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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

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Chapter 12

Chemical Kinetics

(화학반응속도론)


Kinetics
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
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)


Concentration

[NO2]

Time



Concentration

[NO2]

[NO]

[O2]

Time


Calculating rates
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.


Concentration

D[NO2]

Dt

Time


Concentration

D[NO2]

D t

Time


Defining rate
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
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.


2 no 2 2 no o 2
) 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 no 2 2 no o 21
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
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.



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



The method of initial rates
The method of Initial Rates ( the rate law must be..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
An example the rate law must be..

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



B. Calculations the rate law must be..

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 the rate law must be..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
An exercise the rate law must be..

  • 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.



12.4 The Integrated Rate Law the rate law must be..

(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]nDt

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


Integrated first order rate law single reactant
Integrated First-Order Rate Law the rate law must be.. (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 the rate law must be..

    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.



Half life of a first order reaction
Half-life of a First-Order Reaction the rate law must be..(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


  • t the rate law must be..1/2 = 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 single reactant
Integrated Second-Order Rate Law the rate law must be..(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 2
Second Order Half Life the rate law must be..(2차 반응 반감기)

  • [A] = [A]0 /2 at t = t1/2


Example for the reaction 2c 4 h 6 g c 8 h 12 g
Example the rate law must be..: For the reaction 2C4H6(g) C8H12(g)

  • A plot of ln[C4H6] versus t.

  • A plot of 1/[C4H6] versus t.


  • (b) the rate law must be..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 single reactant
Integrated Zero-Order Rate Law the rate law must be..(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 the rate law must be..:

  • 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
More Complicated Reactions the rate law must be..(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 bro 3 br h 2
Rate = the rate law must be..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 bro 3 br h 21
Rate = the rate law must be..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
12.5 Summary of Rate Laws the rate law must be..(속도식 요약)


12 6 reaction mechanisms
12.6 Reaction Mechanisms the rate law must be..(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 example1
An example the rate law must be..

  • 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.


  • Each of the two reactions is called an the rate law must be..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 the rate law must be.. 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
12.7 Collision theory the rate law must be..(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) the rate law must be..→ 2NO(g)+ Br2(g)

Potential

Energy

Reactants

(반응물)

Products

(생성물)

Reaction Coordinate


Potential the rate law must be..

Energy

Activation Energy Ea

(활성화 에너지)

Reactants

Products

Reaction Coordinate


Activated complex or transition state the rate law must be..(활성화물 또는 전이상태)

Potential

Energy

Reactants

Products

Reaction Coordinate


Potential the rate law must be..

Energy

}

Reactants

DE

Products

Reaction Coordinate


Br---NO the rate law must be..

Potential

Energy

Br---NO

Transition State

2BrNO

2NO + Br

2

Reaction Coordinate


Arrhenius
Arrhenius the rate law must be..

  • 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 the rate law must be..

    (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
Problems the rate law must be..

  • 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
Arrhenius Equation BrNO molecules. Orientations (a) and (b) can lead to a reaction, but orientation (c) cannot.

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 BrNO molecules. Orientations (a) and (b) can lead to a reaction, but orientation (c) cannot.

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
12.8 Catalysts ( BrNO molecules. Orientations (a) and (b) can lead to a reaction, but orientation (c) cannot.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
How Catalysts Work BrNO molecules. Orientations (a) and (b) can lead to a reaction, but orientation (c) cannot.

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

  • New pathway has a lower activation energy.


How catalysts work1
How Catalysts Work BrNO molecules. Orientations (a) and (b) can lead to a reaction, but orientation (c) cannot.

  • More molecules will have this activation energy.


How catalysts work2
How Catalysts Work BrNO molecules. Orientations (a) and (b) can lead to a reaction, but orientation (c) cannot.

  • 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 catalysts example the hydrogenation of ethylene

H BrNO molecules. Orientations (a) and (b) can lead to a reaction, but orientation (c) cannot.

H

H

H

H

H

H

H

H

H

Heterogenous Catalystsexample: the hydrogenationof ethylene

Heterogenous Catalystsexample: the hydrogenationof ethylene

  • Hydrogen bonds to surface of metal.

  • Break H-H bonds

Pt surface


H BrNO molecules. Orientations (a) and (b) can lead to a reaction, but orientation (c) cannot.

H

C

C

H

H

H

H

H

H

Pt surface


H BrNO molecules. Orientations (a) and (b) can lead to a reaction, but orientation (c) cannot.

H

H

H

C

C

H

H

H

H

  • The double bond breaks and bonds to the catalyst.

Pt surface


H BrNO molecules. Orientations (a) and (b) can lead to a reaction, but orientation (c) cannot.

H

H

H

C

C

H

H

H

H

  • The hydrogen atoms bond with the carbon.

Pt surface


H BrNO molecules. Orientations (a) and (b) can lead to a reaction, but orientation (c) cannot.

H

C

C

H

H

H

H

H

H

Pt surface


An example BrNO molecules. Orientations (a) and (b) can lead to a reaction, but orientation (c) cannot.

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


Homogenous catalysts
Homogenous Catalysts BrNO molecules. Orientations (a) and (b) can lead to a reaction, but orientation (c) cannot.

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

  • Enzymes regulating the body processes. (Protein catalysts)


Catalysts and rate
Catalysts and rate BrNO molecules. Orientations (a) and (b) can lead to a reaction, but orientation (c) cannot.

  • 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

Concentration of reactants


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