Physical Chemistry Examinations [1 ed.] 9781614705819, 9781606923450

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Copyright © 2009. Nova Science Publishers, Incorporated. All rights reserved. Ibhadon, A.O.. Physical Chemistry Examinations, Nova Science Publishers, Incorporated, 2009. ProQuest Ebook Central,

Copyright © 2009. Nova Science Publishers, Incorporated. All rights reserved. Ibhadon, A.O.. Physical Chemistry Examinations, Nova Science Publishers, Incorporated, 2009. ProQuest Ebook Central,

Copyright © 2009. Nova Science Publishers, Incorporated. All rights reserved.

PHYSICAL CHEMISTRY EXAMINATIONS

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Ibhadon, A.O.. Physical Chemistry Examinations, Nova Science Publishers, Incorporated, 2009. ProQuest Ebook Central,

Copyright © 2009. Nova Science Publishers, Incorporated. All rights reserved. Ibhadon, A.O.. Physical Chemistry Examinations, Nova Science Publishers, Incorporated, 2009. ProQuest Ebook Central,

Copyright © 2009. Nova Science Publishers, Incorporated. All rights reserved.

PHYSICAL CHEMISTRY EXAMINATIONS

A.O. IBHADON

Nova Science Publishers, Inc. New York

Ibhadon, A.O.. Physical Chemistry Examinations, Nova Science Publishers, Incorporated, 2009. ProQuest Ebook Central,

Copyright © 2010 by Nova Science Publishers, Inc. All rights reserved. No part of this book may be reproduced, stored in a retrieval system or transmitted in any form or by any means: electronic, electrostatic, magnetic, tape, mechanical photocopying, recording or otherwise without the written permission of the Publisher. For permission to use material from this book please contact us: Telephone 631-231-7269; Fax 631-231-8175 Web Site: http://www.novapublishers.com NOTICE TO THE READER The Publisher has taken reasonable care in the preparation of this book, but makes no expressed or implied warranty of any kind and assumes no responsibility for any errors or omissions. No liability is assumed for incidental or consequential damages in connection with or arising out of information contained in this book. The Publisher shall not be liable for any special, consequential, or exemplary damages resulting, in whole or in part, from the readers‘ use of, or reliance upon, this material.

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Independent verification should be sought for any data, advice or recommendations contained in this book. In addition, no responsibility is assumed by the publisher for any injury and/or damage to persons or property arising from any methods, products, instructions, ideas or otherwise contained in this publication. This publication is designed to provide accurate and authoritative information with regard to the subject matter covered herein. It is sold with the clear understanding that the Publisher is not engaged in rendering legal or any other professional services. If legal or any other expert assistance is required, the services of a competent person should be sought. FROM A DECLARATION OF PARTICIPANTS JOINTLY ADOPTED BY A COMMITTEE OF THE AMERICAN BAR ASSOCIATION AND A COMMITTEE OF PUBLISHERS. LIBRARY OF CONGRESS CATALOGING-IN-PUBLICATION DATA Ibhadon, A. O. Physical chemistry examinations / A.O. Ibhadon. p. cm. ISBN:  (eBook) 1. Chemistry, Physical and theoretical--Examinations, questions, etc. I. Title. QD456.I24 2009 541.076--dc22 2008036859

Published by Nova Science Publishers, Inc. † New York

Ibhadon, A.O.. Physical Chemistry Examinations, Nova Science Publishers, Incorporated, 2009. ProQuest Ebook Central,

CONTENTS

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Preface

vii

Chapter 1

Chemical Kinetics

Chapter 2

Collision Theory and Free Radical Mechanism

25

Chapter 3

Chemical Thermodynamics (First Law)

53

Chapter 4

The Second Law of Thermodynamics

83

Chapter 5

Surface Chemistry

107

Chapter 6

Units and Physical Constants in Physical Chemistry

133

Index

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1

135

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PREFACE Physical Chemistry Examinations was written to meet the needs of students who find studying physical chemistry a difficult and at times a frustrating experience. I have observed over several years of teaching this subject at the undergraduate level, that students are unrelenting in their desire for so called ‗worked examples‘ and answers to ‗typical examination‘ questions. This text has been prepared primarily to meet this need and assist students at the early years of the undergraduate degree programme in chemistry and chemical engineering understand this subject better and perhaps appreciate the relevance and beauty of the entire subject as a whole. Worked examples have been provided in every chapter as well as exercises and ‗typical examination questions‘. Care has been taken to ensure that worked examples are explained on a step by step basis to avoid confusion and misunderstanding on the part of the reader. S.I units have been explained throughout and a reasonably good grounding in mathematics is assumed throughout this book. This text is also aimed at the physical chemistry lecturer/teacher who will discover that students are the same everywhere in their desire for ‗worked examples‘ and solutions to ‗typical examination questions‘. Physical Chemistry Examinations is thus a useful companion to the physical chemistry lecturer and teacher as well his students. It is not a replacement for a more comprehensive text on the subject and students of physical chemistry are advised to avail themselves of such texts. I am grateful to Suzanne Handyford Styring formerly a lecturer at the University of Hull and whose lecture notes were helpful in preparing this text. Alex Ibhadon CChem FRSC Cleveland, England 2009

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Chapter 1

CHEMICAL KINETICS 1.0. LEARNING OBJECTIVES/OUTCOMES

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A good understanding of this chapter, the exercises as well as the worked examples, should make it easier for students to interpret data on rates of reaction in terms of the theories relating e to: • • • • • •

Order of a reaction, rate constants and rate equations. Van‘t Hoff method and integrated rate equations Half-life of a reaction and its determination Molecularity of a reaction and elementary reactions Arrhenius equation, activation energy and frequency factor Maxwell Boltzman distribution of reaction energies

Students should be able to investigate experimentally, the factors that can influence the rates of chemical reactions. To achieve this objective, students need to note the following: • • • •

Methods of experimentation Evaluation of rate constants for zero, first and second order reactions Variation of rate of reaction with temperature, pressure and concentration Calculation of activation energy of a reaction

Students should be able to interpret kinetic data for appropriate reactions involving complex rate equations in terms of simple consecutive processes

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2 •

•

Free radical reactions and steady-state theory Interpret kinetic data on appropriate gas-phase reactions in terms of a unimolecular mechanism Linderman theory of unimolecular reactions Finally, students should be able to evaluate kinetic evidence in the establishment of a mechanistic interpretation of chemical reactions.

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1.1.WHAT IS CHEMICAL KINETICS Chemical kinetics is the study of the rates(speed) at which chemical reactions take place. The rate of reaction is measured by the rates at which reactants are consumed and products produced. Reactants are the starting materials and products are what are formed at the end of the reaction, that is, the outcome or end result of the reaction. Chemical kinetics also includes a study of how rate changes in presence of a catalyst and the mechanism by which the reaction takes place. Rate is important in simple and complex reactions whether these take place in the laboratory or on an industrial scale. Consider the pharmaceutical industry and the range of products available (drugs, personal and skin care products), kinetics is involved in the manufacturing processes that result in the production of various products including drugs and medicines, figure 1.1 It is important to note that all chemical reactions in industry and laboratory do not necessarily follow the same steps shown in figure 1.1

Figure 1.1. Typical Chemical Processes in a Manufacturing Plant.

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Chemical Kinetics

3

Why Study Reaction Kinetics (Rates of a Reaction)? The reasons why we study the rates of reactions are many and some of these are given below: • •

•

To predict how quickly a reaction will reach equilibrium To determine the dependence of the rate of reaction on experimental variables such as temperature, pressure, presence of a catalyst and the nature of the reaction surface This enables reactions to be optimised by choosing the right conditions especially in industrial processes where optimization is very important. To understand the mechanism of the reaction, that is, the sequence of elementary steps through which the reaction takes place . An example is the reaction between hydrogen and bromine, the mechanism of which is provided below in a series of initiation, propagation and termination steps.

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H2 + Br2 → 2HBr The stoichiometric equation (such as the above reaction), provides information about the mass and mole balance. It is an expression for the equilibrium constant but doesn‘t tell you how the reaction occurs or how fast the reaction takes place or whether the reaction goes through a series of steps as in the following example.

Br2 → 2Br (aq) Br. + H2 → HBr + H. H.+ Br2 → HBr + Br. Br. + Br. → Br2

initiation step propagation termination step

The stoichiometric equation is the sum of the elementary steps in the mechanism and it is the mechanism that explains what happens in the reaction and is determined experimentally. Generally, in industry as well as in the laboratory, the rate of a reaction is measured by using a number of methods including: a.

Changes in colour, pH, conductivity or refractive index change during reaction

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b. Change in concentration with time of a reactant or product during a reaction.

1.2. KINETIC ANALYSIS ENABLES US TO 1. Determine the stoichiometry of the reaction. 2. Collect data on the change in concentrations of reactants and products with time at a constant temperature (isothermal process). The rate of a reaction varies with temperature(see the Arrhenius equation)

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Rate of reaction The rate of a reaction is defined as the rate of change of the concentration of a designated species. This is usually obtained from the slope (dy/dx) of a plot of concentration (y axis) in mol L-1 (or Torr if the reactants/products are gases) as a function of time (x axis) in s. The rate of reaction is in mol L-1 s-1 .( or Torr s-1). You can see from figure 1. 2 that the rate of reaction changes with time. It is only possible, therefore, to determine the rate of reaction at a specific time. We can therefore define rate of consumption of a reactant (R) = -d[R]/dt – negative slope(a). rate of production of a product (P) = d[P]/dt – positive slope. (Note that while the concentration of the product is increasing with time, the concentration of the reactant is decreasing with time). Note that different species in a chemical reaction are consumed and produced at different rates. For example, consider the production of ammonia gas from nitrogen and hydrogen represented by the following equation N2(g) + 3H2(g) → 2NH3 (g) Note from the above equation that the rate of formation of NH3 is twice the rate of removal of N2. The rate of removal of H2 is three times the rate of removal of N2. This is explained by the stoichiometry of the reaction.

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How does the rate of removal of H2 compare with the rate of production of NH3? It is therefore important to specify which species of the reaction is being used to determine the rate. Consider the following example.

Figure 1.2. Concentration as a function of time for a chemical reaction.

Exercise 1.1. From the reaction (stoichiometric equation) shown below, work out the rate of consumption of NOBr and the rate of formation of Br2, given that the rate of formation of NO is 1.6 x 10-4 mol L-1 s-1 2NOBr (g) → 2NO (g) + Br2(g)

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If the stoichiometry of a reaction is known and the rate of change of one of the species is known, then the rate of change in concentration of the other species can be determined.

1.3. RATE LAWS AND RATE CONSTANTS It is observed that the measured rate of a reaction is proportional to the molar concentrations of the reactants raised to a power. For example, for the reaction, A + B → products,

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the rate of a reaction (R) between A and B may be written as R = k[A]2[B]. This would have been found experimentally and shows that the rate varies with the square of [A] and varies with [B]. In this reaction, k is known as the rate constant. It is dependent on the reaction being studied as well as the temperature, but independent of concentration

Rate Equation This is an equation that expresses the rate of a reaction in terms of the concentrations of reactants or products in the overall reaction and is determined experimentally. The rate equation is used for: • • •

When combined with the rate constant, the rate equation can be used to predict the rate of reaction from the composition of the mixture. Can be used to predict the concentrations of the reactants and products. Is a guide to the mechanism of the reaction

Reaction Order The order of a reaction with respect to each species(reactant, product) in a reaction is the power or exponent to which the concentration of that species is raised in the rate equation and is determined experimentally. For the reaction A + B = products, the rate of reaction, R = k[A]2[B] then the reaction is second order in A and first order in B. The overall order of the reaction is the sum of the orders in the rate equation, so that for R = k[A]2[B], the overall order of reaction is third order. Orders of reaction need not be integral numbers – they can be fractional, zero or even negative numbers. For example, in the decomposition of ozone 2O3(g)  3O2 (g)

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Chemical Kinetics

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The rate equation is given as R = - d[O3]/dt = k[O3]2[O2]-1 The reaction is second order with respect to O3 and has an order of –1 for O2 and is therefore first order overall ( +2 -1). Let us consider the example given below NO2(g) + CO (g)  NO (g) + CO2(g) The rate equation is = -d [NO2]/dt = k [NO2]2 This means that the reaction is second order overall, second order with respect to NO2 and zero order with respect to CO, that is, the reaction is independent of CO concentration. The rate equation may not have the form k[A]x[B]y[C]z … in which case the reaction does not have an order. Consider the reaction between bromine and hydrogen H2 + Br2  2HBr,

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The rate of formation of HBr = k [H2][Br2]3/2/([Br2]+k‘[HBr]) The reaction is first order in H2 but has an indefinite order in Br2 and HBr and therefore an overall order for the reaction cannot be determined.

How Are the Unit(S) of K Determined? The units of k depend on the overall order of reaction. Units of rate of reaction = mol L-1 s-1 or Torr s-1 Rate of reaction = k [reactants]n therefore units of k = mol L-1 s-1 or (mol L-1)n

Torr s-1 = (mol L-1)-(n-1)s-1 or (Torr)-(n-1)s-1 or (Torr )n

For example, n 0 1 2 1/2

Units of k / concentration Mol L-1 s-1 s-1 mol-1Ls-1 mol1/2L-1/2s-1

Units of k / pressure Torr s-1 s-1 Torr-1s-1 Torr 1/2 s-1

n = the order of reaction.

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1.4. DETERMINATION OF THE RATE EQUATION AND ORDER The rate of a reaction cannot be measured directly. It is either measured as the loss of reactants and /or gain in products. These can be measured in a number of ways including, for example, taking samples at different times as in the reaction between dimethyl ketone and iodine CH3COCH3 + I2  CH3COCH2I + HI The rate of the reaction is determined by following(measuring) the loss of reactant by thiosulphate titration and product gain by AgNO3 titration and plot of concentration against time. Other ways of measuring the rate of reaction include: •

Following changes in colour, conductivity, viscosity, pressure with time.

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The way the rate varies with concentration may be used to determine the order of reaction and there are a number of methods that may be used to do this. Let us consider two situations: a. One reactant is involved in the reaction such that A = products and b. two reactants are involved in the reaction as in A + B = products (i) For the reaction A  products Rate = -d[A]/dt , that is loss of reactant A Rate equation = -d[A]/dt= R = k[A]n

(1.1)

where n = order of reaction The above cases can be treated using a number of methods which we shall soon learn

1.4.1. The Van’t Hoff Method The order of the reaction is calculated from the gradient of a plot of the log of the rate vs the log of the concentration of A. From equation (1) and taking logs,

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Chemical Kinetics

9

we obtain log R = log k + nlog[A]. A simple plot of log R vs log[A] is a straight line of gradient n(order of reaction) and intercept log k. The advantage of the van‘t Hoff method is that it enables the order of a reaction to be determined in a straight forward manner.

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Figure 1.3. Variation of Reactant and Product Concentration with time.

1.4.2. The Initial Rates Method The rate is measured at the start of the reaction for several different initial concentrations. Consider a reaction with the rate equation represented as , Rate equation (ro)=-d[A]/dt = k[A]no

(1.2)

where ro = initial rate The value for ro is found by drawing a tangent to the curve at t = 0 and calculating the slope as illustrated in figure 1.4 A series of experiments are carried out at different initial concentrations of A so that the data obtained is a table of initial [A] with initial rates as well as rate equations,

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Figure 1.4. The Initial Rates Method.

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The method of initial rates used in determining the order of a reaction (ro)1 = kan1

(1.3)

(ro)2 = kan2

(1.4)

Divide one equation by the other, such that (ro)1/(ro)2 = ka1n/ ka2n = (ro)1/(ro)2 = (a1/ a2)n For example, using the following data [A]o/mol L-1 1.96 x 10-2 2.57 x 10-2

ro/mol min-1 3.14 x 10-4 4.11 x 10-4

The initial concentration of A is increased by a factor of 2.57 x 10-2/1.96 x 10= 1.31 The initial rate of reaction is increased by a factor of 4.11 x 10-4/3.14 x 10-4 = 1.31 The answer we have obtained indicates that the rate is directly proportional to the initial rate, and hence the reaction is first order. We could have also determined the order mathematically as 1.31 = 1.31n, therefore n =1. You have to take the log of both sides of the equation as log 1.31 = n log 1.31 and then determine n from this equation (where n is the order of the reaction). 2

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Exercise 1.2. Calculate the order of reaction from the following data. [A]o/mol L-1 0.0167 0.0569

ro/mol min-1 3.61 x 10-2 4.20 x 10-1

Solution The initial concentration of A is increased by a factor of 0.0167/0.0569 = 2.93 x 10-1 The initial rate of reaction is increased by a factor of 3.61x 10-2/4.20 x 10-1 = 8.62 x 10-2 . Therefore 8.62 x 10-2 = (2.93 x 10-1)n . To solve this expression we need to take logs of both sides of the equation to give

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ln 8.62 x 10-2 = n ln 2.93 x 10-1 Rearrange the equation and solve for n. You will find that n = 2, which means that the reaction is second order. The above method does have some advantages as well as disadvantages: Advantages: There is no product interference; the method is simple and direct. Disadvantages: It is difficult to measure the initial rate accurately and the initial rates method cannot be used for a reaction in which the maximum rate is not reached immediately.

Exercise 1.3. How does the initial rate vary for a zero order reaction and a third order reaction? Use the equation (ro)=-d[A]/dt = k[A]no

(1.6)

to explain your answer.

