Chemical KineticsChapter 14

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화학 반응 속도론

• 14.1 반응속도• 14.2 속도법칙• 14.3 반응물의 농도와 시간과의 관계• 14.4 활성화에너지와 속도상수와의

의존성• 14.5 반응 메카니즘과 속도법칙• 14.6 촉매

Chemical Kinetics

Thermodynamics – does a reaction take place?

Kinetics – how fast does a reaction proceed?

Reaction rate is the change in the concentration of a reactant or a product with time (M/s).

A B

rate = -[A]t

rate = [B]t

[A] = change in concentration of A over time period t

[B] = change in concentration of B over time period t

Because [A] decreases with time, [A] is negative.

A B

rate = -[A]t

rate = [B]t

time

Br2 (aq) + HCOOH (aq) 2Br- (aq) + 2H+ (aq) + CO2 (g)

time

393 nmlight

Detector

[Br2] Absorption3

93 n

m

Br2 (aq)

Br2 (aq) + HCOOH (aq) 2Br- (aq) + 2H+ (aq) + CO2 (g)

average rate = -[Br2]t

= -[Br2]final – [Br2]initial

tfinal - tinitial

slope oftangent

slope oftangent slope of

tangent

instantaneous rate = -d[Br2]dt

rate [Br2]

rate = k [Br2]

k = rate[Br2]

= rate constant

= 3.50 x 10-3 s-1

2H2O2 (aq) 2H2O (l) + O2 (g)

PV = nRT

P = RT = [O2]RTnV

[O2] = PRT1

rate = [O2]t RT

1 Pt=

measure P over time

2H2O2 (aq) 2H2O (l) + O2 (g)

Reaction Rates and Stoichiometry

2A B

Two moles of A disappear for each mole of B that is formed.

aA + bB cC + dD

dt

d

ddt

d

cdt

d

bdt

d

arate

]D[1]C[1]B[1]A[1

dt

drate

]A[

2

1

dt

drate

]B[

Write the rate expression for the following reaction:

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g)

dt

d

dt

d

dt

d

dt

drate

]OH[

2

1]CO[]O[

2

1]CH[ 2224

The Rate Law

The rate law expresses the relationship of the rate of a reaction to the rate constant and the concentrations of the reactants raised to some powers.

aA + bB cC + dD

Rate = k [A]x[B]y

reaction is xth order in A

reaction is yth order in B

reaction is (x +y)th order overall

F2 (g) + 2ClO2 (g) 2FClO2 (g)

rate = k [F2]x[ClO2]y

Double [F2] with [ClO2] constant

Rate doubles

x = 1

Quadruple [ClO2] with [F2] constant

Rate quadruples

y = 1

rate = k [F2][ClO2]

F2 (g) + 2ClO2 (g) 2FClO2 (g)

rate = k [F2][ClO2]

Rate Laws

• Rate laws are always determined experimentally.

• Reaction order is always defined in terms of reactant (not product) concentrations.

• The order of a reactant is not related to the stoichiometric coefficient of the reactant in the balanced chemical equation.

1

Determine the rate law and calculate the rate constant for the following reaction from the following data:S2O8

2- (aq) + 3I- (aq) 2SO42- (aq) + I3

- (aq)

Experiment [S2O82-] [I-]

Initial Rate (M/s)

1 0.08 0.034 2.2 x 10-4

2 0.08 0.017 1.1 x 10-4

3 0.16 0.017 2.2 x 10-4

rate = k [S2O82-]x[I-]y

Double [I-], rate doubles (experiment 1 & 2)

Double [S2O82-], rate doubles (experiment 2 & 3)

k = rate

[S2O82-][I-]

=2.2 x 10-4 M/s

(0.08 M)(0.034 M)= 0.08/M•s

y = 1x = 1

rate = k [S2O82-][I-]

First-Order Reactions

A product rate = -d[A]dt

rate = k [A]

k = rate[A]

= 1/s or s-1M/sM

=d[A]dt

= k [A]-

[A] is the concentration of A at any time t

[A]0 is the concentration of A at time t=0

[A] = [A]0exp(-kt) ln[A] = ln[A]0 - kt

Decomposition of N2O5

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

The reaction 2A B is first order in A with a rate constant of 2.8 x 10-2 s-1 at 800C. How long will it take for A to decrease from 0.88 M to 0.14 M ?

