**Learning Outcomes:a) Define reaction rateb) Explain the graph of concentration against time in relation to reaction rate. c) Write differential rate equation.aA + bB cC + dDDifferential rate equation:rate =

d) determine reaction rate based on differential rate equation of a reaction.10 REACTION KINETICSLECTURE 1

**Chemical kinetics is the study of the rates of chemical reactions. -TimeOptimum yieldOptimum conditions

*control over reaction, *obtain products economically,*using optimum conditions

Importantindustrial processREACTION KINETICS

** Reaction Rate (R)The change in the concentration of reactant or product with time (M/s) General equation:

rate = (concentration) t OR

The ve sign indicates [A] decreases with time.The +ve sign indicates [B] increases with time.10 .1 Reaction Rate

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*ConcentrationTimeProductReactantGraph of concentration of reactant and product versus timeFNH

FNH

**

**The average rate is the rate over a period of time.The rate of reaction at a given time (specified 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 (at the point concerned)

Reaction rate

**Reaction: H2O2(aq) H2O(l) + O2(g) Reaction rates are obtained from the slopes of the straight lines; An average rate from the purple line. The instantaneous rate at t =300 s from the red line.The initial rate from the blue line.Rate of reactionblueredpurple

**instantaneous rate = rate at a specific time

31.bin

**The differential rate equationRate =

A differential rate equation enables the relationship between the rate of disappearance of reactants andthe formation of products.

Simple example: aA aB

Where:[ ] concentration of reactant or product (mol L-1)t time (s @ min @ hr etc)a and b are the stoichiometric coefficients

*Consider the reaction,

aA + bB cC + dD

Rate = A, B, C and D represent subtances in (g) or (aq) a, b, c and d are stochiomieric coefficients

**Example: The formation of NH3,The differential rate equationN2(g) + 3H2(g) 2NH3(g)The differential rate equation is;Rate = The equation means that the rate of disappearanceof N2 is 1/3 the rate of disappearance of H2 and 1/2the rate of formation of NH3.

** Consider the reaction, 2HI H2 + I2, determine the rate of disappearance of HI when the rate of I2 formation is 1.8 x 10-6 M s-1.Example 1:Rate =Rate == 1.8 10-6Solution:= -2 1.8 10-6 = - 3.6 10-6 M s-1

NOTE:d [HI] = - 3.6 10-6 M s-1 @ dt

-d[HI] = 3.6 10-6 M s-1 @ dt

Rate of disappearance oh HI is 3.6 10-6 M s-1FNH*

FNH

Write the rate expressions for the following reactions in terms of the disappearance of the reactants and the appearance of the products:(a) I- (aq) + OCl- (aq) Cl- (aq) + OI- (aq) (b) 3O2 (g) 2O3 (g)(c) 4NH3 (g) + 5O2 (g) 4NO (g) + 6H2O (g)*EXERCISE 1:

**EXERCISE 2: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.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)

**Consider the reaction,

2NO(g) + O2(g) 2NO2(g).

Suppose that at a particular time during the reaction nitric oxide (NO) is reacting at the rate of 0.066 Ms-1 At what rate is NO2 being formed?At what rate is molecular oxygen reacting?Exercise 3:

**Consider the reaction,

N2(g) + 3H2(g) 2NH3(g)

Suppose that at a particular moment during the reaction molecular hydrogen is reacting at the rate of 0.074 M s-1a) At what rate is ammonia being formed?b) At what rate is molecular nitrogen reacting?Exercise 4:

*Learning Outcomes:e) Define i. rate law.ii. Order of reaction. f) Write rate law with respect to order of reaction

h) Determine the order of reaction involving a single reactant using i. initial rate method ii. the units of rate constants, kLecture 2

*FNH*

k is called rate constantx and y is called order of reaction

The Rate Law (Rate equation)The 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.Rate [reactant]Rate = k [A]x[B]y

NOTE!The values of x and y can only be determined experimentally not from the stoichiometry of the balance chemical equation

FNH

*Order of reaction is defined as the power to which the concentration of a reactant is raised in the rate equation.