1.4.3. Integrated Rate Equations Integrated rate equations (functional and graphical methods) are often used to demonstrate that the order of reaction is first order, second order, etc. It is a test and is not used to determine the order of reaction. The value of n has been

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determined from other methods and is substituted into equation (1.1). The rate equation, which is a differential equation, can now be integrated as kt = f[A]. [A] = (a-x) (1.7) where a = initial concentration and x = concentration of products. We shall now consider the integrated rate equations for a zero, first, second and third order reactions.

Zero Order Reactions (N = 0) dx/dt = k(a-x)0 = k,

(1.8)

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because (a-x)0 = 1. Rearranging the above equation yields dx = kdt and if we integrate this equation, we obtain x + c = kt, where c = integration constant. To determine the integration constant, we insert the initial conditions that at t = 0, x = 0. Therefore c = 0 and kt = x, hence a plot of [x] against t will yield a straight line, figure 1.5, of gradient k if the reaction is zero order. Let us now consider the following cases.

Figure 1.5. Plot for a Zero order reaction.

First Order Reactions( N = 1) dx/dt = k(a-x)

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(1.9)

Chemical Kinetics

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Rearrangement of the above equation yields equation [1.10] which on integration yields equation [1.11] dx/(a-x) = kdt (1.10) -ln(a-x) + c = kt

(1.11)

To determine the integration constant, we insert the initial conditions that at t = 0, x = 0, therefore, ln a + c = 0 and hence c = ln a and ln a-ln(a-x) = kt or ln [a/(a-x)] = kt Therefore a plot of the term on the left of equation [1.12] against time ln [a/(a-x)] vs t

(1.12)

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will give a straight line of gradient -k if the reaction is first order. This is illustrated in figure 1.6

Figure 1.6.

Integrated Rate Equations for Second Order Reactions(N = 2) Typically, second order reactions can be represented as dx/dt = k(a-x)2

(1.13)

Equation [1.13] can be rearranged to give

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14 -dx/(a-x)2 = kdt (1.14) which on integration, yields 1/(a-x) + c = kt (1.15)

integrate -xndx  {-xn+1/(n+1)}+ c for n At t = 0, x = 0

-1

c= -1/a

1/(a-x) – 1/a = kt

(1.16)

Rearrangement of equation [1.16] yields x/[a(a-x)] = kt and hence x/(a-x) = a kt

(1.17)

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Plot of x/(a-x) vs t will give a straight line graph of gradient ak if the reaction is second order.

Figure 1.7.

Integrated Rate Equation for Third Order Reactions( N=3) The integrated rate equation for a third order reaction can be stated as -dx/dt = k(a-x)3

(1.18)

Rearrangement of the above equation gives

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-dx/(a-x)3 = kdt Integration of equation (19) gives integrate -xndx  {-xn+1/(n+1)}+ c for n

(1.19)

-1

1/(2(a-x)2 )+ c = kt

(1.20)

At t = 0, x = 0 therefore c = -1/(2a2) 1/(2(a-x)2 )– 1/(2a2 ) = kt

(1.21)

rearrangement of equation (21) gives 1/(a-x)2 – 1/a2 = 2kt

(1.22)

Plot of 1/(a-x)2 vs t is a straight line graph of gradient 2k, figure 1.8, and intercept of –1/a2 if the reaction is third order. This is illustrated in figure 1.8

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Generally for n

-1, dx/dt = k(a-x)n and integration gives

(n+1)kt = 1/((a-x)n+1) – 1/(a-x)n+1 . The plot of 1/ ((a-x)n+1 vs t is a straight line of gradient (n+1)k and the intercept is = – 1/(a-x)n+1

Half-Life Method The half-life is the time taken for the concentration of a reactant or product to fall to half its initial value. This means that when x = a/2 t = t1/2 Substitute a/2 for x in the integrated rate equation and re-arrange to give an equation for t1/2. Summary Order n=0, n=1 n=2 n=n

Integrated rate equation for half-life x = kt a/2 = kt1/2 ln(a/(a-x)) = kt 1/(a-x) – 1/a = kt 1/(a-x)n-1 – 1/an-1 = (n-1)kt 1/an-1}k

Equation t1/2 = a/2k t1/2 = ln 2/k t1/2 = 1/ak t1/2 = {2n-1-

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Figure 1.8.

1.5. MORE COMPLEX REACTIONS

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For the reaction, A+B = products, the reaction may have an overall order of 2 or more. We will look at the integration of the rate equation if the reaction is second order. For a second order reaction, dx/dt = k(a-x)(b-x) Rearrangement of the equation [1.23] gives kdt = dx/(a-x)(b-x) Express equation [1.24] in terms of partial fractions to obtain,

(1.23) (1.24)

kdt = Adx/(a-x)+ Bdx/(b-x)

(1.25)

where A = 1/(b-a) and B = -1/(b-a)

(1.26)

Integration gives kt = -A ln((a-x)-B ln(b-x) + c At t = 0, x = 0 and therefore c = A ln a+ B ln b So that kt = Alna/(a-x)+ Bln b/(b-x)

(1.27) (1.28) (1.29)

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Rearrange and substitute for A and B to give kt = 1/(b-a) ln {a(b-x)/(b(a-x))}

(1.30)

(b-a)kt = ln {a(b-x)/(b(a-x))}

(1.31)

(b-a)kt = (ln a/b)+ln{(b-x)/(a-x)}

(1.32)

A plot of ln{(b-x)/(a-x)}vs t is again a straight line of gradient (b-a)k and intercept ln a/b . If the rate equation takes the form dx/dt = k(a-x)n(b-x)m then it is probably best to use the isolation method.

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1.5.1. Isolation Method The determination of the rate equation in chemical reactions is simplified by the isolation method. This method ensures that the concentrations of all the reactants except one are present in excess. If a species is present in a large enough excess, then it may be assumed that the concentration of that species will remain constant throughout the reaction. The methods described so far may then be used to determine the order of reaction. For example, the initial rate method At t = 0 (ro) = kanbm (1.33) Taking logs of equation (1.33) gives ln(ro) = ln k + n ln a + m ln b If A is present in excess and [B] is varied then n ln a remains constant and a plot of ln(ro) vs ln b is a straight line of gradient m and intercept of ln k + n ln a. If B is present in excess and [A] varied then m ln b remains constant and a plot of ln(ro) vs ln a is a straight line of gradient n and intercept of ln k + m ln b.

Exercise 1.4. 1. The gas phase decomposition of HI is second order and has a rate constant of 30 L mol-1 min-1 at 443 K. a. How long will it take for the [HI] to fall from 0.01 to 0.005 M? b. Calculate [HI] after 10 min.

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18 2HI = H2 + I2 Rate = dx/dt = k[HI]2

Solution a.

Use the integrated rate equation for a second order reaction

1/(a-x) – 1/a = kt where a = 0.01M, (a-x) = 0.005 M and k = 30 L mol-1 min-1 substitute into the integrated rate equation above to yield t = 3.3 min. or use t1/2 = 1/ak = 3.3 min b. Using the integrated rate equation as in (a) above t = 10, a = 0.01 therefore (a-x) = 2.5 x 10-3 M so that [HI] after 10 min is = 2.5 x 10-3 M

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2. The inversion of sucrose to glucose and fructose occurs in dilute acid solution according to the following equation. The concentration of sucrose was followed with time and shown in the table below Sucrose + H+ = glucose + fructose [sucrose]/M Time/min

0.316 0

0.274 39

0.238 80

0.190 140

0.146 210

Work out : i. the reaction order ii. write down the rate equation iii. work out the concentration of sucrose after 175 min.

Solution i. First order reaction R = k[sucrose] ii. First order plot of ln a/(a-x) vs t or ln(a-x) vs t to obtain k from gradient. iii. Using the integrated rate equation, if t =175 min and a = 0.316 then (a-x) = 0.167

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Chemical Kinetics

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Summary We have discussed rates of reaction and defined what the rate of a reaction is. We are now familiar with rate equation, rate constants and how to determine the units of the rate constant, k. We have also learnt that the order of a reaction can be determined experimentally using a number of methods including: a. the Van‘t Hoff method, for example, for rate equation R = k[A]n ; b. the Initial rate method and in this case we can integrate the above equation as ln R = ln k + nln[A] . A plot of ln R vs ln [A] is made and the tangent to the curve at t = 0. and ro gives the order of the reaction. To do this, a series of experiments are carried out and we can then use the equation (ro)1/(ro)2 = (a1/ a2)n.

(1.34)

Taking logs of equation[1.34] we obtain

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ln(ro)1/(ro)2 = n ln(a1/ a2),

(1.35)

from which the order of the reaction can be determined. In addition to the two methods summarised above, we can also determine the order of a reaction using: i. Half-life method ii. Compare initial rate with half-life. iii. Function method – test of the order of reaction—We need to know what graphs to plot for a zero, first and second order reaction and how to obtain k from these plots. For more complex reactions, we can use the Isolation method to determine the order of the reaction, for example, R = k(a-x)n(b-x)m We have stated that the above experiments should be carried out at constant temperature as the rate of reaction is dependent on the temperature at which the reaction is carried out. We are now going to look at why this is the case.

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1.6. EFFECT OF TEMPERATURE ON RATE OF REACTION The plot of the rate constant of a reaction(k against T as shown below in fgure 1.9 is known as an Arrhenius plot. The plot shows that the rate constant increases exponentially with temperature. Arrhenius found that ln k was proportional to 1/T. This can be expressed as follows: k = Ae-Ea/RT Arrhenius equation (equation 1.36)

Figure 1.9.

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where a plot of ln k as a function of 1/T gives a straight line of gradient = –Ea/R and intercept A. A is the pre-exponential factor (units same as k). Ea is the activation energy, ( J mol-1) and T is the temperature in K.

Exercise 1.5. The Determination of the Arrhenius Parameters The rate of the second-order decomposition of acetaldehyde (ethanal, CH3CH0) was measured over the temperature range 700-1000 K, and the rate constants are reported below. Determine Ea and A. T/K K/(Lmol-1s-1)

700 0.011

730 0.035

760 0.105

790 0.343

810 0.789

840 2.17

910 20.0

1000 145

Solution The data can be analysed by plotting ln(k/Lmol-1s-1) against 1/(T/K) to obtain a straight line plot. The slope of this line is (-Ea/R)/K and the intercept at 1/T = 0 is lnA). We can draw up the following table:

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Chemical Kinetics 103(K/T) 1.43 (x-axis) ln(k/Lmol-1s-1) -4.51 (y-axis)

21

1.37

1.32

1.27

1.23

1.19

1.101.009

-3.35

-2.25

-1.07

-0.24

0.77

3.00

4.98

Now plot ln k against 1/T. The least-squares best fit of the line has a slope of 2.21x104 and an intercept of 25.857. Therefore, the activation energy is Ea = (2.21 x 104K) x (8.3145 J mol-1) = 183.75 kJ mol-1 A = e 25.857 = 1.69x1011 L mo1-1s-1

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In figure 1.10, note that you are required to plot the reciprocal of the temperature and not just the temperature. This is a common mistake that students make in examinations (they plot lnk against T) and at times, query the data provided if they think they are right. Note that A (the frequency or pre-exponential factor) has the same units as k , the rate constant. The slopes and intercepts of graphs are always dimensionless, and care must be taken to relate the value from the graph to the physical quantity by noting how the data have been plotted.

Figure 1.10. Determination of Arrhenius Parameters.

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Exercise 1.6. Determination of Arrhenius Parameters Determine A and Ea from the following data: T/K k/(Lmol-1s-1)

300 7.9 x 106

350 3.0 x 107

400 7.9x107

450 1.7x108

500 3.2x108

[Ans: A = 8x1010 L mo1-1s-1, Ea = 23 kJ mol-1]

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Comments Treat as in exercise 1.5. Note that the activation energy, Ea, is given by the slope of the plot of ln k against 1/T and this means that the higher the activation energy, the stronger the temperature dependence of the rate constant, that is, the steeper the slope. A high activation energy means that the rate constant depends strongly on temperature. If a reaction has zero activation energy, its rate is independent of temperature. In some cases the activation energy is negative, which indicates that the rate decreases as the temperature is raised. We shall see later that this type of behaviour is characteristic of a reaction that has a complex mechanism. Significance of Arrhenius Parameters What do these Arrhenius parameters mean and what is their significance? For gas-phase reactions, we can ask the following questions: i. What is needed for a reaction to occur in the first place? ii. Collisions – unless molecules come together they can‘t react. The more frequently they come together, the faster they react, that is, an increased collision rate increases the rate of reaction. This is why the preexponential or frequency factor, A iii. Does every collision result in a reaction? The answer is NO, because every reaction needs a certain critical energy to enable the molecules to react, the activation energy, Ea We should now be able to explain why the rate constant is dependent on temperature. At higher temperatures, molecules have a higher kinetic energy i. therefore the collision rate is increased ii. more molecules with enough energy to exceed the critical energy required for a reaction to occur.

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Figure 1.11. Boltzmann‘s Distribution sketch.

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The fraction of molecules with energy exceeding the activation energy is given by e-Ea/RT (1.37) This is illustrated by the Boltzmann‘s distribution, figure 1.11. Is the collision frequency dependent on anything else other than temperature? The answer is yesthe collision frequency also depends on the concentration of the reactants. This will have an effect on the number of collisions. For a reaction between A and B, a doubling of the concentration of either species will lead to a doubling of the collision frequency, that is, the collision frequency is proportional to the concentrations of A and B. The pre-exponential factor , A, is related to the concentration of reactants and the rate at which they collide.

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Chapter 2

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COLLISION THEORY AND FREE RADICAL MECHANISM The collision theory is a theory of reaction rates developed from the kinetic theory of gases. Consider a reaction between two molecules. For molecules to react they must first collide. The collision frequency is the number of collisions made by one molecule per unit time(the units of collisions is m-3s-1.). For the reaction between hydrogen and iodine to form hydrogen iodide, consider that the I2 molecules are stationary and single H2 molecule is moving. The factors affecting collision frequency (Z) are velocity (related to the temperature, T), the concentration of molecules and size of molecules. This will be explained in more detail later. The numbers of successful collisions that lead to a reaction depend on whether the molecules collide with enough energy to exceed the activation energy so that the rate of reaction = Z e-Ea/RT

(2.1)

Also rate of reaction = k nAnB

(2.2)

where nA and nB are the concentrations in molecules(m-3). If equations (2.1) and (2.2) are combined, we obtain equation [2.3] k = Z/nAnB e-Ea/RT.

(2.3)

We can replace Z/nAnB by Zo where Zo is the collision number (units collisions m3 molecule-1 s-1).

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EXERCISE 2.1. Given the equation below and the following information, calculate k and compare it with the value determined experimentally.

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H2 + C2H4 = C2H6 Zo = 7.3 x 1011 L mol-1 s-1 Ea = 180 kJmol-1, T = 400 K; R = 8.314 J mol-1 K-1; kexpt = 3.8 x 10-18 Units (which is the experimentally determined value of k) The value obtained by calculation is k calc = 2.2 x 10-12 You can see that the two values are quite different and therefore the rate constant and rate are not just dependent on the number of energetically favourable collisions but another factor is involved. Let us go back to the reaction above. When a hydrogen molecule collides with an ethene molecule a double bond is broken and two C-H bonds are made. How does this happen? What actually happens in the actual collision? You can see from this that the way in which the hydrogen molecule approaches the ethene molecule is important. Not only do the molecules need to collide with enough energy to exceed the activation energy but also the molecules must have the right orientation with respect to one another, this reduces the number of successful collisions which lead to a reaction. The collision number needs to be multiplied by another factor, the steric factor P, to bring the theoretical values close to the experimental values. If we now return to k = A e-Ea/RT, then the pre-exponential factor A = ZoP Generally P is between 0 and 1 Comparison of collision number and pre-exponential factor A Reaction 2NOCl 2NO + 2Cl H2 + C2H4  C2H6

Zo/L mol-1s-1 5.9 x 1010 7.3 x 1011

A/L mol-1s-1 9.4 x 109 1.24 x 106

Ea/kJ mol-1 102 180

P 0.16 1.7 x 10-6

The steric factor P is calculated by dividing A (determined experimentally) by Zo which is calculated from the kinetic theory of gases. The collision theory works best for small simple molecules, that is, there is good agreement between Zo and A. The reason for this is that for larger molecules: i.

orientation is very important and more difficult.

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Collision Theory and Free Radical Mechanism

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ii. There are more bonds, therefore the energy from the collision may be dissipated throughout the whole molecule.

2.0. COLLISION FREQUENCY, MEAN FREE PATH AND COLLISION NUMBER Collision frequency (Z1)-the rate at which a single molecule collides with other molecules. Total collision frequency (Z) – total number of collisions in one second by all molecules. The inverse of Z1 = average time that a molecule spends in flight between two collisions ( normally about 1 nanosecond). The mean free path ( ) is the average distance a molecule travels between collisions.(usually several hundred molecular diameters.) The mean free path and collision frequency are related by c=

Z1

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where c = mean velocity = (8RT/ M)

(2.4)

The molecules may not actually physically touch to cause a collision, therefore it is necessary to define the collision diameter, .

Collision Diameter and Cross-Section Collision diameter = d = 2r Collision cross-section = d2 = Consider the situation shown in figure 2.1: All molecules except one are stationary. The molecule that is moving has a speed of c ms-1.

