PART 11 – Chemical KineticsPART 11 Chemical Kinetics

Reference: Chapter 18 in bookReference: Chapter 18 in book

Rate of Chemical EquationsRate of Chemical Equations

1. Representation of Reaction Rate:nΔ n n

v = = (unit: mol·s-1 )tnΔΔ n n

t t2 1

2 1

−−

vC = (unit: mol·L-1·s-1C Ct t2 1

2 1

−−

or mol·L-1·min-1, mol·L-1·h-1)

vP = (unit: atm·s-1 or kpa·s-1 , mmHg·s-1)P Pt t2 1−

2

P ( p g )t t2 1−

2 N O ( ) 4 NO ( ) + O ( )2 N2O5 (g) → 4 NO2 (g) + O2 (g)

v = Δ[N O ] / Δt v ’ = Δ[NO ] / Δt v ’’ = Δ[O ] / Δtvc = – Δ[N2O5] / Δt vc = Δ[NO2] / Δt vc = Δ[O2] / Δt

vP = – ΔPN2O5 / Δt vP’ = ΔPNO2 / Δt vP

’’ = ΔPO2 / ΔtP P P

Most of reactions do not have constant rate. Above

3

are all for average reaction rates.

Example: H2O2 → H2O + ½O2Example: H2O2 → H2O + ½O2

[H O ] h ith ti C 0 80 l L 1[H2O2] changes with time: C0 = 0.80 mol·L-1

After 20 min: C = 0 40 mol·L-1After 20 min: C1 = 0.40 mol·L

v1 = – (0.4 – 0.8)/20 = 0.02 mol·L-1·min-1v1 (0.4 0.8)/20 0.02 mol L min

After another 20 min: C2= 0.20 mol·L-12

v2 = – (0.2 – 0.4)/20 = 0.01 mol·L-1·min-1

After another 20 min: C3= 0.10 mol·L-1

4v3 = – (0.1 – 0.2)/20 = 0.005 mol·L-1·min-1

Cv Δ=

__

tv Δ=

dCv =dt

2 N2O5 (g) → 4 NO2 (g) + O2 (g)2 5 (g) 2 (g) 2 (g)

vN2O5 = – d[N2O5] / dt vNO2 = d[NO2] / dt

5vO2 = d[O2] / dt

2 N2O5 (g) → 4 NO2 (g) + O2 (g)2 N2O5 (g) → 4 NO2 (g) + O2 (g)

v = – d[N2O5] / 2·dt = d[NO2] / 4·dt = d[O2] / dtv d[N2O5] / 2 dt d[NO2] / 4 dt d[O2] / dt

Therefore, for a general chemical reaction:

aA + bB → gG + hHaA + bB → gG + hH

v = –d[A] / a·dt = –d[B] / b·dt = d[G] / g·dt = d[H] / h·dtv d[A] / a dt d[B] / b dt d[G] / g dt d[H] / h dt

6

Rate LawsRate Laws

A bB G hHaA + bB gG + hH

Net rate: v = vforward – vreverse

At initial moments: Reactant → Product

−d[A] / a·dt = k · [A]m · [B ]n Differential Rate Laws

k is rate constant (a function of temperature)k is rate constant (a function of temperature)

m, n are orders of reaction (m is the order of reaction for A,

7n is the order of reaction for B )

Question:Question: A reaction at room temperature is as follows:

S O 2 SO 2S2O82- + 3 I- → 2 SO4

2- + I3-

The rates under different initial concentrations are:The rates under different initial concentrations are:

[S2O82-] [I-] v (–d[S2O8

2-] / dt )

1. 0.076 0.060 2.8 x 10-5 mol/L min

2 0 038 0 060 1 4 10 5 l/L i2. 0.038 0.060 1.4 x 10-5 mol/L min

3. 0.076 0.030 1.4 x 10-5 mol/L min

Write the rate laws for this reaction:

8–d[S2O8

2-] / dt = k [S2O82-]m [I- ]n

Solution:Solution: By simple observation, we can see:

m = 1, n = 1 Total order of this reaction: m + n = 2

– d [S2O82-] / dt = k [S2O8

2- ]1 [I- ]1

k = 2 8 x 10-5 / (0 076)(0 060 ) = 6 14 x 10-3 mol-1·L·s-1k = 2.8 x 10 / (0.076)(0.060 ) = 6.14 x 10 mol L s

