Vanessa N. Prasad-PermaulCHM 1046
Valencia Community College
Introduction 1. Chemical kinetics is the study of reaction rates
2. For a chemical reaction to be useful it must occur at a reasonable rate
3. It is important to be able to control the rate of reaction
4. Factors that influence rate
a) Concentration of reactants (molarity)b) Nature of reaction, process by which the reaction takes place c) Temperatured) Reaction mechanism (rate determining step)
e) Catalyst
Reaction rate
Reaction rate- Positive quantity that expresses how the conc. of a reactant or product changes with time
2N2O5(g) 4NO2(g) + O2(g)
1. [ ] = concentration, molarity, mole/L2. [N2O5] decreases with time3. [NO2] increases with time4. [O2] increases with time5. Because of coefficients the concentration of reactant and
products does not change at the same rate6. When 2 moles of N2O5 is decomposed, 4 moles of NO2
and 1 mole of O2 is produced
2N2O5(g) 4NO2(g) + O2(g)
-[N2O5] = [NO2] = [O2] 2 ½
-[N2O5] because it’s concentration decreases, other positive because they increase
Rate of reaction can be defined by dividing by the change in time, t
rate = - [N2O5] = [NO2] = [O2] t 2t ½ t
Reaction Rates
Generic Formula: aA + bB cC + dD
rate = - [A] = - [B] = [D] = [C] at bt dt ct
EXAMPLE 13.1: CONSIDER THE REACTION OF NO2 WITH F2 TO GIVE NO2F
2NO2(g) + F2(g) 2NO2F(g)
HOW IS THE RATE OF FORMATION RELATED TO THE RATE OF REACTION OF F2?
NO2F = [NO2F] F2 = -[F2] t t
[NO2F] = -[F2] 2t t
EXERCISE 13.1: CONSIDER THE REACTION OF NO2 WITH F2 TO GIVE NO2F
2NO2(g) + F2(g) 2NO2F(g)
HOW IS THE RATE OF FORMATION RELATED TO THE RATE OF REACTION OF NO2?
EXAMPLE 13.2: CALCULATE THE AVERAGE RATE OF DECOMPOSITION OF N2O5
N2O5(g) 2NO2(g) + 1/2O2(g)
N2O5 = -[N2O5] t
-(0.93x10-2M – 1.24x10-2M) = -0.31x10-2M = 5.2x10-6M/s 1200s – 600s 600s
TIME [N2O5]600s 1.24 x10-2M1200s 0.93x10-2M
EXERCISE 13.2: IODIDE ION IS OXIDIZED BY HYPOCHLORITE ION IN BASIC SOLUTION.
CALCULATE THE AVERAGE RATE OF REACTION OF I- DURING THIS TIME INTERVAL:
I-(aq) + ClO-
(aq) Cl-(aq) + IO-
(aq)
TIME [I-]2.00s 0.00169M
8.00s 0.00101M
Reaction Rate and Concentration The higher the conc. of starting reactant the more rapidly
areaction takes place
1. Reactions occur as the result of collisions between reactant molecules2. The higher the concentration of molecules, the greater the # of collisions in unit time and a faster reaction 3. As the reactants are consumed the concentration decreases, collisions decrease, reaction rate decreases4. Reaction rate decreases with time and eventually = 0, all reactants consumed 5. Instantaneous rate, rate at a particular time6. Initial rate at t = 0
Rate Expression and Rate Constant Rate expression / rate law:
rate = k[A]
where k = rate constant, varies w/ nature and temp.
[A] = concentration of A
Rate law: an equation that related the rate of a reaction
to the concentration of reactants (and catalyst) raised
to various powers
Order of rxn involving a single reactant
General equationrate expressionA products rate = k[A]m
where m=order of reactionm=0 zero orderm=1 first orderm=2 second order
m, can’t be deduced from the coefficient of the
balanced equation. Must be determined experimentally!