Figure 2.1. Collision Diameter and Cross-section.

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For a collision to occur, the centre of the molecules must be within the cylinder shown above. The volume swept out by one molecule in 1 s = c m3. If there are no molecules per unit volume, then the number of collisions made by 1 molecule in 1 s = c no That is Z1 =

c no

(2.5)

If we now consider a more real situation where all molecules are moving, then we must consider not the mean speed but the mean relative speed = 2 c. So that Z1 = 2 c.

no

(2.6)

This is the number of collisions made by 1 molecule in 1 second(Units collisions s-1) We can now determine the mean free path ( ) of one molecule as,

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= distance travelled per second = no. of collisions per second = c/ 2 c no = 1/ 2 no

speed Z (2.7) (2.8)

To calculate the total number of collisions made by all molecules per second we need to multiply by no/2 so that: Z = (1/ 2) c.

no2 (units collisions s-1 m-3)

(2.9)

If we now consider two dissimilar molecules A and B Collision diameter = dAB = rA + rB Collision cross section =

AB

= d2AB.

(2.10) (2.11)

The collision frequency for a single A molecule with B molecules present (concentration = nB) is given below and is a modification of equation (2.1) i.e. ZA =

AB

c nB

and therefore the total number of collisions per second is given by

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(2.12)

Collision Theory and Free Radical Mechanism ZAB =

AB

c nB nA

where c = (8RT/

) and

29 (2.13)

= (MAMB/MA+MB) = reduced mass (kg)

and c is the mean relative speed of A and B. Note- this expression contains terms relating to speed, concentration and size of molecules, so don‘t get confused!.

Summary The following equations apply to collision theory of gases: Rate = Z e-Ea/RT

(2.15)

where Z = total collision frequency. For a simple second order reaction of A + B = products K = Zo e-Ea/RT

(2.16)

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where Zo = collision number = Z/nAnB

2.1. COLLISION FREQUENCY, COLLISION NUMBER AND MEAN FREE PATH = 1/ 2

no

(2.17)

For a single species gas, Z = (1/ 2) c.

no2 (Units of collisions is s-1 m-3)

where c = (8RT/ M)

(2.18) (2.19)

For a gas containing two molecules that are not similar ZAB =

AB

c nB nA (units collisions s-1 m-3)

where c = (8RT/

(2.20)

)

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so that collision frequency depends on the speed, concentration (which is in turn dependent on P and T), and size of molecules. Remember how collision frequency varies with Z/nAnB = Zo

2.05 Pre-Exponential Factor, Collision Number and Steric Factor k = A e-Ea/RT Arrhenius equation

(2.21)

where the pre-exponential factor A = ZoP(collision theory modified with steric factor.)

Exercise 2.2. (a) Calculate Z1, below.

and Z for oxygen at 100kPa and 298K using the data

= 0.4 nm2 = 0.4 x 10-18 m2

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(b) Use the collision theory of gases to calculate the theoretical value for the second order rate constant for the following reaction at 650 K. H2 + I2 = 2HI where

= 0.38 nm2

= 1.98 x 10-3 kg mol-1 and Ea = 171 kJmol-1

REACTION MECHANISM 2.1. Reaction Order and Molecularity We have learnt that reaction occur via a sequence of elementary steps, this constitutes the mechanism of the reaction. Each of these steps typically only involves one or two species. The molecularity of an elementary reaction is the number of species coming together to react. These species are all present in the rate determining step of the reaction. For example, in a unimolecular reaction, a

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single molecule shakes itself apart, or re-arranges itself, for example, the isomerisation of cyclopropane to give propene. In a bimolecular reaction, a pair of molecules/atoms collide and this results in a reaction. Such as the reaction between H and Br2. Reaction order is obtained experimentally from the rate equation. Molecularity refers to an elementary reaction in the mechanism and it is possible to write a rate equation for an elementary reaction. Let us consider the following examples of reactions A unimolecular reaction is first order For example, for the reaction, A = products, we can state that: d[A]/dt = -k[A].

(2.22)

that is, the rate of decomposition of a unimolecular reaction is proportional to its molar concentration. A bimolecular reaction is second order. For example, for the reaction, A + B = products, the rate equation is given by

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d[A]/dt = -k[A][B]

(2.23)

that is, the rate of reaction is proportional to the rate at which the reactant species meet (which is proportional to the concentrations). Therefore if an elementary reaction is bimolecular we can write down the rate law. For example, the mechanism for the following reaction is thought to be a single bimolecular reaction. CH3I + CH3CH2O →- CH3OCH2CH3 + ISo that the rate equation for this reaction given by rate = k[CH3I][CH3CH2O-]

(2.24)

This has been confirmed experimentally. However, a rate equation for another reaction which is also second order does not signify that the reaction is a simple bimolecular process. The reaction may be complex and hence may take place in more than one step.

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2.2. Testing/Proposing the Reaction Mechanism Let us consider the following examples (1) 2NO2 + F2  2FNO2

(2.25)

The experimentally determined rate equation is R = k[NO2][F2] This shows that the reaction is not termolecular and that the reaction above is NOT an elementary reaction (otherwise the rate equation would be R = k[NO2]2[F2]) This being the case, we can propose a mechanism as follows: Possible mechanism: NO2 + F2  FNO2 + F Slow step

(2.26)

NO2 + F  FNO2 Fast step

(2.27)

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This mechanism predicts that the rate of reaction will be determined by the slowest step or the rate determining step. Rate equation for step 1 is R = k1[NO2][F2]

(2.28)

Reactions involving radicals (such as F) tend to be fast. The reason reaction 1 is the rate determining step is because F2 has a very strong bond and therefore has a high dissociation energy. Reaction (2.26) thus forms the basis of the rate equation (2)The decomposition of ozone can be represented as 2O3  3O2 The experimentally determined rate equation is R = k[O3]2/[O2] (2.29) This shows that O2 inhibits the reaction (n=-1 for O2) and we therefore need a mechanism that shows how O2 may inhibit the reaction. It may inhibit the reaction by reforming O3. For example, the proposed mechanism: O3  O2 + O

equilibrium k1 forward reaction, fast

O + O3  2O2

k-1 backward reaction, slow

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Therefore the rate equation R = k2[O][O3]. This is not what is observed experimentally. We can determine an expression for the [O] using the reaction at equilibrium. K = [O2][O]/[O3] Therefore

[O] = K[O3]/[O2]

substitute for [O] in the rate equation, to give R = k2K[O3]2/[O2] (2.30) Therefore the proposed mechanism agrees with the experimentally determined rate equation.

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Consecutive Reactions Some reactions proceed through the formation of an intermediate. For example, for the reaction, A B  C each reaction is unimolecular The variation of the concentrations of A, B and C may be monitored with time. -d[A]/dt = ka[A]

(2.31)

d[B]/dt = ka[A]-kb[B]

(2.32)

d[C]/dt = kb[B]

(2.33)

The reaction profiles for A, B and C depend on the relative values of ka and kb For example, for kb = 2ka ka = 10 kb ; kb = 100 ka This final example shows that the maximum value for [B] is reached very early relative to the loss of A and then stays effectively constant, that is, its rate of change is zero. This is important for steady state analysis Therefore d[B]/dt = ka[A]-kb[B] = 0 or ka[A] = kb[B] so that

d[C]/dt = ka[A]

(2.34)

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This means that if the intermediate B is very reactive relative to A then the rate of formation of C = rate of removal of A. This means that the rate of removal of A is the rate determining step. The rate of production of C depends on how fast B is produced and not the rate at which B is being converted into C. If the reaction intermediate is a radical then the reaction proceeds via a chain reaction.

2.3. Free Radical Reaction Chain Mechanisms

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At the beginning of this chapter, an example of a free radical chain mechanism such as the reaction between hydrogen and bromine H2 + Br2 was given. A free radical is any molecule with an odd number of electrons. The step in which radicals are generated is the initiation step in which molecules dissociate into molecules/atoms with odd numbers of electrons either due to a violent collision in a thermolysis reaction or as a result of an absorption of a photon in a photolysis(which means light splitting and thermo means heat). Therefore this is a high energy reaction. For example, for the reaction, Cl2 = 2Cl∙ H2 + O2 = HO2 + H∙ Steps in which products are formed and radicals are regenerated are called propagation steps. Radicals already present attack other reactant molecules to give a product and a new radical as in the following reactions: For example, for the reaction, CH3 + CH3CH3  CH4 + CH3CH2 Br∙ + H2  HBr + H∙ CH3CHO + CH3  CH4 + CO∙ + CH3∙ Steps in which products are formed but no radicals are generated are called termination steps. In termination, radicals combine and the chain is ended. For example, CH3CH2∙ + CH3CH2∙ = CH3CH2CH2CH3 Br∙ + Br∙ + M∙ = Br2 + M∙ Ibhadon, A.O.. Physical Chemistry Examinations, Nova Science Publishers, Incorporated, 2009. ProQuest Ebook Central,

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where M is a third body that is needed to remove the excess energy from the collision so that the molecule (Br2) does not fall apart. Radicals may also be removed by an inhibition step. Radicals may also be removed by reacting with the walls of a vessel or reacting with another foreign radical. For example, for the reaction, CH3CH2 + R  CH3CH2R CH3CH2 + NO  CH3CH2NO∙. For example, CH3CHO  CH4 + CO d[CH4]/dt = k [CH3CHO]3/2 Some ethene is also produced(detected.)

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Proposed mechanism for the reaction is as follows: CH3CHO  CH3 + CHO CH3CHO + CH3  CH4 + CH3CO CH3CO  CH3 + CO. CH3 + CH3  CH3CH3.

initiation step propagation step propagation step termination step

From the mechanism, d[CH4]/dt = kb[CH3CHO][CH3], how do we determine the concentration of CH3? The solution is to write the rate equation for each step. Initiation Propagation Propagation Termination

ratea = ka[CH3CHO] rateb = kb[CH3CHO][CH3] ratec = kc[CH3CO] rated = kd[CH3]2

(2.35) (2.36) (2.37) (2.38)

How do we test the mechanism from these rate equations? We can do this by; a. Integrate the rate equations numerically. b. Make an approximation.

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The approximation is made using the steady state approximation. This assumes that during the major part of the reaction, the rates of change of concentrations of all reaction intermediates are very small. That is, d[intermediate]/dt 0. This means that we can state that d[CH3]/dt = -d[CH3CO] = 0. That is, the rate of formation of CH3 and CH3CO = rate of removal of CH3 and CH3CO. Therefore d[CH3]/dt = ka[CH3CHO]-kb[CH3CHO][CH3]+kc[CH3CO]-kd[CH3]2=0 (2.39) therefore ka[CH3CHO]+ kc[CH3CO] = kb[CH3CHO][CH3]+ kd[CH3]2 (2.40) d[CH3CO]/dt = kb[CH3CHO][CH3] – kc[CH3]

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therefore kb[CH3CHO][CH3] = kc[CH3]

(2.41)

Equating (2.40) to (2.41), gives, ka[CH3CHO]= kd[CH3]2

(2.42)

[CH3] = (ka/kd)1/2[CH3CHO]1/2

(2.43)

Substitute for [CH3] in d[CH4]/dt = kb[CH3CHO][CH3]

(2.44)

so that d[CH4]/dt = kb(ka/kd)1/2 [CH3CHO]3/2. (2.45) This agrees with the experimentally determined rate equation that d[CH4]/dt = k [CH3CHO]3/2

(2.46)

Unimolecular Reactions Finally, we will now study unimolecular reactions A number of reactions follow first order kinetics For example, A  products d[A]/dt = k[A] (2.47)

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How does the molecule aquire its energy to react? It aquires energy from collisions. However these are bimolecular events so how can they be first order? First order reactions are commonly called unimolecular reactions because they involve an elementary step that is unimolecular in which the reactant molecule changes into a product molecule. The Lindeman Hinshelwood theory explained unimolecualr reactions as follows. It was proposed that the reactant molecule A became energeticlly excited by a collision with another A molecule. A + A  A + A A may either i.

d[A]/dt = ka[A]2

(2.48)

lose its excess energy by colliding with another molecule.

A + A  A + A

-d[A]/dt = k-a [A] [A] (2.49)

ii. Shake itself apart to form product P. This is the unimolecular step.

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A  P

-d[A]/dt = kb[A]

(2.50)

If this unimolecular step is the rate determining step, then the overall reaction will be first order. This can be demonstrated using the steady state analysis. d[A]/dt = ka[A]2 - k-a [A] [A] - kb[A] = 0 We can re-arrange equation [2.51] as in terms of [A]

(2.51)

[A] = ka[A]2/(kb+k-a[A])

(2.52)

The rate equation for the rate of formation of P is given by P = d[P]/dt = kb[A] = kbka[A]2/(kb+k-a[A])

(2.53)

For the reaction to be first order, kb needs to be >ka[A], then the reaction will switch to a second order reaction since, d[P]/dt = kbka[A]2/kb = d[P]/dt = kb[A] = ka[A]2

(2.55)

The reason this happens is that at low pressures the rate determining step is now the bimolecular formation of A d[P]/dt = kbka[A]2/(kb+k-a[A]) If this is re-arranged, d[P]/dt = keff[A] where keff = kbka[A]/(kb+k-a[A]) (2.56) The expression for the effective rate constant can be re-arranged into a straight line equation of the form. 1/keff = k-a/kakb + 1/ka[A] (2.57)

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Therefore a plot of 1/keff vs 1/[A] should give a straight line if it predicts this switch in order from first to second,

Figure 2.2.

Exercise 2.3. Function plots for zero, first, second, and third order reactions

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First Order Reactions

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First order reaction Time/min

[A]/moll-1

ln[A]

0.0

1.3

0.26

5.0

1.08

0.08

10.0

0.9

-0.11

15.0

0.75

-0.29

20.0

0.62

-0.48

25.0

0.52

-0.65

30.0

0.43

-0.84

Figure 2.3.

The slope of this line = -3.7 x 10-2 min-1 and therefore k = 3.7 x 10-2 min-1 Similar plots for second and zero order reactions can be plotted From the data above a concentration against time plot can be drawn from which an initial rate can be determined, figure 2.4

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Figure 2.4.

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2.4. The Kinetic Theory of Gases We should now consider briefly the theory behind gaseous reactions. The kinetic theory of gases predicts properties such as pressure from the mass and size of the molecules within the gas. There are three basic assumptions: i.

A gas consists of molecules of mass m and diameter d which are in ceaseless random motion. ii. The size of the molecules is negligible compared to the distance that they travel between collisions. iii. The molecules do not interact, except during collisions. Pressure is experienced, for example on the walls of a container, because molecules collide with those walls. These collisions exert a force on the wall and because the collisions are so numerous, the pressure exerted by the gas is constant. The pressure exerted by the gas can be calculated by calculating the force exerted by the molecules per unit area of wall as they collide. Force = mass x acceleration, acceleration is rate of change of velocity. If we imagine one molecule with velocity u ms-1 colliding with one wall of a container then the change in momentum on each collision = 2mu .

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For a wall of area A, all molecules within a distance of u m will collide with the wall in 1 s. Therefore all molecules within the volume A u will collide with the wall in 1 s. If there are N molecules per unit volume then A u m3 contains NAu molecules. On average, half the molecules will be travelling up and half down, therefore the average number of collisions per second is given by, s = ½ NAu

(2.58)

Total momentum change per s = ½ NAu.2mu = NAmu2.

(2.59)

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Therefore force = NAmu2 And pressure = force / area = Nmu2 pressure exerted on wall ABCD. In a similar way the force exerted on walls CDEF, and BCFG are NAmv2 and NAmw2. Total force exerted on all 6 walls = 2NAm(u2+v2+w2)

(2.60)

So that the total pressure = 2Nm(u2+v2+w2)

(2.61)

Since c2 = u2+v2+w2 and the average velocities can be considered to be equal since the molecular motion is completely random so that c2 = 3u2 Or 1/3 c2 = u2 and pressure (P) = 1/3 Nmc2 N = nNA/V so that P = 1/3nNA/Vmc2 Or PV = 1/3 nNAmc2 = 1/3 nMc2, since mNA = M molar mass (kg mol-1) From PV=nRT (remember the ideal gas equation!) nRT = 1/3 nNAmc2

(2.62)

Therefore c =

(2.63)

3RT/M

where c = root mean square velocity.(r.m.s)

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Exercise 2.4. Calculate the rms velocity of O2 molecules ( molar mass 32 g mol-1) and N2 molecules ( 28 g mol-1) at 25oC. Ans = c (O2) = 482 m s-1; c(N2) = 515 m s-1 2.4.1. Mean Speed This is found by multiplying each speed by the fraction of molecules that have that speed. Integrate c = 8RT/ M. If we now calculate the mean velocity of O2 and N2 molecules. c (O2) = 444 m s-1 c (N2) = 475 m s-1 You can see from these equations that the speed at which the molecules travel depends on the temperature.

