chapter 12 chemical kinetics ( 화학반응속도론 ). kinetics l 반응속도의 이해 l...

Chapter 12

Chemical KineticsChemical Kinetics

(( 화학반응속도론화학반응속도론 ))

Kinetics

반응속도의 이해

자발적인 반응이 빠른 반응을 의미하지는 않는다 .

예 ) 다이아몬드가 자발적으로 흑연이 되는 반응은 매우 느리다 .

반응 메커니즘의 이해 – 반응의 진행 단계 이해

12.1 Reaction Rate ( 반응속도 )

Rate( 속도 ) = Conc. of A at t2 -Conc. of A at t1

t2- t1

=[A] t

Change in concentration per unit time

For the reaction

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

As the reaction progresses the concentration of NO2 goes down.

Concentration

Time

[NO[NO22]]

As the reaction progresses the concentration of NO goes up.

Concentration

Time

[NO][NO]

[NO[NO22]]

As the reaction progresses the concentration of O2 goes up 1/2 as fast.

Concentration

Time

[O[O22]]

[NO][NO]

[NO[NO22]]

Calculating Rates (속도계산법 )

Average rates ( 평균속도 ) are taken over long intervals.

Instantaneous rates ( 순간속도 ) are determined by finding the slope of a line tangent to the curve at any given point because the rate can change over time.

Average slope method

Concentration

Time

[NO[NO22]]

tt

Instantaneous slope method.

Concentration

Time

[NO[NO22]]

tt

Defining Rate ( 속도의 정의 )

We can define rate in terms of the disappearance of the reactant or in terms of the rate of appearance of the product.

In our example

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

-[NO2] = [NO] = 2[O2] t t t

** Choice of rate law depends on what data is easiest to collect.

12.2 Rate Laws (반응속도식 )

Reactions are reversible ( 가역적 ). As products accumulate they can begin to turn

back into reactants ( 역반응 ). Early on the rate will depend on only the amount

of reactants present. We want to measure the reactants as soon as

they are mixed. This is called the initial rate method.

The concentration of the products do not appear in the rate law because this is an initial rate.

The order must be determined experimentally,

** can’t be obtained from the equation.

Two key points

You will find that the rate will only depend on the concentration of the reactants.

Rate = k[NO2]n

This is called a rate law ( 속도식 ) expression.

k is called the rate constant ( 속도상수 ).

n is the order ( 차수 ) of the reactant -usually a positive integer.

예 ) 2 NO2 2 NO + O2

The rate of appearance of O2 can be said to be.

Rate' = [O2] = k'[NO2] t

Because there are 2 NO2 for each O2

Rate = 2 ⅹ Rate‘

So k[NO2]n = 2ⅹk'[NO2]n

So k = 2 ⅹ k'

2 NO2 2 NO + O2

Types of Rate Laws ( 속도식의 종류 )

Differential rate law ( 미분속도식 ) - describes how rate depends on concentration.

Integrated rate law ( 적분속도식 ) - describes how concentration depends on time.

For each type of differential rate law there is an integrated rate law and vice versa.

Rate laws can help us better understand reaction mechanisms.

The first step is to determine the form of the rate law (especially its order).

Must be determined from experimental data. For the reaction

2N2O5 (aq) 4NO2 (aq) + O2(g)The reverse reaction won’t play a role.

12.3 Determining the Form of the Rate Laws

( 속도식의 유형 결정 )

19

0

0.2

0.4

0.6

0.8

1

1.2

0 200

400

600

800

1000

1200

1400

1600

1800

2000

To find rate we have to find the slope at two points.

We will use the tangent method.

0

0.2

0.4

0.6

0.8

1

1.2

0 200

400

600

800

1000

1200

1400

1600

1800

2000

At .90 M the rate is (.98 - .76) = 0.22 =- 5.4x 10 -4 (0-400) -400

0

0.2

0.4

0.6

0.8

1

1.2

0 200

400

600

800

1000

1200

1400

1600

1800

2000

At .45 M the rate is (.52 - .31) = 0.22 =- 2.7 x 10 -4 (1000-1800) -800

23

Since the rate at twice the concentration is twice as fast the rate law must be..

