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– Molecular Orientation– Temperature– Arrhenius Equation

• Reaction Mechanisms– Elementary Steps, Reaction Order, Intermediates

• Catalysts

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

[M]

time

[M]

t

for reaction A B

Reaction Rates (graphical):

• Average Rate = [M] t

Rates

• for A B

- [A] = [B] t t

Rate of the disappearance of A is equal in magnitude but opposite in sign to the rate of the appearance of B

[M]

time

tangent at time, t

t

• Average Rate-- mol (or concentration) over a period of time, t

• Instantaneous Rate-- slope of the tangent at a specific time, t

• Initial Rate-- instantaneous rate at t = 0

Average Rate =

[A]final time - [A]initial time tfinal - tinitial

for A B

[M]

time

tangent at time, t

t

Instantaneous Ratetime, t = slope of the tangent at time = t

Stoichiometry

4PH3 => P4 + 6H2

- 1[PH3] = + 1 [P4] = + 1 [H2] 4 t 1 t 6 t

- [PH3] = + 4 [P4] = + 2 [H2]

t t 3 t

General Relationship

Rate =

- 1 [A] = - 1 [B] = + 1 [C] = + 1 [D] a t b t c t d t

aA + bB cC + dD

Conditions which affect rates

• Concentration

– concentration rate

• Temperature

– temperature rate

• Catalyst

– substance which increases rate but itself remains unchanged

• aA + bB xX rate law

rate = 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

Rate Equations:

Examples:

• 2N2O5 => 4NO2 + O2

rate = k[N2O5] 1st order

• 2NO + Cl2 => 2NOCl

rate = k[NO]2[Cl2] 3rd order

• 2NH3 => N2 + 3H2

rate = k[NH3]0 = k 0th order

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 Rate

Exponent Value [conc] rate

0 double same1 double double2 double x 43 double x 84 double x 16

Problem:

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

Expt. # [NO] [H2] rate

1 6.4x10-3 2.2x10-3 2.6x10-5

2 12.8x10-3 2.2x10-3 1.0x10-4 3 6.4x10-3 4.5x10-3 5.0x10-5

rate = k[NO]2[H2]

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. Time1st and 2nd order integrated rate equations

• First Order: rate = - [A] = k [A] t

• ln [A]t = - kt A = reactant

[A]0

• or ln [A]t - ln [A]0 = - kt

1st ord. decay.lnk

Conversion to base-10 logarithms:

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

to

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

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

• Second Order: rate = - [A] = k[A]2

t• 1 - 1 = kt

[A]t [A]0

• second order rxn with one reactant:

rate = k [A]2

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

b = slopea = y interceptx = time

Graphical Methods

• Equation for a Straight Line

• y = bx + a

• ln[A]t = - kt + ln[A]0 1st order

• 1 = kt + 1 2nd order

[A]t [A]0

time

[H2O2]

First Order: 2H2O2(aq) 2H2O(l) + O2(g)

ln [H2O2]

time

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

First Order: 2H2O2(aq) 2H2O(l) + O2(g)

1/[NO2]

time

slope, b = +k

Second Order: 2NO2 2NO + O2

0.020 M

0.010 M

t1/2

[M]

time

t1/2 = 0.693 k

0.005 M

t1/2

Half-Life of a 1st order process:

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

• 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

Activation Energy, Ea

En

erg

y

Reaction Progress

Reactant

Product

H reactionEa forward rxn.Ea reverse rxn.

Molecular Theory (Collision Theory)

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

• increase temperature, inc. reaction rates

• rxn rates are 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

1/T

ln

k

slope = -Ea/R

Graphical Determination of Ea

rearrange eqtn to give straight-line eqtn y = bx + a ln k = -Ea 1 + ln A

R T

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

slope = -6373 = -Ea/R

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

ln k

1/T

3.14 3.25 3.35 x 10-3

-3.2

0 -2

.50

-

1.8

5

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

k = 0.0985 min-1

Problem:

The energy of activation for

C4H8(g) => 2C2H4(g) is 260 kJ/mol at 800 K and k = 0.0315

secFind k at 850 K.

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

k at 850 K = 0.314 sec-1

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 Equations

• Molecularity Rate Lawunimolecular 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

2O3 => 3O2

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

O3 + O => 2O2 rate = k’[O3]

[O]

2O3 + O => 3O2 + O

O is an intermediate

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

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]

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

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

1: O3(g) O2(g) + O(g) fast

equil. rate1 = k1[O3]rate2 = k2[O2][O]

2: O(g) + O3(g) 2O2(g) slowrate3 = 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

Reaction Mechanisms & Equilibria

k3

k1

k2

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

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

k1

k2 k3

k4

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

• Facilitates the progress of a reaction bylowering the overall activation energy

– homogeneous

– heterogeneous

En

erg

y

Reaction Progress

Ea

Ea

Hrxn

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

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)

Catalyzed Reaction:

Uncatalyzed Reaction:O3(g) <=> O2(g) + O(g)

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

Ea uncatalyzed rxn

Ea catalyzed rxn

ClO + O2 + O

Cl + O3 + O

Cl + O2 + O2