Prepared by PhD Falfushynska H.

1. The Arrhenius Postulations 2. Collisions Theory and Molecular orientations 3. Van-Hoff Rule. 4. How to find the “Activation Energy” 5. Types of photochemical reactions

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, E a is the activation energy, R is the gas constant ( 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 E a are specific to a given reaction. Temperature and Rate ln k = ln A – E a / R T

From k = A e – Ea / R T, calculate A, E a, k at a specific temperature and T. The reaction: 2 NO 2 (g) -----> 2NO(g) + O 2 (g) The rate constant k = 1.0e-10 s -1 at 300 K and the activation energy E a = 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) = E a / R T

13 The reaction: 2 NO 2 (g) -----> 2NO(g) + O 2 (g) The rate constant k = 1.0e-10 s -1 at 300 K and the activation energy E a = 111 kJ mol -1. What are A, k at 273 K and T when k = 1e-11? Use the formula derived earlier: A = k e Ea / R T = 1e-10 s -1 exp ( J mol -1 / (8.314 J mol -1 K – 1 *300 K)) = 2.13e9 s -1 k = 2.13e9 s -1 exp (– J mol -1 ) / (8.314 J mol -1 K –1 *273 K) = 1.23e-12 s -1 T = E a / [R* ln (A/k)] = J mol -1 / (8.314*46.8) J mol -1 K -1 = 285 K

If t 2  t 1 Temperature coefficient:

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 the “Activation Energy”(E a ) …? Expt.  Table  Find E a by Graphical Method k = Ae (-E /RT) a ln k = lnA – EaEaEaEa RT i.e. “ln k” vs “1/T” should give a straight line with slope = -E a /R

Find E a by Experiment … (1) At T 1, find “k 1 ” by Differential / Integrated Rate Eqn At T 2, find “k 2 ” by Differential / Integrated Rate Eqn TT1T1 T2T2 T3T3 T4T4 kk1k1 k2k2 k3k3 k4k4 1/T1/T 1 1/T 2 1/T 3 1/T 4 ln kln k 1 ln k 2 ln k 3 ln k 4

ln k 1/T ln A slope = - E a /R Find E a by Experiment … (2) 1/T1/T 1 1/T 2 1/T 3 1/T 4 ln kln k 1 ln k 2 ln k 3 ln k 4 ln k = – + lnA EaEaEaEaR 1 T ** E a must be +ve.!! 

Slope = K K = -E a /(8.314 J K -1 mol -1 ) E a = J mol -1 E a = J mol -1  E a = 96.9 kJ mol -1 From the graph, ln k =  k = 0.40 s -1 1 st order: k = ln(2) / t 1/2  t 1/2 = 1.73 s  t 1/2 = 1.73 s

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

INTERACTION OF LIGHT AND MATERIALS: a)X 2 * → X 2 + M* (excess energy transferred to the surrounding) b) X 2 * → X 2 + h (fluorescence or phosphorescence) c) X 2 * + Y → chemical reaction (excess energy supplies the activation energy of the reaction)

(energy of the photon supplies the „dissociation heat”) a) Photodissociation b) Photosynthesis: when a larger molecule is formed from simple ones c) Photosensitized reactions: when an excited molecule supplies activation energy for the reactants

Photodissociation Photolysis of hydrogen bromide Overall:

Note: 1 photon absorbed, 2 molecules of HBr dissociated:

Notes: M absorbs energy released in the reaction

Note: Quantum yield is about 10 6 (explosion)

Non-Rotating Annular Photochemical 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