1
CHEMICAL KINETICS 25-05-2015
Contact: Phone: +92-303-7807073 E-mail:
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4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 4.10
4.11 4.12 4.13 4.14
By: Muhammad Asif Lecturer (Physical Chemistry) Govt. College Sahiwal, Pakistan
Introduction ……………………………………………………………………… Rate of reaction ……………..…………….……………………………………… Rate equation and rate constant…………….……………………………………. Order and molecularity of reactions …………..…………………………………. Pseudo order reactions …………………………………………………………… Zero order reactions …………………………….……………………………….. First order reactions …………………………….……………………………….. Second order reactions ……………………..……………………………………. Third order reactions …………………………………………………………….. Elementary and complex reactions ……………………………………………… 4.10.1 Opposing reactions ………………………………………………………. 4.10.2 Consecutive reactions ……………………………………………………. 4.10.3 Parallel reactions ……….…………..……………………………………. 4.10.4 Chain reactions …………………………………………………………… Thermal reactions ……………………………...…………………………………. Photochemical reactions ………………..………………………………………… Effect of temperature on rate of reaction …..……………..……………………… Theories of reaction rates ………………………………………………………… 4.14.1 Collision theory …………………………………………………………... 4.14.2 Transition state theory …………………………………………………….
02 02 03 04 05 06 07 09 13 23 23 34 36 37 37 40 44 48 48 52
2 4.1
INTRODUCTION Chemical kinetics is the branch of chemistry that deals with the rates of chemical
reactions. The chemical reactions can broadly be divided into three classes: 1. Fast reactions e.g., the reaction between aqueous sodium chloride and silver nitrate. 2. Moderate reactions e.g., hydrolysis of an ester. 3. Slow reactions e.g., rusting of iron. However, the chemical reactions can occur at a variety of rates from very fast to very slow. For example, the paper on which this book is printed reacts extremely slowly with the atmosphere oxygen, whereas explosion is extremely fast reaction. The two extreme classes of reactions cannot be studied conveniently. Reactions which proceed with measureable rates are the subjects of study by physical chemists. In 1864 C. M. Guldberg and P. Waage pioneered the development of chemical kinetics by formulating the law of mass action, which states that the speed of a chemical reaction is proportional to the quantity of the reacting substances. Experimentally it has been found that the rate of a chemical reaction depends on the nature of the reacting species, the temperature, the pressure, the concentration of the reacting species and the presence of catalyst. The reactions may be classified kinetically as being either homogenous, if they take place in one phase only, or heterogeneous if two or more phases are involved in the processes. The reactions may be elementary or complex. An elementary reaction occurs in a single step while a complex reaction occurs in two or more steps. 4.2
RATE OF REACTION The rate of a chemical reaction is the change in the concentration of reactants or
products in a unit time or a given time.
Rate
Change in concentration c Time period of change t
This equation gives the average velocity of the reaction during the time of observation. By shortening the time of observation the average velocity approaches more and more closely to the actual velocity of the reaction. So in chemical kinetics the rate of reaction is not represented by the average velocity but by the instantaneously velocity of the reaction that is when the time of observation approaches zero then, Rate
d[c] dt
3 The brackets [ ] represent the concentration in moles per decimeter cube. Thus rate of reaction is expressed in units of moles per decimeter cube per second ( mol dm3 sec1 ). For gas phase reactions, pressure units are used in place of molar concentration. Consider the following general reaction,
A+B
C+D
The rate at which reaction proceeds can be measured in terms of the rate at which one of the reactants disappears or one of the products appears. dx d[A] d[B] d[C] d[D] dt dt dt dt dt
Where [A], [B], [C] and [D] represent the concentrations of A, B, C and D in moles per decimeter cube at time t. The negative sign indicates that the concentrations of A and B decreases with increasing time. The plus sign indicates that the concentrations of C and D increase with increasing time. 4.3
RATE EQUATION AND RATE CONSTANT According to law of mass action, the rate of a chemical reaction is proportional to the
product of the molar concentrations of the reactants raised to the power equal to the number of molecules of each species taking part in the reaction. Consider the reaction,
aA + bB
cC + dD
dx [A]a [B]b dt dx = k[A]a [B]b dt
Where, k is constant of proportionality known as the velocity constant or rate constant or specific rate constant. This expression shows how the reaction rate is related to concentration and is known as rate equation, rate expression or rate law. If the concentration of A and B are kept unity, then dx = k[1]a [1]b = k dt
Thus rate constant may be defined as the rate of reaction when molar concentration of each of the reactants is unity (1 mol dm−3). The value of k varies from reaction to reaction and also varies with temperature for a given reaction. The value of k for a reaction does not change with time. The units of k depend on the order of reaction.