[A]0 = 0.88 M [A] = 0.14 M

ssMM

kkt 66

108.214.088.0

ln]A[]A[

ln]Aln[]Aln[

12

0

0

]Aln[]Aln[ 0 kt

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

First-Order Reactions

The half-life, t½, is the time required for the concentration of a reactant to decrease to half of its initial concentration.

t½ = t when [A] = [A]0/2

What is the half-life of N2O5 if it decomposes with a rate constant of 5.7 x 10-4 s-1?

t½ln2k

=0.693

5.7 x 10-4 s-1= = 1200 s = 20 minutes

How do you know decomposition is first order?

units of k (s-1)

kkkt

693.02ln2/]A[]A[

ln 0

2/1

A product

First-order reaction

# of half-lives [A] = [A]0/n

1

2

3

4

2

4

8

16

13.3

13.3

Second-Order Reactions

A product rate = -d[A]dt

rate = k [A]2

k = rate[A]2

= 1/M•sM/sM2=

d[A]dt

= k [A]2-

[A] is the concentration of A at any time t

[A]0 is the concentration of A at time t=0

1[A]

=1

[A]0

+ kt

t½ = t when [A] = [A]0/2

t½ =1

k[A]0

Zero-Order Reactions

A product rate = -d[A]dt

rate = k [A]0 = k

k = rate[A]0

= M/sd[A]dt

= k-

[A] is the concentration of A at any time t

[A]0 is the concentration of A at time t=0

t½ = t when [A] = [A]0/2

t½ =[A]0

2k

[A] = [A]0 - kt

Summary of the Kinetics of Zero-Order, First-Orderand Second-Order Reactions

Order Rate LawConcentration-Time

Equation Half-Life

0

1

2

rate = k

rate = k [A]

rate = k [A]2

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

1[A]

=1

[A]0

+ kt

[A] = [A]0 - kt

t½ln2k

=

t½ =[A]0

2k

t½ =1

k[A]0

Exothermic Reaction Endothermic Reaction

The activation energy (Ea ) is the minimum amount of energy required to initiate a chemical reaction.

A + B AB C + D++

Temperature Dependence of the Rate Constant

k = A • exp( -Ea/RT )

Ea is the activation energy (J/mol)

R is the gas constant (8.314 J/K•mol)

T is the absolute temperature

A is the frequency factor

lnk = -Ea

R1T

+ lnA(Arrhenius equation)

R

Eslope a

Reaction Mechanisms

a series of simple elementary steps or elementary reactions.

The sequence of elementary steps that leads to product formation is the reaction mechanism.

2NO (g) + O2 (g) 2NO2 (g)

N2O2 is detected during the reaction!

Elementary step: NO + NO N2O2

Elementary step: N2O2 + O2 2NO2

Overall reaction: 2NO + O2 2NO2

+

2NO (g) + O2 (g) 2NO2 (g)

Elementary step: NO + NO N2O2

Elementary step: N2O2 + O2 2NO2

Overall reaction: 2NO + O2 2NO2

+

Intermediates are species that appear in a reaction mechanism but not in the overall balanced equation.

An intermediate is always formed in an early elementary step and consumed in a later elementary step.

The molecularity of a reaction is the number of molecules reacting in an elementary step.

• Unimolecular reaction – elementary step with 1 molecule

• Bimolecular reaction – elementary step with 2 molecules

• Termolecular reaction – elementary step with 3 molecules

Unimolecular reaction A products rate = k [A]

Bimolecular reaction A + B products rate = k [A][B]

Bimolecular reaction A + A products rate = k [A]2

Rate Laws and Elementary Steps

Writing plausible reaction mechanisms:

• The sum of the elementary steps must give the overall balanced equation for the reaction.

• The rate-determining step should predict the same rate law that is determined experimentally.

The rate-determining step is the slowest step in the sequence of steps leading to product formation.

Sequence of Steps in Studying a Reaction Mechanism

The experimental rate law for the reaction between NO2 and CO to produce NO and CO2 is rate = k[NO2]2. The reaction is believed to occur via two steps:

Step 1: NO2 + NO2 NO + NO3

Step 2: NO3 + CO NO2 + CO2

What is the equation for the overall reaction?

NO2+ CO NO + CO2

What is the intermediate?

NO3

What can you say about the relative rates of steps 1 and 2?

rate = k[NO2]2 is the rate law for step 1 so step 1 must be slower than step 2

CH2

CH2

CH2

CH2

CH2 CH22CH2

CH2

CH2

CH2

CH2

CH2

CH2

CH2

• •

CH2 CH22

Chemistry In Action: Femtochemistry

catalyst

k = A • exp( -Ea/RT ) Ea k

uncatalyzed catalyzed

ratecatalyzed > rateuncatalyzed

Ea < Ea‘ 13.6

CatalyzedUncatalyzed

In heterogeneous catalysis, the reactants and the catalysts are in different phases.

In homogeneous catalysis, the reactants and the catalysts are dispersed in a single phase, usually liquid.

