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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 O2 goes up 1/2 as fast.

Concentration

[NO2]

[NO]

[O2]

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.

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.

- 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 .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

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.

(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 (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.

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

- 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.

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

12.6 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)

(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 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.

Activated complex or transition state(활성화물 또는 전이상태)

Potential

Energy

Reactants

Products

Reaction Coordinate

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/Tis 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.

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.

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

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

Rate

Concentration of reactants

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