4 Anyhow it is not essential that the powers in the rate equation are always equal to the coefficients of balance chemical equations. So we can say that the rate equation of above reaction may be, dx = k[A]m [B]n dt
The power or exponent of a concentration term, m or n in the rate equation is usually in small whole number integers (0, 1, 2, 3) or may be fractional. Here n is called the order of reaction with respect to A, and m is the order of the reaction with respect to B. The sum (n + m) is called the overall order of the reaction. Rate equation for an elementary reaction can be predicted from balanced chemical equation but it cannot be predicted for complex reactions (involving more than one steps). Therefore it is written from experimental facts for complex reactions. Actually rate equation is an experimental expression. The order of the reaction is completely independent of the reaction stoichiometry. 4.4
ORDER AND MOLECULARITY OF REACTIONS The order of a reaction is defined as the total number of atoms, ions or molecules
whose concentration changes during a chemical reaction. It can be calculated from an experimentally determined rate equation. It is the sum of exponents of all the concentration terms in the differential form of rate equation. A reactant whose concentration does not affect the rate of reaction is not included in the rate equation. The concentration of such a reactant has the power 0 e.g., [A]0 =1. For a reaction maximum order is three and the minimum is zero. However, the order of reaction may be in minus with respect to a particular species in the reaction; such a species acts as an inhibitor and rate of reaction is inversely proportional to its concentration. The reactions may be classified according to the order. If (m + n) in the rate equation is: m+n=0
the reaction is zero order reaction
m+n=1
the reaction is first order reaction
m+n=2
the reaction is second order reaction
The order of reaction provides valuable information about the mechanism of a reaction and the knowledge of mechanism of a given reaction allows us to control that reaction. The order of reaction should not be confused with molecularity which is defined as the total number of atoms, ions or molecules which take part in a reaction as given by the balanced chemical equation. Molecularity and order are identical for elementary reactions. Most chemical reactions are complex reactions which occur in a series of steps.
5 Each step is an elementary reaction. The stepwise sequence of elementary reactions that convert reactants to products is called mechanism of the reaction. In any mechanism, some of the steps will be fast, others will be slow. The rate of the reaction is determined by the slowest step known as the rate determining step. Hence, molecularity of a reaction can also be defined as the total number of atoms, ions or molecules taking part in the rate determining step. The term unimolecular, bimolecular and termolecular (or trimolecular) indicate the number of particles reacting in a single elementary process. No 1 2 3 4 5
6 7
4.5
Order of Reaction It may be equal to zero. It can be in fractions. It is at most equal to three. It can only be determined experimentally. It is the sum of all the exponents of the concentration terms in the rate equation. It helps in determining the mechanism of a reaction. In case of complex reactions order is determined by the slowest step of the reaction.
Molecularity It can never be equal to zero. It can never be in fractions. It can be more than three. It can be obtained from a simple balanced chemical equation. It is the sum of number of molecules of the reactants taking part in a single step chemical reaction. It gives no idea about the mechanism of a reaction. In case of complex reactions each step of the reaction has its own molecularity i.e., molecularity has no significance for complex reactions.
PSEUDO ORDER REACTIONS A reaction in which one of the reactants is present in large excess shows an order
different from the actual order. The experimental order which is not the actual one is referred as the pseudo order. Since for elementary reactions molecularity and order are identical, pseudo reactions may also be called pseudo molecular reactions. Let us consider a reaction,
A+B
Products
In which the reactant B is present in a large excess. Since it is an elementary reaction, its rate equation can be written as; Rate = k [A] [B] As B is present in large excess its concentration remains practically constant in the course of reaction. Thus the rate law can be written as; Rate = k′ [A]
6 Where the new rate constant k′ = k [B]. Thus the actual order of the reaction is second order but in practice it will be first order. Therefore, the reaction is said to have a pseudo first order. Examples: 1. Hydrolysis of an ester: methyl acetate upon hydrolysis in aqueous solution using a mineral acid as a catalyst forms acetic acid and methyl alcohol. H
CH3COOCH3 + H2O
CH3COOH + CH3OH
Here a large excess of water is used and the rate law can be written as Rate = k [CH3COOCH3] [H2O] Rate = k′ [CH3COOCH3] The reaction is actually second order but in practice it is found to be first order. Thus it is a pseudo first order reaction. 2. Hydrolysis of sucrose: sucrose upon hydrolysis in the presence of a dilute mineral gives glucose and fructose. 4.6
ZERO ORDER REACTION The reaction in which rate is independent of the concentration of the reactants is
called zero order. Let us take a substance A which decomposes into products. Its initial concentration is ‘a’ moles dm−3. Let after time ‘t’ seconds the amount left behind is (a-x) moles dm−3 and that converted into product is ‘x’ moles dm−3. Then
A When t = 0 a When t = teq (a-x) Hence, the rate of reaction is given by,
Products 0 x
dx k(a x) dt dx k dt
dx kdt Integrating,
dx k dt x kt C The value of integration constant C can be found by applying the initial conditions of the reaction when t = 0, x = 0. That is
x kt
……… (1)
7 This equation (1) is similar to the equation for a straight line. Thus, if we plot a graph between x on y-axis and t on x-axis, we should get a straight line passing through the origin with slope k.
