Chapter 12
Sponsored Links
This presentation is the property of its rightful owner.
1 / 76

Chapter 12 PowerPoint PPT Presentation

  • Uploaded on
  • Presentation posted in: General

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 -Conc. of A at t 1 t 2 - t 1

Download Presentation

Chapter 12

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

Chapter 12

Chemical 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

    =D[A] Dt

  • Change in concentration per unit time

  • For thereaction

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

  • As the reaction progresses the concentration of NO2 goes down.




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





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






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





  • Instantaneous slope method.



D t


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)

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

    Dt Dt Dt

    ** 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 NO2 2 NO + O2

  • 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.

2 NO2 2 NO + O2

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

    Rate' = D[O2] = k'[NO2]Dt

  • Because there are 2 NO2 for each O2

  • Rate = 2 ⅹ Rate‘

    So k[NO2]n= 2ⅹk'[NO2]n

    So k = 2 ⅹk'

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

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

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

  • We will use the tangent method.

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

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

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

  • Rate = -D[N2O5] = k[N2O5]1 = k[N2O5] Dt

  • We say this reaction is first order (1차 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.

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 (2차 반응) with respect to A.

Rate = k[A]2

  • 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.

  • Basic rate law for the reaction:

  • Rate = k[NH4+]n[NO2-]m

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-]

  • D. Calculate k, the rate constant (속도상수)

  • Rate is known.

  • b. Both concentrations are known.

  • c. Exponents are known.

  • 1.35ⅹ10-7 mol/L·s = k(0.100 M)(0.0050 M)

  • k = 1.35ⅹ10-7 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


  • 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]nDt

  • 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 (-D[N2O5]/Dt) = 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.

  • General form

    Rate = - D[A] / Dt = 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.

A plot of ln[N2O5] versus time.

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

  • Often expressed in a ratio

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

  • 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

A plot of [N2O5] versus time for the

decomposition reaction of N2O5

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

  • Rate = -D[A] / Dt = 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

Second Order Half Life(2차 반응 반감기)

  • [A] = [A]0 /2 at t = t1/2

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

  • A plot of ln[C4H6] versus t.

  • A plot of 1/[C4H6] versus t.

  • (b)의 그림으로부터 = 이차반응 속도식

    속도 = -D[C4H6] / Dt = k[C4H6]2

  • 반응속도상수

    = (b) 그래프의 기울기 = -D(1/[C4H6]) / Dt

    = (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 aplatinum surface.

More Complicated Reactions(두 종류 이상의 반응물)

  • BrO3- + 5 Br- + 6H+ 3Br2 + 3 H2O

  • For this reaction we found the rate law to be

    Rate = k[BrO3-][Br-][H+]2

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

Rate = k[BrO3-][Br-][H+]2

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

  • [BrO3-]0= 1.0ⅹ10-3 M

  • [Br-]0 = 1.0 M

  • [H+]0 = 1.0 M

  • As the reaction proceeds [BrO3-] changes noticeably.

  • [Br-] and [H+] don’t

Rate = k[BrO3-][Br-][H+]2

  • This rate law can be rewritten

    Rate = k[BrO3-][Br-]0[H+]02

    Rate = k[Br-]0[H+]02[BrO3-]

    Rate = k’[BrO3-]

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

  • k =k’ [Br-]0[H+]02

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

    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)


    NO3(g) + CO(g) NO2(g) + CO2(g) (fast)

  • NO3 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.

Ex)2BrNO(g) → 2NO(g)+ Br2(g)







Reaction Coordinate



Activation Energy Ea

(활성화 에너지)



Reaction Coordinate

Activated complex or transition state(활성화물 또는 전이상태)





Reaction Coordinate







Reaction Coordinate





Transition State


2NO + Br


Reaction Coordinate


  • 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


  • 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/Tis a straight line.

An example

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.

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.

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 E.











Heterogenous Catalystsexample: the hydrogenationof ethylene

Heterogenous Catalystsexample: the hydrogenationof ethylene

  • Hydrogen bonds to surface of metal.

  • Break H-H bonds

Pt surface











Pt surface











  • The double bond breaks and bonds to the catalyst.

Pt surface











  • The hydrogen atoms bond with the carbon.

Pt surface











Pt surface

An example

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

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

  • Rate increases until the active sites of catalyst are filled.

  • Then rate is independent of concentration


Concentration of reactants

  • Login