– d[S2O82-]/dt = 6.14 x 10-3[S2O8

2- ]·[I- ]

d[I- ]/dt = 3 (– d[S O 2-] / dt) = 3 x 6 14 x 10-3[S O 2- ]·[I- ]d[I ]/dt = 3 (– d[S2O8 ] / dt) = 3 x 6.14 x 10 [S2O8 ] [I ]

9

About the rate constant k:

Physical meaning: the reaction rate when concentration of each reactant is 1 mol/Lconcentration of each reactant is 1 mol/L.

Unit: s-1 (1st order reaction)

mol-1·L·s-1 (2nd order reaction)

mol-2·L2·s-1 (3rd order reaction)

Some other reactions have more complicated rate laws, for instance, for H2 (g) + Br2 (g) 2 HBr (g) ,

the rate law is:

[HBr]]][Brk[H 21

22=v

10][Br[HBr]k1

2

′+

Differential rate law Describe the relationshipDifferential rate law — Describe the relationshipbetween reaction rate and concentration of species.

Integrated rate law — Describe howconcentration of each species is changing with timep g g

Example: A (Reactant) → B (Product)

zero order: −d[A] / dt = k [A]0fi t d d[A] / dt k [A]1first order: −d[A] / dt = k [A]1second order: −d[A] / dt = k [A]2

11

Zero Order First Order Second Order

−d[A]/dt = k[A]0 −d[A]/dt = k[A] −d[A]/dt = k[A]2

[A] [A]

∫ ∫−=[A]

[A]

t

00dtkd[A] ∫ ∫−=

[A]

[A}

t

00dtk

[A]d[A] ∫ ∫−=

[A]

[A}

t

020

dtk[A]d[A]

[A] = −k t + [A]0 ln[A] = − kt + ln[A]0 1/[A]= kt + 1/[A]0

12[A]= [A]0·e-kt

Question:

(CH3)3CBr + H2O → (CH3)3COH + HBr

If we know this reaction is a 1st order reaction, and the rate constant k is 0.0518 h-1, the initial concentration of (CH3)3CBr is 0.1039 mol/L.

Please calculate the concentration and conversion yield of (CH3)3CBr after 10 hours of reaction?

Solution:Solution:

C0 = 0.1039 mol/L k = 0.0518 h-1 t = 10 h

First order reaction: ln C/C0 = – kt

ln (C /0.1039) = – 0.0518 x 10 C = 0.0619 mol/L

13

( )

Conversion yield α: (0.1039 – 0.0619) / 0.1039 = 0.402

Half-Life (t1/2) — The time needed for consuming ( 1/2) g50% of reactants.

Half-life of 1st order reaction:a e o o de eac o

ln C/C0= – kt C = ½ C0

ln ½ C0/ C0 = – k t1/2 ln 2 = k t1/2 t1/2= ln 2 / k

t 0 693 / k (f 1 t d ti )t1/2 = 0.693 / k (for 1st order reaction)

N t f 1st d ti t d t d d i iti lNote: for 1st order reaction, t1/2 does not depend on initial concentration (C0)！ 100% → 50% → 25% → 12.5%

14

t1/2 t1/2 t1/2

Question:The dissociation of cyclobutane is a 1st order reaction. At

750 K, 25% of cyclobutane dissociated in 80 sec.

How much time does it need to dissociate 75% of cyclobutane at this temperature?

S l tiSolution: First work out the rate constant k:

ln (C / C0) = – k * 80 ln 0.75/1 = – k * 80

k = 3 6 x 10-3 s-1k = 3.6 x 10-3 s-1

ln (1 – 0.75) /1 = – 3.6 x 10-3 s-1 t

15ln 0.25 /1 = – 3.6 x 10-3 s-1 t t = 385 sec

Half-life of a 2nd order reaction:

1/[A] – 1/[A]0 = kt

[A] = ½ [A] 2/[A] 1/[A] = k t[A] = ½ [A]0 2/[A]0 – 1/[A]0 = k t1/2

1/[A]0 = k t1/2 1=t

02/1 k[A]=t

That is, for a 2nd order reaction, t1/2 depends on the initial concentration of reactants ([A]0).