Rearranging Equation (1) for k, k
x t
……… (2)
This equation (2) gives the rate constant for zero order reaction. The units of k are mole dm−3sec−1. Half-Life It is defined as the time required to reduce the concentration of reactants to half of its original value. Therefore, when t = t1/2 then x = a/2. Substitute this value in equation (2),
t1/2
a 2
k
a 2k
Examples 1. Thermal decomposition of HI on gold surface,
2HI
Gold Surface
H2 + I2
2. Decomposition of ammonia in the presence of tungsten,
2NH3
Tungston
N2 + 3H2
3. Photochemical reactions are usually zero order. 4. The reactions which are catalyzed by enzymes are also zero order. 4.7
FIRST ORDER REACTIONS The reaction in which the concentration of only one molecule is changed is called
a first order reaction. Let us take a substance A which decomposes into products. Its initial
8 concentration is ‘a’ moles dm−3. Let after time ‘t’ seconds the amount left behind is (a-x) moles dm−3 and that converted into product is ‘x’ moles dm−3. Then When t = 0 When t = teq
A a (a-x)
Products 0 x
Hence, the rate of reaction is given by, dx k(a x) dt
dx kdt (a x) dx
(a x) k dt ln(a x) kt C
……… (1)
The value of integration constant C can be found by applying the initial conditions of the reaction when t = 0, x = 0. That is ln(a 0) k(0) C
ln a C Therefore Eq.(1) becomes, ln(a x) kt ln a ln a ln(a x) kt
kt a log a x 2.303
……… (2)
This equation is similar to the equation for a straight line. Thus, if we plot a graph between
log
a on y-axis and t on x-axis, we should get a straight line passing through the origin (a x)
with slope k/2.303. Rearranging Equation (2) for k, k
2.303 a log t ax
……… (3)
This equation (3) gives the rate constant k of first order reaction. The units of k are sec−1. Half-Life When t = t1/2 then x = a/2. Substitute this value in equation (3),
9
a 2.303 log a k a (2) 2.303 0.693 log2 k k
t1/2 t1/2
Examples 1. Thermal decomposition of hydrogen peroxide is a first order reaction, H2O2
H2O + O
2. Decomposition of nitrogen pentoxide in CCl4 solution is a first order reaction, 2NO2 + 1/ 2 O2
N2O5
3. The hydrolysis of an ester in the presence of a mineral acid as catalyst,
CH3COOC2H5 + H2O 4.8
H
CH3COOH + C2H5OH
SECOND ORDER REACTIONS The reaction in which two molecules undergo a chemical change is called a
second order reaction. Let us consider two substances ‘A’ and ‘B’ which react to give the products. The reaction can be carried out by; a) By taking equal concentrations of A and B b) By taking different concentrations of A and B a) Second order reaction with equal concentrations of reactants Let the initial concentrations in moles dm−3 are a moles dm−3 and after time ‘t’ the concentrations left behind are (a-x) for both A and B. Then
A+B a + a (a-x) + (a-x)
When t = 0 When t = teq
Hence, the rate of reaction is given by, dx k(a x)(a x) dt dx k(a x) 2 dt
dx kdt (a x) 2 dx
(a x) (a x)
2
2
k dt dx k dt
Products 0 x
10
(a x)21 (1) kt C 2 1 1 kt C (a x)
……… (1)
The value of integration constant C can be found by applying the initial conditions of the reaction when t = 0, x = 0. That is
1 1 k0 C C (a 0) a Therefore equation (1) becomes,
1 1 kt (a x) a 1 1 kt (a x) a x kt a(a x) x akt (a x)
……… (2)
This equation is similar to the equation for a straight line. Thus, if we plot a graph between
x on y-axis and ‘t’ on x-axis, we should get a straight line passing through the origin (a x) with slope ‘ak’. Rearranging Equation (2) for k,
1 x k t a(a x)
……… (3)
This equation gives the rate constant k for second order reaction with same initial concentrations of the reactants. The units of k are dm3 mol−1 sec−1. Half-Life When t = t1/2 then x = a/2. Substitute this value in equation (3),
t1/2
a 1 1 2 a k a(a 2 ) ka
Examples Saponification of an ester with a strong base is a second order reaction in solution,
CH3COOC2H5 + NaOH
CH3COONa + C2H5OH
11 b) Second order reaction with different concentrations of reactants Let the initial concentrations of reactants A and B are ‘a’ and ‘b’ moles dm−3 respectively, and ‘x’ moles dm−3 of the reactants are disappeared in time ‘t’. Then
A+B a + b (a-x) + (b-x)
When t = 0 When t = teq
Products 0 x
Hence, according to rate law, dx k(a x)(b x) dt
dx kdt (a x)(b x) dx
(a x)(b x) k dt
……… (1)
In order to integrate the L.H.S., we have to do the partial fractions first by using partial fraction method,