• Haber synthesis of ammonia

• Ostwald process for the production of nitric acid

• Catalytic converters

• Acid catalysis

• Base catalysis

N2 (g) + 3H2 (g) 2NH3 (g)Fe/Al2O3/K2O

catalyst

Haber Process

Gerhard Ertl

N2(g) → N2(adsorbed) N2(adsorbed) → 2N(adsorbed) H2(g) → H2(adsorbed) H2(adsorbed) → 2H(adsorbed) N(adsorbed) + 3H(adsorbed)→ NH3(adsorbed) NH3(adsorbed) → NH3(g)

Haber 법으로 매년 1 억톤의 질소 비료 생산 .세계 천연가스의 3–5% 를 Haber 법의 수소 생산에 사용됨 .세계 에너지 소비량의 ~1–2%.이 비료 때문에 지구 인구 1/3 이 생명을 유지하고 있음 .

Haber 법의 발견으로 칠레에서는 대규모 실업자를 발생시켰음 .1918 년 노벨 화학상 수상

Fritz Haber (1868 –1934)

Ostwald Process

Hot Pt wire over NH3 solutionPt-Rh catalysts used

in Ostwald process

4NH3 (g) + 5O2 (g) 4NO (g) + 6H2O (g)Pt catalyst

2NO (g) + O2 (g) 2NO2 (g)

2NO2 (g) + H2O (l) HNO2 (aq) + HNO3 (aq)

Catalytic Converters

CO + Unburned Hydrocarbons + O2 CO2 + H2Ocatalytic

converter

2NO + 2NO2 2N2 + 3O2

catalyticconverter

Enzyme Catalysis

uncatalyzedenzyme

catalyzed

rate = [P]t

rate = k [ES]

Michaelis-Menten MechanismE + S ES → E + P

tion ApproximaStateSteadyESESSEES

:0][][]][[][

211 kkkdt

d

k1

k-1

k2

][

][][

]][[

][][

][02

022 S

SE

SE

SES

P

MMM

M

Kk

KK

Kkk

dt

dV

][][

][

])[][

1(]][[

][][][][

,]][[]][[

][])[(]][[

0

0

1

21

21

1211

ES

E

ESSE

EESEE

SESE

ESESSE

M

M

MM

MM

K

KKK

k

kkK

Kkk

kkkk

Michaelis-Menten Mechanism

][,][

][][02maxmax E

S

SPkV

KV

dt

dV

M

maxmax

1

][

11

VV

K

VM

S

화학반응속도론 요약

속도의 표현

aA + bB cC + dD

dt

d

ddt

d

cdt

d

bdt

d

arate

]D[1]C[1]B[1]A[1

화학반응속도론 요약속도 법칙은 실험으로 정한다 => 반응차수와 속도상수

Order Rate LawConcentration-Time

Equation Half-Life

0

1

2

rate = k

rate = k [A]

rate = k [A]2

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

1[A]

=1

[A]0

+ kt

[A] = [A]0 - kt

t½ln2k

=

t½ =[A]0

2k

t½ =1

k[A]0

화학반응속도론 요약속도 상수와 온도 - Arrhenius equation

k = A • exp( -Ea/RT )

Ea is the activation energy (J/mol), 활성화 상수

R is the gas constant (8.314 J/K•mol)

T is the absolute temperature

A is the frequency factor, 잦음율

화학반응속도론 요약반응메카니즘과 기본 단계

단분자반응 A products rate = k [A]

이분자반응 A + B products rate = k [A][B]

이분자반응 A + A products rate = k [A]2

화학반응속도론 요약속도 상수와 촉매 , 효소 – 활성화 에너지의 변화Haber process

효소 반응 속도

퀴즈 1-1

350℃ 에서 일차반응의 속도상수가 4.60×10-4 s-

1 이다 . 활성화에너지가 104 kJ/mol 일 때 , 속도상수가 8.80×10-4 s-1 이 되는 온도를 계산하시오 .

R(gas constant) = 8.314 J/K•mol

(1) 644 K (2) 356 K (3) 371 K(4) 답 :________K문제와 답의 풀이를 쓰지 않으면 점수가 없다 .

퀴즈 1-2

온도가 295 K 에서 305 K 가 될 때 속도상수가 두 배가 되는 반응이 있다 . 이 반응의 활성화에너지를 계산하시오 .

R(gas constant) = 8.314 J/K•mol

(1) 52 kJ/mol (2) 105 kJ/mol (3) 190 kJ/mol(4) 답 :________ kJ/mol

답의 풀이를 쓰지 않으면 점수가 없다 .

퀴즈 1-3

일차반응에 대하여 반응물의 농도가 처음의 1/8 로 줄어드는 데 얼나마 걸리겠는가 ? 반감기 (t1/2) 로 답을 표시하고 , 또한 속도상수 k 로 답을 표시하시오 .

(1) 8 t1/2 (2) 4 t1/2 (3) 3 t1/2 (4) 답 :____ t1/2

(A) ln(8)/k (B) ln(4)/k (C) ln(2)/k (D) ln(2) k

답의 풀이를 쓰지 않으면 점수가 없다 .