Order of rxn involving a single reactant
Rate of decomposition of species A measured at 2different conc., 1 & 2
rate2 = k[A2]m rate1 = k[A1]m
By dividing we can solve for m, to find the order of the
reaction
Rate2 = [A2]m Rate1 [A1]m (Rate2/Rate1) = ([A2]/[A1])m
EXAMPLE 13.3: BROMIDE ION IS OXIDIZED BY BROMATE ION IN ACIDIC SOLUTION
5Br -(aq) + BrO3
-(aq) + 6H+
(aq) 3Br2(aq) + 3H2O(l)
Rate = k[Br -][BrO3-][H+]2
WHAT IS THE ORDER OF REACTION WITH RESPECT TO EACH REACTANT SPECIES? WHAT IS THE OVERALL ORDER OF REACTION?
Br - IS FIRST ORDERBrO3
- IS FIRST ORDERH+ IS SECOND ORDER
THE REACTION IS FOURTH ORDER OVERALL (1+1+2)=4
EXERCISE 13.3: WHAT ARE THE REACTIO ORDERD WITH RESPECT TO EACH REACTANT SPECIES FOR THE FOLLOWING REACTION? WHAT IS THE OVERALL REACTION?
NO2(g) + CO(g) NO(g) + CO2(g)
RATE = k[NO2]2
Example 1
CH3CHO(g) CH4(g) + CO(g)
[CH3CHO] .10 M .20 M .30 M .40 M
Rate (mol/L s) .085 .34 .76 1.4
Using the given data determine the reaction order
Example 1
Rate2 = [A2]m Rate1 [A1]m
4 = 2m log 4 m = 2
log 2
second order rate = k[CH3CHO]2
once the order of the rxn is known, rate constant can be calculated, let’s calculate the rate constant, k
Example 1
rate = k[CH3CHO]2 rate = .085 mol/L sconc = .10 mol/L
k = rate = 0.085 mol/L s = 8.5 L/mol s
[CH3CHO]2 (0.10 mol/L)2
now we can calc. the rate at any
concentration, let’s try .55 M
Example 1
rate = k[CH3CHO]2
rate = 8.5 L/mol s [.55]2 = 2.6 mol/ L s
Rate when [CH3CHO] = .55 M is 2.6 mol/L s
Order of rxn with more than 1 reactant
aA + bB prod
rate exp: rate = k [A]m [B]n
m = order of rxn with respect to An = order of rxn with respect to B
Overall order of the rxn is the sum, m + n
Key
When more than 1 reactant is involved the order can be determined by holding the concentration of 1 reactant constant while varying the other reactant. From the measured rates you can calculate the order of the rxn with respect to the varying reactant
Example 2
(CH3)3CBr + OH- (CH3)3COH + Br-
Exp. 1 Exp. 2 Exp. 3 Exp. 4 Exp. 5
[(CH3)3CBr] 0.5 1.0 1.5 1.0 1.0
[OH-] 0.05 0.05 0.05 0.10 0.20
Rate (mol/L s) 0.005 0.01 0.015 0.01 0.01
Find the order of the reaction with respect to both reactants, write the rate expression, and find the overall order of the reaction
Reactant concentration and time
Rate expression rate = k[A]
Shows how the rate of decomposition of A changes with concentration
More important to know the relation between concentration and time
Using calculus: Integrated rate equations relating react conc. to time
For the following rate law, what is the overall order
of the reaction?Rate = k [A]2 [B]
1. 1
2. 2
3. 3
4. 4
Zero Order
Zero order: A Products
rate = k
[A]0 – [A] = kt t½ = [A]0
2k
m = 0: zero order - rate is independent of theconcentration of reactant. Doubling theconcentration has no effect on rate.
First Order
First Order: A Products
rate = k[A]
ln [A]o/[A] = kt t½ = .693/k
[A]o = original conc. of A [A] = Conc. of A at time, t
k = first order rate constant ln = natural logarithmm = 1: first order - rate is directly proportional to the concentration of the reactant. Doubling the concentration increases the rate by a factor of 2.
Second Order
Second order: A Products
rate = k[A]2
1 – 1 = kt t½ = 1
[A] [A]0 k[A]0
m = 2: second order - the rate is to the square of the concentration of the reactant. Doubling the concentration increases the rate by a factor of 4.