1 A B (a x)(b x) (a x) (b x)
……… (2)
1 A(b x) B(a x)
……… (3)
To get A, put (a-x) = 0 or x = a in equation (3), 1 A(b a)
A
1 1 b a (a b)
To get B, put (b-x) = 0 or x = b in equation (3), 1 B(a b)
B
1 (a b)
Substituting the value of A and B into equation (2),
1 1 1 (a x)(b x) (a b)(a x) (a b)(b x) Now substituting this value in equation (1),
1
1
(a b)(a x)dx (a b)(b x)dx k dt 1 dx 1 dx k dt (a b) (a x) (a b) (b x)
12
1 1 [ ln(a x)] [ ln(b x)] kt C (a b) (a b) 1 1 ln(a x) ln(b x) kt C (a b) (a b) 1 ln(a x) ln(b x) kt C (a b) 1 (a x) ln kt C (a b) (b x)
……… (4)
The value of integration constant C can be found by applying the initial conditions of the reaction when t = 0, x = 0. That is
1 (a 0) ln k0 C (a b) (b 0) 1 a ln C (a b) b Therefore equation (4) becomes,
1 (a x) 1 a ln kt ln (a b) (b x) (a b) b 1 (a x) 1 a ln ln kt (a b) (b x) (a b) b 1 (a x) a ln ln kt (a b) (b x) b
1 b(a x) ln kt (a b) a(b x) 2.303 b(a x) log kt (a b) a(b x) log
b(a x) (a b)k t a(b x) 2.303
……… (5)
This equation is similar to the equation for a straight line. Thus, if we plot a graph between L.H.S. of this equation (5) on y-axis and time ‘t’ on x-axis, we should get a straight line passing through the origin with slope
(a b)k . 2.303
Rearranging Equation (5) for k,
k
2.303 b(a x) log t(a b) a(b x)
……… (6)
13 This equation (6) gives the rate constant k for a second order reaction when the initial concentrations of the reactants are different. The dimensions of k are again dm3 mol−1 sec−1. Half-Life The half-life method cannot be used for reactions where the concentrations of A and B are different, since A and B will have different times for half reaction. Examples 1. H2 + I2
2HI
2. 2HI
H2 + I2
3. 2O3
3O2
4. 2NO2 4.9
N2 + 2O2
THIRD ORDER REACTIONS The reaction in which only three molecules undergo a chemical change is called a
third order reaction. Four different cases of 3rd order reaction are discussed here. Case (I): when the initial concentration of all the reactants is same Consider the following reaction in which A, B and C reactants are converting into products in an elementary reaction. All reactants have same initial concentration (a). After time (t) the concentration of each reactant becomes equal to a – x as given below;
A + B + C
Products
When t = 0
a
a
a
0
When t = t
a–x
a–x
a–x
x
Rate equation for such a reaction can be written as dx k[A][B][C] dt
dx k(a x)(a x)(a x) dt dx k(a x)3 dt
This is the differential form of rate law for a third order reaction in which concentration of all the reactants is same. Integrating the above equation after separating variables,
dx
(a x) (a x)
3
3
k dt dx k dt
14
(a x) 31 kt c (3 1)(1) (a x) 2 kt c 2 1 kt c 2(a x) 2
……… (1)
When t = 0, x = 0, putting these values in equation (1), the value of ‘c’ can be calculated c
1 2a 2
Hence, equation (1) becomes
1 1 kt 2 2 2(a x) 2a 1 1 2kt 2 2 (a x) a
……… (2)
This is the integrated form of rate law for third order reaction with same initial concentrations of reactants. This equation is similar to the equation for a straight line. Therefore if we plot a graph between
1 on y-axis and t on x-axis, we should get a (a x) 2
straight line with slope equal to 2k and intercept 1/ a 2 . Also from equation (2),
kt
1 1 2 2 2(a x) 2a
kt
a 2 (a x) 2 2(a x) 2 a 2
a 2 (a 2 x 2 2ax) kt 2(a x)2 a 2
kt
a 2 a 2 x 2 2ax 2(a x)2 a 2
2ax x 2 kt 2(a x) 2 a 2 kt
x(2a x) 2(a x) 2 a 2
k
1 x(2a x) 2a 2 t (a x) 2
……… (3)
15 This equation (3) gives the rate constant k. Units of k can be determined as follows,
k
1 (mol dm3 )(mol dm3 ) (mol dm3 )2 sec (mol dm3 ) 2
k mol2dm6 sec1
Half-Life When t = t1/2 then x = a/2 = 0.5a. Substitute this value in equation (3),
t
1 x(2a x) 2a 2 k (a x) 2
t1/2
1 0.5a(2a 0.5a) 3 1.5 2 2 2 2 2a k (a 0.5a) 2ka ka
……… (4)
It is evident from the above equation that the half-life period is inversely proportional to the square of the initial concentration of the reactant. In general for a reaction of nth order, the half-life is found to be, t1/2
1 a n 1
(where n is the order of reaction)
Thus, half-life period of any order reaction is inversely proportional to the initial concentration raised to the power one less than the order of that reaction. Examples: The reaction between NO and H2 to give N2 and H2O follows the third order reaction kinetics. The reaction is,