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TUTORIAL QUESTIONS AND ANSWERS CHAPTERS ONE AND TWO Question 1 The equations below relate to the following mechanism for the decomposition of nitrogen(V) oxide. Derive a rate expression for the decomposition 2N205(g) → 4N02(g) + 02(g) (overall reaction)

(2.64)

The mechanism of this reaction is as follows: N205 → N02 + N03 ka

(2.65)

N02 + N03 → N205 k1a

(2.66)

N02 + N03 → N02 + 02 + N0 kb

(2.67)

N0 + N205 → 3N02 kc

(2.68)

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Solution to Question 1 The first task is to identify the intermediates and write expressions for their net rates of formation. All net rates of change of the concentrations of intermediates are set equal to zero and the resulting equations solved algebraically. The intermediates are N0 and N03; the net rates of change of their concentrations are: (a) d[NO] = kb[NO2] [NO3] – kc[NO] [N2O5] dt [NO] = kb[NO2] [NO3]/kc[N2O5]

0

[NO3] = ka[N2O5] -k1a[NO2] [NO3] – kb[NO2][NO3] dt [NO3] = ka[N2O5]/ k1a[NO2]+ kb[NO2]

0

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(b) the net rate of change of concentration of N205 is d[N2O5] = -ka[N2O5] + k1a[NO2][NO3] – kc[NO][N2O5] dt Insert [NO] and [NO3] into the above equation to obtain the expression required. d[N2O5] = - 2kakb[N2O5] dt k1a + kb For the solution to question one, we used the assumption of the steady state approximation. The steady state approximation,(SSA) assumes that after an initial induction period, an interval during which the concentration of the intermediate, I, rises from zero, and during the major part of the reaction, the rates of change of concentrations of all reaction intermediates are negligibly small, such that, d[I]/dt = 0 2. The variation in the partial pressure of azomethane with time was followed at 600K and the results are given below

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i. show that the decomposition is first order in azomethane ii. determine the rate constant at 600K CH3N2CH3(g)  CH3CH3(g) + N2(g) t/s 0 1000 2000 3000 4000 p/(10-2 Torr) 8.20 5.74 3.98 2.76 1.94

Solution to Question 2 Azomethane decomposes in accordance with the following equation CH3N2CH3(g)  CH3CH3(g) + N2(g)

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For a first order plot of ln[(A)]/[(A)]o against time, a straight line should be obtained. The partial pressure of the gas is proportional to the concentration, that is, ln [P/Po] plotted against time, should be a straight line of slope, = - k The data provided is t/s

0

1000

2000

3000

4000(x-axis)

p/(10-2 Torr)

8.20

5.74

3.98

2.76

1.94

ln[P/Po]

1

-0.356

-0.72

-1.08

-1.44(y-axis)

The plot is a straight line, figure 2.5 confirming a first order reaction and slope is - 6x10-4 s-1. Therefore from the slope, the rate constant is 6x10-4 s-1

Figure 2.5.

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Question 3. For the reaction A  Products (a) Show that ln(A) = ln(Ao) - kt (b) Write an equation for the half-life of the reaction (c)Write an equation for the relaxation time (d) Show how the rate constant may be determined from the reaction profile

Solution to Question Three (a) For a first order reaction, -d[A]/dt = k[A]

(2.69)

that is -d[A]/[A] = kt

(2.70)

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if the concentration of A is [A]1 at t1 and [A]2 at t2, then integration of equation [2.70] yields

(2.71) that is, ln[A]1/[A]2 = k(t2-t1)

(2.72)

If the initial concentration is represented by [A]o and t1 is taken to be zero then ln[A]o/[A] = kt; and [A] = [A]oe-kt

(2.73)

and ln[A] = ln[A]o -kt (which is the required expression)

(2.74)

(b)The half life is the time required for half of the reactants to disappear and is given by

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k = 1/t1/2 ln1/1/2 = 0.693/t 1/2 or t 1/2 = 0.693/k

(2.75)

(c) The relaxation time of a first order reaction is equal to the reciprocal of the first order rate constant, that is, = 1/k

(2.76)

It is clear from the equation, ln[A]1/[A]2 = k(t2-t1), that to determine the rate constant for a first order reaction, the ratio of the concentration at two reaction times would need to be determined. This ratio is proportional to the rate constant(k)

Question 4

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The rate of the second-order decomposition of ethanal (CH3CH0) was measured over the temperature range 700-1000 K, and the rate constants obtained are given below. Apply the Arrhenius equation to this date and determine Ea and A. T/K K/(Lmol-1s-1)

700 0.014

730 0.036

760 0.105

790 0.344

815 0.789

840 2.20

920 20.0

1000 146

Suggest what a high value of the activation energy might indicate in terms of the dependence of the rate constant on temperature

Solution to Question 4 The data is analysed by plotting ln(k/Lmol-1s-1) against 1/(T/K) to give a straight line. The slope of this line is (-Ea/R)/K and the intercept at 1/T = 0 is lnA), see figure 1.10. The table below is drawn up as shown below 103K/T

1.43

1.37

1.32

1.27

1.22

1.19

1.08

ln(K/Lmol-1s-1)

4.27

-3.32

-2.25

-1.06

-0.23

0.788

2.99

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1.00 (x-axis) 4.98 (y-axis)

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A plot of ln k against 1/T is made. The plot has a slope of –2.1717 x 104 and intercept 26.523. Therefore, Ea = (2.115 x 104K) x (8.3145 J K-1mol-1) = 181.12 kJmol-1 (Note the units of activation energy, Ea) as in the above example

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A = e26.523 L mol-1s-1 = 3.302 x 1011 L mol-1s-1 (Note the units of A)

It is important to note that A has the same units as k. The slopes and intercepts of graphs are always dimensionless, and care must be taken to relate the numerical value to the physical quantity by noting how the data have been plotted. In practice, A is obtained from one of the midrange data values rather than by using a lengthy extrapolation. Note that the fact that Ea is given by the slope of the plot of ln k against 1/T means that the higher the activation energy, the stronger the temperature dependence of the rate constant (that is, the steeper slope). A high activation energy indicates that the rate constant depends strongly on temperature. If a reaction has zero activation energy, its rate is independent of temperature. In some cases the activation energy is negative, which indicates that the rate decreases as the temperature is raised. We shall see that such behaviour is a signal that the reaction has a complex mechanism.

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Question 5 Determine A and Ea from the following data: T/K K/(Lmol-1s-1)

300 7.9 x 106

350 3.0 x 107

400 7.9 x 107

450 1.7x108

500 3.2x108

[Answer: A = 8x1010 L mo1-1s-1, Ea = 23 kJ mol-1]

REVISION QUESTIONS: CHAPTERS ONE AND TWO STUDENTS ARE ENCOURAGED TO WORK THROUGH THESE QUESTIONS

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1 (a) At 791K, the half-life for the decomposition of a sample of gaseous acetaldehyde initially at 363 Torr was 410 s. When the pressure was 169 Torr the half-life was 880 s. Determine the order of the reaction from the information provided. (b) The rate constant for the first order decomposition of N2O5 in the reaction 2N2O5 = 4NO2 + O2 is k = 3.38 x 10-5 s-1 at 25oC. (i) What is the half-life of N2O5 (ii) Work out the pressure of N2O5 remaining after (iii) 10s and (iv) 10 min if the initial pressure of N2O5 was 500 Torr. Show your workings (2a) The reaction 2A = P is second order with k = 3.50 x 10-4 L mol-1 s-1. work out the time required for the concentration of A to change from 0.260 mol L1 to 0.011 mol L-1. Show your workings (b)The rate constant for the decomposition of a certain substance is 2.80 x 10-3 L mol-1 s-1 at 30 oC and 1.38 x 10-2 L mol-1s-1 at 50 oC. Calculate the Arrhenius parameters of the reaction. (c) Determine A and Ea from the following data: T/K k/(Lmol-1s-1)

300 7.9 x 106

350 3.0 x 107

400 7.9 x 107

450 1.7x108

500 3.2x108

3. The decomposition of a liquid phase reaction, 2A  B was followed by a spectrophotometric method with the following results. Determine the order of the reaction and its rate constant.

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Collision Theory and Free Radical Mechanism t/min [B]/mol L-1

0 0

10 0.089

20 0.153

30 0.200

40 0.230

49

0.312

4. The reaction, CH3CH2NO2 + OH-  CH3CHNO2- + H2O was carried out at 273K with an initial concentration of each reactant of 5.00.x 10-3 mol L-1. The OH- concentration fell to 2.6 x 10-3 after 5 min, 1.70 x 10-3 after 10 min and 1.30 x 10-3 after 15 min. Show the reaction is second order and calculate k. 5.A possible mechanism for the hydrogenation of ethene in the presence of Hg is shown below. Hg + H2  Hg + 2H H + C2H4  C2H5 C2H5 + H2  C2H6 + H H + H  H2.

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Determine the rate of formation of C2H6 in terms of rate constants and concentrations of Hg, H2 and C2H4. Assume that H and C2H5 reach steady-state concentrations. 3D

6.The following initial rate data were obtained for the reaction 2A + B  C +

Initial [A]/mol L-1 0.127 0.254 0.254

Initial [B] / mol L-1 0.346 0.346 0.692

Initial –d[A]/dt /mol L-1s-1 1.64 x 10-6 3.28 x 10-6 1.31 x 10-5

a.

Determine the order of reaction with respect to [A] and [B] and hence write the rate law for the reaction. b. Calculate the value for the rate constant (k). c. Calculate the rate of disappearance of A if [A] = 0.100 mol L-1 and [B] = 0.200 mol L-1. d. Calculate the rate of formation of D under the conditions of (c). e. For a bimolecular elementary process, what three factors determine the rate of the formation of products? 7. Determine the rate equation from the following data

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50 Run number 1 2 3 4 5 6 7 8

[H2SeO3]o/10-4M

[H+]o/10-2M

[I-]o/10-1M

0.712 2.4 7.2 0.712 0.712 0.712 0.712 0.712

2.06 2.06 2.06 12.5 5.18 2.06 2.06 2.06

3.0 3.0 3.0 3.0 3.0 3.0 9.0 15.0

Initial rate (ro) /10-7Ms-1 4.05 14.6 44.6 173 28 4.05 102 508

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It will have the form Rate = k[H2SeO3]a[H+]b[I-]c. Therefore what you are trying to find is a, b and c. (8).The decomposition of an organic nitrile produced the following data. Determine the order of the reaction and the rate constant. Time/103s

0

2.22

4.00

6.00

8.00

10.00

12.00

[nitrile]/M

1.10

0.86

0.67

0.52

0.41

0.32

0.25

(9).The kinetics of the decomposition of NO2(g)  NO(g) + 1/2O2(g) was investigated by measuring the change in concentration of NO2 with time. Using the data below, determine the order of reaction and the rate constant for this reaction. [NO2]/M

0.020

0.015

0.013

0.012

0.011

0.010

0.009

0.008

Time/min

0

0.50

0.75

1.00

1.25

1.50

2.00

2.50

(10). A certain reaction is found to be first order in A, second order in B and third order in C. What will be the effect on the rate of reaction of doubling (i) [A], ii) [B] (iii) [C]. (11).For the reaction described by the rate laws below, state the order with respect to each species and the overall reaction order: i) Rate = k[A][B]2 ii) Rate = k[A]2 (iii) Rate = k[A][B]/[C]

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(12). Determine the rate law, and calculate the rate constant for the reaction A + B → C using the following initial rate data. Initial [A]/M 0.395 0.482 0.482

Initial [B]/M 0.284 0.284 0.482

Initial –d[A]/dt/Ms-1 1.67 x 10-5 2.04 x 10-5 5.88 x 10-5

(13). The following data were obtained at 320oC for the reaction SO2Cl2  SO2 + Cl2. Determine the order of reaction using the initial rate and the rate at the half-life. Confirm the order of reaction using a Function Plot and calculate the rate constant. Time/hours [SO2Cl2]/M

0 1.2

1.00 1.109

2.00 1.024

3.00 0.946

4.00 0.874

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(14).The half lives for the decomposition of a compound in solution at different initial concentrations were measured at 330K and the results are shown in the table below. Determine the order of reaction. What initial concentration would give a half-life of 10s. Initial conc/10-3 M T1/2/s

0.5 4280

1.10 885

2.48 174

4.63 50

7.63 22

(15). How does molecularity differ from order? Under what circumstances can reaction order be predicted from a knowledge of molecularity? (16).Rank the following bimolecular processes in order of decreasing steric factor p (i) O3 + NO  NO2 + O2( ii) CH3+ CH3  C2H6( iii) I+ + I-  I2 (iv) CH3CH2CH2CH2COOH + CH3OH  CH3CH2CH2COOCH3 + H2O. (17) .If the activation energy for a reaction is 198 kJ mol-1 and k = 5.00 x 10-6 s , at what temperature is k = 5.00 x 10-5 s-1? -1

(18).Use the collision theory of gases to calculate the theoretical value for the second order rate constant for the reaction below at 650 K. (Hint: You need to calculate Zo first).

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52 H2 + I2  2HI

Where AB = 0.38 nm2, = 1.98 x 10-3 kg mol-1.and Ea = 171 kJmol-1 If the experimentally determined rate constant is 3.62 x 10-3 L mol-1 s-1 and Ea = 171 kJ mol-1. Calculate A and compare with Zo. (19).Explain what is meant by the steady state approximation as applied to reaction chain mechanisms. The decomposition of ethanal to give methane and CO has been proposed to proceed by the following mechanism. CH3CHO  CH3 + CHO CH3CHO + CH3  CH4 + CO + CH3 2 CH3  C2H6.

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Determine the rate equation from this mechanism and express your rate with respect to CH4, labelling the initiation, propagation and termination steps. (20) List the methods for determining the order of a reaction.

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Chapter 3

CHEMICAL THERMODYNAMICS (FIRST LAW) In this chapter, we should be considering the basics of chemical thermodynamics including the first law. In this regard, the following learning outcomes are important as they will help you to evaluate your level of understanding of the chapter as a whole and the topics covered.

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GENERAL LEARNING OUTCOMES With respect to this chapter, students should be able to: Compare and correlate thermochemical data and calculate values of thermochemical constants as they relate to: • •

Heat, internal energy, work, enthalpy, First Law. Enthalpy change, Hess‘ Law, bond energies/enthalpies.

Calculate the expected variation of thermochemical data as conditions vary; • •

Variation with temperature. Kirchoff‘s Law.

CONCEPT OF ENERGY, WORK AND HEAT We start our discussion of thermodynamics by considering the concept of energy as well as the relationship between energy and work. Thermodynamics

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A.O. Ibhadon

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literally means heat dynamics(heat movement). Heat is associated with energy and without energy we would not exist. We consume food to enable us to live, work, play and do other things. Energy can be defined as the capacity to do work or produce heat. It can be converted from one form to another, but it cannot be created or destroyed* . Energy may take the form of kinetic energy (body in motion), potential energy (position), heat energy (measured by temperature), electrical energy, chemical energy, etc. There is a natural tendency for other forms of energy to eventually be turned into heat energy. Chemical and physical processes are accompanied by energy changes. Thus mechanical energy produces friction and this in turn produces heat. Electrical energy produces current through a resistance wire and this in turn produces heat. The study of these energy changes is called thermodynamics. J.P.Joule – ― when a definite amount of any form of energy, particularly mechanical and electrical, is converted into heat, a perfectly definite number of calories (joules) of heat is always produced.‖ Therefore there is a relationship between heat and other forms of energy. Heat: This is the energy change in a system as a result of a temperature difference Adiabatic wall: is a boundary that does not permit energy transfer as heat A diathermic wall: is a boundary that permits energy transfer as heat(eg steel, glass) Work: is said to be done when an object is moved against an opposing force System: part of the universe we are interested in. Surroundings: part of the universe we are not interested in Boundary separates the system and the surroundings. Isolated system: neither substances or energy can leave or enter, for example, Bomb Calorimeter Closed system : only heat can leave or enter. Not matter can leave or enter, for example, a sealed container Open system: heat and matter can leave or enter, for example, a laboratory beaker State: solid, liquid, gaseous states of the system is dependent on pressure, volume and temperature. Intensive variables: are independent of the amount of material present, for example, temperature, density, surface tension, etc. The boiling point of 500 ml and 100ml of water is the same

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Extensive variable: are dependent upon the amount of material present, for example, volume, mass, internal energy, enthalpy. Function of state: dependent upon the pressure, volume and temperature of the system. Examples include: total or internal energy is denoted by the symbol U(E). Enthalpy, H, Entropy, S. and Gibb‘s free energy, G. Thermodynamic equation of state: expresses a quantity in terms of two variables and usually applies to any substance

3.0. INTERNAL OR TOTAL ENERGY (U)

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The total energy (U) is the sum of the potential energy (P.E.) and kinetic energy (K.E.) of the system. The kinetic energy results from different types of movements such as translation, vibration, rotation. All these are temperature dependent. The potential energy of the system results from the position of the molecules with respect to one another and will therefore be dependent on volume and pressure.

Figure 3.1. The separation between two Gaseous molecules.

The molecules attract and repel one another (van der Waals forces), figure 3.1. shows the total P.E. (solid line and the energy due to attraction (-ve) and repulsion (+ve) (dotted lines). The repulsive forces are more important the closer the molecules are, and the attractive forces are more important the further the molecules are apart. The point at which the potential energy = 0 is the collision diameter. The total energy U of the system is therefore a function of state: it is dependent only on the pressure, volume and temperature of the system. It is difficult to determine an actual value for U but this is not important as thermodynamics deals with changes in energy.

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THE FIRST LAW OF THERMODYNAMICS Also known as the conservation of energy. This law states that *Whenever one kind of energy is produced, an exactly equivalent amount of other kinds of energy must disappear. Therefore the first law of thermodynamics states that ―the total energy of a system and its surroundings must remain constant, although it may be changed from one form to another. OR the energy of the universe is constant.

