hydrophilic and hydrophobic colloids chapter 14, chemical ...mathews/chem122wi03/... · chapter 14,...

1

Chapter 14, Chemical Kinetics

(continued)

Quizzes next week will be on Chap 14through section 14.5.

Last week we covered the following material:

Review Vapor Pressure with two volatile components

13.6 ColloidsHydrophilic and Hydrophobic ColloidsRemoval of Colloidal Particles

Chapter 14 Chemical Kinetics

14.1 Reaction RatesRates in Terms of ConcentrationsReaction Rates and Stoichiometry

14.2 The Dependence of Rate on ConcentrationsReaction OrderUnits of Rate ConstantsUsing Initial Rates to Determine Rate Laws

14.3 The Change of Concentration with TimeFirst-Order ReactionsHalf-LifeSecond-Order Reactions

14.4 Temperature and RateThe Collision ModelActivation EnergyThe Orientation FactorThe Arrhenius Equation

14.5 Reaction MechanismsElementary StepsMultistep MechanismsRate Laws of Elementary StepsRate Laws of Multistep MechanismsMechanisms with and Initial Fast Step

Week Five Chemical Kinetics (cont)

Results for a reaction of A → B

C4H9Cl(aq) + H2O (l) C4H9OH (aq) + HCl (aq)

tOHHC

tClHCRate

∆∆

+=∆

∆−=

][][ 9494

Note the signs!

2

In fact, the instantaneous rate corresponds to d[A]/dt

Consider the reaction 2 HI(g) H2(g) + I2(g)

It’s convenient to define the rate as

tI

tH

tHIrate

∆∆

+=∆

∆+=

∆∆

−=][][][

21 22

And, in general for

aA + bB cC + dD

tD

dtC

ctB

btA

aRate

∆∆

=∆∆

=∆∆

−=∆∆

−=][1][1][1][1

Sample exercise 14.2

The decomposition of N2O5 proceeds according to the equation

2 N2O5 (g) 4 NO2 (g) + O2 (g)

If the rate of decomposition of of N2O5 at a particular instant in a vessel is 4.2 X 10-7 M/s, what is the rate of appearance of (a) NO2; (b) O2 ?

tO

tNO

tONRate

∆∆

+=∆

∆+=

∆∆

−=][

11][

41][

21 2252

i.e. the rate of the reaction is 2.1 x 10-7 M/s

the rate of appearance of NO2 is 8.4 x 10-7 M/s

and the rate of appearance of O2 is 2.1 x 10-7 M/s

2 N2O5 = 4 NO2 + O2 (g)

at T = 45 oC in carbon tetrachloride as a solvent

Time ∆t {N2O5] ∆[N2O5] - ∆[N2O5]/ ∆tmin min mol/L mol/L mol/L-min

0 2.33

184 2.08

319 1.91

526 1.67

867 1.35

1198 1.11

1877 0.72

2 N2O5 = 4 NO2 + O2 (g)


Time ∆t {N2O5] ∆[N2O5] - ∆[N2O5]/ ∆tmin min mol/L mol/L mol/L-min

0 2.33184 0.25 1.36 x 10-3

184 2.08135 0.17 1.26 x 10-3

319 1.91207 0.24 1.16 x 10-3

526 1.67341 0.32 0.94 x 10-3

867 1.35331 0.24 0.72 x 10-3

1198 1.11679 0.39 0.57 x 10-3

1877 0.72

The information on the previous slide is a bit of a nuisance, since the instantaneous rate keeps changing—and you know how much we like constant values or linear relationships!

So let’s try something rather arbitrary at this point.

Let’s divide the instantaneous, average rate by

[N2O5] and/or [N2O5]2

3

2 N2O5 = 4 NO2 + O2 (g)


{N2O5] [N2O5] ∆[N2O5] - ∆[N2O5]/ ∆t Avg rate Avg ratemol/L avg mol/L mol/L-min /[N2O5]av /[N2O5]av

2

2.332.21 0.25 1.36 x 10-3 6.2 x 10-4 2.8 x 10-4

2.082.00 0.17 1.26 x 10-3 6.3 x 10-4 3.2 x 10-4

1.911.79 0.24 1.16 x 10-3 6.5 x 10-4 3.6 x 10-4

1.671.51 0.32 0.94 x 10-3 6.2 x 10-4 4.1 x 10-4

1.351.23 0.24 0.72 x 10-3 5.9 x 10-4 4.8 x 10-4

1.110.92 0.39 0.57 x 10-3 6.2 x 10-4 6.7 x 10-4

0.72

Notice the nice constant value!!!

It’s convenient to write this result in symbolic form:

Rate = k [N2O5]

where the value of k is about 6.2 x 10-4

so that when [N2O5] = 0.221,

Rate = (6.2 x 10-4 )(0.221)

= 1.37 x 10-4 which is the ‘average rate’we started with

In fact, we really should take into account the 2 in front of theN2O5, in accordance with the rule we developed earlier.

This leads us to the general concept of Reaction Order

When Rate = k [reactant 1]m [reactant 2]n

we say the reaction is m-th order in reactant 1

n-th order in reactant 2

and (m + n)-th order overall.