2NO + 2H2
N2 + 2H2O
The rate law for this reaction is given by Rate = k [H2][NO]2 Case (II): when the initial concentration of two reactants is same and that of the third one is different Consider the following reaction in which the reactants A and B have the same initial concentrations equal to ‘a’ but reactant C has different concentration equal to ‘c’ as given below,
A + B + C
Products
When t = 0
a
a
c
0
When t = t
a–x
a–x
c–x
x
Rate equation for such a reaction is given by
16 dx k[A][B][C] dt dx k(a x)(a x)(c x) dt
dx k(a x) 2 (c x) dt
This is the differential rate equation for a third order reaction in which concentration of two reactants are same and that of the third one is different. Integrating the above equation after separating variables,
dx
(a x) (c x) k dt
……… (1)
2
Using partial fraction, we can write
1 A B C 2 2 (a x) (c x) (a x) (a x) (c x)
……… (2)
1 A(a x)(c x) B(c x) C(a x) 2
……… (3)
For calculating the value of B, put (a – x) = 0 or x = a that is
1 A(a a)(c a) B(c a) C(a a)2 1 B(c a)
or
B
1 (c a)
For calculating the value of C, put (c – x) = 0 or x = c that is
1 A(a c)(c c) B(c c) C(a c)2
1 C(a c)2
or
C
1 (a c) 2
For calculating the value of A, consider equation (3),
1 A(a x)(c x) B(c x) C(a x) 2 1 A(ac ax cx x 2 ) B(c x) C(a 2 x 2 2ax) 1 Aac Aax Acx Ax 2 Bc Bx Ca 2 Cx 2 2Cax
Comparing coefficients of x2, we get
0 AC
A C
or
A
1 (a c) 2
Putting the values of A, B and C in equation (2),
17
1 1 1 1 2 2 2 (a x) (c x) (a c) (a x) (c a)(a x) (a c) 2 (c x) Hence equation (1) becomes
1
1
(a c) (a x) (c a)(a x) 2
2
1 dx k dt 2 (a c) (c x)
1 dx 1 dx 1 dx k dt 2 2 2 (a c) (a x) (c a) (a x) (a c) (c x)
1 ln(a x) 1 (a x) 21 1 ln(c x) kt c 2 (a c) 1 (c a) (2 1)(1) (a c) 2 1 1 1 1 1 ln(a x) ln(c x) kt c 2 (a c) (c a) (a x) (a c) 2 1 1 1 ln(a x) ln(c x) kt c 2 (a c) (c a) (a x) 1 (a x) 1 1 ln kt c 2 (a c) (c x) (c a) (a x)
……… (4)
When t =0, x = 0, putting these values in above equation, the value of c can be calculated,
1 (a 0) 1 1 ln k0 c 2 (a c) (c 0) (c a) (a 0) 1 a 1 ln c 2 (a c) c a(c a) Putting the value of c in equation (4),
1 (a x) 1 1 1 a 1 ln kt ln 2 2 (a c) (c x) (c a) (a x) (a c) c a(c a) 1 (a x) 1 1 1 a 1 ln ln kt 2 2 (a c) (c x) (c a) (a x) (a c) c a(c a) 1 (a c)2
(a x) a 1 1 1 ln (c x) ln c (c a) (a x) a(c a) kt
1 (a c)2
c(a x) a (a x) ln a(c x) a(a x)(c a) kt
1 (a c)2
c(a x) a a x ln a(c x) a(a x)(c a) kt
1 (a c)2
c(a x) x ln a(c x) a(a x)(c a) kt
18
1 (a c)2
ln
c(a x) x ln a(c x) a(a c)(a x) kt
c(a x) x(a c) (a c)2 kt a(c x) a(a x)
……… (5)
This is the integrated rate equation for a third order reaction in which concentration of two reactants are same and that of the third one is different. This equation is similar to the equation for a straight line passing through the origin, from the slope of which k can be calculated if initial concentrations are known that is,
Slope (a c)2 k
k
Slope (a c) 2
Left hand side of equation (5) is a dimensionless quantity. Hence units of k comes out to be, 1
mol / dm k sec 3 2
1 mol2dm6 sec k k mol2dm6 sec1
The half-life period cannot be determined for such type of a reaction. Case (III): when the initial concentration of all the reactants is different Consider the following reaction in which the initial concentration of reactants A, B and C are a, b and c respectively.
A + B + C
Products
When t = 0
a
b
c
0
When t = t
a–x
b–x
c–x
x
Rate equation for such a reaction can be written as dx k[A][B][C] dt dx k(a x)(b x)(c x) dt
Integrating the above equation after separating variables,
dx
(a x)(b x)(c x) k dt
……… (1)
Using partial fraction, we get
1 A B C (a x)(b x)(c x) (a x) (b x) (c x)
……… (2)
19 1 A(b x)(c x) B(a x)(c x) C(a x)(b x)
For calculating the value of A, put (a – x) = 0 or x = a that is 1 A(b a)(c a) B(a a)(c a) C(a a)(b a) 1 A(b a)(c a)
A
1 1 (b a)(c a) (a b)(c a)
For calculating the value of B, put (b – x) = 0 or x = b that is 1 A(b b)(c b) B(a b)(c b) C(a b)(b b) 1 B(a b)(c b)
B
1 1 (a b)(c b) (a b)(b c)
For calculating the value of C, put (c – x) = 0 or x = c that is 1 A(b c)(c c) B(a c)(c c) C(a c)(b c) 1 C(a c)(b c)
C
1 1 (a c)(b c) (c a)(b c)
Putting the values of A, B and C in equation (2),
1 1 1 1 (a x)(b x)(c x) (a b)(c a)(a x) (a b)(b c)(b x) (c a)(b c)(c x) Hence equation (1) becomes
1
1
1
