11.0 REACTION KINETICSObjectives: 1. Define reaction rate, average rate, instantaneous rate and initial rate. 2. Determine the reaction rate based on a differential equation.

REACTION KINETICSChemical kinetics is the study of the rates of chemical reactions, the factors that affect these rates, and the reaction mechanisms by which reactions occur.Important

industrial process -Time -Optimum yield -Optimum conditions control over reaction, obtain products economically, using optimum conditions

Rate of reactionReaction rate is the change in the concentration of a reactant or a product with time. Unit of rate (mol L-1 s-1)

1 rate w TimeExample;

A p B d[A] = change in concentration of A dt = period of time d[B] = change in concentration of B

- d[A] rate = dt d[B] rate = dt

Because [A] decreases with time, d[A] is negative.

A

B

time [B]

rate = -

d[A] dt d[B] dt

rate =

[A]

Rate of reaction The average rate is the rate over a period of time. The rate of reaction at a given time is called an instantaneous rate of reaction. The instantaneous rate at the beginning of a reaction is called the initial rate of reaction. Instantaneous rate is determined from a graph of concentration vs time by drawing a line tangent to the curve at that particular time.

Rate of reactionReaction: H2O2(aq) p H2O(l) + O2(g) Reaction rates are obtained from the slopes of the straight lines;purple

An average rate from the purple line. The instantaneous rate at =300 s from the red line. t

blue

red

The initial rate from the blue line.

Br2 (aq) + HCOOH (aq)

2Br- (aq) + 2H+ (aq) + CO2 (g)

instantaneous rate = rate at a specific time [Br2]final [Br2]initial d[Br2] average rate = =dt tfinal - tinitial

The differential rate equationA differential rate equation enables the relationship between the rate of disappearance of reactants and the formation of products. Consider the reaction, aA + bB p cC + dD

Rate =

1 d[A] 1 d[B] 1 d[C] 1 d[D] ! ! ! a dt b dt c dt d dt

a,b,c and d are the stoichiometric coefficients

The differential rate equationExample: The formation of NH3, N2(g) + 3H2(g) p 2NH3(g) The differential rate equation is; Rate =

d[N 2 ] 1 d[H 2 ] 1 d[NH 3 ] ! ! dt 3 dt 2 dt

The equation means that the rate of disappearance of N2 is 1/3 the rate of disappearance of H2 and 1/2 the rate of formation of NH3.

Example 1: Consider the reaction, 2HI p H2 + I2, determine the rate of disappearance of HI when the rate of I2 formation is 1.8 x 10-6 M s-1. Solution:

1 d[HI] d[H 2 ] d[I 2 ] ! ! Rate = 2 dt dt dtd[I 2 ] = 1.8 v 10-6 dt 1 d[HI] d[I 2 ] ! Rate = 2 dt dt

d[HI] = 2 v 1.8 v 10-6 = 3.6 v 10-6 M s-1 dt

EXERCISE 1: Hydrogen gas produced nonpolluting product is water vapour when react in O2 due to this reaction has been used for fuel aboard the space shuttle, and may be used by Earth-bound engines in the near future. 2H2(g) + O2(g) 2H2O(g)

Express the rate in terms of changes in [H2], [O2] and [H2O] with time. When [O2] is decreasing at 0.23 mol L-1 s-1, at what rate is [H2O] increasing? (0.46 mol L-1 s-1)

Exercise 2: Consider the reaction, NO(g) + O2(g) p2NO2(g). Suppose that at a particular time during the reaction nitric oxide (NO) is reacting at the rate of 0.066 M s-1 a) At what rate is NO2 being formed? b) At what rate is molecular oxygen reacting?

Exercise 3: Consider the reaction, N2(g) + 3H2(g) p 2NH3(g) Suppose that at a particular moment during the reaction molecular hydrogen is reacting at the rate of 0.074 M s-1 a) At what rate is ammonia being formed? b) At what rate is molecular nitrogen reacting?