16

([ ]0)

Elementary Reactiony

a. Elementary Reaction & Non-elementary Reactiona e e ta y eact o & o e e e ta y eact o

NO2 + CO → NO + CO2 (Elementary reaction)2 2 ( y )

H2O2 + 2H+ + 2Br- → 2H2O + Br2 (Non-elementary)2 2 2 2 ( y)

(1) H2O2 + H+ + Br- → HOBr + H2O( ) 2 2 2

(2) HOBr + H+ + Br- → Br2 + H2O

A non-elementary reaction comprises of a few steps of

17

y p pelementary reactions.

b. Rate Law of an Elementary Reaction:

aA + bB → cC + dD

d[C] / dt k [A]a [B]bd[C] / dt = k · [A]a · [B]b

c Equilibrium of Elementary Reactions:c. Equilibrium of Elementary Reactions:

aA + bB cC + dDaA bB cC dD

v forward = k forward · [A]a · [B]b v reverse = k reverse · [C]c · [D]d

At equilibrium: vforward = vreverse , kforward [A]a [B]b = kreverse [C]c [D]d

[C]c · [D]d / [A]a · [B]b = kforward / kreverse = Kequilibrium

N t I d d h th ll ti h ilib i ll

18

Note: Indeed, when the overall reaction reaches equilibrium, all its elementary reactions reach each own equilibrium.

Temperature Dependence of ReactionTemperature Dependence of Reaction

Rate

Normally, when temperature increases, reaction rate y p

increase. (Although many exceptions exist.)

19

k A Ea / RT Arrhenius Equationk = A· e-Ea / RT Arrhenius Equation

Ea is the Activation Energy (kJ/mol).

ln k = – Ea / RT + ln A

ln k2 / k1 = – Ea / [R(1/T2 – 1/T1)]

Calculate rate in different temperatures:

)11()ln( 2

TTRE

kk a −=

20

)()(211 TTRk

Question:

The dissociation of acetic acid is a 1st order reaction, with activation energy (Ea) = 144 kJ/mol. At 557 K, k = 3.3 xwith activation energy (Ea) 144 kJ/mol. At 557 K, k 3.3 x 10-2 s-1. If we want to dissociate 90% of acetic acid in 10 min, how should we set the temperature?how should we set the temperature?

S l tiSolution:

ln C/C0 = – k2 t ln (0.1C0/C0) = – k210 x 600 2 ( 0 0) 2

k2 = 3.8 x 10-3s-1

2 1T1 = 557 K k1 = 3.3 x 10-2 s-1

ln (3.8 x 10-3 / 3.3 x 10-2) = (144x103/8.314)(T2 – 557) / T2x 557

21T2 = 521 K

Physical Meaning of Activation Energya. Theory of the Transition State

Ea+ = Activation energyEa+ Activation energy for the forward reaction

Ea = Activation energyEa- Activation energy for the reverse reaction

ΔE = Ea+ – Ea

22

ΔE Ea+ Ea-

b. Collision Theory of Bi-molecular Reaction (optional)

v Z Z = Z0(A)(B) v = Z0(A)(B)v ∝ Z Z = Z0(A)(B) v = Z0(A)(B)

f = Effective Collision # / Total Collision # = e –Ec /RT

v = Z·f = Z·e –Ec/RT v = Z ·f ·P0 ( )( ) E /RTv = Z0 (A)(B) · P · e –Ec/RT

(A) = (B) = 1 mol / L v = k

k = Z0 · P · e –Ec/RT k = A· e –Ea/RT

Ea = Lowest translational energy to induce an effective collision.