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Consider the process below: Frictionless cylinder piston.

Figure 3.2.

The change in the total energy due to the input of heat (q) and work done (w) U = q + w.

(3.1)

where U = U1- U2 Or dU = dU1-U2 = large change, d = very small change. Convention: A number gives the magnitude of the change and the sign indicates the direction of flow of energy

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Figure 3.3.

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Applying this to the system above, heat is put into the system therefore q = +ve, work is done by the system, therefore energy is lost by the system and w = ve. In this case the work done by the piston is the work of expansion. The work of expansion = -P V (3.2)

Figure 3.4.

Therefore U = q -P V or dU = q - PdV.

(3.3)

Exercise: 3.1. A balloon is inflated to its full extent by heating the air inside it. The volume of the balloon expands from 4.00 x 106 L to 4.50 x 106 L by the addition of 1.3 x108 J of energy as heat. Assuming the balloon expands against a constant pressure of 1 atm, calculate U(the change in internal energy) for the process. U = q + w. 3.3 a

Solution to Exercise 3.1. energy added as heat =q = +1.3 x 108 J

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work done by the balloon = w = -P V where P = 1 atm. convert to Nm-2(Pa) = 1.01x105Nm-2. and V = (4.5 – 4.0)x106 L = 0.50 x 106L = 0.50 x 103 m3 Therefore w = - 1.01x105Nm-2 x 0.50 x103 m3 w = -5.1 x 107Nm = -5.1 x 107 J Therefore U = +1.3 x 108 - 5.1 x 107 J = 8 x 107 J The net increase in the energy is due to the fact that more energy is added than is used in expansion.

ADIABATIC PROCESSES In an adiabatic expansion, no heat gained or lost by the gas, the process is thus adiabatic and the first law of thermodynamics is reduced to equation [3.4]

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dU = dw

(3.4)

For pressure – volume work only, dU = -PextdV and for an adiabatic expansion, dV is positive while dU is negative. If the expansion is opposed by an external pressure, Pext, work is done on the surroundings at the expense of the internal energy of the gas. For any adiabatic process, dU = dw = -PextdV (reversible or irreversible) if PV work is the only type of work involved. If the external pressure is 0, it means adiabatic expansion into a vacuum, and hence no work is done and there is no change in internal energy(this would apply for all gases). If the expansion is opposed by an external pressure, work is done on the surrounding and the temperature drops as internal energy is converted to work. In this case, integration of equation [3.5] dU = -PextdV yields

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(3.5)

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(3.6) And U = U2-U1 = w where w is the work done on the gas Since for an ideal gas, the internal energy is a function only of temperature dU = CvdT (3.7)

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Thus when an ideal gas expands adiabatically against an external pressure, the temperature drop is related to the change in internal energy (see equations below). If Cv is independent of temperature for an ideal gas at the temperature of interest, then

(3.8) And U = U2 – U1 = Cv(T2 – T1)

(3.9)

Since q = 0, then U = w and

(3.10) This relation applies to the adiabatic expansion of an ideal gas with Cv independent of temperature whether the process is reversible or irreversible. If the gas expands, the final temperature T2 will be lower than the initial temperature T1 and the work done on the gas is negative. If the gas is compressed adiabatically, it will heat up.

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Exercise 3.2. An ideal monatomic gas at 1 bar and 273.15K is allowed to expand adiabatically against a constant pressure of 0.315 bar until it doubles its volume. a. b. c. d.

What is the change in molar volume? How much work is done on the gas in this process? What is the final temperature? What is the change in the molar internal energy of the gas?

The value of Cv is 3/2 R. When an adiabatic expansion is carried out reversibly, the equilibrium pressure is substituted for the external pressure, and so for an ideal gas

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(3.11)

If the heat capacity is independent of temperature, then

(3.12)

R (3.13) Cv lnT2/T1 = RlnV1/V2

(3.14)

The above equation is a good approximation only if the temperature range is small enough so that Cv does not change very much. Since Cp - Cv = R, equation [3.14] may be written as T2/T1 = (V1/V2) -1

(3.15)

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where = Cp/Cv. It is also possible to obtain the following alternative form of the above equation: T2/T1 = (P2/P1)( -1)/

(3.16)

P1V1 = P2V2 (obvious from gas law)

(3.17)

Thus when a gas expands adiabatically to a larger volume and a lower pressure, the volume is smaller than it would be after an isothermal expansion to the same final pressure. (( Note that other units of pressure apart from the bar, include atmospheres and Torr.)

3.1. MEASUREMENT OF INTERNAL ENERGY( U)

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3.1.1. Reactions at Constant Volume The internal energy change of a system can be measured easily if the system is unable to do work. For example, in the frictionless cylinder, work of expansion is done by the system. If heat was added to the system, but expansion was not allowed to take place (i.e. V=0), then U = qv

(3.18)

The apparatus used for this type of experiment is a bomb calorimeter. The internal energy change may also be calculated directly from reactions in which no gases are liberated. Many reactions in chemistry take place at constant pressure Since U = q -P V or dU = q - PdV.

(3.19)

We can rearrange equation 3.17 to obtain qp = U + P V or qp = dU + PdV

(3.20)

This makes it possible for us to define the term enthalpy (H)

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3.1.2. Enthalpy (H) The enthalpy is defined as H = U + PV

(3.21)

From equation 3.19 we can state that H= U + P V

(3,21a]

at constant pressure Where U = internal energy, P is the pressure of the system, and V is the volume of the system. Since U, P and V are state functions so is H. If we consider a reaction carried out at constant pressure, where only pressure-volume work is allowed, and if we rearrange the first law as U = q+w, then

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qp = U + P V

(3.22)

where qp = heat transfer at constant pressure. Therefore, H = qp, that is, at constant pressure, where only work of expansion is allowed, the change in the enthalpy of the system = energy flow as heat. Are q and w state functions? No: they depend on the path of change. So what does a path of change mean?

3.2. PATH OF CHANGE What is meant by the path of change? Let us consider the following analogy. If a person, of mass m, walks from an area of (height h1) up to another area of (height h2). His/her P.E. changes as P.E U = mg(h2-h1). The change in potential energy is given by mg(h2-h1). This is Path 1. If that same person walks from the an area of height h1 and then through several other areas and then back to the area of height h2, the change in the P.E is still mg(h2-h1). = Path 2. This means that the change in P.E is independent of the path of change.

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Exercise 3.3. Does that person do the same amount of work (w) and generate the same amount of heat (q) while walking along both paths? The answer is no, that is, the work done and the heat generated depend on the path taken. Therefore q and w are not thermodynamic functions of state. The path in a chemical sense means how the reaction or change is carried out. For example, you can carry out a reaction along a path of constant pressure, or along a path of constant volume or somewhere in between, where pressure and volume change. The heat depends on the path that is taken, but is fixed for a given path. This is why we can say qv = U (constant volume path) and qp = H (constant pressure path)

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3.3. EXOTHERMIC AND ENDOTHERMIC REACTIONS The enthalpy change, H = -ve, the reaction is exothermic, heat is lost to the surroundings therefore there is a decrease in enthalpy of the system. The enthalpy change , H = +ve, the reaction is endothermic, heat is acquired from the surroundings therefore there is an increase in the enthalpy of the system.

Exercise 3.4. (1) Fe(s) + 2HCl(aq)  FeCl2(aq) + H2(g) Heat evolved at 291K = 86.9kJ Use U = qp -P V We can use the ideal gas equation to replace P V by nRT and so V = nRT = P V = nRT

(3.23)

So that U = qp - nRT

(3.24)

qp = -86.9 kJ (-ve; heat is lost from the system)

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The gas causes the volume change, therefore n = change in the number of moles of gaseous molecules (1-0). nRT = 1 x 8.314 x 291 U = -86.9 x103 - 1 x 8.314 x 291 = -89.318 kJ This is the heat released at constant volume H = -86.9 kJ This is the heat released at constant pressure. Why is U H ? The reason is because there is a volume change and work has been done. Less heat is released at constant pressure because some energy has gone into doing work.

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2) C6H6(l) +7 ½ O2(g) = 6CO2(g) + 3H2O(l) The reaction was carried out at 291 K in a bomb calorimeter and the heat evolved qv = 3269 kJ. Calculate H. U = qv = -3269kJ Use H = U +P V = U + nRT = -3269x103 + (-3/2 RT) = -3273kJ In the above example, more energy is released at constant pressure than at constant volume as energy is acquired by the system because work is done on the system. Enthalpy, like internal energy, is an extensive variable- it depends on a number of variables such as temperature, purity of reactants/products (number of moles present) and pressure. In order to compare reactions we need to define standard conditions. Standard conditions are pure, unmixed substances at a pressure of 1 atm. For example, the standard state of H2 at 1 atm is a gas, the standard state of calcium carbonate is the pure solid at 1 atm, the standard states of the various states of methanol are the pure solid, pure liquid or pure vapour at 1 atm. The actual state depends on the temperature. The temperature is not part of the definition of a standard state, the data for all substances is reported at a specific temperature. Most data is reported at 25oC (298.15K). The standard enthalpy of reaction Ho is

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the enthalpy of reaction for the conversion of reactants in their standard states into products in their standard states. For example, 2H2(g) + O2(g) = 2H2O(l)

Ho = -572 kJ at 25oC

This shows that the enthalpy of the system decreases by 572 kJ when 2 moles of pure H2 at 1 atm react with 1 mole of pure oxygen at 1 atm and 2 moles of pure liquid water at 1 atm are formed Standard enthalpies of reaction covers a wide range of reactions such as combustion and formation. These are enthalpies of chemical change. Enthalpies of physical change include vaporisation, fusion, sublimation etc. Enthalpies of atomic and molecular change include ionisation, atomisation, electron gain, electron affinity and dissociation, bond enthalpy. These enthalpies have the units kJmol-1.

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The mol-1 depends on what enthalpy is being determined. For example, 2H2(g) + O2(g) = 2H2O(l) Ho = -572 kJ at 25oC We could write that the reaction above had a Ho = -572 kJmol-1 at 25oC. Alternatively, H2(g) + 1/2O2(g) = H2O(l) Ho = -286 kJmol-1 at 25oC What does the mol-1 stand for. We need to look at the definitions for the standard enthalpies of combustion, formation etc to answer that question. For the reactions above we may say that the mol-1 refers to per mole of reaction as written. The standard enthalpies of reactions can be combined together to obtain the enthalpy of another reaction, see Hess‘s Law. Hess‘s law is a recognition that enthalpy is a function of state and says that the value of Ho for a direct path between reactants and products must be equal to the value of Ho for an indirect path.

SUMMARY The changes in internal energy and enthalpy occur in any kind of process. We have discussed U and H, q and w and path of change.

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Thermodynamic functions of state have been defined. U and H are most easily determined when the process under investigation is carried out at constant volume and constant pressure, respectively.

Exercise 3.5. Calculate the standard enthalpy of formation of CH3OH(l) at 255C from the value of –128.12KJ that would be obtained for the enthalpy change for the reaction in which 1 mol of this substance was formed from hydrogen and carbon monoxide. The standard enthalpy of formation of carbon monoxide is –110.52KJ.

Solution to Exercise 3.5 The solution to exercise 3.5 is an example of an organised procedure for doing calculations that involve the thermodynamic properties of substances of chemical reaction. Set up the calculation as

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2H2(g)+CO(g ) H0s,kJ

0

CH3OH(l) H0 = -128.14kJ -110.52 H0s(CH3OH)

Then from H0 = H0s(CH3OH) – [2 H0s(H2) + H0s(CO)] or -128.14 = H0s(CH3OH) – (0-110.52) this leads to: H0s(CH3OH) = -238.66 kJ/mol

3.4. VARIATION OF ENTHALPY WITH TEMPERATURETHE KIRCHOFF’S EQUATION The heat change associated with any process, physical or chemical, usually varies with temperature.

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For the reaction A=B

H = HB-HA

(3.25)

To determine .the variation of H with temperature, the expression above is differentiated with respect to temperature at constant pressure to give the expression

(3.26)

(3.27)

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(3.28)

(3.29) Where Cp is the heat capacity at constant pressure and Cp is the difference in the heat capacities of the final (B) and initial (A) states. Integration of equation (3.23) between the temperature limits T1 and T2 gives the expression below. H1

d( H) H2

H2

H1

T2 T1

C p dT

(3.30)

H1 and H2 are the enthalpies of reaction, at constant pressure, at temperatures T1 and T2, respectively. A simple assumption is made that Cp is constant and independent of temperature over a limited range.(100K) Equation 3.30 now becomes H2- H1 = Cp(T2-T1) 

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(3.31)

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Exercise 3.6 Worked example Using the standard enthalpy of formation of gaseous water at 25oC. Estimate its value at 100oC given the molar heat capacities below. H2(g) + ½ O2(g) H2O(g) Hoform = 241.82 kJmol-1 at 298K H2O (g) H2(g) O2(g)

Cp

Cp/J K-1 mol-1 33.58 28.84 29.37

= 1xCp(H2O(g)) –[ 1xCp(H2) + ½ Cp(O2)] = -9.95 J K-1

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T = +75K H2- H1 = Cp(T2-T1) H2 = H1+ Cp(T2-T1) H2 = - 241.82 x103+ (-9.95x75) H2 = -242.57 kJmol-1 This calculation shows that the reaction enthalpy at 100oC is only slightly different from that at 25oC. The reason for this is because Cp is small. If Cp were a larger value then the enthalpy of reaction would be more sensitive to temperature changes. We can carry out similar calculations for determining the variation of with temperature using the expression in equation [3.31]

(3.32) Where Cv is the heat capacity at constant volume.

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Exercise 3.7. 1. When 3.0 mol of O2 is heated at constant pressure of 3.25 atm, its

temperature increases from 260K to 285K. Given that the molar heat capacity of O2 at constant pressure is 29.4J mol-1 K-1, calculate q, H, and U. 2. The standard enthalpy of formation of gaseous H20 at 298 K is –241.82 kJ mon-1. Estimate its value at 1000C given the following values of the molar heat capacities at constant pressure: H20(g) : 33.58 JK-1mol-1; H2(g) : 28.84 JK-1mol-1; 02(g) : 29.37 JK-1mol-1. Assume that the heat capacities are independent of temperature.

3.5. HEAT CAPACITY AND UNITS

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The internal energy of a system can be increased by increasing the temperature. The rise in temperature is proportional to the input of heat(q) to the system. This means that T or

q T = q/C

where C = heat capacity(has units of JK-1.) (3.33)

This shows that the temperature of a system with a large heat capacity will have a small increase for a certain input of energy, compared to a system with a small heat capacity with the same input of energy. If we make heat capacity intensive an intensive property it becomes molar heat capacity and the units are J mol-1K-1. On the other hand, the units of specific heat capacity are J g-1K-1. Heat capacity is dependent on whether a system is maintained at the same volume or the same pressure, that is, Cp and Cv. (constant pressure heat capacity and constant volume heat capacity. Cpand Cv have similar values for solids and liquids (negligible volume change with T) but differ appreciably for gases. For perfect gases Cpand Cv are related by Cp= Cv + nR.

(3.34)

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Cp is larger than Cv because some energy is used to do work of expansion in a constant pressure system. This means a constant pressure system needs more energy input to achieve the same temperature rise as a constant volume system

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Relationship between U, H and C. U = Cv T

(3.35)

H = Cp T

(3.36)

For example, if we consider a liquid in equilibrium with its vapour closed in a frictionless piston in a constant temperature water bath, figure 3.2, a decrease in the external pressure Pext so that Pext< V.P. will cause evaporation. A liquid evaporating to a gas requires heat. For this evaporation to be thermodynamically reversible there must be no increase on the temperature of the system relative to the surroundings. If the evaporation process is slow enough, the heat required will be acquired from surroundings at a quick enough rate so that the system is kept in thermal equilibrium with its surroundings. The evaporation process must therefore be carried out in a series of infinitesimal decreases in Pext. A large decrease in Pext would result in rapid evaporation, the energy required for this could not be obtained rapidly enough (from the surroundings) to prevent the temperature of the system from decreasing relative to the surroundings. If we consider the work done by the system as w = (V.P).dV, for a reversible change Pext = V.P. and for an irreversible change, Pext wirrev.

3.6. REVERSIBLE AND IRREVERSIBLE CHANGES We have learnt that q and w depend on the path taken during a change in a system. How do q and w vary with the speed of change? In thermodynamics it is important that all changes are carried out infinitesimally slowly, so that the system is always in temperature and pressure

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equilibrium with its surroundings. If a change is carried out under these conditions then the change is said to be thermodynamically reversible. There is a relationship between thermodynamic reversibility and maximum work. If we expand a gas (pressure Pi) against an external pressure (Pe) in a frictionless piston at constant T then w = -PedV

(3.37)

For an expansion to occur Pe must be less than Pi Or Pi = Pe + dP

(3.38)

We stated that for a change to be thermodynamically reversible it must be in temperature and pressure equilibrium with its surroundings. (T constant)

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i.e. in the example given Pi Pe so that dP is infinitesimally small ( P). To carry out an expansion in a reversible way it must be carried out infinitely slowly so that it can be considered that the system is in equilibrium with its surroundings at all times. If the expansion is carried out at a finite speed then dP is much larger ( P). this means that Pi > Pe. (Piston moves up spontaneously) and the expansion is irreversible. If we now consider the work of expansion (w), for a volume change of dV and an internal pressure of Pi, the maximum work occurs for a reversible change. w = -PedV; Pe has its maximum value for a reversible expansion. How does q vary for a reversible process? Consider the first law. dU = q + w

(3.39)

dU = q – PdV (for expansion)

(3.40)

dU is fixed for a given process (state function) therefore since wrev>wirrev, then qrev>qirrev.

