slide1 l.
Download
Skip this Video
Loading SlideShow in 5 Seconds..
Chapter 14 PowerPoint Presentation
Download Presentation
Chapter 14

Loading in 2 Seconds...

play fullscreen
1 / 51

Chapter 14 - PowerPoint PPT Presentation


  • 251 Views
  • Uploaded on

Chapter 14. Chemical Kinetics. Overview:. Reaction Rates Stoichiometry, Conditions, Concentration Rate Equations Order Initial Rate Concentration vs. Time First Order Rxns. Second Order Rxns. Graphical Methods. Cont’d. Molecular Theory Activation Energy Concentration

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about 'Chapter 14' - mirra


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
slide1

Chapter 14

Chemical Kinetics

overview
Overview:
  • Reaction Rates
    • Stoichiometry, Conditions, Concentration
  • Rate Equations
    • Order
    • Initial Rate
    • Concentration vs. Time
    • First Order Rxns.
    • Second Order Rxns.
  • Graphical Methods
cont d
Cont’d
  • Molecular Theory
    • Activation Energy
    • Concentration
    • Molecular Orientation
    • Temperature
    • Arrhenius Equation
  • Reaction Mechanisms
    • Elementary Steps, Reaction Order, Intermediates
  • Catalysts
reaction rates
Reaction Rates
  • What Affects Rates of Reactions?
    • Concentration of the Reactants
    • Temperature of Reaction
    • Presence of a Catalyst
    • Surface Area of Solid or Liquid Reactants
reaction rates graphical
Reaction Rates (graphical):
  • Average Rate = D[M]Dt

[M]

for reaction A  B

D[M]

time

Dt

rates
Rates
  • for A  B- D[A] = D[B] Dt DtRate of the disappearance of A is equal in magnitude but opposite in sign to the rate of the appearance of B
slide7
Average Rate--D mol (or concentration) over a period of time, Dt
  • Instantaneous Rate-- slope of the tangent at a specific time, t
  • Initial Rate-- instantaneous rate at t = 0

[M]

tangent at time, t

t

time

slide8

Average Rate =

[A]final time - [A]initial timeDtfinal - Dtinitial

for A  B

stoichiometry
Stoichiometry

4PH3 => P4 + 6H2

- 1D[PH3] = + 1D[P4] = + 1D[H2] 4Dt 1Dt 6Dt

- D[PH3] = + 4D[P4] = + 2D[H2] Dt Dt 3Dt

general relationship
General Relationship

Rate = - 1 D[A] = - 1 D[B] = + 1 D[C] = + 1 D[D] aD t b D t cD t dD taA + bB  cC + dD

conditions which affect rates
Conditions which affect rates
  • Concentration
    • concentration  rate
  • Temperature
    • temperature  rate
  • Catalyst
    • substance which increases rate but itself remains unchanged
rate equations
Rate Equations:
  • aA + bB  xX rate lawrate = k[A]m[B]n
  • m, n are orders of the reactants
    • extent to which rate depends on concentration
    • m + n = overall rxn order
  • k is the rate constant for the reaction
examples
Examples:
  • 2N2O5 => 4NO2 + O2 rate = k[N2O5] 1st order
  • 2NO + Cl2 => 2NOClrate = k[NO]2[Cl2] 3rd order
  • 2NH3 => N2 + 3H2rate = k[NH3]0 = k0th order
determination of rate equations
Determination of Rate Equations:

Data for: A + B => C

Expt. # [A] [B] initial rate

1 0.10 0.10 4.02 0.10 0.20 4.03 0.20 0.10 16.0

rate = k[A]2[B]0 = k[A]2 k = 4.0 Ms-1 = 400 M-1s-1 (0.10)2 M2

exponent values relative to d rate
Exponent Values Relative to DRate

Exponent Value [conc] rate

0 double same 1 double double 2 double x 4 3 double x 8 4 double x 16

problem
Problem:

Data for: 2NO + H2 => N2O + H2O

Expt. # [NO] [H2] rate

1 6.4x10-3 2.2x10-3 2.6x10-52 12.8x10-3 2.2x10-3 1.0x10-43 6.4x10-3 4.5x10-3 5.0x10-5 rate = k[NO]2[H2]

units of rate constants
Units of Rate Constants
  • units of rates M/s
  • units of rate constants will vary depending on order of rxn M = 1 (M)2 (M) for s sM2 rate = k [A]2 [B]
  • rate constants are independent of the concentration
concentration vs time 1st and 2nd order integrated rate equations
Concentration vs. Time1st and 2nd order integrated rate equations
  • First Order: rate = - D[A] = k [A]D t
  • ln [A]t = - kt A = reactant [A]0
  • or ln [A]t - ln [A]0 = - kt
conversion to base 10 logarithms
Conversion to base-10 logarithms:

ln [A]t = - kt [A]0

tolog [A]t = - kt [A]0 2.303

problem21
Problem:

The rate equation for the reaction of sucrose in water is, rate = k[C12H22O11]. After 2.57 h at 27°C, 5.00 g/L of sucrose has decreased to 4.50 g/L. Find k. C12H22O11(aq) + H2O(l) => 2C6H12O6

ln 4.50g/L = - k (2.57 h) 5.00g/L

k = 0.0410 h-1

concentration vs time
Concentration vs. Time
  • Second Order:rate = - D[A] = k[A]2Dt
  • 1 - 1 = kt [A]t [A]0
  • second order rxn with one reactant: rate = k [A]2
problem23
Problem:

Ammonium cyanate, NH4NCO, rearranges in water to give urea, (NH2)2CO. If the original concentration of NH4NCO is 0.458 mol/L and k = 0.0113 L/mol min, how much time elapses before the concentration is reduced to 0.300 mol/L?NH4NCO(aq) => (NH2)2CO(aq) rate = k[NH4NCO]

1 - 1 = (0.0113) t

(0.300) (0.458)

t = 102 min

graphical methods
Graphical Methods
  • Equation for a Straight Line
  • y = bx + a
  • ln[A]t = - kt + ln[A]01st order
  • 1 = kt + 1 2nd order[A]t [A]0

b = slopea = y interceptx = time

first order 2h 2 o 2 aq 2h 2 o l o 2 g26
First Order:2H2O2(aq) ® 2H2O(l) + O2(g)

slope, b = -1.06 x 10-3 min-1 = - k

ln [H2O2]

time

second order 2no 2 2no o 2
Second Order: 2NO2 ® 2NO + O2

1/[NO2]

slope, b = +k

time

half life of a 1 st order process
Half-Life of a 1st order process:

0.020 M

t1/2 = 0.693k

[M]

0.010 M

0.005 M

t1/2

t1/2

time

problem29
Problem:

The decomposition of SO2Cl2 is first order in SO2Cl2 and has a half-life of 4.1 hr. If you begin with 1.6 x 10-3 mol of SO2Cl2 in a flask, how many hours elapse before the quantity of SO2Cl2 has decreased to 2.00 x 10-4 mol?

SO2Cl2(g) => SO2(g) + Cl2(g)

temperature effects
Temperature Effects
  • Rates typically increase with T increase
  • Collisions between molecules increase
  • Energy of collisions increase
  • Even though only a small fraction ofcollisions lead to reaction
  • Minimum Energy necessary for reactionis the Activation Energy
molecular theory collision theory
Molecular Theory(Collision Theory)

Activation Energy, Ea

DH reactionEa forward rxn.Ea reverse rxn.

Energy

Reactant

Product

Reaction Progress

activation energy
Activation Energy
  • Activation Energy varies greatly
    • almost zero to hundreds of kJ
    • size of Ea affects reaction rates
  • Concentration
    • more molecules, more collisions
  • Molecular Orientation
    • collisions must occur “sterically”
the arrhenius equation
The Arrhenius Equation
  • increase temperature, inc. reaction rates
  • rxn rates are a to energy, collisions, temp. & orient
  • k = Ae-Ea/RT

k = rxn rate constant

A = frequency of collisions

-Ea/RT = fraction of molecules with energy necessary for reaction

graphical determination of ea
Graphical Determination of Ea

rearrange eqtn to give straight-line eqtn

y = bx + a

ln k = -Ea1 + ln A RT

slope = -Ea/R

ln k

1/T

problem35
Problem:

Data for the following rxn are listed in the table. Calculate Ea graphically, calculate A and find k at 311 K.

