A Radioactivity Example

14C 14N + e- 6 7

Carbon-14 is unstable. It decays very slowly by a process called beta-decay.

A beta particle is an electron. It is ejectedfrom the nucleus. In the process a neutronbecomes a proton.

The rate that Carbon-14 decaysis an inherent property of the nucleus.Thus the rate of decay is proportional to the number of 14C atoms present.

It is a perfect example of a first orderreaction.

-d [14C]

d t= k [14C]

d[A]k [A]=-

d t

d t-= [A]d[A]

k

kd[A] [A]

= - d t

t[A]t

[A]0 0

kd[A] [A]

= - d t

t[A]t

[A]0 0

ln[A]t - ln[A]0 = -k t

ln[A]t

[A]0= -k t [A]t = [A]0 e-kt

[A]t = [A]0 e-kt

when [A]t = 0.5 [A]0

e-kt=0.5

ln 0.5 = -k t

0.693/k = t1/2

N

O

O ON

O

O

N

O

O

+ N

O

OO

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

Decomposition of N2O5 in CCl4 solution

Rate determining step is:

First order reaction

d t- = k [A]2d[A]

kd[A]

[A]2 = - d t

0[A]0

[A]t t

d t-= [A]2d[A]

k

0[A]0

[A]t t

d t-= [A]2d[A]

k

k t=[A]0[A]t

11 -

Second order reaction

k t=[A]0[A]t

11+

First order reaction

ln[A]t = -k t + ln[A]0

2 C4H6(g) C8H12(g)

2

What is the order of the reaction?

k t=[A]0[A]t

11+

ln[A]t = -k t + ln[A]0

2nd order

1st order

Reaction is Second Order

Zero order d[A]k [A]0=-

d tk =

d t-=d[A] k

kd[A] = - d t

t[A]t

[A]0 0

[A]t = - k t + [A]0

Zero order

[A]t = - k t + [A]0

The rate does not depend on anyreactant!

For this to be true something elsemust be determining the rate.

A catalyst for example

2N2O (g) 2N2 (g) + O2 (g)Pt

The rate depends only on the surface areaof the Pt not on the concentration of N2O

Second order reaction

k t=[A]0[A]t

11+

First order reaction

ln[A]t = -k t + ln[A]0

[A]t = - k t + [A]0

Zero order reaction

Consider the famous reaction.

A + B C + D

Assume the reaction in both directions is first order in each reactant

Rateforward = kf [A][B]

Ratereverse = kr [C][D]

The two reactions will compete.

A + B C + DRateforward = kf [A][B]

Ratereverse = kr [C][D]

High [A] or [B] favor fast forward reaction

High [C] or [D] favor fast reverse reaction

At some set of concentrations therates will be equal.

When the forward and reversereactions occur at the same ratethere will be no change in concentrations.

kr [A][B]

A + B C + D

Rateforward = kf [A][B] = Ratereverse = kr [C][D]

This is equilibrium.

The forward and reverse ratesare the same at equilibrium.

kf [A][B] = kr [C][D]

kf [C][D]= = Kequil

kr [A][B]

A + B C + D

kf [A][B] = kr [C][D]

kf [C][D]= = Kequil

The equilibrium constant is determinedby the relative forward and reverse rates.

The rate law does not necessarilyreflect the stoichiometry of the reaction.

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

Rate = k[NO2]2

What mechansim can explain this?

NO2(g) + NO2(g) NO3(g) + NO(g)

NO3(g) + CO(g) NO2(g) + CO2(g)

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

NO2(g) + NO2(g) NO3(g) + NO(g)

NO3(g) + CO(g) NO2(g) + CO2(g)

Over all reaction

Elementary steps

slow

fast

Rate = k[NO2]2

Reactions that start with a fast equilibrium

2O3(g) 3O2(g)

[O3]2

[O2]Rate = k

How can you explain this? Whyis [O2] in the rate?