The order of a reactant is not related to the stoichiometric coefficients of the reactants in the balanced chemical equation. It can be only determined experimentally

ORDER OF REACTIONRate = k [A]x[B]y Order of reaction with respect to A is x Order of reaction with respect to B is y Overall order of reaction = (x + y)

The exponents x, y, can be integers, fractions or decimal or negative values.

* Example Overall reaction order: 2 + 1 = 3The reaction order respect to NO : 2The reaction order respect to O2 : 1 Solution:

*

Solution: The order of reaction with respect to CH3CHO : 3/2 The overall order of reaction : 3/2Solution:The order of reaction with respect to H2O2 : 1 The order of reaction with respect to I- : 1 The order of reaction with respect to H+ : 0 overall order of reaction: 1+1= 2

**The order of reactionFNH

FNH

Determining the Order of ReactionThere are 5 method to determine the order of reaction: Using rate lawUsing the unit of the rate constant.Using initial rate mwethodUsing half-life method (concentration vs time graph)Using linear graph method*

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* 1. Determining the order of reaction from rate law For reaction

Rate = k [A]xi) If x = 0Rate is not dependent on [A]Rate = k [A]0

Rate = kTherefore this reaction is zero order with respect to A

*

ii) If x= 1Rate = k [A]1Assume [A]i = 1.0M,Rate1 = k (1.0M)If the [A] is doubled from 1.0M to 2.0M,

Rate2 = 2 Rate1Doubling the [A] will double the rate of reaction.Therefore this reaction is first order with respect to A

*iii) If x = 2Rate = k[A]2Assume [A]i = 1.0 M,Rate1 = k (1.0 M)2If the [A] is doubled from 1.0 M to 2.0 M, Rate2 = 4 Rate1Doubling [A], the rate will increase by a factor of 4 (by fourfold)Therefore the reaction is second order with respect to A

* The units of rate constant, ki) zero order Rate = k [A]0 Rate = k unit k =unit rate= mol L-1 s-1 or Ms-1 2. Determining the order of reaction from unit of rate constant

*ii) First orderRate = k [A]1iii) Second orderRate = k [A]2

*

The rate equation, order of reaction and unit of k for general equation: A + B C show in table below:-

Note : Reaction order is always defined in term reactant(not product) concentrations

The rate equationOrder of reactionUnit of kR=k[A]0[B]0R=k0Ms-1R=k[A]1R=k[B]11s-1R=k[A]1[B]1R=k[A]2R=k[B]22M-1s-1R=k[A]2[B]13M-2s-1

*Example :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-]1b) Rate = k [S2O82-]1[I-]1Unit k

*

EXAMPLE: 3. Determination of the orders of reaction rate using initial rate method

*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. ;Do a similar calculation for the other reactant (s).Solution:The reaction is first order with respect to O2

*To find the order with respect to NO, we compare experiment 3 and 1, in which [O2] is held constant and [NO] is doubled:The reaction is second order with respect to NO

*The results of the kinetic studies are given below.Exercise 1:ClO2(aq) + 2OH- (aq) productsExplain what is meant by the order of reaction.Referring 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

exp[ClO2]M[OH-]MInitial rate, Ms-110.04210.01858.21 1 0-320.05220.01851.26 1 0-230.04210.02851.26 1 0-2

*When [A] is doubled, rate also doubles. But doubling the [B] has no effect on rate.When [A] is increased 3x, rate increases 3x, and increasing of [B] 3x causes the rate to increase 9x.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.Exercise 2:A + B CWrite rate law for this equation,

*Use the following data to determine the individual and overall reaction orders:

**Learning Outcomes:Write the integrated rate equation for zero, first and second order reactions.iii)To define half-life,t1/2

i) Perform calculation using the integrated rate equations.LECTURE 3Integrated Rate Law and Half life

*Is used to derive the rate law for zero order, first order and second order reaction.

