15 February 2012. Objective : You will be able to: define “kinetics” and identify factors that affect the rate of a reaction. write rate expressions for balanced chemical reactions. Agenda. Do now Kinetics notes Reaction Rates Demonstrations Rate constant and reaction rates problems.
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Homework: p. 602 #2, 3, 5, 7, 12, 13, 15, 16, 18: Thurs.
A B
rate =
D[A]
D[B]
rate = 
Dt
Dt
A B
rate =
D[A]
D[B]
rate = 
Dt
Dt
Chemical Kinetics
Reaction rate is the change in the concentration of a reactant or a product with time (M/s).
D[A] = change in concentration of A over
time period Dt
D[B] = change in concentration of B over
time period Dt
Because [A] decreases with time, D[A] is negative.
Br2(aq) + HCOOH (aq) 2Br(aq) + 2H+(aq) + CO2(g)
time
393 nm
Detector
light
redbrown
t1< t2 < t3
D[Br2] aD Absorption
Br2(aq) + HCOOH (aq) 2Br(aq) + 2H+(aq) + CO2(g)
slope of
tangent
slope of
tangent
slope of
tangent
[Br2]final – [Br2]initial
D[Br2]
average rate = 
= 
Dt
tfinal  tinitial
instantaneous rate = rate for specific instance in time
2A B
aA + bB cC + dD
rate = 
=
=
rate = 
= 
D[C]
D[B]
D[A]
D[B]
D[D]
D[A]
rate =
1
1
1
1
1
Dt
Dt
Dt
Dt
Dt
Dt
c
d
a
2
b
Reaction Rates and Stoichiometry
Two moles of A disappear for each mole of B that is formed.
D[CO2]
=
Dt
D[CH4]
rate = 
Dt
D[H2O]
=
Dt
D[O2]
= 
1
1
Dt
2
2
Write the rate expression for the following reaction:
CH4(g) + 2O2(g) CO2(g) + 2H2O (g)
rate
k =
[Br2]
rate a [Br2]
rate = k [Br2]
= rate constant
= 3.50 x 103 s1
Consider the reaction:
Suppose that, at a particular moment during the reaction, molecular oxygen is reacting at the rate of 0.024 M/s.
Homework Quiz: N2(g) + 3H2(g) → 2NH3(g)
Suppose that at a particular moment during the reaction, hydrogen is reacting at the rate of 0.074 M/s.
Homework: p. 602 #15, 16, 18, 19, 20: Mon after break
Hint: Use pressure just like concentration.
Diagnostic test (Tues after break)
Consider the reaction:
4PH3(g) P4(g) + 6H2(g)
Suppose that, at a particular moment during the reaction, molecular hydrogen is being formed at the rate of 0.078 M/s.
aA + bB cC + dD
The Rate Law
The rate law is a mathematical relationship that shows how rate of reaction depends on the concentrations of reactants
Rate = k [A]x[B]y
x and y are small whole numbers that
relate to the number of molecules of A
and B that collide and are determined
experimentally!
aA + bB cC + dD
The Rate Law
Rate = k [A]x[B]y
Reaction is xth order in A
Reaction is yth order in B
Reaction is (x +y)th order overall
Rate = k [A]1[B]2
F2(g) + 2ClO2(g) 2FClO2(g)
rate = k [F2][ClO2]
Write the reaction rate expressions for the following in terms of the disappearance of the reactants and the appearance of products:
Consider the reaction
N2(g) + 3H2(g) 2NH3(g)
Suppose that at a particular moment during the reaction molecular hydrogen is reacting at a rate of 0.074 M/s.
Homework: Diagnostic test
revisit/correct p. 603 #15, 16, 18
F2(g) + 2ClO2(g) 2FClO2(g)
1
Rate Laws
rate = k [F2][ClO2]
Determine the rate law and calculate the rate constant for the following reaction from the following data:
S2O82(aq) + 3I(aq) 2SO42(aq) + I3(aq)
Determine the rate law and calculate the rate constant for the following reaction from the following data:
S2O82(aq) + 3I(aq) 2SO42(aq) + I3(aq)
rate
k =
2.2 x 104 M/s
=
[S2O82][I]
(0.08 M)(0.034 M)
rate = k [S2O82]x[I]y
y = 1
x = 1
rate = k [S2O82][I]
Double [I], rate doubles (experiment 1 & 2)
Double [S2O82], rate doubles (experiment 2 & 3)
= 0.08/M•s
2NO(g) + 2H2(g) N2(g) + 2H2O(g)
From the following data collected at this temperature, determine (a) the rate law, (b) the rate constant and (c) the rate of the reaction when [NO] = 12.0x103 M and [H2] = 6.0x103 M
Consider the reaction X + Y Z
From the following data, obtained at 360 K,
Consider the reaction A B.
