Chapter 12

Chemical Kinetics (

화학반응속도론

)

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 t 2 t 2 - t 1 = D [A] D t -Conc. of A at t 1  Change in concentration per unit time  For the reaction 2NO 2 (

g

) 2NO(

g

) + O 2 (

g

)

t a i o n C o n c t r e n

 As the reaction progresses the concentration of NO 2 goes down.

[NO 2 ]

Time

t a i o n C o n c t r e n

 As the reaction progresses the concentration of NO goes up.

[NO 2 ] [NO]

Time

t a i o n C o n c t r e n

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

[NO [O 2 ] 2 ] [NO]

Time

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.

t a i o n C o n c t r e n

 Average slope method D [NO D t 2 ]

Time

t a i o n C o n c t r e n

 Instantaneous slope method.

D [NO D t 2 ]

Time

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 2NO 2 (

g

) 2NO(

g

) + O 2 (

g

) D [NO 2 ] = D [NO] = 2 D [O 2 ] D t D t D 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 .

Two key points

 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.

예

) 2 NO 2 2 NO + O 2  You will find that the rate will only depend on the concentration of the reactants.

 Rate =

k

[NO 2 ]

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.

2 NO 2 2 NO + O 2  The rate of appearance of O 2 Rate' = D [O 2 ] =

k'

[NO 2 ] D t can be said to be.

 Because there are 2 NO 2 for each O 2  Rate = 2 ⅹ Rate‘ So

k

[NO 2 ]

n

So

k

= 2 ⅹ = 2 ⅹ

k'

[NO 2 ]

n k'

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.

12.3 Determining the Form of the Rate Laws ( 속도식의 유형 결정 )

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

 Must be determined from experimental data.

 For the reaction 2N 2 O 5 (aq) 4NO 2 (aq) + O 2 (g) The reverse reaction won’t play a role.

19

1.2

1 0.8

0.6

0.4



0.2

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

We will use the tangent method.

1.2

1 0.8

0.6

0.4

At .90 M the rate is (.98 - .76) = 0.22 =- 5.4x 10 0.2

(0-400) -400 -4 0

1.2

1 0.8

0.6

0.4

0.2

At .45 M the rate is (.52 - .31) = 0.22 =- 2.7 x 10 (1000-1800) -800 -4 0

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

 Rate = D [N 2 O 5 ] =

k

[N 2 O 5 ] 1 D t =

k

[N 2 O 5 ]  We say this reaction is first order (1 차 반응 ) in N 2 O 5

.

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

23

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.

An example

A. Determining the value of

n

( 반응 차수 ) 1. Example [A] Rate (mol/L·s) 1.00

M

8.4 ⅹ 10 -3 0.50

M

2.1 ⅹ 10 -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

a. In experiment 1 and 2, the concentration of NH 4 + held constant.

b. In experiment 2 and 3, the concentration of NO 2 held constant.

is is c. Basic rate law for the reaction: Rate =

k

[NH 4 + ]

n

[NO 2 ]

m

26

B. Calculations a. Experiment 1 and 2 (1) Rate doubles when concentration of NO 2 (2)

m

= 1 (first order) doubles b. Experiment 2 and 3 (1) Rate doubles when concentration of NH 4 + (2)

n

= 1 (first order) doubles 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

[NH 4 + ][NO 2 ] 27

D. Calculate

k

, the rate constant ( 속도상수 ) a. Rate is known.

b. Both concentrations are known.

c. Exponents are known.

1.35

ⅹ 10 -7 mol/L·s =

k

(0.100

M

)(0.0050

M

)

k

= 1.35

ⅹ 10 -7 mol/L∙s (0.100

M

)(0.0050

M

) 28

An exercise

 For the reaction BrO 3 + 5 Br + 6H + 3Br 2 + 3 H 2 O  The general form of the Rate Law is Rate =

k

[BrO 3 ]

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 = D [A]

n

D t  We will only work with

n

= 0, 1, and 2

Integrated First-Order Rate Law ( 적분 일차반응 속도식 ) (single reactant)

   For the reaction 2N 2 O 5 4NO 2 + O 2 We found the rate ( D [N 2 O 5 ]/ D t) =

k

[N 2 O 5 ] 1 If we integrate this equation with respect to time, we get the integrated rate law ln[N 2 O 5 ] = -

k

t + ln[N 2 O 5 ] 0 ln is the natural log [N 2 O 5 ] 0 is the initial concentration.

 General form Rate = D [A] / D t =

k

[A]  Integrated first-order rate law  ln[A] = -

k

t + ln[A] 0 In the form y = mx + b A plot of ln[N 2 O 5 ] versus time.

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

ln

   0  

= kt

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 = t 1/2 ln    0 0 2    = kt 1 2  ln(2) =

k

t 1/2 or t 1/2 = ln(2) /

k

 t 1/2 = 0.693/

k

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

A plot of [N 2 O 5 ] versus time for the decomposition reaction of N 2 O 5  An easy way to find

k

Integrated Second-Order Rate Law ( 적분 이차반응 속도식 ) (single reactant)

 Rate = D [A] / D t =

k

[A] 2  integrated rate law   1/[A] =

k

t + 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 time t you can calculate [A] at any

Second Order Half Life (2 차 반응 반감기 )

 [A] = [A] 0 /2 at t = t 1/2 1 [ A ] 0 2 = kt 1 2 + 1 [A] 0 1 [ A] 0 = kt 1 2 2 [ A] 0 1 [ A] 0 = kt 1 2 t 1 1 2 = k[A] 0

Example: For the reaction 2C

4

H

6

(g) C

8

H

12

(g)

(a) A plot of ln[C 4 H 6 ] versus

t

.