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SUMMARY OF CHAPTER THREE The first law of thermodynamics is the law of the conservation of energy. We have discussed enthalpy change at constant volume and at constant pressure, the relationship between heat capacity and enthalpy of a system as well as expansion of gases. We have discussed reversible and irreversible changes as well as the application of the Kirchoff‘s Equation. In chapter three, we should be discussing the second law of thermodynamics.

Revision Questions in Chapter Three 1) Write notes on heat capacity . 2) Define the standard enthalpies of vaporisation, fusion, sublimation, ionisation, electron gain, electron affinity, dissociation, bond enthalpy, combustion, formation. 3) Give an example of the use of Hess‘s law.

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For the combustion reactions 2Mg(s) + O2(g) = 2MgO(l) CH4(g) + 2O2(g) = CO2(g) + 2H2O(l)

Ho = -1204 kJ Ho = -890 kJ

Determine the standard enthalpy of combustion, Ho com , in kJmol-1 4) For the reaction 2H2(g) + O2(g) = 2H2O(l)

Ho = -572 kJ at 25oC

Calculate the standard enthalpy of formation Ho form 5) The mass of a typical sugar cube is 1.5 g. Calculate the energy released as heat when a cube is burned in air. To what height could you climb on the energy a cube provides assuming 25% of the energy is available for work and you weigh 65kg?

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6) A system expands at constant 1.00 atm pressure from a volume of 3.00L to one of 7.00 L while absorbing 13.0 kJ of heat. Determine H and U for the process. Comment on the difference between the two values. 7) A system expands from a volume of 2.5 L to one of 5.50 L against a constant opposing pressure of 4.8 atm. Calculate a) the work done by the system on the surroundings and b) the work done on the system by the surroundings. Express your answer in J. 8) N2(g) + 1½ H2(g) NH3(g) Ho = -46.11 kJmol-1 at 298K Use the data below to calculate Hoform at 400 K.

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NH3 (g) H2(g) N2(g)

Cp/J K-1 mol-1 36.00 28.83 29.13

9) The standard enthalpy of combustion of propane gas is –2220 kJmol-1 at 298K and the standard enthalpy of vaporisation of the liquid is +15 kJmol-1 at 298K. Calculate (a) the standard enthalpy and (b) the standard internal energy of combustion of the liquid at 298K. 10( a). For the reaction A = B, show how the standard molar enthalpy of B can be obtained from heat capacity and enthalpy measurements. (b). For a closed system, doing non - expansion work, at constant composition, show that dG = VdP -SdT (c) Show that w = Cv(T2-T1) for an ideal gas (d). The standard enthalpy of formation of gaseous H20 at 298 K is –241.82 kJ mon-1. Estimate its value at 1000C given the following values of the molar heat capacities at constant pressure: H20(g) : 33.58 JK-1mol-1; H2(g) : 28.84 JK-1mol-1; 02(g) : 29.37 JK-1mol-1. Assume that the heat capacities are independent of temperature. State any assumptions made in your calculation.

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11. Calculate the standard enthalpy of formation of CH3OH(l) at 255C from the value of –128.12KJ that would be obtained for the enthalpy change for the reaction in which 1 mol of this substance was formed from hydrogen and carbon monoxide. The standard enthalpy of formation of carbon monoxide is –110.52KJ. Assume that the standard enthalpy of formation of hydrogen gas from its elements is zero. 12. The enthalpy change H when one mole of water freezes at 273K is – 6.00kJ. Calculate the enthalpy change when water freezes at 253K. Cp for water is 75.3 JK-1mol-1 and for ice is 37.6 JK-1mol-1. 13. For a system that does only PV work, show that dG = VdP -SdT 14. For n moles of a perfect gas, show that G = nRTlnPB/PA

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15. Sketch the variation of free energy with temperature and show how Hvap can be obtained from the Clapeyron and Classius –Clapeyron Equations. 16. At 273.16K the enthalpy change on fusion of water is 6.0kJ/mol and the corresponding volume change is -1.6x10-6 m3mol-1. Estimate the temperature at which ice will melt at 1000atm pressure(take 1 atm = 105Nm-2) 17. Derive the van‘t Hoff Isochore The equilibrium constant Kp for the dissociation of bromine into atoms Br2 → Br.+Br. is 6x10-12 at 600K and 1x10-7 at 800K. Calculate the standard free energy change for the reaction at these temperatures and the standard enthalpy change assuming this is constant in the temperature range 600-800K. Also calculate the standard entropy change associated with the reaction.

3.8.2. Tutorial Questions and Answers 1 (a) Explain briefly the meaning of: (i) Thermodynamic state function (ii) A thermodynamic equation of state

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(b) State the Gibbs- Helmholtz equation ( c) For an isolated system, in the absence of non-expansion work and at constant composition, show that: dG = Vdp - SdT state any assumptions you have made (d) Show that standard reaction enthalpies at different temperatures may be obtained from heat capacity and reaction enthalpy measurements. State any assumptions you have made (e) The standard enthalpy of formation of gaseous H20 at 298 K is –244.82 kJ mol-1. Determine its value at 100 oC given the following values of the molar heat capacities at constant pressure: H20(g) : 33.58 JK-1mol-1; H2(g) : 28.84 JK-1mol-1;

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02(g) : 29.37 JK-1mol-1. Assume that the heat capacities are independent of temperature. (f) State two factors that could determine the direction of a spontaneous change in a chemical or physical system

Question 2 (a) Explain briefly the meaning of the term ‗adiabatic expansion or compression‘ of a gas (b) Write the equation for the First law of thermodynamics for an adiabatic process (c) Show that when an ideal gas expands adiabatically against an external pressure, the temperature drop is proportional to the change in internal energy of the gas, that is,

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Sate any assumptions you have made in (c0 and explain the meaning of the term(U) in the equation (d) An ideal monatomic gas at 1 bar and 273.15K is allowed to expand adiabatically against a constant pressure of 0.320 bar until its volume doubles, Calculate i. ii. iii. iv.

the change in molar volume the work done on the gas in this process the final temperature the change in the molar internal energy of the gas

Assume that

v is 3/2 R.( R is the gas Constant)

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Solution to Question 1 a(i) A thermodynamic function of state(same as thermodynamic state function) is independent of path of change – depends on initial and final states, depends only on the pressure, volume and temperature of the system (ii) A thermodynamic equation of State Is an equation that expresses a quantity in terms of two variables, for example, temperature and pressure and applies to any substance or system. An example is the following equation where the coefficient is the internal pressure of the system T=

T( P/ T)v-P

(3.41)

(b) The Gibbs – Helmholtz equation This equation can be expressed(stated) as [( / T(G/T)]p = -H/T2

(3.42)

where G, T and H mean what?

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Chemical Thermodynamics (First Law) (c)show that dG = Vdp -SdT.

77 (3.43)

The Gibbs function can be defined as G = H- TS. If the system undergoes a change of state, G changes because H, T and S also change. (G,S, H, T, V, U, P) have their usual meaning) dG = dH -TdS – SdT

(3.44)

and H = U+PV

(3.45)

dH = dU+pdV+Vdp

(3.46)

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For a closed system doing non-expansion work, dU can be replaced by the fundamental equation dU = TdS-PdV, resulting in dG = (TdS-pdV)+pdV+Vdp-TdS-SdT

(3.47)

= Vdp –SdT

(3.48)

(d) Standard molar enthalpies for several substances have been determined and these data are available. However in the absence of these data, standard reaction enthalpies at different temperatures may be estimated from reaction enthalpy and heat capacities at some other temperature, as illustrated below. The heat change associated with a chemical or physical process such as the reaction given below varies with temperature For the reaction, A = B, the enthalpy change is given by H = HB-HA Change in G, H, S or U is always product minus reactant To determine .the variation of H with temperature, the expression above is differentiated with respect to temperature at constant pressure to give the expressions below.

(3.49)

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(3.50)

(3.51)

(3.52) Where Cp is the heat capacity at constant pressure and Cp is the difference in the heat capacities of the final (B) and initial (A) states. Integration of equation (1) between the temperature limits T1 and T2 gives the expression below. d( H) = H2- H1 =

CpdT

(3.49)

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(Insert limits T1 and T2 on the left side of equ. 3.49) H1 and H2 are the enthalpies of reaction, at constant pressure, at temperatures T1 and T2, respectively. A simple assumption is made that Cp is constant and independent of temperature over a limited range.(100K) Equation 2.49 gives the required expression and can also be written as H2- H1 = Cp(T2-T1)

(3.50)

(e) If H is independent of temperature in the range T1 to T2 , then H(T2) = H(T1) + Cp (T2 – (T1) The reaction is H2(g) + 1/202(g) → H20(g) and Cp = Cp,(H20,g) – {Cp (H2,g) + 1Cp(02,g)}

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= 33.58-{(28.84+14.685)} = 33.58 –43.525 = -9.95JK-1 mol-1 ( note the difference in units between heat capacity and units for H,G, U or S) It follows from the above that ø fH (373

K) = -241.82 kJ mol-1 +[ (75 K) x (-9.945)] J K-1mol-1 = -242.6kJmol-1 (note the units of enthalpy and molar heat capacity) The assumption is that the heat capacities are independent of temperature, over a limited range of temperature. (f) Spontaneous changes lead generally to a dispersal of energy in a more chaotic form and generate entropy. They are determined mainly by the entropy(S),the free energy (G) and enthalpy(H).

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Solution to Question 2 (a) In an adiabatic process, no heat is lost or gained by the gas (system) (b) Thus the first law of thermodynamics becomes, dU = dw The term U is the internal(total) energy of a system, and is made up of the potential and kinetic energies. (c) For pressure –volume work only, dU = -PextdV Integration of the above equation yields

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U 2

(3.51) And U = U2-U1 = w

(3.52)

where w is the work done on the gas Since for an ideal gas, the internal energy is a function only of temperature dU = CvdT

(3.53)

If Cv is independent of temperature for an ideal gas at the temperature of interest, then U

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2

(3.54) U = U2 – U1 = Cv(T2 – T1)

(3.55)

Since q = 0, then U = w and

(3.56) is the required expression Thus when an ideal gas expands adiabatically against an external pressure, the temperature drop is related to the change in internal energy d (i)V1 = RT/P1 = (0.083 145 L bar K-1 mol-1) (273.15 K)/(1 bar)

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= 22.71 L mol-1 V = 2V1 - V1 = 22.71 L mol-1 (ii)w = -Pext V = -(0.315 x 105Pa) (22.711 x 10-3m3mol-1) = -715.4 J mol-1 (iii)w = Cv T T = (-715.4 J mol-1)/3/3(8.3145 J K-1 mol-1) = -57.4 K T2 = T1 + T = 273.15 K – 57.4 K = 215.8 K (iv) U = Cv T = 3/2(8.3145 J K-1 mol-1) (-57.4 K) = -715.4 J mol-1 * volumes in above calculation are molar volumes

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Question 3 - Using Kirchhoff’s Law The standard enthalpy of formation of gaseous H20 at 298 K is –241.82 kJ mon-1. Estimate its value at 1000C given the following values of the molar heat capacities at constant pressure: H20(g) : 33.58 JK-1mol-1; H2(g) : 28.84 JK-1mol-1 and 02(g) : 29.37 JK-1mol-1. Assume that the heat capacities are independent of temperature.

Solution to Question 3 When rHøp is independent of temperature in the range T1 to T2, the integral in eqn 2.49 evaluates to rCøp (T2 - T1). Therefore, ø rH (T2)

=

ø rH (T1)

+ rCøp (T2 – (T1)

(3.57)

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To proceed, write the chemical equation, identify the stoichiometric coefficients, and calculate rCøp from the data. The reaction is 1 ( ) → H 0( ) H2(g) + 1/2O 2 g 2 g

So that rCøp = Cøp,m(H20,g) – {Cøp,m (H2,g) + 1/2 Cøp,m(02,g)} = -9.94 JK-1mol-1 It then follows that kJ mol-1 + (75 K) x (-9.94 J K-1mol-1)

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ø fH (373 K) = -241.82 ø -1 fH = -242.6 kJ mol

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Chapter 4

THE SECOND LAW OF THERMODYNAMICS

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In this chapter, students are expected to be able to evaluate the significance of free energy and free energy change in relation to the following themes after reading and working through the exercises in this chapter. Evaluate the significance of free energy and free energy change • • • •

Second Law and concept of entropy S for phase changes and temperature changes S for chemical reactions, comparison and interpretation Free energy and spontaneous change.

Relate equilibrium constant to the concept of free energy change and calculate the values of equilibrium constants from thermodynamic data. • • •

Gibbs-Helmholtz equation Variation of G with temperature, pressure and concentration in chemical equilibrium Reaction Isotherm

Use the reaction isotherm to interpret the variation in ln(Kp) with temperature. Investigate suitable equilibrium systems in order to determine H from variation of equilibrium constant with temperature. • •

Second Law and concept of entropy S for phase changes and temperature changes

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S for chemical reactions, comparison and interpretation free energy and spontaneous change. Reaction Isotherm

THE SECOND LAW In thermodynamics, reactions can be classified as spontaneous or nonspontaneous. The first law deals with changes in internal energy and enthalpy of a system accompanying physical and chemical processes. It states that energy is conserved during that change (or the energy of the universe is constant). The first law gives no indication as to whether a reaction is feasible (spontaneous) or not. There are no explanations as to why some processes occur while others don‘t. The second law of thermodynamics tries to provide an explanation as to spontaneous changes and the driving force for spontaneous change. The driving force for spontaneous change or entropy can be determined quantitatively.

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DEFINITIONS 4.0. Thermodynamically Reversible The change is carried out infinitesimally slowly so that at each stage the system is in pressure and temperature equilibrium with its surroundings. Consider the work of expansion done by a gas against an external pressure of Pe in a frictionless piston, figure 4.1

Figure 4.1. Work of Expansion.

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Work of expansion done by the gas expanding against an external pressure of Pe is given by Pe V

(4.1)

The expansion is carried out in a number of infinitesimally small steps. This means that the external pressure Pe is only infinitesimally ( ) smaller than Pi And this means that Pi=Pe+ P

(4.2)

so that Pi Pe and the system is in pressure equilibrium with its surroundings. Pe is therefore at the maximum value it can be and still allow the gas to expand. This means that the work of expansion done by the gas is also at its maximum value.

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4.1. Thermodynamically Irreversible The change is carried out at a finite speed. If we consider the expansion of the gas above then Pi=Pe+ P [4.3] And Pi>Pe so that the work of expansion done by the gas for the same volume change is less than the work of expansion done reversibly. Universe – system + its surroundings, the universe can be considered to be a global isolated system and relationship between system and surrounding is shown

Figure 4.2. Thermodynamic System.

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schematically in figure 4.2. It is useful now to define spontaneous and non – spontaneous changes.

Spontaneous Changes Are thermodynamically irreversible. They do not require work to cause the change to take place. Non Spontaneous Changes have no natural tendency to occur, they can be made to occur by doing work. Carnot in (1824) and Clausius in (1860), investigated the efficiency of steam engines. It was found that heat could not be converted completely into work (100% efficiency). If all the heat is not converted to work where does it go? (Remember that the first law of thermodynamics says that energy must be conserved). The second law explains this. We must first and foremost examine some spontaneous changes- be these physical or chemical.

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4.2. Spontaneous Changes

Figure 4.3. Spontaneous Change.

For example, a gas will spontaneously expand to fill a vessel, it does not spontaneously contract to a smaller volume. A hot object will cool to the temperature of the surroundings, it does not get hotter. Some chemical reactions occur spontaneously. For example, in the following reaction, Zn(s) + CuSO4 (aq) = ZnSO4 (aq) + Cu (s)

heat evolved.

A ball dropped from a height will fall to the floor and bounce a few times before eventually coming to rest. These are all common-sense spontaneous

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changes. We could have predicted that all the above processes would be spontaneous. We haven‘t explained how these changes are bought about. What determines the direction of spontaneous change? For example, chemical changes having a positive H Hg2Cl2 (s) + H2(g)  2Hg(l) + 2HCl(g)

(1)

SOCl2 (l) + H2O(l)  SO2(g) + 2HCl(g)

(2)

Physical changes having positive H Salt(s) + water(l) = aqueous solution (aq)

(3)

Chemical changes having negative H

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H2(g)+O2(g) = 2H2O(l) Consider exothermic reactions. If the energy of a system does decrease, the energy of the surroundings must increase (first law of thermodynamics). The increase in the energy of the surroundings is just as spontaneous as the decrease in the energy of the system. For every spontaneous change occurring to the system (closed or open), an opposite spontaneous change is generally happening to the surroundings. We cannot just look at the system we must also consider the surroundings. First law of thermodynamics states that energy is not created or destroyed but converted from one form to another, i.e. the energy is distributed differently after a spontaneous change has occurred. It is this that is the driving force for a spontaneous change. How does this distribution of energy bring about a spontaneous change? In all the examples given above the spontaneous changes are always accompanied by a reduction in the quality of energy. (Quality meaning less available for work and energy in a more dispersed, chaotic form). Finally, then, we have an explanation for spontaneous processes. There is a natural tendency of the universe towards greater chaos.

