Chapter 14 Kinetics Chapter 10 provided an introduction to kinetics and equilibrium. In this chapter...

Chapter 14Kinetics

Chapter 10 provided an introduction to kinetics and equilibrium. In this chapter we expand the quantitative treatment of chemical kinetics.

The Forces That Control a Chemical Reaction

● A large negative value of ΔG° for a reaction means that the products are thermodynamically favored over the reactants.

● It does not mean the reactants will convert to products.


● Why would the reactants not convert to products if the products are more stable?


● Why would the reactants not convert to products if the products are more stable?

The rate of conversion may be too slow to observe.

● Calculating and measuring this rate is part of the field of chemical kinetics.

Chemical Kinetics

● What is the rate of a reaction?● What factors influence the rate?● What is the mechanism of the reaction?

Chemical Kinetics

t

X

tt

XXRate

initialfinal

initialfinal

• Sucrose(aq) + H3O+(aq) → glucose(aq) + fructose(aq)

• Rate = k[sucrose]

• As reaction proceeds, the rate decreases.

Is the Rate of Reaction Constant?

● In excess base, phenolphthalein turns pink. Over time the pink color disappears.

● A solution is prepared which is initially 0.00500 M in phenolphthalein and 0.61 M in OH-(aq).

● The slow decrease in phenolphthalein concentration is shown in the next three slides.


Table 14.1


Figure 14.2

Instantaneous Rates of Reaction

Figure 14.3

Instantaneous Rates of Reaction

● Instantaneous rate of reaction:Rate = - dX/dt

Minus sign used for reactants so that rate > 0● Initial instantaneous rate of reaction is

measured at the moment of mixing

Rate Laws and Rate Constants

● In the phenolphthalein case, the rate is proportional to the current phenolphthalein concentration.

Rate = k[phenolphthalein]

k, the constant of proportionality, is called the rate constant.

The equation is called the rate law.

The Rate Law Versus the Stoichiometry of a Reaction

● The rate law for a reaction can’t be predicted from the stoichiometry of the reaction; it must be determined experimentally.


● CH3Br + OH- → CH3OH + Br-

Rate = k[CH3Br][OH-]

● (CH3)3CBr + OH- → (CH3)3COH + Br-

Rate = k’[(CH3)3CBr ]


● The expression for the rate will depend on the stoichiometric coefficients and whether reactants or products are being considered.

● For this reaction

2N2O5 → 4NO2 + O2

several equivalent expressions can be written.


2N2O5 → 4NO2 + O2

dt

d

dt

d

dt

dRate

)NO(

4

1)ON(

2

1)O( 2522

Order and Molecularity

● Molecularity Unimolecular: single molecule consumed

Bimolecular: two molecules consumed


● ClNO2 + NO → ClNO + NO2

Bimolecular

● N2O5 ⇄ NO2 + NO3

Forward reaction is unimolecular

Reverse reaction is bimolecular


● Reactions are also classified by their order.● ClNO2 + NO → ClNO + NO2

Rate = k[ClNO2][NO]– Overall second order reaction. First order in each

reactant

● N2O5 ⇄ NO2 + NO3

Rate = k[N2O5]– First order reaction


● Zero order?● Yes, reactions occurring on metal surfaces

can be zero-order:

)(O)(N2)(ON2 222 ggg Pt

Rate = k[N2O]0 = k’


● Molecularity describes what happens at the molecular level.

● The order of a reaction describes what happens at the macroscopic level.

● We deduce the molecularity from the experimentally determined rate law.

A Collision Theory of Chemical Reactions

● Chemical reactions occur as a result of collisions between molecules.

● The rate of a reaction depends on the frequency of the collisions.

● Consider again

ClNO2 + NO → ClNO + NO2


Figure 14.4


● If this picture is accurate, then the rate of the reaction should depend on the concentration of each reactant.

This is observed experimentally.


● Now consider again

(CH3)3CBr + OH- → (CH3)3COH + Br-

● Experimentally, the rate is determined to depend only on the (CH3)3CBr concentration.

No collisions?


● A multistep reaction is proposed to account for the experimental observations. 1. (CH3)3CBr → (CH3)3C+ + Br-

2. (CH3)3C+ + H2O → (CH3)3COH2+

3. (CH3)3COH2+ + OH- → (CH3)3COH + H2O

● The first step is slowest and is the rate-limiting step.

● The steps are the mechanism.

The Mechanisms of Chemical Reactions

● Main goal of chemical kinetics: determine a chemical reaction mechanism.

● Proposed mechanisms lead to predicted rate laws.

● These predicted laws are compared to the experimentally determined law.

Some mechanisms can be eliminated as possibilities.


● Mechanisms often propose intermediates.● In this mechanism

1. (CH3)3CBr → (CH3)3C+ + Br-

2. (CH3)3C+ + H2O → (CH3)3COH2+

3. (CH3)3COH2+ + OH- → (CH3)3COH + H2O

(CH3)3C+ and (CH3)3COH2+ are proposed

intermediates.


● What intermediate is proposed in this mechanism?

● What is the predicted rate law for this mechanism? The observed rate law is rate=k[NO]2[O2].

2NO ⇄ N2O2 (fast)

N2O2 + O2 → 2NO2 (slow)


● Does the agreement of the predicted and experimental rate laws prove the mechanism is correct?


● Does the agreement of the predicted and experimental rate laws prove the mechanism is correct?

No, only that the mechanism is consistent with the observed rate law.