Rate = -[N2O5] = k[N2O5]1 = k[N2O5] t

We say this reaction is first order (1 차 반응 ) in

N2O5.

The only way to determine order is to run the experiment.

The method of Initial Rates (초기속도법 )

This method requires that a reaction be run several times.

The initial concentrations of the reactants (반응물의 초기 농도 ) are varied.

The reaction rate is measured just after the reactants are mixed.

Eliminates the effect of the reverse reaction.

A. Determining the value of n ( 반응 차수 ) 1. Example

[A] Rate (mol/L·s) 1.00 M 8.4 10ⅹ -3

0.50 M 2.1 10ⅹ -3

☞ Doubling the concentration of species A quadruples the rate of the reaction. Therefore, the reaction is of second order (2 차 반응 ) with respect to A. Rate = k[A]2

An example

26

a. In experiment 1 and 2, the concentration of NH4+ is

held constant.b. In experiment 2 and 3, the concentration of NO2

- is held constant.

c. Basic rate law for the reaction: Rate = k[NH4

+]n[NO2-]m

27

B. Calculations

a. Experiment 1 and 2 (1) Rate doubles when concentration of NO2

- doubles (2) m = 1 (first order) b. Experiment 2 and 3 (1) Rate doubles when concentration of NH4

+ doubles (2) n = 1 (first order)

C. Overall reaction order ( 전체반응차수 )

a. Overall reaction order is the sum of m and n (1) m +n = 2 so overall the reaction is second order Rate = k[NH4

+][NO2-]

28

D. Calculate k, the rate constant ( 속도상수 )

a. Rate is known.b. Both concentrations are known.c. Exponents are known.

1.35 10ⅹ -7 mol/L·s = k(0.100 M)(0.0050 M)

k = 1.35 10ⅹ -7 mol/L∙s (0.100 M)(0.0050 M)

An exercise

For the reaction BrO3-

+ 5 Br- + 6H+ 3Br2 + 3 H2O

The general form of the Rate Law is

Rate = k[BrO3-]n[Br-]m[H+]p

We use experimental data to determine the values of n, m, and p.

Now we have to see how the rate changes with concentration

12.4 The Integrated Rate Law ( 적분속도식 )

Expresses the reaction concentration as a function of time.

Form of the equation depends on the order of the rate law (differential).

Changes Rate = [A]n t

We will only work with n = 0, 1, and 2

Integrated First-Order Rate Law ( 적분 일차반응 속도식 )

(single reactant) For the reaction 2N2O5 4NO2 + O2

We found the rate (-[N2O5]/t) = k[N2O5]1

If we integrate this equation with respect to time, we get the integrated rate law

ln[N2O5] = - kt + ln[N2O5]0

ln is the natural log

[N2O5]0 is the initial concentration.

A plot of ln[N2O5] versus time.

General form

Rate = - [A] / t = k[A] Integrated first-order

rate law

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

In the form y = mx + b

y = ln[A], m( 기울기 ) = -k

x = t, b( 절편 ) = ln[A]0

A graph of ln[A] vs time is a straight line.

By getting the straight line you can prove it is first order

Often expressed in a ratio

lnA

A = kt0

Half-life of a First-Order Reaction( 일차반응의 반감기 )

• The time required to reach half the original concentration.

• If the reaction is first order

[A] = [A]0/2 when t = t1/2

ln

A

A = kt0

01 2

2

ln(2) = kt1/2 or t1/2 = ln(2) / k

A plot of [N2O5] versus time for the decomposition reaction of N2O5

t1/2 = 0.693/k

The time to reach half the original concentration does not depend on the starting concentration.

An easy way to find k

Integrated Second-Order Rate Law( 적분 이차반응 속도식 )

(single reactant)

Rate = -[A] / t = k[A]2 integrated rate law

1/[A] = kt + 1/[A]0

y= 1/[A], m( 기울기 ) = k, x= t, b( 절편 ) = 1/[A]0 A straight line if 1/[A] vs t is graphed Knowing k and [A]0 you can calculate [A] at any

time t

[A] = [A]0 /2 at t = t1/2

021

0 [A]

1 + kt =

2]A[1

2100

kt = A][

1 -

A][

2

0

1 = 21 k[A]

t1

[A] = kt

01 2

Second Order Half Life(2 차 반응 반감기 )

Example: For the reaction 2C4H6(g) C8H12(g)

(a) A plot of ln[C4H6] versus t.