11.1 RATE LAW

Objectives:At the end of the lesson the students should be able to: 1. define rate law and write the rate equation 2. define the order of reaction and the rate constant 3. calculate the order with respect to a certain reactant from experimental data 4. determine the overall order of a reaction from experimental data 5. calculate the value and determine the units of rate constants

The Rate LawThe rate law expresses the relationship of the rate of a reaction to the rate constant and the concentrations of the reactants raised to some powers. aA + bB cC + dD

Rate

[reactant]

Rate = k [A]x[B]yreaction order is x with respect to A reaction order is y with respect to B Overall reaction order is (x + y)The exponents x, y, can be integers, fractions or decimal or negative values. k is called rate constant

Rate LawThe values of x and y can only be experimentally. determined

Reaction order is usually defined in terms of reactant concentrations. The order of a reactant is not related to the stoichiometric coefficients of the reactants in the balanced chemical equation. F2 (g) + 2ClO2 (g) 2FClO2 (g)1

rate = k [F2][ClO2]

The units of rate constant, kA Products Rate, r = k [A]x i) The reaction is zero order Rate = k [A]0 Rate = k unit k = unit rate = mol L-1 s-1 or M s-1

ii) First order

Rate = k [A]1

rate k= [A] Unit k = M s-1 M

= s-1 iii) Second order Rate = k [A]2 rate k= [A]2 M s-1 M2 = M-1 s-1

Unit k =

Example :

S2O82- + 3I-

2SO42- + I3-

The above reaction is first order with respect to iodide ions and to thiosulphate ions. a) Write the rate of equation for the reaction. b) What is the unit of rate constant, k? Solution : a) Rate = k [S2O82-]1[I-]1 b) Rate = k [S2O82-]1[I-]1 rate k= [S2O82-]1[I-]1 Ms-1 M2

Unit k =

=

M-1 s-1

The order of reactionFor reactionA Products Rate = k [A]x i) If x = 0 Rate = k [A]0 Rate = k Rate is not dependent on [A] Therefore this reaction is zero order with respect to A

ii) If x= 1

Rate = k [A]1

Assume [A]i = 1.0M Rate = k (1.0M) If the [A] is doubled from 1.0M to 2.0M, Rate = k (2.0M) = 2k(1.0M) hence Rate = 2k[A] Doubling the [A] will double the rate of reaction. Therefore this reaction is first order with respect to A

iii) If x = 2

Rate = k[A]2

Assume [A]i = 1.0 M Rate = k (1.0 M)2 If the [A] is doubled from 1.0 M to 2.0 M, Rate = k (2.0 M)2 = 4k(1.0 M) hence Rate = 4k[A] Doubling [A], the rate will increase by a factor of 4. Therefore the reaction is second order with respect to A

Example Determining Reaction Order from Rate Law For each of the following reactions, determine the reaction order with respect to each reactant and the overall order from the given rate law.(a)

2NO(g) + O2(g)

2NO2(g); rate = k[NO]2[O2]

The reaction order respect to NO : 2 The reaction order respect to O2 : 1 overall reaction order = 3

(b) CH3CHO (g)

CH4(g) + CO(g);

rate = k[CH3CHO]3/2 Solution: The reaction order with respect to CH3CHO : 3/2 The reaction order (overall) : 3/2 (c) H2O2(aq) + 3I-(aq) + 2H+(aq) rate = k[H2O2][I-] Solution: The reaction of order with respect to H2O2 : 1 The reaction of order with respect to I- : 1 and zero order in H+, while overall order is 2. I3-(aq) + 2H2O(l);

Determination of the orders of reaction rate;O2(g) + 2NO(g)Initial Reactant Concentrations (molL-1) Experiment O2 1 2 3 4 5 1.10x10-2 2.20x10-2 1.10x10-2 3.30x10-2 1.10x10-2 NO 1.30x10-2 1.30x10-2 2.60x10-2 1.30x10-2 3.90x10-2

2NO2(g)Initial Rate (M s-1) 3.21x10-3 6.40x10-3 12.8x10-3 9.60x10-3 28.8x10-3

Solution: O2(g) + 2NO(g) rate = k [O2]m[NO]nCompare 2 experiments in which the concentration of one reactant varies and the concentration of the other reactant(s) remains constant.