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A.O. Ibhadon

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4.3. The Dispersal of Energy Let us consider again the example of the bouncing ball(produce your own sketch) The ball does not rise as high after each bounce because of inelastic losses. WHY? The potential energy is converted to kinetic energy, but this kinetic energy is not converted completely back to potential energy, as the ball does not rise to its initial height. What happens to this energy? It has not been lost (first law) it must have been converted to another form of energy. What actually happens is that some of the kinetic energy is converted into thermal motion in the molecules in the floor. This energy does not remain localised but is dissipated throughout the floor. (Heat moves from hot bodies to cold bodies and not the reverse). Eventually the ball will come to rest so that all the potential energy it started with has been converted into thermal motion in the molecules in the floor and has been dissipated throughout the floor. The floor doesn‘t get hot as it is infinitely large. The reverse will never happen spontaneously. A ball at rest on the floor will not start to bounce spontaneously. From the examples of the chemical reactions we can see that for exothermic reactions, heat is lost to the surroundings and dissipated. The endothermic reactions are spontaneous because there is a decrease in the order of the system and therefore the energy is more chaotically distributed. Solid  Liquid  More random, more disorder  (1) Solid +gas = liquid + gas (2) liquid + liquid = gas (3) solid + liquid = liquid

Gas

How do we explain disorder and chaos in thermodynamics? How can we quantify the disorder of a system?

4.4. The Meaning of Entropy(S) Entropy is the measure of the dispersal of matter and energy. If matter and/or energy disperses chaotically then the entropy increases. We will be discussing the changes in entropy of the universe that occur during a spontaneous change. Entropy is a state function, that is it depends only on P, V

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and T and the change in entropy S depends only on the final and initial states and is independent of the path taken. The second law states that the entropy of an isolated system (universe) increases in a spontaneous change. Suniverse >0 where Suniverse = Ssystem+ Ssurroundings We need to mention reversible and irreversible processes. Spontaneous processes are irreversible and they generate entropy. Reversible processes occur without degrading the quality of energy, i.e. they do not generate entropy, but they may transfer entropy from one part of the universe to another. If a process is reversible, therefore, maximum work may be done, as there is no degradation of the quality of the energy. We can relate the dispersal of energy (and therefore the change in entropy) to the heat transferred during a spontaneous process, since the transfer of energy as heat makes use of the chaotic motion of atoms in the surroundings.

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4.4.1. Definition of Entropy

Figure 4.4.

The weight falls, which operates a generator, which produces heat to the infinitely large reservoir. We can determine the change in entropy of the reservoir. Sreservoir will depend on its temperature and how much heat it receives from the falling weight. The transfer of energy to a high temperature body (one which already contains molecules with a lot of thermal motion) will result in a smaller change in entropy than the same transfer of energy to a body at a lower temperature.(the molecules have much less thermal motion), that is, Sreservoir

1/Treservoir.

(4.4)

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Heat produces greater chaotic motion of molecules. The more heat the reservoir receives from the falling weight, the greater the increase in Sreservoir i.e. Sreservoir

-q

(4.5)

(definition q = heat absorbed by the falling weight, therefore heat given out by falling weight = -q) Or Sreservoir = -q/Treservoir

(4.6)

What about the change in the entropy of the falling weight? The process is spontaneous and therefore Suniverse > 0 .

(4.7)

That is Sfalling weight > - Sreservoir

(4.8)

Then,

(4.9)

Sfalling weight>q/T

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for a spontaneous change.

FOR A THERMODYNAMICALLY REVERSIBLE PROCESS Stotal = Ssystem+ Ssurroundings=0 therefore Ssystem= q/T

(4.10) (4.11)

The second law of thermodynamics may now be given by the mathematical equation below. This shows that the entropy of a substance is related to the heat transferred to it and its temperature. S

q/T

(>irreversible, = reversible)

(4.12)

Spontaneously changing systems experience greater disorder than they would if the changes were made under equilibrium conditions. We can use this expression to show that the cooling of a hot body are caused by an increase in entropy.

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4.3. Spontaneous Cooling Consider the transfer of heat, q from a hot body at temperature Th to the cold surroundings at temperature Tc.

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Figure 4.5. Spontaneous Cooling.

Ssurroundings = q/Tc

(4.13)

Ssystem = -q/Th

(4.14)

Stotal = q/Tc – q/Th = q(1/Tc-1/Th) > 0

(4.15)

When the temperature of the surroundings = temperature of the system, Stotal = 0 and the system is in thermal equilibrium with its surroundings. We can calculate the change in entropy for phase changes and temperature changes.

4.4. Entropy of Phase Transitions We have already said that solids are more ordered than liquids and liquids are more ordered than gases That is Ssolid < Sliquid < ,Sgas

(4.16)

If there is a phase change or a transition from one phase to another then there should be a corresponding change in entropy. The entropy change can be calculated as shown below if the phase change is carried out at constant pressure and the system is at equilibrium with its surroundings. Ssystem = Htrans/Ttrans

(4.17)

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where Htrans = enthalpy of the phase transition. An example is the enthalpy of vapourisation, fusion, sublimation, etc. It should be noted that and Ttrans is the temperature at which the transition takes place so that the system is in thermal equilibrium with its surroundings, for example, the change of ice to water occurs at a temperature of 0oC, while the change of water to steam occurs at a temperature of 100oC. The two phases present are in equilibrium with one another. For exothermic phase transitions the change in the entropy is negative. (that is freezing, condensing leads to more ordered systems). Melting and vapourising are endothermic and there is a corresponding increase in the entropy of the system. For example, calculate the change in the entropy for the melting of ice and evaporation of water. The two answers are : Hfus = 6 kJ mol-1

Hvap = 40.7 kJmol-1

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Entropy change is much larger for the change from liquid to gas than for the change from solid to liquid. Can you explain why?

4.4.1. Entropy, Temperature and the Third Law The third law states that the entropy of every pure, crystalline substance at absolute zero is zero.

(4.18) This equation means that if you know the entropy of a system at a temperature Ti and you know the input of energy to change its temperature to Tf along a reversible path (the system is then at equilibrium with its surroundings during the transition), then its entropy can be calculated at Tf. We will consider the entropy change occurring at constant pressure and constant volume. At constant pressure the heat absorbed by the system = enthalpy change H. The change in enthalpy with temperature = heat capacity.Cp We can therefore write qrev= CpdT. Therefore

(4.19)

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(4.20) Similarly, at constant volume

(4.21)

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Cp and Cv are independent of temperature over a short temperature range so that the entropy after heating the system at constant pressure and constant volume can be calculated using the 2 equations below. S(Tf) = S(Ti) + Cpln(Tf/Ti)

(4.22)

S(Tf) = S(Ti) + Cvln(Tf/Ti)

(4.23)

Calculate the change in entropy when the temperature of 1.0 mol of H2 is raised from 20oC to 30oC at constant volume.(Cv = 20.44 J mol-1 K-1) S = 0.69 J K-1.

4.4.2. Efficiencies of Thermal Processes Thermodynamics grew out of the study of ways of improving the efficiencies of heat engines. 4.4.3. Production of Heat From Work. For an engine to be of any use, the production of work from heat must be spontaneous. If the engine extracts heat from the hot sink at temperature Th and converts it entirely to work then the corresponding change in entropy in the hot sink is given by S =-qh/T that is heat is withdrawn from the hot sink therefore the entropy of the hot sink decreases. Work does not generate disorder therefore the total change in entropy is negative and the reaction is non spontaneous. If a cold sink is present, some heat can be discarded into it, figure 3.7.

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Figure 4.6. Efficiency of Thermal Processes.

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Figure 4.7.

The entropy lost by the hot sink = -qh/Th

(4.24)

The entropy gained by the cold sink =qc/Tc

(4.25)

The overall entropy change = qc/Tc - qh/Th

(4.26)

If qc > (Tc/Th ) qh then spontaneous.

S will be positive and the process will be

4.6. Gibbs Free Energy (G) When determining whether a change will be spontaneous or not we need to consider the changes in entropy of the system and the surroundings. Stotal = Ssystem + Ssurroundings

(4.27)

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If the process occurs at constant pressure and at the same temperature of the surroundings then this equation becomes: Stotal = Ssystem- Hsystem /T

(4.28)

-q = - Hsystem = T Ssurroundings

(4.29)

Since

at constant pressure. The total entropy change is expressed in terms of the properties of the system alone provided that the system is undergoing the change at constant temperature and pressure. Chemists are generally interested in reactions occurring at constant temperature and pressure. For a spontaneous change

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Stotal = Ssystem- Hsystem/T >0

(4.30)

Or T Ssystem > Hsystem

(4.31)

Or Hsystem -T S 3 ( the activity of the various sites) Some adsorption sites may be more active than others and adsorption will occur preferentially at these sites. This will cause the adsorption to become less exothermic as increases. Another reason why H is not constant is that there is likely to be interactions between adsorbed molecules. If this interaction is repulsive then this will cause H to decrease. This effect will become more prominent at high . Despite these criticisms, a large number of experimental isotherms do fit the Langmuir isotherm well especially those relating to N2, O2, Ar on charcoal at – 183oC.

5.3. Freundlich Adsorption Isotherm Originally this was an empirical observation, but later it was shown that this was consistent with Hads decreasing exponentially with increasing coverage such that V = kP1/n V = volume of gas adsorbed, P = pressure, k and n are constants.

(5.8)

The plot log V vs Log P is a straight line of gradient 1/n and intercept log k. This is consistent with Hads decreasing exponentially with increasing (surface coverage). Freundlich Adsorption Isotherm is obeyed by adsorption from solution at low concentrations.

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5.4. Temkin Adsorption Isotherm There is a linear variation of H with surface coverage( ) = k ln cP

(5.9)

where k and c are constants. Assumes linear variation of H with surface coverage, .

5.5. B.E.T Adsorption Isotherm

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The B.E.T isotherm deals with multilayer adsorption on non-porous solids and represents an extension of Langmuir isotherm. It was derived form balancing the rates of evaporation and condensation for various layers. The assumptions of the BET isotherm include: 1. Surface is homogeneous surface and all sites are equivalent 2. No lateral interactions within the film. 3. Each adsorption site that is filled forms a site for multilayer adsorption i.e no limit to adsorption therefore no plateau and no limiting value to adsorption. 4. All layers above the monolayer are all equivalent and like a bulk liquid. This model is essentially a Langmuir adsorbed layer and multi-layers of liquid like adsorbate which are all energetically equivalent.

Figure 5.7.

V/Vm = cx/((1-x)(1-x+cx))

B.E.T Isotherm equation

(5.10)

Where x = P/Po = relative pressure.

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C = exp ((E1-E)/RT)

(5.11)

The shape of the BET isotherm depends on the value of c. This describes the binding energy between the first and subsequent layers. This isotherm is used in calculating the area of solid surfaces.

Exercise 5.1. Using the Langmuir Isotherm The data given below are for the adsorption of C0 on charcoal at 273 K. Confirm that they fit the Langmuir isotherm, and find the constant K and the volume corresponding to complete coverage. In each case V has been corrected t0 1.00 atm. P/Torr 100 V/cm3 13.2 (p/Torr)/(Vcm3) 7.58

200 20.6 9.71

300 26.5 11.32

400 33.5 11.94

500 34.9 14.33

600 44.6 13.45

700(x-axis) 48.1 14.55(y-axis)

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From our discussion, we have shown that Kp +

= Kp

With = V/V , where V is the volume corresponding to complete coverage. This expression can be rearranged into equation [5.12] as

(5.12) Hence, a plot of p/V against p, figure 5.8, should give a straight line of slope 1/V and intercept 1/KV . The data to plot is highlighted in blue in the above table

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Figure 5.8.

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Exercise 5.2. Measuring the Isosteric Enthalpy of Adsorption The data below show the pressures of C0 needed for the volume of adsorption (corrected to 1.00 atm and 273 K) to be 10.0 cm3 Calculate the adsorption enthalpy at this surface coverage. T/K 200 210 220 230 p/Torr 30.0 7.1 45.2 54.0 ln(p/Torr) 3.40 3.61 3.81 3.99 3 10 /(T/K) 5.00 4.76 4.54 4.35 Solution The Langmuir isotherm can be rearranged to

240 63.5 4.15 4.17

250 73.9 4.309(x-axis) 4.00(y-axis)

(5.13) Therefore, when

is constant,

1n K + 1np = constant

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It follows that

(5.14) with d(1/T)/dT = -1/T2, this expression rearranges to

(5.15)

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Therefore, a plot of 1np against 1/T, as in figure 5.9, should be a straight line of slope adH /R. Solution Draw up the following table and plot 1np against 1/T and determine adH from the slope. T/K 200 210 3 10 /(T/K) 5.00 1n(p/Torr) 3.40

220 4.76 3.61

230 4.55 3.81

240 4.35 3.99

250 4.17 4.15

4.00(x-axis) 4.30(y-axis)

Figure 5.9.

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Exercise 5.3. Repeat the calculation using the following data: T/K 200 210 220 230 P/Torr 32.4 41.9 53.0 66.0

240 80.0

250 96.0

[Ans = -9.0 kJ mo1-1]

Exercise 5.4. Using the BET Isotherm The data below relate to the adsorption of N2 on rutile (Ti02) at 75 K. Confirm that they fit a BET isotherm in the range of pressures reported, and determine Vmon and c.

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p/torr Vmm3

1.20 601

14.0 720

45.8 822

87.5 935

127.7 1046

164.4 1146

204.7 1254

At 75 K,p* = 570 Torr. The volumes have been corrected to 1.00 atm and 273 K and refer to 1.00 g of substrate. The BET isotherm can be reorganised into

(5.16) It follows that (c – 1)/cVmon can be obtained from the slope of a plot of the expression on the left against z, figure 5.10 and cVmon can be found from the intercept at z = 0. The results can then be combined to give c and Vmon. Solution Draw up the following table: p/torr 103z 104z/(1 – z) (V/mm3)

1.20 2.11 0.035

14.0 24.6 0.350

45.8 80.4 1.06

87.5 154 1.95

127.7 224 2.76

164.4 288 3.53

204.7 359(x-axis) 4.47(y-axis)

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Figure 5.10.

The least squares best line has an intercept at 0.0406, so 1 = 4.06 x 10-6 mm-3 cVmon Thus 1 = cVmon X 4.06 x 10-6 mm-3 And hence cVmon = 246305.42mm3 The slope of the line is 1.23 x 10-2, so c – 1 = (1.23 x 10-2) x 103 x 10-4 mm-3 = 1.23 x 10-3mm-3 cVmon The solutions of these equations are c = 303 and Vmon = 814mm3. We should now discuss the practical uses of adsorption process. As you can imagine, these are many and varied.

5.6. Practical Uses of Adsorption Processes There are many practical uses and applications of adsorption processes. Some of these are summarised as follows:

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Figure 5.11.

1. Catalysis Mechanisms vary widely and there are many types. Have seen that adsorption at solid surfaces allows reactions to proceed which either wouldn‘t occur or would occur slowly. An example is the hydrogenation of unsaturated compounds. CH3COCH3 + H2  CH3C(OH)CH3 In the presence of a metal catalyst such as Ni or Pt, the hydrogen dissociates as shown below.

2. Determination of Solid Surface Areas Examples include N2 adsorption , BET Ibhadon, A.O.. Physical Chemistry Examinations, Nova Science Publishers, Incorporated, 2009. ProQuest Ebook Central,

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123

3. Adsorption of Impurities Examples include sewage , gas masks, water, gas purification. 4. Surfactants Example water loss from reservoirs in hot climates, washing powders, detergents (solubilisation and wetting). 5. Separation processes. Example, Chromatographic techniques

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SUMMARY OF CHAPTER FIVE Surface chemistry deals with chemistry at surfaces – solid surfaces. We have discussed the isotherms that describe the reactions on these surfaces whether these be chemisorption or physisorption. We have also discussed the practical uses and applications of surface chemistry , such as in catalysis and surfactants. Finally, we have mentioned some of the techniques that are available for the characterisation of solid surfaces. Worked examples have been provided to aid the student in understanding the chapter and preparing for an examination in this subject or topic.

Revision Questions Question 5.1. (a) State two differences between the Langmuir and the BET isotherms (b) List two practical applications of adsorption processes The data below relate tot he adsorption of N2 on titanium dioxide at 75 K. confirm that they fit a BET isotherm in the range of pressures reported, and determine Vmon and c. p/torr Vmm3

1.22 600

16.0 740

46.8 820

86.5 940

128.7 1040

164.6 1140

204.8 1260

At 75 K, p* = 570 Torr. The volumes have been corrected to 1.00 atm and 273 K and refer to 1.00 g of substrate.