(b) A plot of 1/[C4H6] versus t.

(b) 의 그림으로부터 = 이차반응 속도식 속도 = -[C4H6] / t = k[C4H6]2

반응속도상수 = (b) 그래프의 기울기 = -(1/[C4H6]) / t

= (481 – 100)L/mol = 6.14ⅹ10-2L/mol ∙ s

(6200 – 0)s

이차반응의 반감기 = t1/2 = 1/k[C4H6]0

= 1/(6.14ⅹ10-2L/mol ∙ s)(1.000ⅹ10-2mol/L)

Integrated Zero-Order Rate Law( 적분 영차반응 속도식 )

(single reactant)

Rate = k[A]0 = k Rate does not change with

concentration. Integrated [A] = -kt + [A]0

When [A] = [A]0/2, t = t1/2

t1/2 = [A]0/2k

영차반응의 예 : Most often when reaction happens on a surface

because the surface area stays constant. Also applies to enzyme( 효소 ) chemistry.

The decomposition reaction 2N2O(g) 2N2(g) + O2(g) takes place on a platinum surface.

More Complicated Reactions( 두 종류 이상의 반응물 )

BrOBrO33-- + 5 Br + 5 Br-- + 6H + 6H++ 3Br 3Br22 + 3 H + 3 H22OO

For this reaction we found the rate law to beFor this reaction we found the rate law to be

Rate = Rate = kk[BrO[BrO33--][Br][Br--][H][H++]]22

To investigate this reaction rate we need to To investigate this reaction rate we need to control the conditions. control the conditions.

Rate = Rate = kk[BrO[BrO33--][Br][Br--][H][H++]]22

We set up the experiment so that two of the reactants are in large excess.

[BrO[BrO33--]]00= 1.0= 1.0ⅹ1010-3-3 M M

[Br[Br--]]0 0 = 1.0 M= 1.0 M [H[H++]]0 0 = 1.0 M= 1.0 M As the reaction proceeds [BrOAs the reaction proceeds [BrO33

--] changes ] changes

noticeably. noticeably. [Br[Br--] and [H] and [H++] don’t] don’t

This rate law can be rewritten Rate = Rate = kk[BrO[BrO33

--][Br][Br--]]00[H[H++]]0022

Rate = Rate = kk[Br[Br--]]00[H[H++]]0022[BrO[BrO33

--]]

Rate = Rate = k’k’[BrO[BrO33--]]

This is called a pseudo first order (This is called a pseudo first order (유사일차반응유사일차반응 ) rate law.) rate law.

k k == k’k’

[Br[Br--]]00[H[H++]]0022

Rate = Rate = kk[BrO[BrO33--][Br][Br--][H][H++]]22

12.5 Summary of Rate Laws( 속도식 요약 )

12.6 Reaction Mechanisms( 반응 메커니즘 )

The series of steps that actually occur in a chemical reaction.

Kinetics can tell us something about the mechanism.

A balanced equation does not tell us how the reactants become products.

For the reaction 2NO2(g) + CO(g) 2NO(g) + CO2(g) Rate = k[NO2]2

The proposed mechanism is NO2(g) + NO2(g) NO3(g) + NO(g) (slow) ( 속도결정단계 ) NO3(g) + CO(g) NO2(g) + CO2(g) (fast)

NO3 is called an intermediate ( 중간체 ) It is formed then consumed in the reaction.

An example

Each of the two reactions is called an elementary step ( 단일단계 반응 ).

The rate for a reaction can be written from its molecularity( 분자도 ) .

Molecularity is the number of pieces that must come together.

Unimolecular step ( 일분자 반응 ) involves one molecule - Rate is first order.

Bimolecular step ( 이분자 반응 ) requires two molecules - Rate is second order.