(a b)(c a)(a x) (a b)(b c)(b x) (c a)(b c)(c x) dx k dt 1 dx 1 dx 1 dx k dt (a b)(c a) (a x) (a b)(b c) (b x) (c a)(b c) (c x) 1 1 1 ln(a x) ln(b x) ln(c x) kt c (a b)(c a) (a b)(b c) (c a)(b c) 1 (b c) ln(a x) (c a) ln(b x) (a b) ln(c x) kt c (a b)(b c)(c a)
(b c) ln(a x) (c a) ln(b x) (a b) ln(c x) kt c (a b)(b c)(c a)
……… (3)
When t =0, x = 0, putting these values in above equation, the value of c can be calculated,
(b c) ln(a 0) (c a) ln(b 0) (a b) ln(c 0) k0 c (a b)(b c)(c a)
20
(b c) ln a (c a) ln b (a b) ln c c (a b)(b c)(c a)
Putting the value of c in equation (3),
(b c) ln(a x) (c a) ln(b x) (a b) ln(c x) kt (b c) ln a (c a) ln b (a b) ln c
(a b)(b c)(c a) (a b)(b c)(c a) (b c) ln(a x) (c a) ln(b x) (a b) ln(c x) (b c) ln a (c a) ln b (a b) ln c kt (a b)(b c)(c a) (a b)(b c)(c a)
(b c) ln(a x) (c a) ln(b x) (a b) ln(c x) (b c) ln a (c a) ln b (a b) ln c kt (a b)(b c)(c a)
(b c) ln
(a x) (b x) (c x) (c a) ln (a b) ln (a b)(b c)(c a)kt a b c
……… (4)
This is the integrated rate equation for a third order reaction in which the concentrations of all the reactants are different. This is the equation of a straight line passing through the origin, where the left hand side of the equation represents the dependent variable and t is the independent variable. The plot will give a straight line whose slope is given by Slope (a b)(b c)(c a)k
k
Slope (a b)(b c)(c a)
Hence, the value of k can be determined if the initial concentrations of the three reactants are known. Units of k can be determined by rearranging the equation (4), k
k
(b c) ln
(a x) (b x) (c x) (c a) ln (a b) ln a b c (a b)(b c)(c a)t
mol dm 3
mol dm
3 3
sec
k mol2 dm6 sec1
Case (IV): when only two reactants are involved Sometime the rate of a reaction has more dependence on the concentration of one reactant than that of the other. For example, 3rd order reactions, rate of reaction may be directly proportional to the square of concentration of one reactant and concentration of the other reactants. Consider a third order reaction of the type in which two moles of reactant A are reacting with one mole of reactant B to give products. The initial concentration of reactant A is a and that of B is b as given below,
21
2A + B
Products
When t = 0
a
b
0
When t = t
a–2x
b–x
x
Rate equation for such a reaction is given by dx k[A]2 [B] dt dx k(a 2x) 2 (b x) dt
Integrating the above equation after separating variables,
dx
(a 2x) (b x) k dt 2
……… (1)
Using partial fraction, we get
1 A B C 2 2 (a 2x) (b x) (a 2x) (a 2x) (b x)
……… (2)
1 A(a 2x)(b x) B(b x) C(a 2x) 2
……… (3)
For calculating the value of C, put (b – x) = 0 or x = b that is
1 A(a 2b)(b b) B(b b) C(a 2b) 2
1 C(a 2b)2 C
1 (a 2b) 2
For calculating the value of B, put (a – 2x) = 0 or x = a/2 that is a a a a 1 A(a 2 )(b ) B(b ) C(a 2 ) 2 2 2 2 2
a 2b a 1 B(b ) B( ) 2 2
B
2 2 (2b a) (a 2b)
For calculating the value of A, consider equation (3),
1 A(a 2x)(b x) B(b x) C(a 2x) 2 1 A(ab ax 2bx 2x 2 ) B(b x) C(a 2 4x 2 4ax) 1 Aab Aax 2Abx 2Ax 2 Bb Bx Ca 2 4Cx 2 4Cax) Comparing coefficients of x2, we get
1 2A 4C
22
2A 4C A 2C
2 (a 2b) 2
Putting the values of A, B and C in equation (2),
1 A B C 2 2 (a 2x) (b x) (a 2x) (a 2x) (b x) 1 2 2 1 2 2 2 (a 2x) (b x) (a 2b) (a 2x) (a 2b)(a 2x) (a 2b) 2 (b x) Hence equation (1) becomes
2
2
(a 2b) (a 2x) (a 2b)(a 2x) 2
2
1 dx k dt 2 (a 2b) (b x)
1 2dx 1 2dx 1 dx k dt 2 2 2 (a 2b) (a 2x) (a 2b) (a 2x) (a 2b) (b x)
1 1 2 (a 2x) 21 ln(a 2x) ln(b x) k dt (a 2b)2 (a 2b)2 (a 2b) (2 1)(2) (a 2x) 1 1 ln kt c 2 (a 2b) (b x) (a 2b)(a 2x) 1 (a 2b) 2
(a 2x) (a 2b) ln (b x) (a 2x) kt c
……… (4)
When t =0, x = 0, putting these values in above equation, the value of c can be calculated,
1 (a 2b) 2
(a 2 0) (a 2b) ln (b 0) (a 2 0) k0 c
1 (a 2b) 2
a (a 2b) c ln b a
Putting the value of c in equation (4),
1 (a 2b) 2
(a 2x) (a 2b) 1 ln (b x) (a 2x) (a 2b) 2
1 (a 2b)2
(a 2x) (a 2b) a (a 2b) kt ln (b x) (a 2x) ln b a
a (a 2b) kt ln b a
ln
b(a 2x) (a 2b) (a 2b) (a 2b) 2 kt a(b x) (a 2x) a
ln
1 b(a 2x) 1 (a 2b) (a 2b) 2 kt a(b x) (a 2x) a
23
ln
a (a 2x) b(a 2x) (a 2b) (a 2b) 2 kt a(b x) a(a 2x)
ln
2x b(a 2x) (a 2b) (a 2b) 2 kt a(b x) a(a 2x)
ln
b(a 2x) 2x(a 2b) (a 2b)2 kt a(b x) a(a 2x)
……… (5)
This is the equation of straight line passing through the origin. The value of k can be found from the slope if the initial concentrations are known.