EXAMPLE 13.4: IODIDE ION IS OXIDIZED IN ACIDIC SOLUTION TO TRIIODIDE ION I3
-, BY HYDROGEN PEROXIDE
H2O2(aq) + 3I-(aq) + 2H+(aq) I3-(aq) + H2O(l)
a)OBTAIN THE REACTION ORDER WITH RESPECT TO EACH REACTANTb)FIND THE RATE CONSTANT
Initial concentrations
(mol/L)
Initial Rate [mol/(L.s)]
H2O2 I- H+
EXP. 1
0.010 0.010
0.00050
1.15 x 10-6
EXP. 2
0.020 0.010
0.00050
2.30 x 10-6
EXP. 3
0.010 0.020
0.00050
2.30 x 10-6
EXP. 4
0.010 0.010
0.00100
1.15 x 10-6
Rate = k[H2O2]m [I-]n [H+]p
(Rate) 2 = k[H2O2]m [I-]n [H+]p
(Rate) 1 = k[H2O2]m [I-]n [H+]p
Rate constant cancels
2.30 x 10-6 = [0.020]m [0.010]n [0.00050]p
1.15 x 10-6 = [0.010]m [0.010]n [0.00050]p
2= 2m
m = 1doubling the H2O2 concentration doubles the rate
Rate = k[H2O2] [I-]Reaction order = 1, 1, 0
Rate = k[H2O2] [I-]
1.15 x 10-6 mol = k x 0.010 mol x 0.010 mol L.s L L
k = 1.15 x 10-6 mol L.s 0.010 mol x 0.010 mol L L
k = 1.2 x 10-2 L/(mol.s)
EXERCISE 13.4: THE INITIAL-RATE METHODWAS APPLIED TO THE DECOMPOSITION OF NITROGEN DIOXIDE
NO2 (aq) 2NO (g) + O2 (g)
FIND THE RATE LAW AND THE VALUE OF THE RATE CONSTANT WITH RESPECT TO O2 FORMATION
INITIAL NO2 CONCENTRATIO
N (mol/L)
INITIAL RATE OF FORMATION OF
O2 mol/(L.s)EXP. 1 0.010 7.1 x 10-5
EXP. 2 0.020 28 x 10-5
Example 3
The following data was obtained for the gas-phase
decomp. of HI
Is this reaction zero, first, or second order in HI?Hint: Graph each Conc. Vs. time
corresponding to correct [A], ln [A], or 1/[A]
Time (h) 0 2 4 6
[HI] 1.00 0.50 0.33 0.25
[HI] vs. time not linear so it is not zero order
[HI] vs. t
00.5
11.5
0 2 4 6 8
Time (h)
[HI]
ln [HI] vs. time not linear so it is not first order
ln [HI] vs. t
-1.5
-1
-0.5
00 2 4 6 8
Time (h)
ln [H
I]
1/[HI] vs. time is linear so it is second order
1/[HI] vs. t
02
46
0 2 4 6 8
Time (h)
1/[H
I]
Order Rate Expression
Conc-TimeRelation
Half-life Linear Plot
0 Rate = k [A]0 – [A] = kt [A]0
2k[A] vs. t
1 Rate = k[A] ln [A]0 = kt [A]
0.693 k
ln [A] vs. t
2 Rate = k[A]2 1 – 1 = kt [A] [A]0
1 k[A]0
1 vs. t[A]
EXAMPLE 13.5: HOW LONG WOULD IT TAKE FOR THE CONC. OF N2O5 TO DECREASE TO 1.00 x 10-2 mol/L FROM THE INITIAL VALUE?
kt = ln [A]0
[A]t
4.80 x 10-4/s (t) = ln [1.65 x 10-2 mol/L] [1.00 x 10-2 mol/L]
4.80 x 10-4/s (t) = 0.500775
t = 0.500775 4.80 x 10-4/s
t = 1043.28125s * 1min 60sec
t = 17.4min
EXERCISE 13.5:
A) WHAT WOULD THE CONCENTRATION BE OF DINITROGEN PENTOXIDE IN THE EXPERIMENT DESCRIBED IN EXAMPLE 13.5 AFTER 6.00 x 102 s?
B) HOW LONG WOULD IT TAKE FOR THE CONCENTRATION OF N2O5 TO DECREASE TO 10% OF ITS INITIAL VALUE?
