Quiz #4 Next Week (Feb 12-16)

Covers sections 14.5-14.7

During lectures next week, we will complete ALL OFChapter 15 and parts of Chapter 16!

The Second Midquarter Examwill cover

Chapters 14, 15, and 16.1-16.3

Week six, a continuation of

Chapter 14 Chemical Kinetics

14.1 Factors that Affect Reaction Rates14.2 Reaction Rates14.3 Concentration and Rate

14.4 The Change of Concentration with Time

14.5 Temperature and Rate The Collision Model Activation Energy The Orientation Factor The Arrhenius Equation and Activation Energies

14.5 Reaction Mechanisms Elementary Steps; Multistep Mechanisms Rate Laws for Elementary Steps Rate Laws for Multistep Mechanisms

14.7 Catalysis Homogeneous and Heterogeneous Catalysis Enzymes

Note the DRAMATIC effect of temperature on k

Activation EnergyActivation Energy• Consider the rearrangement of acetonitrile:

– In H3C-NC, the C-NC bond bends until the C-N bond breaks and the NC 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 NC portion can continue to rotate forming a C-CN 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).

Activation EnergyActivation Energy• Consider the reaction between Cl and NOCl:

– 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.

Activation EnergyActivation Energy

The Arrhenius EquationThe Arrhenius Equation• 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.

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

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, then we can use

AT

Ek a ln

Rln

121

2 11

Rln

TT

E

k

k a

Look Familiar???

Sample Exercise 14.11

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

Sample Exercise 14.11

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

Sample Exercise 14.11

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

Sample Exercise 14.11

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 K)(8.314 J mol-1 K-1)(1 kJ / 1000 J)

= 1.6 x 102 kJ/mol

or 160 kJ/mol

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

121

2 11

Rln

TT

E

k

k a

To calculate k1 for a temperature of 430.0 K, make substitutionsfor all other parameters:Ea = 1.6 x 102 kJ/mol k2 = 2.52 x 10-5 s-1 T1 = 430.0 K T2 = 462.9 K

18.31

1000

0.430

1

9.462

1

/314.8

/160

1052.2ln

151

kJ

J

KKKmolJ

molkJ

sx

k

218.35

1 1015.41052.2

xe

x

kor

161521 100.1)1052.2)(1015.4( sxsxxkand

Activation Energy-orientation factorActivation Energy-orientation factor

RT

aEAek

Note the “A” term.

Note that termolecular reactions are extremely unlikely !

Multistep Mechanisms and Rate Laws

Overall: NO2 (g) + CO (g) NO (g) + CO2 (g)

with an observed rate law of Rate = k[NO2]2

It appears that at temperatures below 225 oC, the reaction proceeds via two elementary steps:

NO2 + NO2 NO3 + NO (1)NO3 + CO NO2 + CO2 (2)

Which yields two rateexpressions:

Rate1 = k1 [NO2]2

Rate2 = k2 [NO3][CO]

Important requirements:(a) Multi steps must add up to yield overall reaction.(b) may involve reactive intermediates (different from activated complexes).(c) One of these may be the ‘rate-determining’ step—the slower one!

In this case given above, step (1) is the rate-determining step and step (2) is a faster step.

Another example: 2 NO2 + F2 2 NO2F observed rate = k[NO2][F2]

Proposed mechanism:

NO2 + F2 NO2F + F (1) slowF + NO2 NO2F (2) fast

Sum gives overall reaction, and

rate = rate1 = k1[NO2][F2]

Another example: overall 2 NO + Br2 2 NOBr with an observed rate law of rate = k[NO]2[Br2]

The proposed mechanism is:

NO + Br2 == NOBr2 (1) (fast)

NOBr2 + NO 2 NOBr (2) (slow)

Our earlier guideline would suggest we use the slow step to determine therate law, giving rate = k2[NOBr2][NO]but this presents an immediate problem, since we don’t know what experimental quantities to put in for [NOBr2] !

