Kinetics Kotz Chap 15

Reaction:

Hydrolysis of proteins (connected through amide linkages)

In your body, this reaction is mostly controlled by enzymes.

The rate in neutral water is about rate constant 10^-6 / s

(J. Am. Chem. Soc. 1988: 7529), i.e.: ½ life 7 years

By contrast, reactions occur 10³ in vivo at 10⁷ / sec with enzymes (Lehninger)

i.e.: ½ life 10^-5 sec

Some interesting statistics

(Deepak Chopra, Ageless Body, Timeless Mind), 1993

Skin completely replaced each month

Stomach lining each 5 days

Liver 6 weeks

Skeleton 3 months

By the end of this year, 98% of the atoms in your body will have been replaced!

N.B. Thre are only very few "errors" per millions of reactions when amino acids incorporated into protein

Your body does a lot of chemistry: it controls 6 x 10¹² reactions/cell/s

insert graph(From notes 1996)

Thermodynamics tell you about the relative stability / energy of reactants & products H_reaction = H(initial) - H(final)

That is, will a reaction actually occur (if a mechanistic pathway exists)

On the other hand, Kinetics tells you about the reaction rate / reaction pathway

The rates depend not on the stability of the product, but on the height of the barrier to the reaction H^‡ (barrier to reaction)

We are answer the following questions:

I) How to measure the rate of a reaction?

II) How do we interpret the measured reaction rate in molecular terms?

III) Can we make use of rates to understand reaction pathway? (YES)

Typical classes of reactivity

-

- react at the diffusion controlled limit (as fast as molecules can migrate to one another)

Radicals e.g., CH₃ + CH₃  H₃CCH₃

- similarly react at the diffusion controlled limit (as fast as molecules can migrate to one another)

Covalent compounds react very slowly, by comparison

Range of reaction

Millions of years -- femtosecond (10^-12 sec) scale

[Demo

Ionic

I) Fe³⁺ + SCN ® Fe(SCN)²⁺

II) Bu²⁺ + CO₃^2- ® BaCO₃

III) Cu²⁺ + 4NH₃ ® Cu(NH₃)₄²⁺

Covalent

IV) NaI + EtBr  CH₃CH₂I + NaBr(s)

V) 2 H₂ g + O₂ g ® 2 H₂O (note, in the absence of a catalyst, this reaction is very slow)]

Experimental measurement of Reaction Rates.

We shall compare, in this course, only bimolecular (two molecules come together to make products) and unimolecular reaction (a molecule spontaneously does something)

To follow the reaction, can use either

I) the rate at which reagents disappear [NaI] or [CH₃CH₂Br] (reagents)

OR

II) or the rate at which products are formed [NaBr] or [CH₃CH₂I]

NaI + CH₃CH₂Br  NaBr + CH₃CH₂I ()

If you want to measure a rate - how to do it? Need some independent news to determine instantaneous concentration (in this reaction, NaBr is insoluble in the solvent, we can follow the reaction by watching precipitate formation)

Measure: [CH₃CH₂I] + [CH₃CH₂Br]

or rate of appearance of new products [NaBr] or [CH₃CH₂I]

An inaccurate, "back of the envelope" of the instantaneous slope can be had by taking the difference between two adjacent peaks.

C_1,2 (concentration at time t) t_1,2

By convention

-reaction rate of the forward reaction (as drawn) is positive

A more accurate way to look at the rate is to take the slope of the tangent to the curve - this is the instantaneous rate

Typically reactions start fast and then slow down - why?

What is happening

A + B ® C

as [A] ¯ statistical chance of hitting B ¯

The Experimental Rate Law:

Use NaI + EtBr  NaBr(s) + EtI

Rate of reaction may be defined in terms of any of the reagents/products. Since only complete molecules can react (can't have only part molecules, unlike part marks), the reagents and products are related by interger values, for the reaction shown above

The rate = -d[I-] / dt = -d[EtBr] / dt (rate of loss of reagents)

= +d[NaBr] / dt = +d[EtI] / dt (= rate of production of products)

One way to determine the integer relationships between the reagents/products is to run several different experiments keeping one species CONSTANT and we vary the concentration of the other

e.g. [EtBr] constant, & vary [NaI] or vice versa

Keep temperature constant (why?)(see below)

Expt	Conc.		Initial Rate
	[EtBr]	[NaI]	+d[NaBr(s)] / dt
1	1.0 M	1.0 M	3.0 X 10-2 mol/LS
2	1.0 M	0.5 M	1.5 x 10-2
3	2.0 M	1.0 M	6.0 x 10-2

The difference betwen reaction 1 and 2 is that the [NaI] was halved. The difference between 1 and 3 is that [EtBr] was doubled.