EXAMPLE 13.6: SULFURYL CHLORIDE, SOC2Cl2, IS A COLORLESS CORROSIVE LIQUID WHOSE VAPOR DECOMPOSES IN A FIRST-ORDER REACTION TO SULFUR DIOXIDE AND CHLORINE.
SO2Cl2 (g) SO2 (g) + Cl2 (g)
AT 320oC, THE RATE CONSTANT IS 2.20 x 10-5/s. WHAT IS THE HALF-LIFE OF SO2Cl2 VAPOR AT THIS TEMPERATURE?
A) HOW LONG (IN HOURS) WOULD IT TAKE FOR 50.0% OF THE SO2Cl2 TO DECOMPOSE?
B) HOW LONG WOULD IT FOR 75.0% OF THE SO2Cl2?
EXAMPLE 13.6: t½ = 0.693/k
t½ = 0.693 2.20 x 10-5/s
t½ = 3.15 x 104 sThe half-life is the time required for 50.0% of the SO2Cl2 to
decompose. This is 8.75 hours
8.75 hrs x 2 = 17.5 hrs
kt = ln [A]0
[A]t
EXERCISE 13.6: THE ISOMERIZATION OF CYCLOPROPANE TO PROPYLENE IS FIRST ORDER IN CYCLOPROPANE AND OVERALL. AT 1000oC THE RATE CONSTANT IS 9.2/s
A) WHAT IS THE HALF-LIFE?
B)HOW LONG WOULD IT TAKE FOR THE CONCENTRATION OF CYCLOPROPANE TO DECREASE TO 50% OF ITS INITIAL VALUE?
C) TO 25% OF ITS INITIAL VALUE?
Activation Energy
Activation Energy: Ea (kJ) For every reaction there is a certain minimum
energy that molecules must possess for collisions to be effective.
1. Positive quantity (Ea>0)
2. Depends only upon the nature of reaction
3. Fast reaction = small Ea
4. Is independent of temp and concentration
For the following reaction:A + B C
If the concentration of A is doubled, and B is constant, the rate doubles. What is the order of the reaction with respect to A?
1. 02. 13. 2
Activation Energy
Reaction Rate and Temp
Reaction Rate and Temp
1. As temp increases rate increases, Kinetic Energy increases, and successful collisions increase
2. General rule for every 10°C inc. in temp, rate doubles
If I increase the temperature of a reaction from
110 K to 120 K, what happens to the rate of the
reaction?
1. Stay the same
2. Doubles
3. Triples
The Arrhenius Equation
The Arrhenius Equationf = e-Ea/RT
f = fraction of molecules having an En. equal to or greater than Ea
R = gas constant A = constant T = temp in K ln k = ln A –Ea/RT plot of ln k Vs. 1/T linear
slope = -Ea/R
Two-point equation relating k & Tln k2 = Ea [1/T1 – 1/T2]
k1 R
EXAMPLE 13.7: THE RATE CONSTANT FOR THE FORMATION OF HYDROGEN IODIDE FROM THE ELEMENTS IS 2.7 x 10-4L/(mol.s) @600 K AND 3.5 x 10-3L/(mol.s) AT 650 K.
FIND THE ACTIVATION ENERGY Ea
ln k2 = Ea [1/T1 – 1/T2] k1 R
ln 3.5 x 10-3L/(mol.s) = Ea 1 - 1 2.7 x 10-4L/(mol.s) 8.314 J/(mol.K) 600K 650K
Ea = 2.56 * (8.314 J/mol ) 1.28 x 10-4
Ea = 1.66 x 105 J/mol or 166kJ
EXERCISE 13.7: ACETALDEHYDE DECOMPOSES WHEN HEATED. THE RATE OF DECOMPOSITION IS 1.05 x 10-3 L/mol.s AT 759 K AND 2.14 x 10-2 L/mol.s AT 836 K.
CH3CHO(g) CH4(g) + CO(g)
WHAT IS ACTIVATION ENERGY FOR THIS DECOMPOSITION?
WHAT IS THE RATE CONSTANT AT 865 K?
Example 4
For a certain rxn the rate constant doubles when the temp increases from 15 to 25°C.
a) Calc. The activation energy, Ea
b) Calc. the rate constant at 100°C, taking k at 25°C to be 1.2 x 10-2 L/mol s
Reaction Mechanism Reaction Mechanism
Description of a path, or a sequence of steps, by which a
reaction occurs at the molecular level.