2NO2(g)

rate2 rate1

=

k [O2]2m[NO]2n k [O2]1m[NO]1n=

=

[O2]2m [O2]1m

=m

[O2]2 [O2]1

m

6.40x10-3Ms-1 3.21x10-3Ms-1

2.20x10-2mol/L 1.10x10-2mol/L ; ,m=1

2 = 2m

The reaction is first order with respect to O2Do a similar calculation for the other reactant(s).

To find the order with respect to NO, we compare experiment 3 and 1, in which [O2] is held constant and [NO] is doubled: n k [O2]3m [NO]3n [NO]3 Rate 3 = = [NO]1 Rate 1 k [O2]1m [NO]1n n -3Ms-1 12.8 x 10 2.60 x 10-2mol/L = -3Ms-1 1.30 x 10-2mol/L 3.21 x 10 4=2 n ; n=2 The reaction is second order with respect to NO Thus the rate law is : Rate = k[O2][NO]2

Exercise:ClO2(aq) + 2OH- (aq) p products The results of the kinetic studies are given below. exp 1 2 3 [ClO2] M 0.0421 0.0522 0.0421 [OH-] M 0.0185 0.0185 0.0285 Initial rate, Ms-1 8.21 v1 0-3 1.26 v1 0-2 1.26 v1 0-2

a) Explain what is meant by the order of reaction. b) Reffering to the data determine (i) rate law /rate equation (ii) rate constant, k (iii) the reaction rate if the concentration of both ClO2 and OH- = 0.05 M

Exercise: Write rate law for this equation, A + B p C When [A] is doubled, rate also doubles. But doubling the [B] has no effect on rate. ii) When [A] is increased 3x, rate increases 3x, and increasing of [B] 3x causes the rate to increase 9x. iii) Reducing [A] by half has no effect on the rate, but reducing [B] by half causes the rate to be half the value of the initial rate. i)

Exercise: Many gaseous reactions occur in a car engine and exhaust system. One of the gas reaction is given below. NO2(g) + CO(g) NO(g) + CO2(g) Rate = k [NO2]m[CO]nUse the following data to determine the individual and overall reaction orders:Experiment Initial Rate(Ms-1) Initial [NO2](M) Initial [CO](M)

1 2 3

0.0050 0.080 0.0050

0.10 0.40 0.10

0.10 0.10 0.20

Integrated Rate LawObjectives: 1.Write the rate law for zero order, 1st order and 2nd order reaction 2. Define half-life. 3. Draw the respective graphs for the different order reactions 4. Solve quantitative problems.

Integrated rate equationsZero Order Reaction A zero order reaction is a reaction independent of the concentration of reactant. Ap product The rate law is given by rate = k[A]0 rate = k

rate

[A] M

- d[A] = k dt Using calculus, - d[A] = kdt - d [A] = k dt - [A] = kt + c substituting t=0, [A] = [A]0 - [A]0 = k(0)+c c = - [A]0 [A]0 -[A] = kt Unit of k for zero order reaction M s-1

Graphs for zero order reaction

[A]o [A]

[A]o [A] = kt y = mx + c

t [A] [A]o[A] = -k t + [A]o y=m x+c

rate

t

[A]

Half-life (t) Half life (t) is the time required for the concentration of a reactant to decrease to half of its initial value. zero order reaction Substituting t = t1/2 , and [A] = [A]0 into the zero order reaction, gives 2

[A]0 - [A] = kt [A]0 [A]0 = kt1/2 2 Solving for t1/2 gives t1/2 = [A]0 2k

First Order Reactions A first order reaction is a reaction whereby its rate depends on the concentration of reactant raised to the first power.