Slope (a 2b)2 k
k
Slope (a 2b) 2
Units of k can be found as follows,
Units of k
mole1n time1 litre1n
Units of k mol2litre2 time1
4.10
ELEMENTARY AND COMPLEX REACTIONS An elementary reaction is a simple reaction which occurs in a single step. A complex
reaction is that reaction which occurs in two or more steps. Chemistry is comprised of billions of reactions. The majority of reactions with which the chemist deals are not elementary instead they involve two or more elementary steps and are complex. The complex reactions are also known as composite reactions. There are various types of complex reactions which may involve: 1. Reversible or opposing reactions 2. Consecutive reactions 3. Simultaneous or parallel or side reactions 4. Chain reactions 4.10.1 REVERSIBLE OR OPPOSING REACTIONS These reactions may be of the following types: a) First order opposed by first order reaction b) First order opposed by second order reaction c) Second order opposed by first order reaction d) Second order opposed by second order reaction
24 a) First order opposed by first order reaction In such reactions, the forward as well as the backward reaction is first order. This is the simplest case of opposing reactions. Consider the following reaction
k1 A
B k-1
When t = 0
a
0
When t = t
a–x
x
When t = teq
a–xe
xe
Rate equation for such a reaction is given by dx k1 (a x) k 1x dt
……… (1)
If xe is the equilibrium concentration then at the state of dynamic equilibrium, Rf = Rb
and
x = xe
k1 (a x e ) k 1x e k 1
k1 (a x e ) xe
……… (2)
Putting the value of k-1 from equation (2) in equation (1), k (a x e ) dx k1 (a x) 1 x dt xe
(a x e )x dx k1 (a x) dt xe x (a x) (a x e )x dx k1 e dt xe
ax xx e ax x e x dx k1 e dt xe a(x e x) dx k1 dt xe dx k1a (x e x) dt xe
Integrating the above equation after separating variables,
(x
ka dx 1 dt xe e x)
25 ln(x e x)
k1a tc xe
……… (3)
When t =0, x = 0, putting these values in above equation, the value of c comes out to be,
c ln x e Putting the value of c in equation (3), ln(x e x)
k1a t ln x e xe
ln x e ln(x e x) ln
k1a t xe
xe ka 1 t (x e x) x e
k1
xe xe ln at (x e x)
……… (4)
This is the integrated rate equation for opposing reaction of the type first order opposed by first order. b) First order opposed by Second order reaction In such reactions, the forward reaction is first order and backward reaction is second order. Consider the following reaction,
k1 A
B
+
C
k-1 When t = 0
a
0
0
When t = t
a–x
x
x
When t = teq
a–xe
xe
xe
Rate equation for such a reaction is given by dx k1 (a x) k 1x 2 dt
……… (1)
If xe is the equilibrium concentration then at the state of dynamic equilibrium, Rf = Rb
and
x = xe
k1 (a x e ) k 1 (x e )2
k 1
k1 (a x e ) (x e ) 2
……… (2)
Putting the value of k-1 from equation (2) in equation (1),
26 k (a x e ) 2 dx k1 (a x) 1 x dt (x e ) 2
k (a x e ) 2 dx k1 (a x) 1 x dt (x e ) 2
dx k1 (a x)x e2 k1 (a x e )x 2 dt x e2 x e2
dx k1 (a x)x e2 k1 (a x e )x 2 dt
x e2
dx k1ax e2 k1xx e2 k1ax 2 k1x e x 2 dt
x e2
dx k1ax e2 k1ax 2 k1x e x 2 k1xx e2 dt
x e2
dx k1a(x e2 x 2 ) k1xx e (x e x) dt
x e2
dx k1a(x e x)(x e x) k1xx e (x e x) dt
x e2
dx k1 (x e x) a(x e x) xx e dt
x e2
dx k1 (x e x) ax e ax xx e dt
k dt dx 12 (x e x)(ax e ax xx e ) xe
Integrating the above equation,
(x
k dx 12 dt xe e x)(ax e ax xx e )
……… (3)
Using partial fraction method, we get 1 (x e x)(ax e ax xx e )
A B (x e x) (ax e ax xx e )
1 A(ax e ax xx e ) B(x e x) For calculating the value of A, put (xe – x) = 0 or x = xe that is
1 A(ax e ax e x e x e ) B(x e x e ) 1 A(2ax e x e2 )
1 A x e (2a x e )
……… (4) ……… (5)
27 A
1 x e (2a x e )
For calculating the value of B, consider equation (5),
1 A(ax e ax xx e ) B(x e x) 1 Aax e Aax Axx e Bx e Bx Comparing coefficients of x, we get
0 Aa Ax e B B Aa Ax e B A(a x e ) Putting the value of A, (a x e ) x e (2a x e )
B
Putting the values of A and B in equation (4), 1 (x e x)(ax e ax xx e )
(a x e ) 1 x e (2a x e )(x e x) x e (2a x e )(ax e ax xx e )
Hence equation (3) becomes dx
x (2a x )(x e
e
e
x)
(a x e )dx k 12 dt x e (2a x e )(ax e ax xx e ) x e
(a x e )dx k1 1 dx dt x e (2a x e ) (x e x) (ax e ax xx e ) x e2 k 1 ln(x e x) ln(ax e ax xx e ) 12 t c x e (2a x e ) xe