Simplest case- only a single step, collision between two reactant molecules
Reaction Mechanism
“Mechanism” for the reaction of CO with NO2 at high temp (above 600 K)
CO(g) + NO2(g) NO(g) + CO2(g)
“Mechanism” for the reaction of CO with NO2 at low tempNO2(g) + NO2(g) NO3(g) + NO(g) slow CO(g) + NO3(g) CO2(g) + NO2(g) fast
CO(g) + NO2(g) NO(g) + CO2(g) overall
The overall reaction, obtained by summing the individual steps is
identical but the rate expressions are different. High temp: rate = k[CO][NO2]Low temp: rate= k[NO2]2
EXAMPLE 13.8: CCl4 IS OBTAINED BY CHLORINATING METHANE OR AN INCOMPLETELY CHLORINATED METHANE SUCH AS CHLOROFORM CH3Cl. THE MECHANISM FOR THE GAS PHASE CHLORINTATION OF CH3Cl IS
Cl2 2ClCl + CHCl3 HCl + CCl3
Cl + CCl3 CCl4
OBTAIN THE NET (OVERALL) CHEMICAL EQUATION
Cl2 2ClCl + CHCl3 HCl + CCl3
Cl + CCl3 CCl4
Cl2 + CHCl3 HCl + CCl4
EXERCISE 13.8: THE IODIDE ION CATALYZES THE DECOMPOSITION OF AQUEOUS HYDROGEN PEROXIDE. THIS DECOMPOSITION IS BELIEVED TO OCCUR IN TWO STEPS:
H2O2 + I- H2O + IO-
H2O2 + IO- H2O + O2 + I-
WHAT IS THE OVERALL EQUATION REPRESENTING THIS DECOMPOSITION?
NOTE THAT THE IO- IS A REACTION INTERMEDIATE. THE IODIDE ION IS NOT AN INTERMEDIATE; IT WAS ADDED TO THE REACTION MIXTURE.
Reaction Mechanism
Elementary Steps: Individual steps that constitute a reaction mechanism
Unimolecular A B + C rate = k[A]
Bimolecular A + A B + C rate = k[A][A] = [A]2
Termolecular A + B + C D + E rate = k[A][B][C]
The rate of an elementary step is equal to a rate constant, k, multiplied by the concentration of each reactant molecule. You can treat all reactants as if they were first order. If a reactant is second order it will appear twice in the general equation.
EXAMPLE 13.9: WHAT IS THE MOLECULARITY OF EACH STEP IN THE MECHANISM DESCRIBED IN EXAMPLE 13.8
1) Cl2 2Cl2) Cl + CHCl3 HCl + CCl3
3) Cl + CCl3 CCl4
1)UNIMOLECULAR2)BIMOLECULAR3)BIMOLECULAR
EXAMPLE 13.9: THE FOLLOWING IS AN ELEMENTARY REACTION THAT OCCURS IN THE DECOMPOSITION OF OZONE IN THE STRATOSPHERE BY NITROGEN MONOXIDE
NO + O3 O2 + NO2
WHAT IS THE MOLECULARITY OF THIS REACTION?
Reaction Mechanism
Slow Steps- A step that is much slower than any other in a reaction mechanism.
Rate-determining step - The rate of the overall reaction can be taken to be that of the slow step
Step 1: A B fastStep 2: B C slowStep 3: C D fast A D
The rate A D (overall reaction) is approx. equal to the rate of B C the slow step
Deducing a Rate Expression from a Proposed
Mechanism 1. Find the slowest step and equate the rate of
the overall reaction to the rate of that step.2. Find the rate expression for the slowest step. NO2(g) + NO2(g) NO3(g) + NO(g) slow CO(g) + NO2(g) CO2(g) + NO2(g) fast
CO(g) + NO2(g) CO2(g) + NO(g)
Rate of overall reaction = rate of 1st step = k[NO2] [NO2] = k[NO2]2
EXAMPLE 13.10: WRITE RATE EQUATIONS FOR EACH OF THE FOLLOWING ELEMENTARY REACTIONS:
OZONE IS CONVERTED TO O2 BY NO IN A SINGLE STEP
1) O3 + NO O2 + NO2 rate = k[O3][NO]
THE RECOMBINATION OF IODINE ATOMS OCCURS ASFOLLOWS
2) I + I + M M* + I2 rate = k[I]2[M]
A WATER MOLECULE ABSORBS ENERGY; LATER IN THE PROCESSENOUGH ENERGY FLOWS INTO O-H BOND TO BREAK IT
3) H2O H + O H rate = k[H2O]
EXAMPLE 13.10: WRITE RATE EQUATIONS FOR THE FOLLOWING ELEMENTARY REACTION:
NO2 + NO2 O2 + N2O4
EXAMPLE 13.11: OZONE REACTS WITH NITROGEN DIOXIDE TO PRODUCE OXGEN AND DINITROGEN PENTOXIDE.