From the rate law, rate = k[A] To obtain the units of k k = rate [A] unit k = M s-1 M = s-1

Rate = k[A] y = mx + c

Rate Ms-1

[A] ,M

For first order reaction, Rate = k[A] - d[A] dt - d[A] [A] - d[A] [A] = k dt -ln [A] = kt ln[A]0 = k[A] substituting t = 0, [A]=[A]0 = k dt

- ln [A]0 = k(0) + c c = ln[A]0

- ln [A] = kt + c ln [A]0 [A] = kt

Characteristic graphs of 1st order reaction ln[A]o ln[A] =kt

[A]

ln[A] = - kt +ln[A]o [A]0 ln -----[A]

=

kt

ln[A] ln[A]o

t ln[A]o [A]

t

t

Example The reaction 2A B is first order with respect A with a rate constant of 2.8 x 10-2 s-1 at 800C. How long will it take for A to decrease from 0.88 M to 0.14 M ? [A]0 = 0.88 M ln[A]o ln[A] = kt [A] = 0.14 M kt = ln[A]0 ln[A] ln[A]0 ln[A] = t= k ln [A]0 [A] k = 0.88 M 0.14 M 10-2 s-1 = 66 s

ln

2.8 x

Example Decomposition of H2O2 (aq) is first order, given that k = 3.66 x 10-3 s-1 and [H2O2 ]o = 0.882 M, determine a) the time at which [H2O2] = 0.600 M b) the [H2O2 ] after 225 s.

Solution : a) ln [H2O2]0 [H2O2] ln 0.882 0.600 = 3.66 x 10-3 s-1 x t = kt

ln 1.47 = 3.66 x 10-3 s-1 x t t= ln 1.47 3.66 x 10-3

= 105.26 s b) [H2O2]0 [H2O2] 0.882 [H2O2] [H2O2] = 0.387 M

ln

=

kt

ln

=

3.66 x 10-3s-1x 225 s

Exercise, The conversion of cyclopropane to propene in the gas phase is a first order reaction with a rate constant of 6.7 X 10-4 s-1 at 500C. CH2

CH2

CH2

CH3-CH=CH2

a) If the initial concentration of cyclopropane was 0.25 M, what is the concentration after 8.8 minute. (0.18 M) b) How long will it take for the concentration of cyclopropane to decrease from 0.25 M to 0.15 M? (13 min) c) How long will it take to convert 74 percent of the starting material? (33 min)

Half-life of a first-order reaction

The half-life, t, is the time required for the concentration of a reactant to decrease to half of its initial concentration. t=t when [A] = [A]0 2 [A]0 ln [A]0/2 0.693 ln2 t = = = k k k What is the half-life of N2O5 if it decomposes with a rate constant of 5.7 x 10-4 s-1? 0.693 t = ln2 = = 1200 s = 20 minutes -4 s-1 5.7 x 10 k How do you know decomposition is first order? units of k (s-1)

First-order reaction A No. of half-lives 1 2 3 4 t = ln2 k product

[A]o = 8 M 4 2 1 1/2

Example The decomposition of ethane C2H6 to methyl radicals is a first order reaction with a rate constant of 5.36 x 10-4 s-1 at 700o C. C2H6 (g) 2 CH3 (g) Calculate the half life of the reaction in minutes.

Solution t1/2 =

ln 2 k 0.693 5.36 x 10-4 = 1.29 x 103 s = 21.5 min

=

Problem 2 What is the half-life of a compound if 75% of a given sample of the compound decomposes in 60 min? Assume it is first-order reactions kinetics. (t1/2=30 min)

Problem 3 The decomposition of SO2Cl2 is a first-order reaction. SO2Cl2(g) SO2 (g) + Cl2 (g) i) Write the rate differential equation for the reaction. ii) Calculate the value of rate constant, k at 500 K if 5.00 % SO2Cl2 decomposed in 6.75 min. Ans (7.6X10-3) iii) Specify the half-life for the decomposition reaction. Ans (91.20 )

Second Order Reactions A second order reaction is a reaction which rate depends on the concentration of one reactant raised to the second power or on the concentration of two different reactants each raised to the first power.