It can be seen that the rate is directly proporational to both [NaI] and [EtBr].

We can thus write a rate law that shows how the rate is related to concentration.

Reaction Rate = k [NaI]¹ [EtBr]¹ = k [NaI] [EtBr]

k (note small k, not K the equilibrium constant) is the "specific rate constant" a number that is specific to a given reaction at a given temperature. The bigger the number, the faster the reaction. At the same time, the equation shows that the higher the concentration the faster the concentration.

The exponent to the right of the concentration is called the "order" of reaction for that reagent.

i.e. Reaction is first order is NaI, Reaction in first order is EtBr

The overall Reaction Order is the sum of the exponents.

Take sum of exponents in the concentrations in rate law and add

-d[NaI] / dt = k [NaI]¹ [EtBr]¹

Overall order = 2 (1+1) a second reaction order may define the order for each species.

Since we have give the law in terms of disappearing reactants, the rate is preceeded by -

ie. -d [EtBr] / dt

Products appearing are +ve rates

ie. +d [NaBr] / dt

e.g. NaBr appearing

Different proportions

For example:

* We also choose d[ ] / dt term for one mole

ie. If A + 3B ® 4C

-d[A] / dt = - 1/3 × d[B] / dt = 1/4 d[C] / dt

The loss of B goes three time faster than the loss of A. The loss of A is four times faster than the production of C.

How to treat reactions of different order

Rate = -d[D] / dt = k [D] (i.e., a first order reaction)

This reaction actually is unimolecular (first order) (Since the rate is in mol/Ls, the units of k will be /s)

Does stoichimoetry always reflect rate equation??? NO!!!!!!!

Take a different reaction. The stoichimetry shows a 1:1 relationship between reactants, but a 1:2 relationship of products and reactants. What is going on? Look at the initial rates.

We would predict that rate = +d[N₂] / dt = -1/2 d[NO] / dt = k [NO]²[H₂]²

2NO(g) + 2H₂(g) ® N₂(g) + 2H₂O

	[NO]	[H2]	Rate of Decrease of [NO]
1	0.1	0.01	0.062 mol/L×sec
2	0.1	0.04	0.246
3	0.3	0.01	0.558

The rates between 1 and 2 show that the reaction rate is directly proportional to [H₂] not to [H₂]²

The rates between 1 vs 3 show that a tripling of [NO] leads to a 9 fold increase [3]² . Thus, the Rate is proportional to [NO]²

The rate law is then clear

Rate = k[NO]²[H₂]

ie. The rate law is not necessarily the same as the stoichiometry of the reaction - why? Rate depends on mechanism (the observed transformation may involve a complex sequence of steps rather than a "one-shot" transformation)

Another e.g.

CH₃CHO(g) ® CH₄(g) + CO(g)

Our prediction from the stoichiometry is

Rate = k[CH₃CHO]¹ not dependent upon products

entry	[CH3CHO]	rate (mol/L×s)
1	0.1M	0.085
2	0.2M	0.34
3	0.3M	0.76
4	0.4M	1.4

From entry 1 and 2, a doubling of the concentration leads to a quadrupling of the rate, thus the rate law

Rate = -d[CH₃CHO] / dt = k [CH₃CHO]²

Since the units of rate are still mol/L s, the units of k must be L/mol s (L/mol s x mol²/L² = mol/L s)

Here's an ugly one:

CHCl₃(g) + Cl₂(g) ® CCl₄(g) + HCl(g)

Chloroform

Rate law found = -d[CHCl₃] / dt = k [CHCl₃]¹[Cl₂]^1/2

overall order = 3/2

Another cutie:

2 NO(g) + Cl₂(g) ® 2NOCl(g)

nitrosyl chloride

Rate Law: can't write expression down - must determine by experiment

It was found to be

Reaction Rate = -1/2 d[NO] / dt = -d[Cl₂] / dt = k [NO]² [Cl₂]

Order = 2 + 1 = 3

2nd order in [NO]

1st order in [Cl₂]

From this series of reactions we see that the: stoichiometric equation may not relate to overall reaction order. If the rate law does match the stoichiometry often the reaction is occurring in a single "primary reaction step" This is a step in which the reagents shown do go directly to product by direct interaction (see below).