2O3(g) 3O2(g)

Overall Reaction

Elementary Steps

O3 O2 + Ok1

k-1

k2O + O3 2O2

Fast

Slow

k2O + O3 2O2

Forward = Reverse Rate= k1[O3] = k-1[O2][O]

Rate of slow step = k2[O][O3]

k-1

k1O3 O2 + O

[O] = k-1[O2]

k1[O3]

Rate of slow step = = k-1[O2]

k2 k1[O3]2 [O3]2

[O2]k

Cl2 + CHCl3 HCl + CCl4all in the gas phase

Rate = k[Cl2]1/2[CHCl3]This is really getting weird.

Cl2 2Clk1

k-1

Cl + CHCl3 HCl + CCl3k2

k3CCl3 + Cl CCl4

fast

slow

fast

Cl2 2Clk1

k-1

Cl + CHCl3 HCl + CCl3k2

k1[Cl2] = k-1[Cl]2

Rate of slow step = k2[Cl][CHCl3]

[Cl]2 = k1/k-1[Cl2]

[Cl] = (k1/k-1[Cl2])1/2

Rate of slow step = k [Cl2]1/2[CHCl3]

H

+

+HO- CH3Br

HOCH3 Br-

TransitionState

Ea

HO Br

Energy distribution of molecules versus Temperature

Fraction of Collisions with Energy > Ea

-Ea/RTe

As -Ea/RT approaches zero theFraction of Collisions with Energy > Ea

approaches unity.

This would happen at high T or low Ea

Fraction of Collisions with Energy > Ea

-Ea/RTk = Ae

-Ea/RTe

APre-exponential factor.fraction of collisions with propersteric orientation

Arrhenius Equation

-Ea/RTk = Ae

ln(k) = -Ea/RT + ln(A)

Arrhenius Equation

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

ln(k) = -Ea/RT + ln(A)

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

Slope = -Ea/R

Ea = -R (-1.2 x 104)K

Ea = 100 kJ/mol

Ea = - 8.31J/K (-1.2 x 104)K

ln(k1) = -Ea/RT1 + ln(A)

ln(k2) = -Ea/RT2 + ln(A)

ln(k2/ k1) = Ea/R(1/T1 - 1/T2)

This equation allows youto compare rates at different T

ln(k2/ k1) = Ea/R(1/T1 - 1/T2)Assume Ea = 100 kJ/mol and T = 300 K

What temperature wouldbe required to double the rate?

T2 = ? K

ln(k2/ k1) = Ea/R (1/T1 - 1/T2)Assume Ea = 100 kJ/mol and T = 300 K

ln(2) = 105 / R (1/300 - 1/T2)

5.8 x 10-5 = 1/300 - 1/T2

T2 = 305

Catalysis

A catalyst changes the mechansimof a reaction to one with a lower energy of activation.

H = -127kJ/mol

CH3CH=CH2 H2+

CH3CH2CH3

? Ea too high thusno reaction

Try to find a new mechanismwith a lower Ea

How do you dothat? Use acatalyst.

NN

NO

OH

OH

OH

O

RH

H2O

NN

OH

OH

O

RO

HN

O

OH

HH

Amide hydrolysis

R

O

N

H

R

H

R

O

N

H

R

HOH2

O

N

R O

H R

H

H

H

O

N

R O

HH

H H

R

R

O

O H

HNH2R

R

O

O

H

H2O NH2RH2O NH2R

Too slow

R

O

N

H

R

R

O

O

H

H2O NH2RH2O NH2R

Too slow

OHO

R O

N

H

H R

OH

H O

H

R

O

O

H O

H

OH

NH2R

R

O

N

H

R

R

O

O

H

H2O NH2RH2O NH2R

Too slow

Uncatalyzed reactionvery slow.

Acid or base catalyzedreaction faster, but still requires high temperature and longtime and it is unselective.

Enzymes can do itvery fast at bodytemperature.

NN

NO

OH

OH

OH

O

RH

H2O

NN

OH

OH

O

RO

HN

O

OH

HH

Carboxypeptidase-A

Hydrolyzes the lastamide bond on a peptide

Carboxypeptidase-A Enzyme

Zn

Carboxypeptidase-A Enzyme