Integrated Rate Law of a Zero Order Reaction

A zero order reaction is a reaction independent of the concentration of reactant.

A product

The rate law is given byrate = k[A]0rate = k[A] MrateIntegrated rate equations

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*

-d[A] = k dt - d[A] = kdt .(1) intergrating equation (1) - 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] = kt[A] = -k t + [A]oty = mx + c

y = m x + c

[A]rateMk= rate constant can be determined from the slope of the straight line

Half-life (t) Half life (t) is the time required for the concentration of a reactant to decrease to half of its initial value.

*

*

Half life of a zero order reaction

Substituting t = t1/2 , and [A] = [A]0 into the zeroorder reaction, gives 2 [A]0 - [A] = kt [A]0 [A]0 = kt1/2 2 Solving for t1/2 gives

t1/2 = [A]0 2k

*From the rate law, rate = k[A] Intergrated Rate Law of a First Order Reactions

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

Consider the reaction :

A productsy = mx + cRateMs-1[A] ,M

*For first order reaction,- d[A]dt=k[A]- d[A][A]=k dt

- d[A][A]=k dt- ln [A] = kt + csubstituting t = 0, [A]=[A]0- ln [A]0 = k(0) + c c = ln[A]0-ln [A] = kt ln[A]0Rate = k[A]

*Characteristic graphs of 1st order reactionln[A]ln[A]ottln[A]o ln[A] =kt

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

*ln[A]o ln[A] = ktkt = ln[A]0 ln[A]t == 66 s[A]0 = 0.88 M[A] = 0.14 MExample 1:

*Example 2: 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 Mb) the [H2O2 ] after 225 s.Solution :a)=

ln0.8820.6003.66 x 10-3 s-1 x t

* ln 1.47 = 3.66 x 10-3 s-1 x t= 105.26 sb)

ktln0.882

[H2O2]=3.66 x 10-3s-1x 225 s[H2O2] = 0.387 M

*Exercise 1:

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

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

*

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 2Half-life of a first-order reactiont = ln 2 k

*= 1200 s = 20 minutesHow do you know decomposition is first order?units of k (s-1)EXAMPLE 1:

*Example 2: 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:==1.29 x 103 s= 21.5 min

*

Problem 2

The decomposition of SO2Cl2 is a first-order reaction. SO2Cl2(g) SO2 (g) + Cl2 (g)

Write the rate differential equation for the reaction.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 )Problem 1

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)

**Integrated rate law of a Second Order ReactionsA 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 A productWhere

**

Rearrange to:- = kdt (1) d[A]

[A]2

= kt + c (2)

At t =0,

Combining (3) and (2) and rearranging the results gives:1[A]=+ ktIntegrating equation 1[A][A] = [A]o1[A]0= k(0) + c

c = 1 (3)[A]0

**Characteristic graphs for second order reactionrate[A]rate[A]2

**Graphs for second order reaction[A]t

1/[A] M-11/[A]ot

1/[A] 1/[A]otFNH

FNH

**

Half life of a second order reaction

+ ktSubstituting t = t1/2[A]= [A]o1=+ kt1/2t1/2= 1

k[A]022

**Example 1:Iodine atoms combine to form molecular iodine inthe gaseous phase I (g) + I (g) I2(g)This reaction is a second order reaction , with the rate constant of 7.0 x 109 M-1 s-1If 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 :i)

+ kt1=+ (7.0 x 109 x 2 x 60 )= 8.4 x 1011[A] = 1.190 x 10-12 Mii)[I2] = 0.06 M

=7.0 x 109 x 0.060=2.4 x 10-10 s

**[I2] = 0.42 M

7.0 x 109 x 0.42

== 3.4 x 10-10 s

**Lecture 4Objective:h) Determine the order of reaction involving a single reactant using:iii) half-life based on the graph of concentration against time.iv) linear graph method based on the integrated rate equation and rate law.

