-CH3
CH3Cl + Cl  HCl + CH2Cl
CH2Cl + Cl  CH2Cl2
* * * (more)
Write down the rate laws
and describe them as
uni- bi- or ter-molecular
steps yourself, please.
Rate Laws and Mechanisms
A mechanism is a collection of elementary steps devise to explain the the
reaction in view of the observed rate law. You need the skill to derive a
rate law from a mechanism, but proposing a mechanism is task after you
have learned more chemistry
For the reaction, 2 NO2 (g) + F2 (g)  2 NO2F (g), the rate law is,
rate = k [NO2] [F2] .
Can the elementary reaction be the same as the overall reaction?
If they were the same the rate law would have been
rate = k [NO2]2 [F2],
Therefore, they the overall reaction is not an elementary reaction. Its
mechanism is proposed next.
Rate-determining Step in a Mechanism
The rate determining step is the slowest elementary step in a mechanism,
and the rate law for this step is the rate law for the overall reaction.
The (determined) rate law is,
rate = k [NO2] [F2],
for the reaction,
2 NO2 (g) + F2 (g)  2 NO2F (g),
and a two-step mechanism is proposed:
i NO2 (g) + F2 (g)  NO2F (g) + F (g)
ii NO2 (g) + F (g)  NO2F (g)
Which is the rate determining step?
Answer:
The rate for step i is rate = k [NO2] [F2], which is the rate law, this
suggests that step i is the rate-determining or the s-l-o-w step.
Explain rate determining step in a mechanism
and use it to derive the rate law.
2 ‫مثال‬
The decomposition of H2O2 in the presence of I– follow this
mechanism,
i
ii
H2O2 + I–  k1 H2O + IO–
H2O2 + IO–  k2 H2O + O2 + I–
What is the rate law?
slow
fast
Energy
Eai
Eaii
reaction
Solve the problem
The decomposition of H2O2 in the presence of I– follow this mechanism,
i
ii
H2O2 + I–  k1 H2O + IO–
slow
H2O2 + IO–  k2 H2O + O2 + I–
fast
What is the rate law?
Solution
The slow step determines the rate, and the rate law is:
rate = k1 [H2O2] [I –]
Since both [H2O2] and [I –] are measurable in the system, this is the
rate law.
‫مثال ‪( 3‬تمرین ‪ 24‬پایان فصل)‬
‫در واکنش زیر معادله سرعت به صورت‬
‫‪N2O5+NO 3NO2‬‬
‫]‪k1k3[ N 2O5 ][NO‬‬
‫]‪k2 [ NO2 ]  k3[ NO‬‬
‫‪RN 2O5 ‬‬
‫است مکانیسم پیشنهادی به صورت زیر است‬
‫‪k1‬‬
‫‪N 2O5 ‬‬
‫‪NO2  NO3‬‬
‫‪k2‬‬
‫‪NO2  NO3 ‬‬
‫‪N 2O5‬‬
‫‪k3‬‬
‫‪NO  NO3 ‬‬
‫‪2NO2‬‬
‫مرحله سوم مرحله کند واکنش بوده و با توجه به تقریب حالت پایا نشان دهید که مکانیسم‬
‫فوق همان معادله سرعت را تایید می کند‬
‫در صورتی که ‪ k2<<k3‬باشد معادله سرعت چه تغییری مینماید‬
‫راه حل‬
‫‪ -1‬نوشتن معادله سرعت واکنش چون در مسئله مشخص نشده که کدام‬
‫مرحله مرحله کند واکنش است بنابر این مرحله آخر به عنوان مرحله‬
‫تعیین کننده سرعت در نظر گرفته می شود‬
‫سرعت واکنش = ]‪k3[NO][NO3‬‬
‫‪ -2‬تقریب حالت پایا‪ :‬در این تقریب فرض می شود که غلظت حد واسط‬
‫ثابت است و سرعت تولید شدن آن و سرعت مصرف شدنش برابر باشد‬
‫پس‬
‫سرعت تولید ]‪=[NO3‬سرعت مصرف ]‪[NO3‬‬
‫سرعت تولید ]‪K1[N2O5]=[NO3‬‬
‫سرعت مصرف ]‪K2[NO2][NO3]+k3[[NO][NO3]=[NO3‬‬
‫‪ -3‬با برابر قرار دادن سرعت تولید و مصرف حد واسط خواهیم داشت‬
‫] ‪k1[ N 2O5‬‬
‫‪[ NO3 ] ‬‬
‫]‪k2 [ NO2 ]  k3[ NO‬‬
‫باجانشین کردن غلظت حد واسط در رابطه سرعت خواهیم داشت‬
‫]‪k1k3[ N 2O5 ][NO‬‬
‫‪‬‬
‫]‪k2 [ NO2 ]  k3[ NO‬‬
‫‪RN 2O5‬‬
‫درصورتی که ‪ k2<<k1‬خواهیم داشت‬
‫‪k1k3‬‬
‫‪‬‬
‫] ‪[ N 2O5 ]  k[ N 2O5‬‬
‫‪k3‬‬
‫‪RN 2O5‬‬
Deriving rate laws from mechanisms – steadystate approximation
The steady-state approximation
is a general method for deriving
rate laws when the relative
speed cannot be identified.
It is based on the assumption
that the concentration of the
intermediate is constant.