The rate of the reaction is 1.6x102 M/s when the concentration of A is 0.35 M. Calculate the rate constant if the reaction is
Homework: p. 603 #19, 20 (use Excel!), 24, 26
rate = k[A] =
ln[A] = ln[A]0  kt
A product
rate
=
[A]
M/s
D[A]

M
= k [A]
Dt
[A] = [A]0e−kt
ln[A] = ln[A]0  kt
D[A]
rate = 
Dt
FirstOrder Reactions
rate = k [A]
= 1/s or s1
k =
Graphical Determination of k
2N2O5 4NO2 (g) + O2 (g)
The reaction 2A B is first order in A with a rate constant of 2.8 x 102 s1 at 800C. How long will it take for A to decrease from 0.88 M to 0.14 M ?
0.88 M
ln
0.14 M
=
2.8 x 102 s1
ln
ln[A]0 – ln[A]
=
k
k
[A]0
[A]
[A]0 = 0.88 M
ln[A] = ln[A]0  kt
[A] = 0.14 M
kt = ln[A]0 – ln[A]
= 66 s
t =
The conversion of cyclopropane to propene in the gas phase is a first order reaction with a rate constant of 6.7x104 s1 at 500oC.
Homework Quiz:
The conversion of cyclopropane to propene in the gas phase is a first order reaction with a rate constant of 6.7x104 s1 at 500oC.
If the initial concentration of cyclopropane was 0.25 M, what is the concentration after 8.8 minutes?
The rate of decomposition of azomethane (C2H6N2) is studied by monitoring partial pressure of the reactant as a function of time:
CH3N=NCH3(g) → N2(g) + C2H6(g)
The data obtained at 300oC are shown here:
Are these values consistent with firstorder kinetics? If so, determine the rate constant.
Ethyl iodide (C2H5I) decomposes at a certain temperature in the gas phase as follows:
C2H5I(g) → C2H4(g) + HI(g)
From the following data, determine the order of the reaction and the rate constant:
[A]0
ln
t½
[A]0/2
0.693
=
=
=
k
k
ln 2
k
FirstOrder Reactions
The halflife, t½, is the time required for the concentration of a reactant to decrease to half of its initial concentration.
t½ = t when [A] = [A]0/2
What is the halflife of N2O5 if it decomposes with a rate constant of 5.7 x 104 s1?
How do you know decomposition is first order?
[A]0
ln
t½
[A]0/2
0.693
=
=
=
=
k
k
t½
ln 2
ln 2
0.693
=
k
k
5.7 x 104 s1
FirstOrder Reactions
The halflife, t½, is the time required for the concentration of a reactant to decrease to half of its initial concentration.
t½ = t when [A] = [A]0/2
What is the halflife of N2O5 if it decomposes with a rate constant of 5.7 x 104 s1?
= 1200 s = 20 minutes
How do you know decomposition is first order?
units of k (s1)
A product
# of
halflives
[A] = [A]0/n
Firstorder reaction
1
2
2
4
3
8
4
16
C2H6(g) 2CH3(g)
Calculate the halflife of the reaction in minutes.
2N2O5 4NO2(g) + O2(g)
A product
rate
=
[A]2
M/s
D[A]
1
1

M2
= k [A]2
=
+ kt
Dt
[A]
[A]0
t½ =
D[A]
rate = 
Dt
1
k[A]0
SecondOrder Reactions
rate = k [A]2
= 1/M•s
k =
[A] is the concentration of A at any time t
[A]0 is the concentration of A at time t=0
t½ = t when [A] = [A]0/2
Iodine atoms combine to form molecular iodine in the gas phase:
I(g) + I(g) I2(g)
This reaction follows secondorder kinetics and has the high rate constant 7.0x109/M·s at 23oC.
The reaction 2A → B is second order with a rate constant of 51/M·min at 24oC.
The reaction 2A → B is second order with a rate constant of 51/M·min at 24oC.
Homework: p.
A product
rate
[A]0
D[A]

= k
Dt
[A]0
t½ =
D[A]
2k
rate = 
Dt
ZeroOrder Reactions
rate = k [A]0 = k
= M/s
k =
[A] is the concentration of A at any time t
[A] = [A]0  kt
[A]0 is the concentration of A at time t = 0
t½ = t when [A] = [A]0/2
ConcentrationTime Equation
Order
Rate Law
HalfLife
1
1
=
+ kt
[A]
[A]0
=
[A]0
t½ =
t½
t½ =
ln 2
2k
k
1
k[A]0
Summary of the Kinetics of ZeroOrder, FirstOrder
and SecondOrder Reactions
[A] = [A]0  kt
rate = k
0
ln[A] = ln[A]0  kt
1
rate = k [A]
2
rate = k [A]2
+
A + B AB C + D
+
Exothermic Reaction
Endothermic Reaction
The activation energy (Ea ) is the minimum amount of energy required to initiate a chemical reaction.
=a barrier that prevents less energetic molecules from reacting
Arrhenius equation
Eais the activation energy (J/mol)
R is the gas constant (8.314 J/K•mol)
T is the absolute temperature
A is the frequency factor
The rate constants for the decomposition of acetaldehyde:
CH3CHO(g) → CH4(g) + CO(g)
were measured at five different temperatures. The data are shown below. Plot lnk versus 1/T, and determine the activation energy (in kJ/mol) for the reaction. (Note: the reaction is order in CH3CHO, so k has the units of )
The second order rate constant for the decomposition of nitrous oxide (N2O) into nitrogen molecule and oxygen atom has been measured at different temperatures. Determine graphically the activation energy for the reaction.