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FNH* 4. Determination of the orders of reaction rate using half life methodtZero order:PlotSketch a [A] vs t graph

t1/2 = [A]0 2k*A linear graph

FNH

FNH*[A]otx x x[A]o/2[A]o/4[A]o/8First order:

t = ln 2 k*Half-life is constant!

FNH

*[A]0[A]0/2[A]0/4[A]0/8tx2x4xSecond order:

*Half-life is twice the initial half-life.

*First-order reaction12344211/2

Example 1:

**Example 2:

The following results were obtained from an experimental investigation on dissociation of dinitrogen pentoxide at 45oCN2O5(g) 2 NO2(g) + O2(g)time, t/min0102030405060[N2O5] x 10-4 M1761249371533929Plot graph of [N2O5] vs time, determineThe order of the reaction the rate constant k

**Solution :[N2O5] x 10-4 /MTime ( min)180160140801201008060402010203040506070xx

**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

i)ii)k =ln2

20 min= 0.03 min-1

Zero Order:A plot of [A] vs t should be linear

* 5. Determination of the orders of reaction rate using linear graph method

First Order:A plot of ln [A] vs t should be linear

*

Second Order:A plot of 1/ [A] vs t should be linear

*

Example:The following data were obtained from the gas-phase decomposition of hydrogen iodide:

Find the order of HI?

Solution:Strategy; prepare a table listing [HI], ln [HI] and 1/[HI]

*

Time (h)0246[HI] (M)1.000.500.330.25

Time (h)[HI]Ln [HI]1/[HI]01.0001.0020.50-0.692.0040.33-1.103.0060.25-1.394.00

Then, sketch graphs:

*Because a plot of 1/[HI] vs t is linear, the decomposition of HI must be a second order.

**Summary of the Kinetics of Zero-Order, First-Orderand Second-Order Reactions012rate = krate = k [A]rate = k [A]2ln[A] = ln[A]0 - kt[A] = [A]0 - kt

**Zero order1st order2nd orderA productA productA productr = k [A]0r = k [A]1r = k [A]2Unit k = M s-1Unit k = s-1Unit k = M-1 s-1Integrated rate lawIntegrated rate lawIntegrated rate law[A]0 [A] = ktln([A]0 / [A]) = kt1/[A] 1/[A]0 = ktt1/2 = [A]0/2kt1/2 = ln2/kt1/2 = 1/k[A]0

The data refers to the decomposition of N2O4 at 100kPa and 380KExercisea)Determine the order with respect to N2O4b) Calculate half life of the reaction.c) Calculate the initial rate if the initial concentration of N2O4 isi) 1.5x103Mii) 3.1x10-2M*

[N2O4]/M0.60.91.2Initial rate/Ms-13.2x10-34.8x10-36.4x10-3

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**LECTURE 5Explain collision theory Define activation energy.Explain transition state theoryDraw energy profile diagram of a reaction

Objectives:Collision theory and transition state

**Collision Theory is the theory to explain the rate of chemical reactions. It is based on:-

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 for the reaction to occur.

According to this theory, not all collisions between the molecules result in the formation of products.Collision Theory

**

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.

** Activation energy is the minimum energy required to initiate the chemical reaction. Definition of Activation Energy (Ea)

**Importance of OrientationOrientation is unimportantOrientation is important

**Importance of OrientationOrientation is important

**Transition State TheoryThe 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.Activated complex- the species temporarily form by the reactant molecule as the result of the collision before they form product. (also called the transition state)

**Very unstable i.e. 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.

Characteristics of Activated Complex

**ReactantproductA 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 -DH for the reaction.

*Reactant molecules must have enough energy to overcome an energy barrier separating products from reactants, Ea.