The solution comes from an analysis of the reversible fast reaction (1).

rate forward = k 1[NO][Br2] and rate reverse = k -1[NOBr2]

but these exist in a fast, dynamic equilibrium where rate forward = rate reverse

And this gives us the relationship

rate forward = k 1[NO][Br2] = k -1[NOBr2] = rate reverse

and

][][

]][][[]][[

]][[][

22

21

1222

21

12

BrNOk

NOBrNOk

kkNONOBrkRate

finallyand

BrNOk

kNOBr

Quiz #4 Next Week (Feb 12-16)

Covers sections 14.5-14.7

During lectures next week, we will complete ALL OFChapter 15 and parts of Chapter 16!

The Second Midquarter Examwill cover

Chapters 14, 15, and 16.1-16.3

Somewhat more complicated: 2 O3 3 O2

obs rate = k[O3]2[O2] -1

Proposed mechanism:O3 = O2 + O (1) fast, reversible

O + O3 2 O2 (2) slow

tells us rate = k2[O][O3] now what???

assume k1[O3] = k -1[O2][O]

so that !!!

][

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2

3

21

31

O

Ok

Ok

OkO

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2

23

32

3

1

12

32

2

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O

OkO

O

O

k

kkrate

becomesOOkrateand

O

Ok

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OkO

Cl2 + CHCl3 HCl + CCl4 rate = k[Cl2]1/2[CHCl3]

Proposed mechanism:

Cl2 = 2 Cl (1) fast, reversible

Cl + CHCl3 HCl + CCl3 (2) slow

CCl3 + Cl CCl4 (3) fast

Evaluate the rate constant using this mechanism.

CatalysisA catalyst changes the rate of a chemical reaction

without being consumed.

• Homogeneous catalysis: catalyst and reaction are in the same, single phase.

• Heterogeneous catalysis: catalyst and reaction are in different phases. Often the catalyst is a solid in contact with gaseous or liquid reactions.

• Enzymes: In living systems, usually a large molecule which catalyzes a specific reaction, an enzyme-substrate complex.

• Homogeneous: catalyst and reaction are in same phase:• Hydrogen peroxide decomposes very slowly:

2H2O2(aq) 2H2O(l) + O2(g)

• In the presence of the bromide ion, the decomposition occurs rapidly:– 2Br-(aq) + H2O2(aq) + 2H+(aq) Br2(aq) + 2H2O(l).

– Br2(aq) is brown.

– Br2(aq) + H2O2(aq) 2Br-(aq) + 2H+(aq) + O2(g).

– Br- is a catalyst because it can be recovered at the end of the reaction and it makes the reaction rate faster.

• Generally, catalysts operate by lowering the activation energy for a reaction.

• Catalysts can operate by increasing the number of effective collisions.

• That is, from the Arrhenius equation: catalysts increase k be increasing A or decreasing Ea.

• A catalyst may add intermediates to the reaction.

• Example: In the presence of Br-, Br2(aq) is generated as an intermediate in the decomposition of H2O2.

• When a catalyst adds an intermediate, the activation energies for both steps must be lower than the activation energy for the uncatalyzed reaction.

Heterogeneous Catalysis• The catalyst is in a different phase than the reactants and

products. • Typical example: solid catalyst, gaseous reactants and

products (catalytic converters in cars).• Most industrial catalysts are heterogeneous.• First step is adsorption (the binding of reactant molecules

to the catalyst surface).• Adsorbed species (atoms or ions) may have increased

reactivity, but they are always easily available.• Reactant molecules are also adsorbed onto the catalyst

surface and may migrate to active sites.

– Consider the hydrogenation of ethylene:

C2H4(g) + H2(g) C2H6(g), Ho = -137 kJ/mol.– The reaction is slow in the absence of a catalyst.

– In the presence of a metal catalyst (Ni, Pt or Pd) the reaction occurs quickly at room temperature.

– First the ethylene and hydrogen molecules are adsorbed onto active sites on the metal surface.

– The H-H bond breaks and the H atoms migrate about the metal surface.

– When an H atom collides with an ethylene molecule on the surface, the C-C bond breaks and a C-H bond forms.

– When C2H6 forms it desorbs from the surface.