Termolecular step ( 삼분자 반응 ) requires three molecules - Rate is third order

Termolecular steps are almost never heard of because the chances of three molecules coming into contact at the same time are miniscule.

12.7 Collision theory( 충돌 모형 )

Molecules must collide to react. Concentration affects rates because

collisions are more likely. Must collide hard enough. Temperature and rate are related. Only a small number of collisions

produce reactions.

Potential Energy

Reaction Coordinate

Reactants

( 반응물 )

Products

( 생성물 )

Ex)2BrNO(g) → 2NO(g)+ Br2(g)

Potential Energy

Reaction Coordinate

Reactants

Products

Activation Energy Ea

( 활성화 에너지 )

Potential Energy

Reaction Coordinate

Reactants

Products

Activated complex or transition state ( 활성화물 또는 전이상태 )

Potential Energy

Reaction Coordinate

Reactants

ProductsE}

Potential Energy

Reaction Coordinate

2BrNO

2NO + Br

Br---NO

Br---NO

2

Transition State

Arrhenius

Said the at reaction rate should increase with temperature.

At high temperature more molecules have the energy required to get over the barrier.

The number of collisions with the necessary energy increases exponentially.

Number of collisions with the required energy

( 활성화 에너지 이상의 에너지를 갖는 충돌수 )

= ze -Ea/RT

z = total collisions ( 총 충돌수 ) e is Euler’s number (opposite of ln) Ea = activation energy ( 활성화 에너지 ) R = ideal gas constant ( 기체상수 ) T is temperature in Kelvin

Problems

Observed rate is less than the number of collisions that have the minimum energy.

( 모든 분자 충돌이 화학반응을 일으키지는 않는다 .)

Due to Molecular orientation ( 충돌순간의 분자배향 )

written into equation as p the steric factor (입체인자 ).

Several possible orientations for a collision between two BrNO molecules. Orientations (a) and (b) can lead to a

reaction, but orientation (c) cannot.

Arrhenius Equation

k = zpe -Ea/RT = Ae -Ea/RT

A is called the frequency factor ( 잦음율 ) = zp

ln k = -(Ea/R)(1/T) + ln A

ln k vs 1/T is a straight line.

Plot of ln(k) versus 1/T for the reaction 2N2O5(g) 4NO2(g) + O2(g). The value of the activation energy for this reaction can be obtained from the slope of the line, which equals -Ea/R.

An example

12.8 Catalysts ( 촉매 )

Speed up a reaction without being used up in the reaction.

Enzymes ( 효소 ) are biological catalysts. Homogenous Catalysts ( 균일 촉매 ) are in the

same phase as the reactants. Heterogeneous Catalysts ( 불균일 촉매 ) are in

a different phase as the reactants.

How Catalysts Work Catalysts allow reactions to

proceed by a different mechanism - a new pathway.

New pathway has a lower activation energy.

More molecules will have this activation energy.

How Catalysts Work

How Catalysts Work

Catalysts allow reactions to proceed by a different mechanism - a new pathway.

New pathway has a lower activation energy.

More molecules will have this activation energy.

Do not change E.

Pt surface

HH

HH

HH

HH

Hydrogen bonds to surface of metal.

Break H-H bonds

Heterogenous Catalystsexample: the hydrogenation of ethylene

HH

Heterogenous Catalystsexample: the hydrogenation of ethylene

Pt surface

HH

HH

C HH C

HH

Pt surface

HH

HH

C HH C

HH

The double bond breaks and bonds to the catalyst.

Pt surface

HH

HH

C HH C

HH

The hydrogen atoms bond with the carbon.

Pt surface

H

C HH C

HH

H HH

The exhaust gases from an automobile engine are passed through a catalytic converter to

minimize environmental damage.

An example

Homogenous Catalysts

Chlorofluorocarbons (CCl2F2: 프레온 -12) catalyze the decomposition of ozone.

Enzymes regulating the body processes. (Protein catalysts)

Catalysts and rate

Catalysts will speed up a reaction but only to a certain point.

Past a certain point adding more reactants won’t change the rate.

Zero Order

Concentration of reactants

Rate

Rate increases until the active sites of catalyst are filled.

Then rate is independent of concentration