[Intermediate]
Rate of producing the
intermediate, Rprod, is
the same as its rate of
consumption, Rcons.
Rprod > Rcons
Be able to apply the steadystate approximation to derive
rate laws
Rprod = Rcons
Rprod < Rcons
time
Steady-state
Let’s assume the mechanism for the reaction.
H2 + I2  2 HI
approximatio
as follows.
n-2
Step (1)
I2 —k1 2 I
Step (1)
2 I —k-1 I2
Step (2)
H2 + 2 I —k2 2 HI
Derive the rate law.
Derivation:
rate = k2 [H2] [I] 2 (‘cause this step gives products)
but I is an intermediate, this is not a rate law yet.
Since k1 [I2]
(= rate of producing I) Steady state
= k-1 [I]2 + k2 [H2] [I]2
(= rate of consuming I)
Thus,
k1 [I2]
[I]2 = ——————
k-1 + k2 [H2]

rate = k1 k2 [H2] [I2] / {k-1 + k2 [H2] }
Steady-state approximation - 3
From the previous result:
k1 k2 [H2] [I2]
rate = ———————
{k-1 + k2 [H2] }
Discussion:
(i) If k-1 << k2 [H2] then {k-1 + k2 [H2]} = k2 [H2] ,
then rate = k1 k2 [H2] [I2] / {k2 [H2] } = k1 [I2] (pseudo 1st order wrt I2)
using large concentration of H2 or step 2 is fast (will meet this condition).
(ii) If step (2) is slow, then k2 << k1,
and if [H2] is not large, we have {k-1 + k2 [H2]} = k-1
and rate = k1 k2 [H2] [I2] / k1 = k2 [H2] [I2]
Steady-state approximation - 4
In an alkaline solution, peroxydisulfate oxidizes sulfite to yield sulfate,
S2O82- + SO32- + 2 OH-  3 SO42- + H2O.
The following mechanism has been proposed:
i
S2O82- + SO32- —k1 S2O72- + SO42ii
S2O72- + H2O —k2 2 SO42- + 2 H+
iii
H+ + OH- —k3 H2O (fast equilibrium to be discussed)
Steady-state approximation follows these steps:
What is or are the intermediates I?
Use which step to give the rate law that may involve [I]?
Express the rates of producing and consuming intermediate(s)
Express [I] of intermediate(s) in terms of [Reactants]
Derive the rate law in terms of [Reactants]
Discuss
See page 607 PHH Text
Catalysis
Energy
A catalyst is a substance that changes the rate
Uncatalyzed rxn
of a reaction by lowing the activation energy, Ea.
It participates a reaction in forming an
intermediate, but is regenerated.
Enzymes are marvelously selective catalysts.
A catalyzed reaction,
NO (catalyst)
2 SO2 (g) + O2 — 2 SO3 (g)
via the mechanism
i
2 NO + O2  2 NO2 (3rd order)
ii
NO2 + SO2  SO3 + NO
Catalyzed rxn
rxn
Catalyzed decomposition of
ozone
R.J. Plunkett in DuPont discovered
carbon fluorine chlorine compounds.
The CFC decomposes in the atmosphere:
CFCl3  CFCl2 + Cl
CF2Cl3  CF2Cl + Cl.
The Cl catalyzes the reaction via the mechanism:
i
O3 + h v  O + O 2,
ii
ClO + O  Cl + O2
iii
O + O3  O2 + O2.
The net result or reaction is
2 O3  3 O2
Scientists sound the alarm, and the CFC is banned now.
Homogenous vs.
heterogeneous catalysts
A catalyst in the same phase (gases and solutions) as the reactants is a
homogeneous catalyst. It effective, but recovery is difficult.
When the catalyst is in a different phase than reactants (and products),
the process involve heterogeneous catalysis. Chemisorption, absorption,
and adsorption cause reactions to take place via different pathways.
Platinum is often used to catalyze hydrogenation
Catalytic converters reduce CO and NO emission.
Heterogeneous catalysts
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Catalyzed reactions:
CO + O2 CO2
2 NO  N2 + O2
Enzymes – selective catalysts
Enzymes are a long protein molecules that fold into balls. They often
have a metal coordinated to the O and N sites.
Molecules catalyzed by enzymes are called substrates. They are
held by various sites (together called the active site) of the enzyme
molecules and just before and during the reaction. After having
reacted, the products P1 & P2 are released.
Enzyme + Substrate ES (activated complex)
ES  P1 + P2 + E
Enzymes are biological catalysts for biological systems.
X-ray 3-D structure of
fumarate reductase. It
reduces fumerate, an
important role in the
metabolism of
anaerobic bacteria,
from Max Planck Inst.
Chemical Kinetics - Summary
Explain how the various factors affect reaction rates.
Define reaction rates, average rates, initial rates and rate constants.
Evaluate rate law from experiments
Properly apply 1st and 2nd differential rate laws and integrated rate laws.
Interpret elementary reactions and mechanisms. Derive rate laws from
a given mechanism. Apply the steady-state method to derive the rate
law of a given mechanism, and discuss the results.
Explain the action of catalysts in terms of chemistry and in terms of
energy of activation.
Fig. 13.21
Fig. 13.23
Fig. 13.26
A) What is the Rate Law For This Reaction?
B) What is the order in respect to each reactant?
C) What is the overall reaction order?
D) What is k, the rate constant?