Homework: Finish correcting and explaining answers to multiple choice: due Weds.
Homework: Complete problem set and
p. 605 #40, 42
Quiz tomorrow
Homework p. 605 #44, 45, 49, 51, 52, 54: Mon.
Importance of Molecular Orientation
effective collision
ineffective collision
2NO (g) + O2 (g) 2NO2 (g)
Elementary step:
NO + NO N2O2
+
Elementary step:
N2O2 + O2 2NO2
Overall reaction:
2NO + O2 2NO2
Reaction Mechanisms
The overall progress of a chemical reaction can be represented at the molecular level by a series of simple elementary steps or elementary reactions.
The sequence of elementary steps that leads to product formation is the reaction mechanism.
N2O2 is detected during the reaction!
2NO (g) + O2 (g) 2NO2 (g)
Mechanism:
Homework: p. 605 #44, 45, 49, 51, 52, 54, 55, 56, 61: Tues.
Elementary step:
NO + NO N2O2
+
Elementary step:
N2O2 + O2 2NO2
Overall reaction:
2NO + O2 2NO2
Intermediates are species that appear in a reaction mechanism but not in the overall balanced equation.
An intermediate is always formed in an early elementary step and consumed in a later elementary step.
Unimolecular reaction
Bimolecular reaction
Bimolecular reaction
A + B products
A + A products
A products
Rate Laws and Elementary Steps
rate = k [A]
rate = k [A][B]
rate = k [A]2
The ratedetermining step is the sloweststep in the sequence of steps leading to product formation.
Step 1:
Step 2:
NO2 + NO2 NO + NO3
NO3 + CO NO2 + CO2
The experimental rate law for the reaction between NO2 and CO to produce NO and CO2 is rate = k[NO2]2. The reaction is believed to occur via two steps:
What is the equation for the overall reaction?
What is the intermediate?
What can you say about the relative rates of steps 1 and 2?
Step 1:
Step 2:
NO2 + NO2 NO + NO3
NO2+ CO NO + CO2
NO3 + CO NO2 + CO2
The experimental rate law for the reaction between NO2 and CO to produce NO and CO2 is rate = k[NO2]2. The reaction is believed to occur via two steps:
What is the equation for the overall reaction?
What is the intermediate?
NO3
What can you say about the relative rates of steps 1 and 2?
rate = k[NO2]2 is the rate law for step 1 so
step 1 must be slower than step 2
2A + 2B → A2B2
Step 1: N2O N2 + O
Step 2 N2O + O N2 + O2
Experimentally the rate law is found to be
rate = k[N2O].
NO2 + F2 → NO2F + F
NO2 + F → NO2F
H2(g) + 2ICl(g) → 2HCl(g) + I2(g)
The rate law for the reaction is
rate = k[H2][ICl]. Suggest a possible mechanism for the reaction.
2H2O2(aq) 2H2O(l) + O2(g)
Can be catalyzed using iodide ions (I)
rate = k[H2O2][I] Why?!
Determined experimentally.
Step 1: H2O2 + I H2O + IO
Step 2: H2O2 + IO H2O + O2 + I
Ea
k
′
Ea< Ea
A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed.
Catalyzed
Uncatalyzed
ratecatalyzed > rateuncatalyzed
Very slow, until you add MnO2, a catalyst. The MnO2 can be recovered at the end of the reaction!
Step 1: HBr + O2 → HOOBr
Step 2: HOOBr + HBr → 2HOBr
Step 3: HOBr + HBr → H2O + Br2
Step 4: HOBr + HBr → H2O + Br2
Homework: Problem Set: Monday
This cycle continually repeats, producing and destroying ozone at the same rate while absorbing harmful ultraviolet light from the sun.
In heterogeneous catalysis, the reactants and the catalysts are in different phases (usually, catalyst is a solid, reactants are gases or liquids).
In homogeneous catalysis, the reactants and the catalysts are dispersed in a single phase, usually liquid.
N2 (g) + 3H2 (g) 2NH3 (g)
Fe/Al2O3/K2O
catalyst
Haber Process
Synthesis of Ammonia
Extremely slow at room temperature. Must be fast and high yield!
Process occurs on the surface of the Fe/Al2O3/K2O catalyst, which weakens the covalent NN and HH bonds.
4NH3(g) + 5O2(g) 4NO (g) + 6H2O (g)
2NO (g) + O2(g) 2NO2(g)
2NO2(g) + H2O (l) HNO2(aq) + HNO3(aq)
PtRh catalysts used
in Ostwald process
Ostwald Process
Pt catalyst
catalytic
CO + Unburned Hydrocarbons + O2
CO2 + H2O
converter
catalytic
2NO + 2NO2
2N2 + 3O2
converter
Catalytic Converters
Enzyme Catalysis
biological catalysts
Binding of Glucose to Hexokinase
Homework: Lab notebook due Monday
$13 for AP Exam due by 8:00 am TOMORROW!!!