DHEaProgress of reactionPotential energyActivated complexEnergy Profile Diagram: Exothermic reaction

**Energy Profile Diagram: Exothermic reactionEaDHEa (reverse reaction)Activated complexTransition state(Forward reaction)CO(g) + NO2(g) CO2(g) + NO(g)FNH

FNH

**

Product

Reactantactivated complex.Ea(reverse reaction)Ea (forward reaction)HEnergy Profile Diagram: Endothermic reaction

**Energy Profile Diagram: Endothermic reactionEa2NOCl 2NO + Cl2

DH

NOTE: Ea for reverse reaction

EXOTHERMICEa(reverse) = Ea(forward) + H

ENDOTHERMICEa(reverse) = Ea(forward) - HFNH*

FNH

**Energy profile diagram and activation energy for : a)Exothermic reaction b) Endothermic reaction

*KMP-SK027**

**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 productsb)The activation energy for the forward and the reverse reactionc)The heat of the reaction

**Factors affecting rate of reactionLECTURE 6

**Objective:To explain the effect of the following factors on the reaction rate.

Concentration or pressureTemperatureCatalystParticle sizeExplain the effect of temperature on reaction rate using Maxwell-Boltzman distributionExplain the effect of catalyst on activation energy based on energy profile diagram.

**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.PARTICLE SIZE:

The rate increases as the smaller the size of reacting particles .CATALYSTS:

Catalysts speed up reactions.Factors affecting rate of reaction PRESSURE

**A) CONCENTRATIONS OF REACTANTS

The frequency of collision increases with the concentration

4 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 discussed

Reaction rate = k [ A ]x [ B ]y Based on this equation,

(A & B = reactants)(x & y = rate order)Reaction rate concentration of reactantsREMINDER!

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

**Pressure

When pressure is applied, volume decreases. Number of molecules per unit volume increases resulting in more collisions per second.

Number of effective collision increases so the rate of reaction also increases.

*KMP-SK027**

**B) TEMPERATUREAs temperature increases, kinetic energy, of molecules increasesSo, more collisions occur in a given timeFurthermore, the higher the kinetic energy, the higher the energy of the effective collisions.So more molecules will have energy greater than activation energy, EaThus, the rate of reaction increases

**Distribution of Kinetic Energies of MoleculesRepresent total number of moleculeswith kinetic energy greater than Ea

**The area under curve represent the total number of molecules in the reaction

So that ,the area under each curve ( T1 and T2 ) is the same, because the total number of molecules used for temperature T1 and T2 are the same.

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**At higher temperature the average energy increases, the fraction of molecules with energy higher than Ea will increases also.

The shaded area under the curve represent the number of molecules with kinetic energy greater than the activation energy

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**

The graph for T1 shows that only a small fraction of molecules has sufficient energy ( the Ea) to react at T1.

At the higher temperature T2, the fraction of molecules that have sufficient energy( equal to or greater than Ea) to react increases.

So that rate of reaction will increases with increasing of temperature

*KMP-SK027**

**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, andCorrect orientation

C) CATALYSTFNH

FNH

**Function of catalyst: To speed up the reaction rate by providing an alternative pathway by lowering the activation energy.

**Addition of a catalyst changes the value of k (rate constant) .

Reaction rate = k [ A ]x [ B ]y(A & B = reactants)(x & y = rate order)

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

FNH

FNH

**rateuncatalyzed < ratecatalyzed13.6When Ea decreases, k increases,

REACTION RATE increasesEa > Ea

** Reaction pathwayEa (without catalyst)

**D) PARTICLE SIZEThe smaller the size of reacting particles, the greater the total surface area exposed for reaction .

The greater surface area causes a greater frequency of collisions,thus increasing the reaction rate.

*Arrhenius equationLECTURE 7

**Objective:a) State Arrhenius equation.

b) Relate temperature and activation energy to the rate constant based on the Arrhenius equation.

c) Determine k, Ea, T and A using Arrhenius equation by calculation and graphical method.