2NO2(g) + O3(g) O2(g) + N2O5(g)
THE PROPOSED MECHANISM IS
O3 + NO2 NO3 + O2 (slow) NO3 + NO2 N2O5 (fast)
WHAT IS THE RATE LAW PREDICTED BY THIS MECHANISM?
RATE = rate = k[O3][NO2]
EXAMPLE 13.11: THE IODIDE-ION-CATALYZED DECOMPOSITION OF HYDROGEN PEROXIDE IS BELIEVED TO HAVE THE FOLLOWING MECHANISM:
H2O2 + I- H2O + IO- (slow)
H2O2 + IO- H2O + O2 + I- (fast)
WHAT IS THE RATE LAW PREDICTED BY THIS MECHANISM?
EXAMPLE 13.12: NITROGEN MONOXIDE CAN BE REDUCED WITH HYDROGEN GAS TO GIVE NITROGEN AND WATER VAPOR
2NO(g) + 2H2(g) N2(g) + 2H2O(g) (OVERALL RXN)
THE PROPOSED MECHANISM IS 2NO N2O2 (fast, eq.)
H2 + N2O2 N3O + H2O (slow) H2 + N2O N2 + H2O (fast)
WHAT IS THE RATE LAW PREDICTED BY THIS MECHANISM?
RATE = rate = k[NO]2[H2]
EXERCISE 13.12: NITROGEN MONOXIDE REACTS WITH OXYGEN TO PRODUCE NITROGEN DIOXIDE
2NO(g) + O2(g) 2NO2(g) (OVERALL RXN)
THE PROPOSED MECHANISM IS
O2 + NO NO3 (fast, eq.) NO + NO3 NO2 + NO2 (slow)
WHAT IS THE RATE LAW PREDICTED BY THIS MECHANISM?
Elimination of Intermediates Intermediate1. A species produced in one step of the
mechanism and consumed in a later step.
2. Concentration too small to determine experimentally
3. Must be eliminated from rate expression
4. The final rate expression must include only those species that appear in the balanced equation for the overall reaction
Example 5
Find the rate expression for the following reaction mechanism
Step1: NO(g) + Cl2(g) NOCl2(g) fast Step2: NOCl2(g) + NO(g) 2NOCl(g) slow 2 NO(g) + Cl2(g) 2NOCl(g)
Example 6
The decomposition of ozone, O3, to diatomic oxygen, O2, is believed to occur by a two-step mechanism:
Step 1: O3(g) O2(g) + O(g) fast
Step 2: O3(g) + O(g) 2 O2(g) slow
2 O3(g) 3 O2(g)
Find the rate expression for this reaction
Which is an intermediate for the following multi step mechanism?
2 A + 2 B C + 2 D
Step 1 2 B EStep 2 E + A D + FStep 3 F + A C + D
1. E2. F3. E & F4. A
Catalysts
Catalysis A catalyst increases the rate of a reaction without
being consumed by it. Changes the reaction mechanism to one with a lower activation energy.
1. Heterogeneous catalysisa) Catalyst is in a different phase from the reaction mixture. Most common, solid catalyst with gas or liquid mixture.b) Solid catalyst is easily poisoned, foreign material deposited on the surface during reaction reduce or destroy its effectiveness.
2. Homogeneous Catalysis a) Same phase as the reactants
Which is the catalyst for the following multi step mechanism?