23

Reaction Mechanism

1 Elementary Reaction1. Elementary Reaction

2. Rate Law and Reaction Mechanism

3 A + 2 B → C + D –d[A] / 3dt = kexp [A] [B]

This reaction has the following mechanism:

(1) A + B → E + F k (slow) rate-determining(1) A + B → E + F k1 (slow), rate-determining

(2) A + E → H k2 (fast)

(3) A + F → G k3 (fast)

(4) H + G + B → C + D k4 (fast)

24

(4) H G B → C D k4 (fast)

Question:

2 NO2 (g) + F2 (g) → 2 NO2F (g)

– d[NO2] / 2dt = kexp [NO2] [F2]

Possible reaction mechanism:Possible reaction mechanism:

(1) NO2 (g) + F2 (g) → NO2F (g) + F (g) k1 (slow)2 2 2

(2) NO2 (g) + F (g) → NO2F (g) k2 (fast)

Because Step (1) is the rate-determining step:

25– d[NO2] / 2dt = k1 [NO2] [F2] kexp = k1

Example: H2 (g) + I2 (g) → 2 HI (g)

–d[H2] / dt = kexp [H2] [I2]

Possible reaction mechanism:

(1) I2 2 I (fast equilibrium） K1 = k1 / k-1

(2) 2 I + H2 → 2 HI (slow) k2

Because Step (2) is rate determining stepBecause Step (2) is rate-determining step, –d[H2] /dt = k2 [H2][I]2

[I]2/[I2] = k1/k-1 = K1 [I]2 = K1 [I2]

26–d[H2] /dt = k2 K1[H2] [I2] kexp = k2 K1

Example: H2 (g) + Br2 (g) → 2 HBr (g)

d[H ] / dt = k [H ] [Br ]1/2–d[H2] / dt = kexp [H2] [Br2]1/2

Reaction Mechanism:

(1) Br2 + M 2Br + M (fast Equil.） K1 = k1/k-1

(2) Br + H2 → HBr + H (slow) k2(2) Br H2 → HBr H (slow) k2

(3) H + Br2 → HBr + Br (fast) k3

Because Step (2) is rate-determining step, so –d[H2] / dt= k2 [H2] [Br]

27

2 [ 2] [ ]

[Br]2 / [Br2] = k1 / k-1 = K1

[Br]2 = K1 [Br2] = k1 / k-1 [Br2]

[Br] = K11/2 [Br2]1/2

Because:

/–d[H2] / dt = k2 [H2] [Br]

Then:Then:

–d[H2] /dt = k2 K11/2 [H2] [Br2]1/2d[H2] /dt = k2 K1 [H2] [Br2]

kexp = k2 K11/2 or : kexp= k2 (k1 / k 1)1/2

28

exp 2 1 exp 2 ( 1 -1)

Steady State Approximation — Assume in the reaction process, the appearance rate of Intermediates equals to the diappeance rate.

Example: 2 NO + O2 → 2 NO2

Possible Reaction Mechanism:

(1) NO + NO → N2O2 k1

(2) N O → NO + NO k(2) N2O2 → NO + NO k-1

(3) N2O2 + O2 → 2NO2 k2

Prove the rate law of this reaction is as following:

29d[NO2] / 2dt = k2 k1 [O2] [NO]2 / (k-1 + k2 [O2])

For intermediate, N2O2, the steady-state approximation:

d[N O ] / dt = d[N O ] / dtd[N2O2] / dt = -d[N2O2] / dt

k1 [NO]2 = k 1 [N2O2] + k2 [N2O2] [O2]k1 [NO] k-1 [N2O2] k2 [N2O2] [O2]

k1 [NO]2 = [N2O2] (k-1+ k2 [O2] )

[N2O2] = k1[NO]2 /(k-1+ k2 [O2] )

d[NO2] / 2dt = k2 [N2O2] [O2]

d[NO2] / 2dt = k2 k1 [O2] [NO]2/(k-1+ k2 [O2] )

30

Question: C H + H → 2 CHQuestion: C2H6 + H2 → 2 CH4

Reaction Mechanism:Reaction Mechanism:

(1) C2H6 2 CH3• (fast equilibrium) k1/ k 1(1) C2H6 2 CH3 (fast equilibrium) k1/ k-1

(2) CH3• + H2 → CH4 + H • k23 2 4 2

(3) H • + C2H6 → CH4 + CH3• k3

Prove the rate law of this overall reaction is:

d[CH4] / 2dt = k2 (k1 / k-1) [C2H6]1/2 [H2]

31

[ 4] 2 ( 1 1) [ 2 6] [ 2]