**ARRHENIUS EQUATION

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

Wherek = rate constantA = constant known as the collision frequency factore = natural log exponentEa = activation energy for the reactionR = universal gas constant (8.314 J mol-1 K-1)T = absolute temperature(in Kelvin)

**ARRHENIUS EQUATION

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

T (K)FNH

FNH

**ARRHENIUS EQUATION DERIVATION

The relationship between k and T is clearer when we further derive the Arrhenius Equation

Natural log both ends(But ln e = 1)ThusSee the linear relationship?y =m x + C

**Graph Representation Of The Arrhenius EquationPlotting a ln k vs 1/T graph would show a clearer relationship between k (Rate constant) and temperature

Where, Ea = Activation Energy R = 8.314 Jmol-1K-1 T = Absolute Temp A = Collision freq. factorA plot of In k versus 1/T gives a straight line with slope m is equal to Ea/R and whose intersect c with the y axis is In A

**

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.

Since A is a constantandRearranging the equations would give

**

DATA: SOLUTION:Exercise: the Activation energyThe decomposition of hydrogen iodide,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.Ea = 1.76 x 105 J/mol = 176 kJ/mol2 HI (g) H2(g) + I2(g)k1 = 9.51 x 10-9 L mol-1 s-1 T1 = 500Kk2 = 1.10 x 10-5 L mol-1 s-1 T2 = 600K

**Example 1:

The rate constants for decomposition of acetaldehyde

2HI(g) H2(g) + I2(g)

were measured at five different temperatures. The data are shown below. Plot ln k versus 1/T, and determinethe activation energy (in kJ/mole) for the reaction.

k (1/M s)T (oC)3.52 x 10-72833.02 x 10-53562.19 x 10-43931.16 x 10-34273.95 x 10-2508

*KMP-SK027**

**SolutionWe need to plot ln k (on the y-axis) versus 1/T( on the x-axis). From the given data we obtain

*

ln k1/T (K-1)14.8601.80 x 10-310.4081.59 x 10-38.4261.50 x 10-36.7591.43 x 10-33.2311.28 x 10-3

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**

Plot of ln k versus 1/T. The slope of the line is calculated from two pairs of coordinates. ln k

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** Slope = (-14.0) (-3.9) (1.75 1.3) x 10-3 K-1 = -2.24 x 104 K

The slope, m = Ea/R. Ea = R (m) = (8.314 J K-1 mol-1) x ( 2.24 x 104 K) = 1.9 x 105 J mol-1 = 190 kJ mol-1

*KMP-SK027**

**An equation relating the rate constants k1 and k2 at temperatures T1 and T2 can be used to calculate the activation energy or to find the rate constant at another temperature if the activation energy is known.

ln k1 = ln A Ea/RT1 ln k2 = ln A Ea/RT2

Subtracting ln k2 from ln k1 gives

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

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

*KMP-SK027**

**Example 10.7.2: The rate constant of a first-order reaction is 3.46 x 10-2s-1 at 298 K. What is the rate constant at 350 K if the activation energy for the reaction is 50.2 kJ/mole?

SolutionGivenk1 = 3.46 x 10-2 s-1k2 = ?T1 = 298 KT2 = 350 K

Subtituting in equation

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

*KMP-SK027**

** ln 3.46 x 10-2 = 50.2 x 103 J/mol 298K 350K k2 8.314 J/K mol (298K)(350K)

Solving the equation givesln 3.46 x 10 -2 = 3.01k2 3.46 x 10 -2 = 0.0493 k2 k2 = 0.702 s 1

*KMP-SK027**

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Changes in conditionReaction rateRate constantIncrease in temperatureIncreaseIncreaseDecrease in temperatureDecreaseDecreaseAddition of catalystIncreaseIncreaseIncrease in pressureIncreaseNo effectDecrease in pressureDecreaseNo effectIncrease concentrationIncreaseNo effectDecrease concentrationDecreaseNo effect

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