Example Ap product Where

Rate = -

d[A] dt

= k[A]2

To obtain the units of k k = rate [A]2 M/s M2 = M-1 s-1 Using calculus, the following expression can be obtained 1 [A] 1 [A]0

Unit k

=

=

+ kt

Characteristic graphs for second order reaction

Rate = k [A]2 rate

[A] rate

[A]2

Graphs for second order reaction

[A]

1 [A]

=

1 [A]o

+ kt

t 1/[A] M-1 1/[A] 1/[A]o

1/[A]o t t

nd 2

order reaction, r =

2 k[A]

If [A] doubles, r2 = k (2[A])2 = k ( 4 [A]2 ) = 4 k [A]2 =4r R will increase by 4 times if [A] doubles

Half life of a second order reaction 1 [A] 1 [A]0 [A]= [A]o 2 1 [A]0 + kt1/2 + kt

=

Substituting t = t1/2 1 [A]0 2 t1/2= 1 k[A]0 =

Detemination of half-life using graph for second order reaction

[A]0t1/2=

1 k[A]0

[A]0/2 [A]0/4 [A]0/8

x

2x

4x

t

Example Iodine atoms combine to form molecular iodine in the gaseous phase I (g) + I (g) p I2(g) This reaction is a second order reaction , with the rate constant of 7.0 x 109 M-1 s1

If the initial concentration of iodine was 0.086 M, i) calculate its concentration after 2 min. ii) calculate the half life of the reaction if the initial concentration of iodine is 0.06 M and 0.42 M respectively.

Solution : 1 i) [A] 1 [A] 1 [A]0 1 [0.086] + (7.0 x 109 x 2 x 60 ) + kt

= =

= 8.4 x 1011 [A] = 1.190 x 10-12 M ii) [I2] = 0.06 M t1/2= 1 k[A]0 = 1 = 7.0 x 109 x 0.060 2.4 x 10-10 s

[I2] = 0.42 M t1/2= 1 k[A]0 1 = 7.0 x 109 x 0.042

= 3.4 x 10-10 s

Using graph Example, The following results were obtained from an experimental investigation on dissociation of dinitrogen pentoxide at 45oC

N2O5(g)

2 NO2(g) + O2(g)

time, t/min [N2O5] x 10-4 Mi) ii)

0 176

10 124

20 93

30 71

40 53

50 39

60 29

Plot graph of [N2O5] vs time, determine The order of the reaction the rate constant k

Solution :

180 160 140 120 80 60 40 20 10 20 30 40 50 60 70 80

[N2O5] x 100 10-4 /M

Time ( min)

i) Based on the above graph, Time taken for concentration of N2O5 to change from 176 x 10-4 M to 88 x 10-4 M is 20 min Time taken for concentration of N2O5 to change from 88 x 10-4 M to 44 x 10-4 M is also 20 min The half life for the reaction is a constant and does not depend on the initial concentration of N2O5

Thus, the above reaction is first order ii) ln2 k= 20 min = 0.03 min-1

Summary of the Kinetics of Zero-Order, First-Order and Second-Order Reactions

Order 0 1 2

Rate Law rate = k rate = k [A] rate = k [A]2

Concentration-Time Equation [A] = [A]0 - kt ln[A] = ln[A]0 - kt 1 1 = + kt [A] [A]0

Half-Life [A]0 t = 2k t = ln2 k t = 1 k[A]0

Zero order A p product r = k [A]0 Unit k = M s-1r [A]

1st order A p product r = k [A]1 Unit k = s-1r [A] r

2nd order A p product r = k [A]2 Unit k = M-1 s-1r

Integrated rate law [A]0 [A] = kt[A] [A]0

Integrated rate law ln([A]0 / [A]) = ktln[A] ln[A]0 [A]

[A]

[A]2

Integrated rate law 1/[A] 1/[A]0 = kt1/[A] 1/[A]0

t1/2 = [A]0/2k [A]

[A]0 - [A]

t ln([A]0 / [A])

t

t

t 1/[A] 1/[A]0

t

t

t1/2 = ln2/kt

t1/2 = 1/k[A]0t

11.2 Collision TheoryObjectivesAt the end of the lesson the students should be able to: 1. explain reaction rates in terms of collision theory. 2. identify factors affecting the effectiveness of collision. 3. define activation energy. 4. Define and state the characteristics of an activated complex

Collision Theory Collision Theory is the theory to explain the rate of chemical reactions. It is based on; Rate w Number of effective collisions time

1- molecule must collide to react 2- molecules must possess a certain minimum kinetic energy (activation energy) to initiate the chemical reaction.