(b) A plot of 1/[C 4 H 6 ] versus

t

.

 (b) 의 그림으로부터 속도 = 이차반응 속도식 = D [C 4 H 6 ] / D t =

k

[C 4 H 6 ] 2  반응속도상수 = (b) 그래프의 기울기 = D (1/[C 4 H 6 ]) / D t = (481 – 100)L/mol = 6.14

ⅹ 10 -2 L/mol ∙ s (6200 – 0)s  이차반응의 반감기 = t 1/2 = 1/

k

[C 4 H 6 ] 0 = 1/(6.14

ⅹ 10 -2 L/mol ∙ s)(1.000

ⅹ 10 -2 mol/L)

Integrated Zero-Order Rate Law ( 적분 영차반응 속도식 ) (single reactant)

 Rate =

k

[A] 0 =

k

 Rate does not change with concentration.

   Integrated [A] = -

k

t + [A] 0 When [A] = [A] 0 /2, t = t 1/2 t 1/2 = [A] 0 /2

k

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 2N 2 O(

g

) 2N 2 (

g

) + O 2 (

g

) takes place on a platinum surface.

More Complicated Reactions ( 두 종류 이상의 반응물 )

 BrO 3 + 5 Br + 6H + 3Br 2 + 3 H 2 O  For this reaction we found the rate law to be Rate =

k

[BrO 3 ][Br ][H + ] 2  To investigate this reaction rate we need to control the conditions.

Rate =

k

[BrO

3 -

][Br

-

][H

+

]

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

[BrO 3 ] 0 = 1.0

ⅹ 10 -3 [Br ] 0 = 1.0 M M [H + ] 0 = 1.0 M As the reaction proceeds [BrO 3 ] changes noticeably. [Br ] and [H + ] don’t

Rate =

k

[BrO

3 -

][Br

-

][H

+

]

2  This rate law can be rewritten Rate =

k

[BrO 3 ][Br ] 0 [H + ] 0 2 Rate =

k

[Br ] 0 [H + ] 0 2 [BrO 3 ] Rate =

k’

[BrO 3 ]  This is called a pseudo first order ( 유사일차반응 ) rate law.



k k’

= [Br ] 0 [H + ] 0 2

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.

An example

 For the reaction 2NO 2 (g) + CO(g) 2NO(g) + CO 2 (g) Rate =

k

[NO 2 ] 2  The proposed mechanism is NO 2 (g) + NO 2 (g) NO 3 (g) + NO(g) (slow) ( 속도결정단계 ) NO 3 (g) + CO(g) NO 2 (g) + CO 2 (g) (fast)  NO 3 is called an intermediate ( 중간체 ) It is formed then consumed in the reaction.

 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.

E n e r g y P o t e n t i a l Ex)2BrNO(g) → 2NO(g)+ Br Reactants (

반응물

) 2 (g) Products (

생성물

)

Reaction Coordinate

E n e r g y P o t e n t i a l Reactants

Reaction Coordinate

Activation Energy E a (

활성화 에너지

) Products

E n e r g y P o t e n t i a l Activated complex or transition state ( Reactants

활성화물 또는 전이상태

) Products

Reaction Coordinate

E n e r g y P o t e n t i a l Reactants

Reaction Coordinate

Products

} D

E

Br---NO E n e r g y P o t e n t i a l 2BrNO Br---NO Transition State

Reaction Coordinate

2NO + Br 2

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) E a = 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

-E a /RT

= Ae

-E a /RT  A is called the frequency factor ( 잦음율 ) = zp  ln

k

= -(E a /R)(1/T) + ln A  ln

k

vs 1/T is a straight line.

An example

Plot of ln(

k

) versus 1/

T

for the reaction 2N 2 O 5 (

g

) 4NO 2 (

g

) + O 2 (

g

). The value of the activation energy for this reaction can be obtained from the slope of the line, which equals -Ea/

R

.

12.8 Catalysts ( 촉매 )

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

Enzymes ( 효소 ) are biological catalysts.

Homogenous Catalysts ( 균일 촉매 ) same phase as the reactants.

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

are in

How Catalysts Work

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

 New pathway has a lower activation energy.

How Catalysts Work

 More molecules will have this activation energy.

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 D E.

H H  Hydrogen bonds to surface of metal.

 Break H-H bonds H H H H Pt surface

H H C H C H H H Pt surface H H

 The double bond breaks and bonds to the catalyst.

H H H H C H C Pt surface H H H

 The hydrogen atoms bond with the carbon.

H H H H C H C Pt surface H H H

H H H C H H C H H H Pt surface

An example

The exhaust gases from an automobile engine are passed through a catalytic converter to minimize environmental damage.

Homogenous Catalysts

 Chlorofluorocarbons (CCl 2 F 2 :

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

t e R a  Rate increases until the active sites of catalyst are filled.

 Then rate is independent of concentration Concentration of reactants