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.