2 A + 2 B C + 2 D
Step 1 2 B + G EStep 2 E + A D + FStep 3 F + A C + D + G
1. E2. G3. E & F4. A
Chapter 12: Kinetics
Hydrogen peroxide decomposes to water and oxygen according to the following reaction
H2O2(aq) H2O + ½ O2(g)
It’s rate of decomposition is measured by titrating samples of the solution with potassium permanganate (KMnO4) at certain intervals.a) Initial rate determinations at 40C for the decomposition give the following data:
[H2O2] Rate (mol/L min)
0.10 1.93 x 10-4
0.20 3.86 x 10-4
0.30 5.79 x 10-4
1. Order of rxn?
2. Rate expression?
3. Calc. k @ 40°C
4. Calc. half-life @ 40°C
b) Hydrogen peroxide is sold commercially as a 30.0% solution. If the solution is kept at 40C, how long will it take for the solution to become 10.0% H2O2?
c) It has been determined that at 50C, the rate constant for the reaction is 4.32 x 10-3/min. Calculate the activation energy for the decomposition of H2O2
d) Manufacturers recommend that solutions of hydrogen peroxide be kept in a refrigerator at 4C. How long will it take for a 30.0% solution to decompose to 10.0% if the solution is kept at 4C?
e) The rate constant for the uncatalyzed reaction at 25C is 5.21 x 10-4/min. The rate constant for the catalyzed reaction at 25C is 2.95 x 108/min.
1) What is the half-life of the uncatalyzed reaction at 25C?
2) What is the half-life of the catalyzed reaction?
1) Express the rate of reaction
2HI(g) H2(g) + I2(g)
a) in terms of [H2]
b) in terms of [HI]
3) Dinitrogen pentaoxide decomposes according to the following equation:
2N2O5(g) 4NO2(g) + O2(g)
a) write an expression for reaction rate in terms of [ N2O5], [NO2], and [O2]
4) What is the order with respect to each reactant and the overall order of the reactions described by the following rate expressions?
a) rate = k1[A]3
b) rate = k2[A][B]
c) rate = k3[A][B]2
d) rate = k4[B]
e) rate = k
5) Complete the following table for the reaction, which is first order in both reactants
A(g) + B(g) products
[A] [B] K (L/mol s) Rate (mol/L s)
.2 .3 1.5
.029 .78 .025
.45 .520 .033
7) The decomposition of ammonia on tungsten at 1100C is zero-order, with a rate constant of 2.5 x 10-4 mol/L min
a) write the rate expression
b) calculate the rate when the concentration of ammonia is 0.080M
c) At what concentration of ammonia is the rate equal to the rate constant?
8) In solution at constant H+ concentration, I- reacts with H2O2 to produce I2
H+(aq) + I-
(aq) + ½ H2O2(aq) ½ I2(aq) + H2O
The reaction rate can be followed by monitoring iodine production. The following data apply:
[I-] [H2O2] Rate (mol/L s)
.02 .02 3.3 x 10-5
.04 .02 6.6 x 10-5
.06 .02 9.9 x 10-5
.04 .04 1.3 x 10-4
a) Order of I-
b) Order of H2O2
c) Calc. k
d) Rate? When [I-] = .01 M [H2O2] = .03 M
9) In the first-order decomposition of acetone at 500C it is found that the concentration is 0.0300 M after 200 min and 0.0200M after 400 min.
H3C-CO-CH3(g) products
Calculatea) The rate constantb) The half-lifec) The initial concentration
10) The decomposition of hydrogen iodide is second-order. Its half-life is 85 seconds when the initial conc. is 0.15M
HI(g) ½ H2(g) + ½ I2(g)
a) What is k for the reaction?b) How long will it take to go from 0.300M to 0.100M?
11) Write the rate expression for each of the following elementary steps:
a) K+ + HCl KCl + H+
b) NO3 + CO NO2 + CO2
c) 2NO2 2NO + O2
12) For the reaction
2H2(g) + 2NO(g) N2(g) + 2H2O(g)
the experimental rate expression is rate = k[NO]2[H2]
The following mechanism is proposed:2NO N2O2 fast
N2O2 + H2 H2O + N2O slow
N2O + H2 N2 + H2O fast
Show that the mechanism is consistent with the rate expression.