3- molecule must collide in the right orientation in order the reaction to occur.

for

Only effective collisions cause formation of product; collisions of molecules with Ea and at correct orientation. The activation energy (Ea) is the minimum energy that must be supplied or required by collisions for a reaction to occur.

The activation energy (Ea) ..is the minimum energy is required to initiate the chemical reaction.

Importance of Orientation

Orientation is unimportant

Orientation is important

Importance of Orientation

Orientation is important

Transition State Theory The configuration of the atoms of the colliding species at the time of the collision is called the transition state. Species formed at transition state is called activated complex.

Characteristics of Activated ComplexVery unstable i.e. It has a short half-life. Its potential energy is greater than reactants or products. The activated complex and the reactants are in chemical equilibrium. It decomposes to form products or reactants.

Potential energy - A reaction profile shows potential energy plotted as a function of the progress of the reaction. - The difference in potential energies between the products and the reactants is -(H for the reaction. - Reactant molecules must have enough energy to overcome an energy barrier separating products from reactants, Ea.

Ea Reactant

(H

product Progress of reaction

A Reaction Profile: exothermic reactionActivated complex Transition state

Ea (Forward reaction) Ea (reverse reaction)

(H

CO(g) + NO2(g)

CO2(g) + NO(g)

A Reaction Profile for endothermic processactivated complex.Ea (reverse reaction) Ea (forward reaction)

Product (H Reactant

A Reaction profile: endothermic reaction

Ea (H

2NOCl

2NO + Cl2

Example:1. For the reaction A + B C + D , the enthalphy change of the forward reaction is + 21 kJ/mol. The activation energy of the forward reaction is 84 kJ/mol. a) What is activation energy of the reverse reaction? b) Sketch the reaction profile of this reaction 2. Draw a potential energy diagram for an exorthermic reaction. Indicate on the drawing: a) Potential energy of the reactants and the products b) The activation energy for the forward and the reverse reaction c) The heat of the reaction

Factors affecting rate of reaction CONCENTRATIONS OF REACTANTS:Reaction rates generally increase as the concentrations of the reactants are increased.

TEMPERATURE: Reaction rates generally increase rapidly as the temperature is increased. CATALYSTS: Catalysts speed up reactions. PARTICLE SIZE: The rate increases as the smaller the size of reacting particles .

A) CONCENTRATIONS OF REACTANTSReaction rate

w

collision time

The frequency of collision increases increases with the concentration4 particle system (2 and 2) 4 collision

A) CONCENTRATIONS OF REACTANTSA concentration of reactants increases, the frequency of collision increases. This would also result in the increase in the quantity of effective collision. Thus the reaction rate increases. 5 particle system (3 and 2) 6 collision

A) CONCENTRATIONS OF REACTANTSThis observation correlates with the RATE LAW that has been previously discussedReaction rate = k [ A ] [ B ]x y

(A & B = reactants) (x & y = rate order)

Based on this equation, Reaction rate w concentration of reactants (depending on its rate order) REMINDER! Only in zero order reactions, the rate of reaction is not dependant upon the concentration of the reactants. Reaction rate = k [ A ]0 = k (constant)

B) TEMPERATURE

As temperature increases, kinetic energy, of molecules increases So, more collisions occur in a given time Furthermore, the higher the kinetic energy, the higher the energy of the effective collisions. So more molecules will have energy greater than activation energy, Ea Thus, the rate of reaction increases