With gases, can use partial pressures

÷ P_NO           ÷P_Cl2

2 NO (g) + Cl₂(g) ® 2 NOCl(g)

conc in terms of partial pressure

Reaction Rate = k [P_NO]²[P_cl]¹ but note change in units for k (not mol L^-1 s^-1 but atm^-2 s^-1)

-d[P_cl]/dt atm s^-1= k [P_NO]²[P_Cl]¹

k -d[P_cl]/dt atm s^-1 / [P_NO]²[P_Cl]¹ atm² atm = atm^-2 × s^-1

First order

For example, for a reaction you think is first order

A ® B + C

Assume first order

-d[A] / dt = k[A]¹

(k is a constant), A₀ = [A] at t=0, A_t = [A] at time t=t

then 

ie. ln[A_t] - ln[A_o] = -k (t - 0)

or

ln [At] / [Ao] = -kt ( y = mx + b)

ie. Plot ln [At] / [Ao] vs t

k = slope - y straight line if it is 1^st order

eg. Hydrolysis of alkyl halide

(H₃C)₃CCl + H₂O ® (H₃C)₃COH + HCl

Alcohol measure rate of HCl appearance (pH meter)

Find +d[HCl] / dt = k [C₄HgCl]

In this case, the plot WAS a straight line, therefore the reaction was first order. What is the mechanism of this reaction? See below.

For first order reactions one can also use the Half-Life as a measure of First Order Reactions

Half life = time for the initial concentration A₀ to be reduced by ½ (i.e., A₀/2)

For a first order reaction t_1/2

ie: ln([A_t]/[A₀]) = -k t

At t = t_1/2 At = A₀/2

So ln[A₀/2]/[A₀] = -k t_1/2 @ t = -t_1/2

ln ½ = -kt_1/2 or t_1/2 = ln2 / k k = 0.693/t_1/2

t ½ therefore gives directly value of k

Note that t_1/2 not is not related to concentration

Who cares about this, you may ask: See problems in Tutorials in the book

Here's one example

Radio Carbon Dating

Carbon 14 decays in a first order process. While objects are living, they absorb ¹⁴C from the atmosphere. This ceases for some reason once something dies and the ¹⁴C concentration starts to decrease

In this case, we know that the number of counts is proportional to the [¹⁴C]

ln Ao/At = kt

t ½ = 5730 years. Thus, by comparison of the [¹⁴C] in living objects, and in the object to be dated, and knowing the t_1/2 we can know k and calculate the concentration.

Temperature Effects

Specific rate constant changes are responsible ie. k T

More or less, for T increase by 10 C there is a change of reaction rate by 2x (faster)

We can understand this in terms of the amount of energy

The theory of reactivity of molecules suggests that, for reaction to occur:

i) molecules must have sufficient energy to overcome an intrinsic reaction barrier (the activation energy - this will be different for every reaction)

ii) molecules must collide with one another for reaction to occur (e.g., bimolecular reactions)

iii) not every collision will lead to successful reaction (some collisions are non-productive)

As the temperature of the environment increases, the number of molecules with more energy than the intrinsic barrier increases - hence, the reaction goes faster.

If number of molecules with E > E_a is n_a (area under the curve)

(Ea = energy required for something (reaction) to occur)

Then

n_a/n_total = e^-Ea/RT exponential function

The Reaction rate = k [A]^x[B]^y....with k = A e^-Ea/RT (called Arrhenius equation)

A = constant with T ("Preexponential Term"). It reflects the frequency of collisions with correct geometry for reaction when the concentration = 1M

E_a = Activation Energy (depends ONLY on reaction) (the factor e^-Ea/RT is always <1 (never have all collisions with E > E_a lead to successful reaction)

Then E_a is Barrier to chemical reaction

Applications

Take k = Ae^-Ea/RT

ln k = Ea/R x 1/T + ln A (constant)

y = M x + b

Plot ln k vs 1/T Þ straight line (especially if only two points)