Be careful—because these orders are NOT related necessarily

to the stoichiometry of the reaction!!!

2 N2O5 = 4 NO2 + O2 (g) Rate = k [N2O5] !!!

CHCl3 (g) + Cl2 (g) CCl4 (g) + HCl(g) Rate = k[CHCl3][Cl2]1/2

H2 (g) + I2 (g) 2 HI (g) Rate = k[H2][I2]

Other reactions and their observed reaction orders

The order must be determined experimentally!!!

We’ll see later that it depends on the Reaction Mechanism,rather than the overall stoichiometry.

Let’s explore the results for the result Rate = k [N2O5]

This can be expressed as Rate = - (∆[N2O5] / ∆ t = - d[N2O5] / dt = k [N2O5]

or, in general for A products

Rate = - ∆[A] / ∆t = d[A] / dt = k [A]

rearrangement and integration some time = t and t =0 gives the result

ln[A]t - ln[A]o = -kt

or ln [A]t = -kt + ln [A]o

or ln ([A]t/[A]o = - kt

This is the expression of concentration vs time

for a First-Order Reaction

An example of the plots of concentration vs timefor a First-Order Reaction

4

The Change of Concentration with TimeThe Change of Concentration with TimeHalfHalf--LifeLife• Half-life is the time taken for the

concentration of a reactant to drop to half its original value.

• That is, half life, t1/2 is the time taken for [A]0to reach ½[A]0.

• Mathematically,

kkt 693.0ln 2

1

21 =−=

The Change of Concentration with TimeThe Change of Concentration with TimeFor a FirstFor a First--Order ReactionOrder Reaction

The identical length of thefirst and second half-life

is a SPECIFIC characteristicof First-Order reactions

Consider now Second-Order Reactions

2][][][ AkdtAd

tARate −==∆∆

−=

∫∫ −=tA

Adtk

AAdt

0

][

][ 20 ][

][

0][1

][1

Akt

A t

+=

Example of Second-Order Plots of conc vs time

5

SecondSecond--Order ReactionsOrder Reactions• We can show that the half life

• A reaction can have rate constant expression of the form

rate = k[A][B],i.e., is second order overall, but has first order dependence on A and B.

[ ]0A1

21 k

t =

t1/2 = 0.693/0.4 = 1.73 sec

t1/2 = (k[A]0)-1 = [(0.4)(1.0)] -1= 2.5 sec

t1/2 = [(0.4)(0.5)] -1= 5.0 sec

General Order of reaction

First Order reactions

Second Order reactions

Integrated form of each

Half lives of each





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MQ1 Results:

Hi 175/175

Lo 36/175

Mean: 125 (71.4 %)

6

Chapter 14, Chemical Kinetics

(continued)







14.3 The Change of Concentration with Time

Mechanisms with and Initial Fast Step


Note the DRAMATIC effect of temperature on kTemperature and RateTemperature and Rate——Collision FrequencyCollision Frequency

The Collision Model The Collision Model egeg HH22 + I+ I22

The Collision ModelThe Collision Model• The greater the frequency of collision the faster

the rate for a given concentration.• Complication: not all collisions lead to products. In

fact, only a small fraction of collisions lead to product. Add an Orientation Factor.

• The higher the temperature, the more energyavailable to the molecules and the faster the rate.

• In order for reaction to occur the reactant molecules must collide in the correct orientation and with enough energy to form products.

Activation EnergyActivation Energy• Arrhenius: molecules must posses a

minimum amount of energy to react. Why?– In order to form products, bonds must be broken

in the reactants.– Bond breakage requires energy.

• Activation energy, Ea, is the minimum energy required to initiate a chemical reaction.

7

Activation EnergyActivation Energy Activation EnergyActivation Energy• Consider the rearrangement of acetonitrile:

– In H3C-N≡C, the C-N≡C bond bends until the C-N bond breaks and the N≡C portion is perpendicular to the H3C portion. This structure is called the activated complex or transition state.

– The energy required for the above twist and break is the activation energy, Ea.

– Once the C-N bond is broken, the N≡C portion can continue to rotate forming a C-C≡N bond.

H3C N CC

NH3C H3C C N

Activation EnergyActivation Energy• The change in energy for the reaction is the

difference in energy between CH3NC and CH3CN.

• The activation energy is the difference in energy between reactants, CH3NC and transition state.

• The rate depends on Ea.• Notice that if a forward reaction is exothermic

(CH3NC → CH3CN), then the reverse reaction is endothermic (CH3CN → CH3NC).

‘‘Effective Collisions’Effective Collisions’• Consider the reaction between Cl and NOCl

to produce Cl2 and NO - maybe:– If the Cl collides with the Cl of NOCl then the

products are Cl2 and NO.– If the Cl collided with the O of NOCl then no

products are formed.• We need to quantify this effect.

