Prepared by PhD Falfushynska H.

1. The Arrhenius Postulations2. Collisions Theory and Molecular

orientations3. Van-Hoff Rule.4. How to find the “Activation Energy”5. Types of photochemical reactions

3

1903 Nobel Prize citation” …in recognition of the extraordinary services he has rendered to the advancement of chemistry by his electrolytic theory of dissociation”

Collisions and Rate － the rate of reaction is much smaller than

calculated collision frequency.A threshold energy (activation energy) － This kinetic energy is changed into

potential energy as the molecules are distorted during a collision, breaking bonds and rearranging the atoms into product molecules.

Particles must collide before a reaction can take place

Not all collisions lead to a reaction: Experiments show that the observed reaction rate is considerably smaller than the rate of collisions with enough energy to surmount the barrier.

The collision must involve enough energy to produce the reaction – Activation energy.

The relative orientation of the reactants must allow formation of any new bonds necessary to products. – Steric effect

The order of each reactant depends on the detailed reaction mechanism.

Chemical reaction speed up when the temperature is increased.

－ molecules must collide to react － an increase in temperature increases the frequency of intermolecular collisions.

The 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 (8.3145 J K-1 mol-1) and T is the temperature in K.

– A is called the frequency factor.

– A is a measure of the probability of a favorable collision.

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

Temperature and RateTemperature and Rate

RTEa

eAk

ln k = ln A – Ea / R T

From k = A e – Ea / R T, calculate A, Ea, k at a specific temperature and T.

The reaction: 2 NO2(g) -----> 2NO(g) + O2(g)

The rate constant k = 1.0e-10 s-1 at 300 K and the activation energy Ea = 111 kJ mol-1. What are A, k at 273 K and T when k = 1e-11?

k = A e – Ea / R T

A = k e Ea / R T

A / k = e Ea / R T

ln (A / k) = Ea / R T

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The reaction:

2 NO2(g) -----> 2NO(g) + O2(g)

The rate constant k = 1.0e-10 s-1 at 300 K and the activation energy Ea = 111 kJ mol-1. What are A, k at 273 K and T when k = 1e-11?

Use the formula derived earlier:

A = k eEa / R T = 1e-10 s-1 exp (111000 J mol-1 / (8.314 J mol-1 K –

1*300 K)) = 2.13e9 s-1

k = 2.13e9 s-1 exp (– 111000 J mol-1) / (8.314 J mol-1 K –1*273 K) = 1.23e-12 s-1

T = Ea / [R* ln (A/k)] = 111000 J mol-1 / (8.314*46.8) J mol-1 K-1 = 285 K

10

1

2

1

2

t

t

t

t

t

k

k

If tIf t22tt11

t

t

k

k 10

Temperature coefficientTemperature coefficient::

oTT

Ckm

v 298

Tn 2298

The shelf-life for the product can be calculated from the rate constant based on an acceptable degree of decomposition. The time taken for 10% loss of activity is given by t90 = 0.105/k1

The Arrhenius equation is used as the basis of a method for accelerating decomposition by raising the temperature of the preparations.

1. Determination of the order of reaction by plotting stability data at several elevated temperatures according to the equations relating decomposition to time for each of the orders of reaction, until linear plots are obtained.

Values of the rate constant k at each temperature are calculated from the gradient of these plots, and the logarithm of k is plotted against reciprocal temperature according to the Arrhenius equation log k = log A – Ea/2.303RT.

A value of k can be interpolated from this plot at the required temperature.

where k1 and k2 are the rate constants at temperatures T1 and T2 respectively. A mid-range value of Ea = 75 kJ mol–1 may be used for these rough estimation

How to find theHow to find the “Activation Energy”(E“Activation Energy”(Eaa)) …? …?