Here's an example

H₂(g) + I₂ (g) ® 2HI (g)

observed rates:

k₁ = 2.7 x 10^-4 L/mol s @ 600 K

k₂ = 3.5 x 10^-3 L/mol s @ 650 K

1) What is E_a

k = Ae^-Ea/RT

Do ratio 2 rates

k₁ / k₂ = Ae^-Ea/R600 / Ae^-Ea/R650

ln (k₁ / k₂ ) = -(Ea/R600)+Ea/R650

ln (k₂ / k₁ ) = Ea/R(1/600  -  1/650)

log (3.5 x 10^-3 / 2.7 x 10^-4) = [Ea / (2.303 x 8.31 J/mol K)] x (1/600 - 1/650)

(2.303 converts ln®log)

E_a = 1.11 x 2.303 x 8.31 J/mol / 1.28 x 10^-4

Another example

The reaction of ^-IO₃ and HSO₃^- gives a purple colour - assume we don't know the reaction products aside from the colour

The rate law was shown to be Rate = k [^-IO₃][HSO₃^-]

Ea is usually LARGE (kJ)

R is small in J

(8.314 J/kmol)

at 15C, colour change took 4.4 s

at 43C, colour change took 3.3 s (the concentrations were the same for both reactions)

To calculate E_a do a 2 point Arrhenius plot (or just take ratio of rate constants)

We know.

k₁ = 1 / 4.4 (rate) x 1 / [IO₃^-] [H₃O₃^-] for T₁ = 15 C = 288 K

k₂ = 1 / 3.3 (rate) x 1 / [IO₃^-] [HSO₃^-] for T₂ = 43 C = 316 K

ln (1 / 4.4) (1 / [IO₃^-] [HSO₃^-]) = E_a/R 1/288 + ln / A

ln (1 / 3.3) (1 / [IO₃^-] [HSO₃^-]) = E_a/R 1/316 + ln / A

Both E_a and A are temperature independant

If take ratio of k₁/k₂

ln 3.3/ 4.4 = E_a/R [1 / 288 - 1 / 316]

Ea = 7.7 (7) x 10³ J/mol

= 7.7 (7) kJ / mol

NO(g) + ½ Cl₂ (g) ® ClNO nitrosyl chloride

We can draw a REACTION COORDINATE. This is a picture of the energy of the molecule(s) as the reactants are converted into products. The energy from the starting materials, via a barrier represented by the activation energy A, leads to (in this case) lower energy products. The DELTA H_react is represented by B. The top of the curve is called the transition state, or activated complex. It has NO finite lifetime (< 10^-13s)

[C......N = O] means transition state

On way to transition state, Cl moves closer starts to form bond.

Cl more away

NO + Cl reactants Cl - N= O product

Reaction Mechanism

If one molecule - UNIMOLECULAR-- if 2 moleucles, BIMOLECULAR

Three are at a time is very rare -- especially in gas phase (termolecular)

Each reaction step is called an elementary step. The series of elementary steps that leads from products to reactants is called the reaction mechanism. FOR AN ELEMENTARY STEP, THE RATE IS DETERMINED BY THE STOICHIOMETRY.

Once know the elementary steps, can deduct rate law.

For unimolecular

A ® B + C Raw Law -d[A] / dt = k(A)

Biomolecular

A + B ® AB -d[A] / dt = k [A][B]

eg. 2NO₂(g) + F₂(g) ® 2 FNO₂

Observed Rate Law -1/2 d[NO₂]/dt = k [NO₂][F₂]

Recall no termolecular steps \ must be sequence of steps

Propose the following mechanism

1. NO₂ + F₂ FNO₂ + F^·      Rxn

2. F^· + NO₂ FNO₂

if assume k₁ << k₂ i.e. Step 1 very slow with respect to Step 2. The observed rate for the overall reaction is the rate of the slowest step, also known as Rate Determining Step.

ie. Subsequent steps don't matter in this case, can use step, to write down rate law

\ Rxn Rate = -d[NO₂]/dt = l[NO₂][F₂]

The two proposed steps are consistent with rate law and called a reaction mechanism. Can't prove by kinetics that this is the reaction mechanism, but can disprove proposed mechanisms by kinetics.

Another Ex

2 NO(g) + O₂(g) ® 2NO₂(g)

Observed Rate Law = -d[O₂] / dt = k[O₂][NO]²

But no termolecular reactions

A Possible Mechanism

1. 