Effective & Ineffective CollisionsEffective & Ineffective Collisions

8

The Arrhenius Equation The Arrhenius Equation ---- FinallyFinally• Arrhenius discovered most reaction-rate data

obeyed the Arrhenius equation:

– k is the rate constant, Ea is the activation energy, R is the gas constant (8.314 J/K-mol) and T is the temperature in K.

– A is called the frequency factor, and is a measure of the probability of a favorable collision (including the collision frequency).

– Both A and Ea are specific to a given reaction.

RTaE

Aek−

=

RTaE

Aek−

=

The Arrhenius EquationThe Arrhenius Equation• If we have a lot of data, we can determine Ea

and A graphically by rearranging the Arrhenius equation:

• If we do not have a lot of data, or if we don’t know A, then we can use

ATEk a lnR

ln +−=

−

−=

121

2 11R

lnTT

Ekk a

Sample Exercise 14.8 For CH3NC CH3CN

Temp. /oC k / (s-1 )

189.7 2.52 x 10-5

198.9 5.25 x 10-5

230.3 6.30 x 10-5

251.2 3.16 x10-5


Temp. (oC) T (K) 1/T (K-1) k (s-1 ) ln k

189.7 2.52 x 10-5

198.9 5.25 x 10-5

230.3 6.30 x 10-5

251.2 3.16 x10-5


Temp. (oC) T (K) 1/T (K-1 x 103) k (s-1 ) ln k

189.7 462.9 2.160 2.52 x 10-5

198.9 472.1 2.118 5.25 x 10-5

230.3 503.5 1.986 6.30 x 10-5

251.2 524.4 1.907 3.16 x10-5

9


Temp. (oC) T (K) 1/T (K-1 x 103) k (s-1 ) ln k

189.7 462.9 2.160 2.52 x 10-5 -10.589

198.9 472.1 2.118 5.25 x 10-5 -9.855

230.3 503.5 1.986 6.30 x 10-5 -7.370

251.2 524.4 1.907 3.16 x10-5 -5.757

Fig 14.18 For CH3NC CH3CN reaction

From the graph we find the slope = -1.9 x 104 K

But this is also equal to - Ea/R

Or Ea = -(slope)(R)

= - ( - 1.9x104 )(8.314 J mol-1 K-1)(1 kJ / 1000 J)

= 1.62 x 102 kJ/mol

or 162 kJ/mol

We can now use these results to calculate the rate constant at any temperature.

−

−=

121

2 11R

lnTT

Ekk a

To calculate k1 for a temperature of 430.0 K, make substitutionsfor all other parameters, after choosing a starting point:

k2 = 2.52 x 10-5 s-1

T2 = 462.9 KAnd T1 = 430.0 K

to obtain k1 = 1.0 x 10-6 s -1

RTaE

Aek−

= Catalysts serve to reduce the activation energy, thus increasing k.

10

Reaction Mechanisms

Involve Elementary Steps – or Reactions

When we know these, we CAN write downdirectly the rate expressions!!!

Furthermore, these elementary steps represent fundamental molecular processes—

and lead to the term molecularity of the elementary reaction.

Elementary StepsElementary Steps• Elementary steps must add to give the

balanced chemical equation.• Intermediate: a species which appears in an

elementary step which is not a reactant or product.

Rate Laws of Elementary StepsRate Laws of Elementary Steps• The rate law of an elementary step is

determined by its molecularity:– Unimolecular processes are first order,– Bimolecular processes are second order, and– Termolecular processes are third order.

Rate Laws of Multistep MechanismsRate Laws of Multistep Mechanisms• Rate-determining step: is the slowest of the

elementary steps.• Therefore, the rate-determining step governs

the overall rate law for the reaction.Mechanisms with an Initial Fast StepMechanisms with an Initial Fast Step• It is possible for an intermediate to be a

reactant.• Consider

2NO(g) + Br2(g) → 2NOBr(g)

Reaction MechanismsReaction MechanismsMechanisms with an Initial Fast StepMechanisms with an Initial Fast Step

2NO(g) + Br2(g) → 2NOBr(g)• The experimentally determined rate law is

Rate = k[NO]2[Br2]• Consider the following mechanism

for which the rate law is (based on Step 2):

NO(g) + Br2(g) NOBr2(g)k1

k-1

NOBr2(g) + NO(g) 2NOBr(g)k2

Step 1:

Step 2:

(fast)

(slow)

Mechanisms with an Initial Fast StepMechanisms with an Initial Fast Step• The rate law is (based on Step 2):

Rate = k2[NOBr2][NO]• The rate law should not depend on the

concentration of an intermediate because intermediates are usually unstable.

• Assume NOBr2 is unstable, so we express the concentration of NOBr2 in terms of NOBrand Br2 assuming there is an equilibrium in step 1 we have

[ ] ]Br][NO[NOBr 21

12

−=

kk

Mechanisms with an Initial Fast StepMechanisms with an Initial Fast Step• By definition of equilibrium (see Chap 15):

k1[NO][Br2] = k-1[NOBr2]• Therefore, the overall rate law becomes

• Note the final rate law is consistent with the experimentally observed rate law.

]Br[]NO[Rate 22

11

2−

=kkk

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