Expt. Expt. Table Table Find E Find Eaa

by by Graphical MethodGraphical Method

k = Aek = Ae(-E /RT)(-E /RT)aa

ln k = lnA – ln k = lnA – EEaa

RTRT

i.e. “i.e. “ln kln k” vs “” vs “1/T1/T” ” should give a should give a straight line with straight line with slope = -Eslope = -Eaa/R/R

Find Find EEaa by Experiment by Experiment … (1)… (1)

At TAt T11, find “k, find “k11” by Differential / Integrated Rate Eqn” by Differential / Integrated Rate Eqn

At TAt T22, find “k, find “k22” by Differential / Integrated Rate Eqn” by Differential / Integrated Rate Eqn

T T1 T2 T3 T4

k k1 k2 k3 k4

1/T 1/T1 1/T2 1/T3 1/T4

ln k ln k1 ln k2 ln k3 ln k4

ln kln k

1/T1/T

ln Aln Aslope = - Eslope = - Eaa/R/R

Find Find EEaa by Experiment by Experiment … (2)… (2)

1/T 1/T1 1/T2 1/T3 1/T4

ln k ln k1 ln k2 ln k3 ln k4

ln k = – + lnAln k = – + lnAEEaa

RR11TT

** E** Eaa must be +ve.!! must be +ve.!!R

Etg a

1/T 0.0037 0.0034 0.0032 0.0030ln k -10.61 -7.65 -5.16 -2.90

Slope = -11658 KSlope = -11658 K

-11658K = -E-11658K = -Eaa/(8.314 J K/(8.314 J K-1-1molmol-1-1))

EEaa = 96924 J mol = 96924 J mol-1-1

EEaa = 96.9 kJ mol = 96.9 kJ mol-1-1

From the graph, ln k = -0.92From the graph, ln k = -0.92

k = k = 0.40 s0.40 s-1-1

11stst order: k = ln(2) / t order: k = ln(2) / t1/21/2

tt1/21/2 = 1.73 s = 1.73 s

Excitation:

X2h *X2

A bonding electron is lifted to a higher energy level (higher orbital)

INTERACTION OF LIGHT AND MATERIALS:

a)X2* → X2 + M* (excess energy transferred to the surrounding)

b) X2* → X2 + h (fluorescence or phosphorescence)

c) X2* + Y → chemical reaction (excess energy supplies the activation energy of the

reaction)

hX2 X + X (photodissociation)

(energy of the photon supplies the „dissociation heat”)

a) Photodissociationb) Photosynthesis: when a larger

molecule is formed from simple onesc) Photosensitized reactions: when

an excited molecule supplies activation energy for the reactants

PhotodissociationPhotolysis of hydrogen bromide

HBr hH + Br (photochemical reaction)

H + HBr H2 + Br

Br + Br Br2

(dark reactions)

Overall:

2HBr h H2 + Br2

Note:

1 photon absorbed, 2 molecules of HBr dissociated:

QUANTUM YIELD = 21 = 2

number of molecules undergoing the processnumber of quanta absorbed=

O2 O + O (<240 nm)h

2O2 + 2O (+M) 2O3 (+M*)

Notes: M absorbs energy released in the reaction

QUANTUM YIELD = 21 = 2

hCl2 < 500 nm 2Cl Photochem. initiation

Cl + H2 HCl + H Dark reactions

H + Cl2 HCl + Cl Chain reaction

H + H + M

H2 + M*

Cl2 + M* Cl + Cl + M

Recombination reactions (chainis terminated)

Note:

Quantum yield is about 106 (explosion)

6CO2 + 6H2O C6H12O6+6O2

carbohydrate

h; chlorophyllseveral steps

Photosensitized reactions. Photosynthesis in plants

Photosynthesis in plants

Non-Rotating AnnularPhotochemical Reactor

# Large Quartz immersion well.# 400 watt medium pressure mercury lamp.# Reactor base and carousel assembly (non rotating), including support rod and immersion well adjustable clamp.# set of sample tube support rings for eight 25mm sample tubes# Only the inner or the outer tubes may be irradiated effectively at one time# UV Screen:- consisting of three black coated consisting of three black coated aluminum sections. A light tight lid, a removable front and back section, that are joined by means of a light tight seal