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(c) List three techniques that may be used for the surface characterisation of materials (d) State one industrial application of the BET isotherm

Solution (a) Differences between the Langmuir and BET Isotherms Langmuir isotherm applies to monolayer adsorption, of equivalent sites and uniform surface. On the other hand, the BET isotherm deals mainly with multilayer adsorption. It can be used to determine the surface area of solids in industry. (b) Application of surface chemistry processes Catalysis, hydrogenation, separation processes, surfactants, etc

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The BET isotherm can be reorganised into

It follows that (c – 1)/cVmon can be obtained from the slope of a plot of z/ (1 – z)V against z, and cVm is obtained from the intercept at z = 0. The results can then be combined to give c and Vm. (see figure 5.10) This is similar to the straight line equation, y = mx+c The following table is drawn up: p/torr 1.22 3 10 z 2.10 104z/(1 – z) 0.026 (V/mm3)

16.0 28.0 0.346

46.8 82.10 1.01

86.5 151.7 1.87

128.7 225.8 2.79

164.6 288.8 3.57

204.8 359.3 4.44

The intercept of the plot is 0.002, so

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The slope of the line is 1.24 x 10-2, so c – 1 = (1.23 x 10-2) x 103 x 10-4 mm-3 = 1.24 x 10-3mm-3 cVm The solutions to these equations are c = 621 and Vmon = 805.15mm3. (Note that Z = P/P*, where P* is 570 torr) (c) Characterisation of the surface of Materials: Any three of: Scanning tunnelling microscopy, Low energy electron diffraction, Electron loss spectroscopy, High vacuum techniques, Auger electron spectroscopy, Ionization techniques Atomic Force Microscopy, Molecular Beam Scattering (d) Surface area determination

Question 5.2. The pressures of CO needed for the volume of adsorption corrected to 1.00 atm and 273 K to be 10.0 cm3 are given below. Determine the adsorption enthalpy at this surface coverage. T/K P/Torr

200 30

210 39.1

220 46.2

228 56.0

240 66.5

260 75.9

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Solution to Question 5.2. Enthalpy of Adsorption Determination The Langmuir isotherm can be rearranged to

Therefore, when It follows that

is constant, 1n K + 1np = constant

with d(1/T)/dT = -1/T2, this expression rearranges to

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Therefore, a plot of 1np against 1/T, figure 5.12, should be a straight line of slope adH/R. The following table is drawn up:

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T/K 200 3 10 /(T/K) 5.00 1n(p/Torr) 3.40

210 4.76 3.65

220 4.54 3.83

230 4.34 4.02

240 4.16 4.19

260 3.84 4.32

103/( T/K) Figure 5.12.

The slope of the graph is –0.8145, so = -6.77 kJmo1-1.

adH

= -(0.8145 x 103 K) x R

Question 5.3. Using The Langmuir Isotherm The data given below are for the adsorption of C0 on charcoal at 273 K. Confirm that they fit the Langmuir isotherm, and find the constant K and the volume corresponding to complete coverage. In each case V has been corrected t0 1.00 atm.

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Surface Chemistry 100 10.2

P/Torr V/cm3

200 18.6

300 25.5

127 400 31.5

500 36.9

600 41.6

700 46.1

Solution to Question 5.3 From Kp + = Kp and with = V/V , where V is the volume corresponding to complete coverage, this expression can be rearranged into

Hence, a plot of p/V against p, figure 5.13, should give a straight line of slope 1/V and intercept 1/KV . The data for the plot are as follows: p/Torr (p/Torr) / (V/cm3)

100 9.80

200 10.8

300 11.8

500 13.6

600 14.4

y = 0.009x + 9.0143

16 15 14 13 12 11 10 100

200

300

700 15.2

= 111 cm3. The intercept at p = 0

K = 1 = 1.00 x 10-3 Torr-1 (111 cm3) x (8.99 torr cm-3)

lnp

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The (least squares) slope is 0.00904, so V is 8.99, so

400 12.7

400

P/Torr 500

600

Temperature

Figure 5.13.

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Comment At high surface coverage, the data would deviate from a straight line. The dimensions of K are 1/pressure.

Question 5.4. Measuring the Isosteric Enthalpy of Adsorption The data below show the pressures of C0 needed for the volume of adsorption (corrected to 1.00 atm and 273 K) to be 10.0 cm3.. Calculate the adsorption enthalpy at this surface coverage. T/K 200 210 220 230 240 250 p/Torr 30.0 37.1 45.2 54.0 63.5 73.9

Solution to Question 5.4.

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The Langmuir isotherm can be rearranged to

Therefore, when is constant, 1n K + 1np = constant It follows from equation 4.13 that

(5.14) with d(1/T)/dT = -1/T2, this expression rearranges to

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Therefore, a plot of 1np against 1/T, figure 5.14, should be a straight line of slope adH /R. Draw up the following table: T/K 200 210 220 230 240 250 103/(T/K) 5.00 4.76 4.55 4.35 4.17 4.00(x-axis) 1n(p/Torr) 3.40 3.61 3.81 3.99 4.15 4.30(y –axis) The slope(of the least squares fitted line) is –0.904, so adH = -(0.902 x 103 K) x R = -7.50 kJmo1-1.

4.4

y = -0.9041x + 7.9195

4.2

lnp

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4 3.8 3.6 3.4 3.2 3 3.9

4.1

4.3

4.5

1034.7 / (T/K) 4.9

5.1

Temperature

Figure 5.14.

Comment the value of K can be used to obtain a value of adG , and then that value combined with adH to obtain the standard entropy of adsorption.

Exercise 5.5. Repeat the calculation using the following data:

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130 T/K 200 210 P/Torr 32.4 [Ans = -9.0 kJ mo1-1]

220 41.9

230 53.0

240 66.0

250 80.0

96.0

5.6. Using the BET Isotherm The data below relate to the adsorption of N2 on rutile (Ti02) at 75 K. confirm that they fit a BET isotherm in the range of pressures reported, and determine Vmon and c. p/torr 1.20 14.0 45.8 87.5 127.7 164.4 204.7 Vmm3 601 720 822 935 1046 1146 1254 At 75 K,p* = 570 Torr. The volumes have been corrected to 1.00 atm and 273 K and refer to 1.00 g of substrate.

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Solution to Question 5.6 Equation 4.16 can be reorganised into

It follows that (c – 1)/cVmon can be obtained from the slope of a plot of the expression on the left against z, and cVmon can be found from the intercept at z = 0. The results can then be combined to give c and Vmon, figure 5.15 We can draw up the following table: p/torr 1.20 3 10 z 2.11 104z/(1 – z) 0.035 (V/mm3)

14.0 24.6 0.350

45.8 80.4 1.06

87.5 154 1.95

127.7 224 2.76

164.4 288 3.53

204.7 359 (x-axis) 4.47 (y - axis)

The least squares best line has an intercept at 0.0406, so 1 = 4.06 x 10-6 mm-3 cVmon

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The slope of the line is 1.23 x 10-2, so c – 1 = (1.23 x 10-2) x 103 x 10-4 mm-3 = 1.23 x 10-3mm-3 cVmon The solutions of these equations are c = 303 and Vmon = 814mm3.

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Figure 5.15.

Comment At 1.00 atm and 273 K, 810 mm3 corresponds to 3.6 x 10-5mo1, or 2.2 x 1019 atoms. As each atom occupies an area of about 0.16 nm2, the surface area of the sample is about 3.5 m2.

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Chapter 6

UNITS AND PHYSICAL CONSTANTS IN PHYSICAL CHEMISTRY Property

Unit

Symbol

Area Specific surface Area

Square meter Square meter per kilogram Moles of charge per cubic meter Moles of charge per kilogram of adsorbent Moles per cubic meter Faraday Coulomb Volt

m2 m2kg-1

Siemens per meter Joule Newton Kilogram per cubic meter Moles per cubic meter Moles per kg of solvent Pascal, Torr, Newton second per square meter Cubic meter

S m-1 J N kg m-3

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Charge concentrations Specific adsorbed charge Concentration Electrical Capacitance Electric Charge Electric Potential difference Electrical Conductivity Energy Force Mass density Molarity Molality Pressure Viscosity Volume

Expression in SI Base Units

mol m-3 mol kg -1 mol m-3 F C V

m-2kg-1S4A2 As m2kgs-3A-1 m-3kg-1S2A2 m2kg s-2 m kg s-2

mol m-3 mol kg -1 Pa Ns m-2 m3

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(Continued). Property

Unit

Symbol

Specific volume

Cubic meter per kilogram J J J J J JK-1 JK-1 Jmol-1 Ө

m3kg -1

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Heat Work Internal energy Enthalpy Gibbs energy Heat Capacity Entropy Chemical Potential Surface Coverage

Expression in SI Base Units

Q W U(E) H G C S Μ m3

Physical Constant Plank‘s Constant

Symbol h

Value 6.6260775x10-34 Js

Faraday‘s Constant

F

9.6485309x104Cmol-

Remarks Energy, wavelength Frequency Electrochemistry

Gas Constant

R

8.314510JK-1

Properties of Gases

Boltzmann‘s Constant

KB

Molar Volume of an Ideal Gas Avogadro‘s Constant

Vo

1.38065812x10-23 JK-1 22.711108Lmol-1

Kinetics, activation energy, temperature Properties of Gases

6.0221367x10-23 mol-1

Moles, molecules, atom

NA

1

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INDEX

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A absolute zero, 104 absorption, 37, 122 acetaldehyde, 23, 53 acid, 20 activation energy, 1, 22, 23, 24, 25, 26, 27, 28, 51, 52, 57, 124, 125, 126, 152 adiabatic, 65, 67, 85, 90 aid, 140 air, 64, 82 alternative, 68 ammonia, 5, 114 application, 81, 108, 122 aqueous solution, 98 assumptions, 44, 83, 84, 85, 129, 133 Atomic Force Microscopy, 142 atoms, 33, 37, 84, 100, 149

B binding, 133 binding energy, 133 boiling, 61, 124 bomb, 69, 72 bonds, 28, 29 bromine, 3, 8, 37, 84

C calcium, 72 calcium carbonate, 72 carbon, 74, 83 carbon monoxide, 74, 83 cast, 116 catalysis, 122, 140 catalyst, 2, 3, 122, 139 CH4, 37, 38, 40, 57, 81, 110, 119 chaos, 98, 99 charcoal, 122, 123, 124, 130, 131, 134, 144 chemical bonds, 123 chemical energy, 60 chemical engineering, ix chemical reactions, 1, 2, 19, 93, 94, 97, 99 chemisorption, 123, 125, 127, 131, 140 CO2, 8, 81, 110, 111, 119, 130 collisions, 26, 27, 28, 29, 30, 31, 32, 40, 44, 45 combustion, 73, 81, 82 components, 112 composition, 7, 83, 84, 112, 114 compounds, 139 condensation, 122, 127, 133 conductivity, 4, 9 conservation, 62, 81 consumption, 5, 6 conversion, 72

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Index

136 cooling, 102 crystalline, 104

D

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decomposition, 7, 19, 23, 34, 35, 47, 48, 51, 53, 55, 56, 57, 111 definition, 72, 101 degradation, 100 degrading, 100 density, 61, 151 desorption, 129 detergents, 139 disorder, 99, 102, 105, 110 dissociation, 35, 73, 81, 84, 126 distribution, 1, 26, 98 drugs, 2 dyeing, 122

flight, 29 flow, 63, 70 food, 60 freezing, 103 friction, 60 fructose, 20 fusion, 73, 81, 84, 103

G gas phase, 19 gases, 5, 27, 29, 31, 33, 44, 57, 65, 69, 78, 81, 103, 115, 126 gel, 123 glass, 60 glucose, 20 graph, 23, 144 grounding, ix

E

H

electron, 73, 81, 142 electron diffraction, 142 electrons, 37 emulsions, 122 endothermic, 71, 99, 103, 110, 118 energy transfer, 60 engines, 96, 105 entropy, 84, 89, 93, 94, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 124, 147 environment, 116 equilibrium, 3, 4, 35, 67, 79, 80, 84, 93, 94, 95, 102, 103, 104, 109, 110, 113, 114, 115, 123, 124, 129, 130 evaporation, 79, 104, 133 examinations, iv, 23 exercise, 24, 74 extrapolation, 52

half-life, 17, 18, 21, 49, 53, 56 heat capacity, 67, 75, 77, 78, 81, 83, 84, 88, 89, 104 heat release, 71 heat transfer, 70, 100, 102 heating, 64, 105 height, 70, 82, 97, 99 helmholtz equation, 84, 86, 93, 117 high temperature, 101, 131 hydrogen, 3, 5, 8, 27, 28, 37, 74, 83, 126, 139 hydrogen gas, 83 hydrogenation, 54, 139, 141

F film, 122, 129, 131, 133

I ice, 83, 84, 103, 104 indication, 94 induction, 48 induction period, 48 industrial, 2, 3, 141 industrial application, 141 industry, 2, 4, 141

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Index inelastic, 99 inhibition, 38 initial state, 100 initiation, 3, 4, 37, 38, 57 integration, 13, 14, 15, 17, 18, 50, 65 interaction, 131 interactions, 123, 129, 131, 133 interface, 122 interference, 12 intermolecular, 129 interval, 48 inversion, 20 iodine, 9, 27 isolation, 19 isothermal, 4, 68 isotherms, 121, 124, 127, 128, 131, 140, 141

137

molecular forces, 123 molecules, 25, 26, 27, 28, 29, 30, 31, 32, 33, 37, 44, 45, 46, 61, 62, 71, 99, 101, 122, 124, 131, 152 momentum, 45 monolayer, 127, 130, 133, 141 motion, 44, 45, 60, 99, 100, 101 movement, 60

N natural, 60, 96, 98 nitrogen, 5, 47 non-uniform, 131

O

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L

oil, 122 optimization, 3 organic, 55 orientation, 28, 29 oxide, 47 oxygen, 33, 73, 116 ozone, 7, 35

laws, 56 learning, 59 learning outcomes, 59 line graph, 16, 17 linear, 132 liquid interfaces, 121 liquid phase, 53 liquid water, 73 liquids, 78, 103 localised, 99 losses, 99 low temperatures, 126

P

M manufacturing, 2 mechanical energy, 60 melt, 84 melting, 104 methane, 57 methanol, 72 microscopy, 142 misunderstanding, ix molar volume, 67, 86, 92 mole, 4, 73, 83

parameter, 128, 130 perfect gas, 78, 83 permit, 60 pharmaceutical, 2 pharmaceutical industry, 2 phase decomposition, 19 phase transitions, 103 photolysis, 37 photon, 37 physical chemistry, ix play, 60 pores, 127 porous solids, 123, 127, 132 powders, 123, 139 power, 7 production, 2, 5, 37, 105

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Index

138 propagation, 3, 4, 37, 38, 57 propane, 82 purification, 139

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R radical reactions, 2 random, 44, 45, 99 reactant, 4, 5, 7, 9, 17, 34, 37, 40, 54, 87 reactants, 2, 4, 5, 7, 8, 9, 19, 26, 50, 72, 73, 111, 114 reaction order, 20, 56, 57 reaction rate, 27 reaction time, 51 reactivity, 122 reading, 93 recognition, 73 refractive index, 4 relationship, 60, 80, 81, 96 relaxation, 49, 50 relaxation time, 49, 50 relevance, ix reservoir, 101 reservoirs, 139 resistance, 60 rutile, 137, 147

S sample, 53, 149 separation, 62, 121, 141 series, 3, 4, 11, 21, 79 sewage, 139 shape, 127, 133 sign, 63 silica, 123 sites, 122, 129, 131, 133, 141 skin, 2 SO2, 56, 98 solid surfaces, 128, 133, 139, 140 solvent, 151 species, 4, 5, 6, 7, 19, 26, 32, 33, 34, 56, 122 specific heat, 78 spectrophotometric, 53

spectrophotometric method, 54 spectroscopy, 142 speed, 2, 30, 31, 32, 46, 79, 80, 96, 126 spontaneity, 110, 111 steady state, 36, 39, 41, 48, 57 steel, 60 steric, 28, 29, 32, 57 stoichiometry, 4, 5, 6 students, ix, 1, 2, 23, 59, 93, 121 substances, 61, 72, 74, 87 sucrose, 20, 21 sugar, 82 surface area, 122, 123, 141, 149 surface chemistry, 121, 140, 141 surface properties, 121 surface tension, 61, 122 surfactants, 140, 141 synthesis, 114 systems, 94, 102, 103, 110, 121

T temperature dependence, 24, 52 thermal equilibrium, 79, 103 thermodynamic function, 70, 86, 107 thermodynamic properties, 74 thermodynamics, 59, 60, 62, 65, 79, 81, 85, 90, 94, 96, 98, 99, 102, 109, 110, 111 thermolysis, 37 thiosulphate, 9 third order, 7, 12, 13, 16, 17, 42, 56 time, 4, 5, 6, 9, 10, 14, 17, 20, 27, 29, 36, 43, 48, 49, 50, 53, 55 titanium dioxide, 141 titration, 9 total energy, 61, 62, 63 transfer, 60, 70, 100, 101, 102 transition, 103, 104 translation, 61 travel, 44, 46

U undergraduate, ix

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Index uniform, 122, 131, 141 universe, 61, 62, 94, 96, 98, 100, 108, 109

V

velocity, 27, 29, 45, 46 vibration, 61 violent, 37 viscosity, 9

W walking, 70 water, 61, 76, 79, 83, 84, 98, 103, 104, 111, 122, 139 wetting, 122, 139

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vacuum, 65, 142 values, 28, 36, 52, 59, 77, 78, 82, 83, 85, 92, 93, 117 van der Waals forces, 62, 123 variables, 3, 61, 72, 86 variation, 36, 48, 59, 75, 77, 84, 88, 94, 116, 132

139

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