2. N₂O₂ + O₂ 2NO₂ -------------- slow

Assume k₂ << k₁/k_-1

That is, propose step 1 is a fast equilibrium

k₁ [NO]² = _k-1[N₂O₂]

ie. [N₂O₂]/ [NO]² = k₁ / k_-1 = K_equi

if the equibrium is faster, then the second step is RDS

Then can write rate law

Rate = k[N₂O₂][O₂], we can substitute the unknown [N₂O₂] by the equilibrium

K_eq = [N₂O₂]/ [NO]²

[N₂O₂] = K_eq [NO]²

The reaction rate = k₂ [N₂O₂] [O₂]

=k₂ K_eq [NO]² [O₂] = k_obs [NO]² [O₂]

k₂ and K_eq are both const @ given T

Our proposed mechanism is thus consistent with this rate law.

Why do we care? What is value of knowing mechanism?

Fundamental understanding of mechanisms allows one to develop new reactions (to make plastics), interfere with reactions (design drugs) etc.

Chain Reactions

Initiate (for example by light) - generate reaactive species

H₂(g)+ Cl₂(g) ® 2HCl(g)

DELTA H_f^o for HCl is ~ 92 kJ/mol. The reaction does not proceed unless we initiate it. Light and heat.

Mechanism

Initiation 1. Cl₂® 2 Cl^·

Propagation 2. Cl^· + H₂® HCl + H^·

3. H^· + Cl₂® HCl + Cl

____________________________________________

NOTE SUM: H₂ + Cl₂ ® 2HCl

is equal to the stoichiometry of the reaction

Termination: 4. Cl^· + Cl^·® Cl₂

5. H^· + Cl^·® HCl

6. H^· + H^·® H₂

all above are very exothermic usually need a third body. E.g. A wall

Other examples:

Initiate

1) 2H₂ + O₂ ® 2H₂O

Heat

Initiation: E.g. A flame ® HO^· Hydroxyl radical

Propagation: E.g. HO^· + H - H ® HO-H + H^·

H^· + O=O ® HO-O^· peroxy rad

HO-O^· + H-H ® HOOH + H^·

HO-OH  2^·OH

Chains increase in number very rapidly - EXPLOSIVE

Another example,

2) Methane & chlorine

CH₄(g) + Cl₂ HCl + CH₃Cl choromethane

Initiation Cl₂ 2Cl^·
 

Prop. 1) Cl^·  + CH₄® CH₃^·  _{+ HCl}

2) CH₃^·  ₊ Cl₂ ® CH₃Cl + Cl^·

Termination you do it...........

3) Polymerization (like dominoes)

Initiate

Propagation

Catalysis

B is the reaction coordinate without a catalyst, A is the reaction coordinate with a catalyst.

TWO CASES:

1) Homogenous catalyst where it is in the same phase as the reaction

eg. Both liquids

2) Heterogeneous catalyst where catalysts and compounds are in different phases

e.g. catalytic converter in auto exhaust catalyst is solid Pt/Pd/Rh on a support reagent reaction are gases

e.g. 2 H₂O₂ + I^_ ® 2 H₂O + O₂ + I^-

Rate Law

Rate = d[O₂] / dt = k[H₂O₂][I-] (NOTE - ONLY 1 H₂O₂ in the rate law I^- is the catalyst)

Proposed Mechanism

1. 

2. IOH + OH-® IO^_ + H₂O

3. IO^_ + HO -- OH  I- + O₂ + H₂O

From Step 1, reaction rate = k[H₂O₂][I-]
 

ii) Heterogeneous catalysis

a) oxidation of acetone on a Cu surface

b) SO₂ oxidation with O₂ + V₂O₅(s)

2SO₂(g) + O₂(g) ® 2SO₃(g) ® H₂SO₄

"Contact" process for H₂SO₄ manufacture

What did we learn?

A + 2B ® 3C

-d[A]/dt= -1/2 d[B]/dt = 1/3 d[C]/dt

Rate law from initial reaction rates and concentration changes, specific rate constant k

For first order reaction; can integrate

ln Ao/At = kt t_1/2 = ln 2/k

B) Temperature effects

Activeation energy Ea

Arrhenius equation k = A e^-Ea/RT

C) Reaction mechanism

Transition states

Intermediates

Elementary Reaction Step

Slow Rate Determining Steps relation to rate law

Proposed Mech

With fast equilibrium steps - steady state

Chain Reactions

Initiation propagation & termination

Catalysis lowering Ea

Homogeneous

Heterogeneous