Mn(CO)5(CH3CN)+ + NC5H5 => Mn(CO)5(NC5H5)+ + CH3CN

ln k k, min-1 T (K) 1/T x 10-3

-3.20 0.0409 298 3.35

-2.50 0.0818 308 3.25

-1.85 0.157 318 3.14

slide36

slope = -6373 = -Ea/REa = (-6373)(-8.31 x 10-3 kJ/K mol) = 53.0 kJ

y intercept = 18.19 = ln A A = 8.0 x 10 7

-3.20 -2.50 -1.85

ln k

k = 0.0985 min-1

3.14 3.25 3.35 x 10-3

1/T

problem37
Problem:

The energy of activation for

C4H8(g) => 2C2H4(g)

is 260 kJ/mol at 800 K and k = 0.0315 sec

Find k at 850 K.

ln k2 = - Ea (1/T2 - 1/T1) k1 R

k at 850 K = 0.314 sec-1

reaction mechanisms
Reaction Mechanisms
  • Elementary Step
    • equation describing a single molecular event
  • Molecularity
    • unimolecular
    • bimolecular
    • termolecular
  • 2O3 => 3O2

(1) O3 => O2 + O unimolecular(2) O3 + O => 2 O2 bimolecular

rate equations39
Rate Equations
  • Molecularity Rate Law

unimolecular rate = k[A] bimolecular rate = k[A][B] bimolecular rate = k[A]2 termolecular rate = k[A]2[B]

    • notice that molecularity for an elementary step is the same as the order
2o 3 3o 2
2O3 => 3O2

O3 => O2 + O rate = k[O3]

O3 + O => 2O2 rate = k’[O3][O]

2O3 + O => 3O2 + O

O is an intermediate

problem41
Problem:
  • Write the rate equation and give the molecularity of the following elementary steps:

NO(g) + NO3(g) => 2NO2(g)

rate = k[NO][NO3] bimolecular

(CH3)3CBr(aq) => (CH3)3C+(aq)+ Br-(aq)

rate = k[(CH3)3CBr] unimolecular

mechanisms and rate equations
Mechanisms and Rate Equations

rate determining step is the slow step --the overall rate is limited by the ratedetermining step

step 1 NO2 + F2 => FNO2 + F rate = k1[NO2][F2] k1 slow

step 2 NO2 + F => FNO2 rate = k2[NO2][F] k2 fast

overall 2NO2 + F2 => 2FNO2 rate = k1[NO2][F2]

problem43
Problem:
  • Given the following reaction and rate law:NO2(g) + CO(g) => CO2(g) + NO(g)rate = k[NO2]2
    • Does the reaction occur in a single step?
    • Given the two mechanisms, which is most likely:NO2 + NO2 =>NO3 + NO NO2 => NO + ONO3 + CO => NO2 + CO2 CO + O => CO2
reaction mechanisms equilibria

k1

k2

Reaction Mechanisms & Equilibria

2O3(g) 3O2(g) overall rxn

1: O3(g) O2(g) + O(g) fastequil. rate1 = k1[O3] rate2 = k2[O2][O]

2: O(g) + O3(g) 2O2(g) slow rate3 = k3[O][O3]

rate 3 includes the conc. of an intermediate and the exptl. rate law will include only species that are present in measurable quantities

k3

substitution method
Substitution Method

at equilibrium k1[O3] = k2[O2][O]

rate3 =k3[O][O3] [O] = k1 [O3] k2 [O2]

rate3 = k3k1 [O3]2 or k2 [O2]

overall rate = k’ [O3]2 [O2]

substitute

problem46

k1

k2

Problem:

Derive the rate law for the following reaction given the mechanism step below:

OCl -(aq) + I -(aq) OI -(aq) + Cl -(aq)

OCl - + H2O HOCl + OH - fast

I - + HOCl HOI + Cl - slow

HOI + OH - H2O + OI - fast

k3

k4

cont d47
Cont’d

rate1 = k1 [OCl -][H2O] =

rate 2 = k2 [HOCl][OH -]

[HOCl] = k1[OCl -][H2O] k2[OH -]

rate 3 = k3 [HOCl][I -]

rate 3 = k3k1[OCl -][H2O][I -] k2 [OH -]

overall rate = k’ [OCl -][I -][OH -]

solvent

catalyst
Catalyst
  • Facilitates the progress of a reaction bylowering the overall activation energy
    • homogeneous
    • heterogeneous
slide49

Ea

Ea

DHrxn

Energy

Reaction Progress

catalysts are used in an early rxn step but regenerated in a later rxn step

slide50

Uncatalyzed Reaction:

O3(g)<=> O2(g) + O(g)

O(g) + O3(g) => 2O2(g)

Catalyzed Reaction:

Step 1: Cl(g) + O3(g) + O(g) => ClO(g) + O2(g) + O(g)

Step 2: ClO(g) + O2(g) +O(g) => Cl(g) + 2O2(g)

Overall rxn: O3(g) + O(g) => 2O2(g)

slide51

Ea uncatalyzed rxn

Ea catalyzed rxn

Cl + O3 + O

ClO + O2 + O

Cl + O2 + O2