– When ethylene and hydrogen are adsorbed onto a surface, less energy is required to break the bonds and the activation energy for the reaction is lowered.

ENZYMES• Enzymes are biological catalysts.• Most enzymes are protein molecules with large molecular

masses (10,000 to 106 ).• Enzymes have very specific shapes.• Most enzymes catalyze very specific reactions.• Substrates are the reactants that undergo reaction at the

active site of an enzyme.• A substrate locks into an enzyme and a fast reaction

occurs.• The products then move away from the enzyme.

Considerable research is currently underway to modifyenzymes to prevent undesirable reactions and/or to prepare new products.

• Only substrates that fit into the enzyme lock can be involved in the reaction.

• If a molecule binds tightly to an enzyme so that another substrate cannot displace it, then the active site is blocked and the catalyst is inhibited (enzyme inhibitors).

• The number of events (turnover number) catalyzed is large for enzymes (103 - 107 per second).

Nitrogenasein legumesconverts

N2 to NH3.

‘Fixation’of N2’

converts itto

compoundsuseful to

plants

Consider the reaction 2 N2O5 4 NO2 + O2

Calculate Ea from the following data:

k/s-1 T/oC

2.0 x 10-5 207.3 x 10-5 302.7 x 10-4 409.1 x 10-4 502.9 x 10-4 60

Formic acid alone,

in the gas phase.

Formic acidin presence of

ZnO.

Sample Problem, page 562-563:

The decomposition of formic acid shown on the previous slide is given by HCOOH (g ) CO2 (g) + H2 (g)

It has been found to be first order at 838 K. see pp 614-615

a) Estimate the half-life and first-order rate constant for the decomposition of pure formic acid and formic acid in the presence of ZnO.

b) What is the effect of ZnO?

c) Suppose we express the concentration of formic acid in mol/L. What effect would that have on the rate constant?

d) The pressure of formic acid at the beginning of the reaction can be read from the graph. Assume constant T, ideal gas behavior, and a reaction volume of 436 cm3. How many moles of gas are in the container at the end of the reaction?

e) The standard heat of formation of formic acid vapor is ΔHof = -378.6

kJ/mol. Calculate ΔHo (see Sec. 5.7) for the overall reaction. Assume the activation energy, Ea , is 184 kJ/mol, sketch an approximate energy profile for the reaction, and label Ea, ΔHo

f , and the transition state.

Chapter 15 Chemical Equilibrium

15.1 The Concept of Equilibrium

15.2 The Equilibrium ConstantThe Magnitude of Equilibrium ConstantsThe Direction of the Chemical Equation and Keq

Other ways to Manipulate Chemical Equations and Keq

Units of Equilibrium Constants– Kp and Kc

15.3 Heterogeneous Equilibria

15.4 Calculating Equilibrium Constants

15.5 Applications of Equilibrium ConstantsPredicting the Direction of Reaction

Calculation of Equilibrium Concentrations

15.6 Le Châtelier’s PrincipleChange in Reactant or Product ConcentrationsEffects of Volume and Pressure ChangesEffect of Temperature ChangesThe Effect of Catalysts

The Equilibrium Constant

• Forward reaction:N2O4 (g) 2 NO2 (g)

• Rate law:Rate = kf [N2O4]

The Equilibrium Constant

• Reverse reaction:2 NO2 (g) N2O4 (g)

• Rate law:Rate = kr [NO2]2

What’s wrong (or not always correct) aboutthis last statement?

The Equilibrium Constant• Therefore, at equilibrium

Ratef = Rater

kf [N2O4] = kr [NO2]2

• Rewriting this, it becomes

kf

kr [NO2]2

[N2O4]=

Always correct

NOT always correct.

The Equilibrium Constant

The ratio of the rate constants is a constant at that temperature, and the expression becomes

Keq =kf

kr [NO2]2

[N2O4]=

The Equilibrium Constant

• To generalize this expression, consider the reaction

• The equilibrium expression for this reaction would be

Kc = [C]c[D]d

[A]a[B]b

aA + bB cC + dD