Distribution of Kinetic Energies of Molecules

Represent total number of molecules with kinetic energy greater than Ea

B) TEMPERATURE

ARRHENIUS EQUATION In 1889, Svante Arrhenius proposed the following mathematical expression for the effect of temperature on the rate constant, k:

k = AeWhere

-Ea

RT

k = rate constant A = constant known as the ollision frequency factor e = natural log exponent Ea = activation energy for the reaction R = universal gas constant (8.314 J mol-1 K-1) T = absolute temperature

B) TEMPERATURE

ARRHENIUS EQUATION The relation ship between the rate constant, k and temperature can be seen in the k vs T graph:

k = Ae

-Ea

RT

1/T (K-1)

B) TEMPERATURE

ARRHENIUS EQUATION - DERIVATION The relationship between k and T is clearer when we further derive the Arrhenius Equation Ea

k ! AeNatural log both ends

RT Ea RT

ln k ! ln( A.e

)

ln k ! ln A ln(e RT ) Ea ln k ! ln e ln A RTThus See the linear relationship ?

Ea

(But ln e = 1)

Ea 1 ln k ! ( ) ln A R T y= m x +C

B) TEMPERATURE

Graph Representation Of The Arrhenius Equation Plotting a ln k vs 1/T graph would show a clearer relationship between k (Rate constant) and temperature Ea 1 ln k ! ( ) ln A R T

Where,Ea = Activation Energy R = 8.314 Jmol-1K-1 T = Absolute Temp A = Collision freq. factor

B) TEMPERATURE If the value of A (collision frequency factor) is not known and the same reaction conducted at two different temperatures.The Arrhenius equation at each temperature can be written and combined to formed the equation shown in the box. Ea 1 ln k1 ! ( ) ln A R T1and

Ea 1 ln k2 ! ( ) ln A R T2

Rearranging the equations would give

ln k1

Ea 1 ( ) ! ln A R T1

ln k2

Ea 1 ( ) ! ln A R T2

Since A is a constant E 1 E 1 ln k1 a ( ) ! ln k2 a ( ) R T1 R T2

ln k1 ln k2 !

Ea 1 E 1 ( ) a ( ) R T2 R T1

k1 Ea 1 1 ln ! ( ) k2 R T2 T1

Exercise: the Activation energyThe decomposition of hydrogen iodide,

2 HI (g)

H2(g) + I2(g)

has rate constants of 9.51 x 10-9 L mol-1 s-1 at 500 K and 1.10x10-5 L mol-1 s-1 at 600 K. Find Ea.DATA:

k1 = 9.51 x 10-9 L mol-1 s-1 k2 = 1.10 x 10-5 L mol-1 s-1

T1 = 500K T2 = 600K

SOLUTION:

k1 Ea 1 1 ln ! ( ) k2 R T2 T1Ea ! R ln( 1 k1 1 )( ) 1 k 2 T2 T1

1.10 v 10 5 1 1 1 )( ) Ea ! (8.314) ln( 9 9.51v 10 600 500Ea = 1.76 x 105 J/mol = 176 kJ/mol

C) CATALYST

A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed. Addition of a catalyst increases the reaction rate by increasing the frequency of effective collision. That is byDecreasing the Ea, and Correct orientation

C) CATALYST Addition of a catalyst changes the value of k (rate constant) .Reaction rate = k [ A ] [ B ]x y

(A & B = reactants) (x & y = rate order)

The catalyst reacts by reducing the Ea and increasing A, thus increasing the k.

Ea 1 ln k ! ( ) ln A R T

C) CATALYST

When Ea decreases, k increases, REACTION RATE increases

rateuncatalyzed < ratecatalyzed

uncatalyzed

catalyzed

Ea > E a

Reaction pathway

D) PARTICLE SIZE The smaller the size of reacting particles, the greater is the total surface area exposed for reaction and consequently the faster the reaction. In the case of heterogeneous systems, in which the reactants are in different phases, the area of contact between the reacting substances will influence the reaction rate considerably.