[CH3•]2 / [C2H6] = k1/ k-1 [CH3

•] = (k1/ k-1) 1/2[C2H6]1/2

Use the steady-state approximation, work out [H•] :

d[H•] / dt = k2 [CH3•] [H2] (H• appearance rate)

-d[H•] / dt = k3 [C2H6] [H•] (H• disappearance rate)

k [CH •] [H ]= k [C H ] [H•]k2 [CH3•] [H2]= k3[C2H6] [H•]

[H•] = k2 [CH3•] [H2] / k3 [C2H6]

d[CH4] / dt = k2 [CH3•] [H2] + k3 [C2H6] [H•]

= k [CH •] [H ] + k [C H ] k [CH •] [H ]/ k [C H ]= k2 [CH3 ] [H2] + k3 [C2H6] k2 [CH3 ] [H2]/ k3 [C2H6]

= 2 k2 [CH3•] [H2]

= 2 k2 [H2] [CH3•] = 2 k2 [H2] (k1/ k-1) 1/2 [C2H6]1/2

d[CH4] / 2dt = k2 (k1/ k 1) 1/2 [H2] [C2H6]1/2

32

d[CH4] / 2dt k2 (k1/ k-1) [H2] [C2H6]

kexp = k2 (k1/ k-1) ½

Question:Question:

For a set of reactions:

A + B → C + D k1

C + D → A + B k-1

C + E → F kC + E → F k2

What kind of relationship should exist among k1、 k-1、k2, and the reagent concentrations can result in the following rate laws:

(a) d[F] / dt = k [A] [B] [E] / [D](a) d[F] / dt = k [A] [B] [E] / [D]

(b) d[F] / dt = k’ [A] [B]

33

Assume intermediate C is in steady-state: d[C]dt = -d[C]dt

k [A][B] = k [C] [D] + k [C] [E]k1[A][B] = k-1 [C] [D] + k2 [C] [E]

k1 [A] [B] = [C] (k-1 [D] + k2 [E])

[C] = k1 [A] [B] / (k-1 [D] + k2 [E])

d[F] / dt = k2 [C] [E]d[F] / dt k2 [C] [E]

d[F] / dt = k2 [E] k1[A][B] / (k-1 [D] + k2 [E])

When k2 [E] << k-1 [D] :

d[F]/dt = k [E] k [A][B] / k [D] = k [A][B][E]/[D]d[F]/dt = k2 [E] k1[A][B] / k-1 [D] = k [A][B][E]/[D]

k = k1k2/k-1

When k2 [E] >> k-1 [D] :

d[F]/dt = k [E] k [A][B] / k [E] = k’ [A][B]

34

d[F]/dt = k2 [E] k1[A][B] / k2 [E] = k [A][B]

k’ = k1

Question: For the following reaction, write the relationship of vforward and vreverse with regard to concentrations of species:vforward and vreverse with regard to concentrations of species:

H3AsO4 (aq) + 3I− (aq) + 2H+ (aq) H3AsO3(aq) + I3− (aq) + H2O (l)

2343

333

][H]][IAsO[H]][IAsO[H+−

−

=K

]][][[dt

]AsOd[H43

33 +−== HIAsOHkv forwardforward dt ff

]][[]][][[33343−+− IAsOHHIAsOHk forward

]][[

][]][[]][[]][][[

2343

33343

−

+−×=

IA OHk

HIAsOHIAsOH

Kv forward

reverse

][][]][[

2333

+−=HIK

IAsOHk forward

35][][]][[

2333

+−

−

=HI

IAsOHkreverse

LAST Homework #11 (due on 6/22, Fri)( )① (Book pg.729—732):

18 37 18 49 18 56 18 66 18 75 18 7918.37, 18.49, 18.56, 18.66, 18.75, 18.79

Final ExamFinal ExamTime: July 5 (Thursday), 8:30—10:30 AM

Location: West Guang-Hua Building 509 （光华西辅楼509）Content: All this semester, Close-book, Bring a calculator

Q & A time: July 3 Tuesday (?), full day

Location: Room 434, Lab of Advanced Materials, Jiang-wan Campus

（江湾校区 先进材料楼 434室）（江湾校区 先进材料楼 434室）

Direction: http://www.chemistry.fudan.edu.cn/m_netteacher/gfzheng/tour.html