Zero-Order Reactions

● Zero-order behavior for a reactant implies a multi-step mechanism.

● What about zero-order in all reactants?

Zero-Order Reactions

● Rate = k● Rate is zero order in all reactants as long as

there is excess NO.● Rate is determined by the availability of sites

on the Pt surface.

)(O)(N)(NO2 22 ggg Pt

Determining the Order of a Reaction From Rates of Reaction

● Consider the decomposition of HI

2HI → H2 + I2

● Record the initial instantaneous rate of reaction for a series of initial concentrations.

● From such a data set, the order of individual reactants can be determined.

Determining the Order of a Reaction From Rates of Reaction

Table 14.2

2HI → H2 + I2

The Integrated Form of Zero-, First-, and Second-Order Rate Laws

● Zero-Order● Rate = -d(X)/dt = k is the differential rate

law. It contains a derivative.● The integrated rate law is found by

integrating the differential rate law.● X(t) –X(0) = kt is the integrated rate law

for a zero-order reaction.


● First-order● -d(X)/dt = k(X)● Integrated form is

ln(X(t)) – ln(X(0)) = -kt


● Second-order in one reagent● -d(X)/dt = k(X)2

● Integrated form is

1/X(t) – 1/X(0) = kt


● Consider radioactive decay (more in next chapter)

● It is a first-order process.● Rate=k(amount of radioactive material)● After a period of time, t1/2, only half of the

radioactive material remains.This time, t1/2, is called the half-life of the radioactive material.


● Using the integrated rate law, the half-life for a first order reaction can be expressed as

1/ 2

ln(1/ 2)t

k


● Each radioactive isotope has a characteristic half-life.

● The half-life for 14C is 5730 years.


Table 14.3

Determining the Order of a Reaction with the Integrated Form of Rate Laws● The order of a reaction can be determined

by finding the functional form of concentration vs. time.

For zero-order, X(t) = -kt + X(0)

For first-order, ln(X(t)) = -kt + ln(X(0))

For second-order, 1/X(t) = kt + 1/X(0)

Determining the Order of a Reaction with the Integrated Form of Rate Laws

Figure 14.7


Figure 14.8


Figure 14.9

Reactions That Are First-Order in Two Reactants

● We have already seen

ClNO2 + NO → ClNO + NO2

with rate = k[ClNO2][NO].

None of these are appropriate:X(t) = -kt + X(0)

ln(X(t)) = -kt + ln(X(0))

1/X(t) = kt + 1/X(0).

Reactions That Are First-Order in Two Reactants

● Make the reaction into a pseudo-first-order reaction.

● If [ClNO2] >>[NO], [ClNO2] will not change much during the reaction: it will be essentially constant.

● Now, rate = k’[NO], where k’=k[ClNO2].A pseudo-first-order reactionln[NO(t)] = -k’t + ln[NO(0)]

The Activation Energy of Chemical Reactions

● Not every molecular collision results in the conversion of reactants to products.

● The molecules must collide with the proper orientation and with enough kinetic energy to overcome the activation energy, Ea.


Figure 14.15


● As the temperature of a reaction increases, more molecules have enough energy to overcome the activation energy.

Figure 14.16

Catalysts and the Rates of Chemical Reactions

● The rate of a chemical reaction can be increased by increasing the temperature.

● It can also be increased by adding a catalyst.


● Catalysts increase the rate of reaction.● They are consumed and regenerated in a

chemical reaction.● Only a small amount is needed.● Catalysts do not affect the equilibrium

position of a reaction (ΔG°).


● Catalysts increase the rate of reaction by providing a different mechanism which has a lower activation energy, Ea.

Ozone decomposes to molecular oxygen with or without a chlorine catalyst.

The mechanism with the chlorine has a lower Ea.


● Mechanism by which Cl atoms catalyze the decomposition of ozone in the stratosphere:

Cl + O3 → ClO +O2

ClO + O → Cl + O2

Overall reaction:

O3 + O → 2O2


● Iodide (I-) catalyzes the decomposition of hydrogen peroxide in water. So does platinum metal and the enzyme catalase.

2H2O2(aq) → 2H2O(l) + O2(g)

● Reaction will occur without a catalyst● The rate is significantly depressed.


Table 14.4


Figure 14.17


Figure 14.18

Determining the Activation Energy of a Reaction

● The rate constant depends on the activation energy, Ea, and the temperature, T.

● The equation is the Arrhenius equation:

122

1 11ln

TTR

E

k

k a

RTEaZeTk /)(

Determining the Activation Energy of a Reaction

● Determining k at 2 or more temperatures allows Ea to be calculated.

Figure 14.19

The Kinetics of Enzyme-Catalyzed Reactions

● Enzymes are proteins which catalyzed reactions in living systems.

● Rate = k[sucrose] for low concentration of sucrose when the reaction is catalyzed by an enzyme.

● At high sucrose concentration, the order goes to zero: rate = k’.

)(glucose)(fructose)(OH)(Sucrose 2 aqaqlaq Enzyme


Figure 14.20


● Explanation is similar to the zero-order behavior seen at metal-surface-catalyzed reactions.

At high concentration of sucrose, the reaction rate is limited by the available sites on the enzyme and independent of the high sucrose concentration.

Chapter 14 Kinetics Chapter 10 provided an introduction to kinetics and equilibrium. In this chapter...

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Transcript of Chapter 14 Kinetics Chapter 10 provided an introduction to kinetics and equilibrium. In this chapter...