(ax e ax xx e ) k1 1 ln 2 t c x e (2a x e ) (x e x) xe
……… (6)
When t =0, x = 0, putting these values in above equation, the value of c comes out to be,
(ax e a0 0x e ) k1 1 ln 2 0c x e (2a x e ) (x e 0) xe c
1 ln a x e (2a x e )
Putting the value of c in equation (6),
(ax e ax xx e ) k1 1 1 ln a ln 2 t x e (2a x e ) (x e x) x e (2a x e ) xe
28
(ax e ax xx e ) k 1 1 ln a 12 t ln x e (2a x e ) (x e x) xe x e (2a x e ) (ax e ax xx e ) k1 1 ln t x e (2a x e ) (x e x)a x e2 k1
xe ax x(a x e ) ln e t(2a x e ) (x e x)a
……… (7)
This is the integrated rate equation for opposing reaction of the type first order opposed by second order. c) Second order opposed by first order reaction In such reactions, the forward reaction is second order and backward reaction is first order. Consider the following reaction,
A
+
k1
B
C
k-1 When t = 0
a
a
0
When t = t
a–x
a–x
x
When t = teq
a–xe
a–xe
xe
Rate equation for such a reaction is given by dx k1 (a x) 2 k 1x dt
……… (1)
If xe is the equilibrium concentration then at the state of dynamic equilibrium, Rf = Rb
and
x = xe
k1 (a x e )2 k 1x e
k 1
k1 (a x e ) 2 xe
……… (2)
Putting the value of k-1 from equation (2) in equation (1),
k (a x e )2 dx k1 (a x)2 1 x dt xe dx k1 (a x) 2 x e k1 (a x e ) 2 x dt xe dx k1 (a x)2 x e (a x e ) 2 x dt x e dx k1 (a 2 x 2 2ax)x e (a 2 x e2 2ax e )x dt x e
29 dx k1 2 a x e x 2 x e 2axx e (a 2 x x e2 x 2ax e x) dt x e
dx k1 2 a x e x 2 x e 2axx e a 2 x x e2 x 2ax e x dt x e
dx k1 2 a x e x 2 x e a 2 x x e2 x dt x e dx k1 2 a x e a 2 x x 2 x e x e2 x dt x e dx k1 2 a (x e x) xx e (x x e ) dt x e dx k1 2 a (x e x) xx e (x e x) dt x e
dx k1 (x e x)(a 2 xx e ) dt x e k dt dx 1 2 (x e x)(a xx e ) xe
Integrating the above equation,
(x
k dx 1 dt 2 xe e x)(a xx e )
……… (3)
Using partial fraction method, we get 1 A B 2 2 (x e x)(a xx e ) (x e x) (a xx e )
……… (4)
1 A(a 2 xx e ) B(x e x)
……… (5)
For calculating the value of A, put (xe – x) = 0 or x = xe that is 1 A(a 2 x e x e ) B(x e x e )
1 A(a 2 x e2 )
A
1 (a x e2 ) 2
For calculating the value of B, consider equation (5), 1 A(a 2 xx e ) B(x e x) 1 Aa 2 Axx e Bx e Bx
Comparing coefficients of x, we get
30
0 Ax e B
B Ax e Putting the value of A, x e (a x e2 )
B
2
Putting the values of A and B in equation (4), x e 1 1 2 2 2 2 2 (x e x)(a xx e ) (a x e )(x e x) (a x e )(a 2 xx e )
Hence equation (3) becomes
(a
2
x e dx k dx 2 1 dt 2 2 x )(x e x) (a x e )(a xx e ) x e 2 e
x e dx k1 1 dx dt (a 2 x e2 ) (x e x) (a 2 xx e ) x e k 1 ln(x e x) ln(a 2 xx e ) 1 t c 2 (a x e ) xe 2
(a 2 xx e ) k1 1 ln t c (a 2 x e2 ) (x e x) xe
……… (6)
When t =0, x = 0, therefore the value of c comes out to be,
(a 2 0x e ) k1 1 ln 0c (a 2 x e2 ) (x e 0) xe 1 a2 ln c (a 2 x e2 ) x e Putting the value of c in equation (6),
(a 2 xx e ) k1 1 1 a2 ln t ln (a 2 x e2 ) (x e x) xe (a 2 x e2 ) x e (a 2 xx e ) 1 1 a 2 k1 ln ln t (a 2 x e2 ) (x e x) (a 2 x e2 ) x e x e
(a 2 xx e )x e k1 1 ln t (a 2 x e2 ) (x e x)a 2 x e k1
xe (a 2 xx e )x e ln t(a 2 x e2 ) (x e x)a 2
……… (7)
31 This is the integrated rate equation for opposing reaction of the type second order opposed by first order. d) Second order opposed by second order reaction In such reactions, the forward reaction is second order and backward reaction is also second order. Consider the following reaction, A
+
k1
B
C
+
D
k-1
When t = 0
a
a
0
0
When t = t
a–x
a–x
x
x
When t = teq
a–xe
a–xe
xe
xe
Rate equation for such a reaction is given by dx k1 (a x) 2 k 1x 2 dt
……… (1)
If xe is the equilibrium concentration then at the state of dynamic equilibrium, Rf = Rb
and
x = xe
k1 (a x e )2 k 1x e2
k1 (a x e ) 2 k 1 x e2
……… (2)
Putting the value of k-1 from equation (2) in equation (1),
k1 (a x e ) 2 2 dx 2 k1 (a x) x dt x e2 dx k1 (a x) 2 x e2 k1 (a x e ) 2 x 2 dt x e2 dx k1 2 (a x)2 x e2 (a x e ) 2 x 2 dt x e
dx k1 2 (a 2 x 2 2ax)x e2 (a 2 x e2 2ax e )x 2 dt x e dx k1 2 2 a x e x 2 x e2 2axx e2 a 2 x 2 x e2 x 2 2ax e x 2 dt x e2 dx k1 2 2 a x e 2axx e2 a 2 x 2 2ax e x 2 dt x e2
dx k1 2 2 2 a x e a 2 x 2 2axx e2 2ax e x 2 dt x e
32 dx k1 2 2 2 a (x e x 2 ) 2axx e (x e x) dt x e
dx k1 2 a (x e x)(x e x) 2axx e (x e x) dt x e2
dx k1 2 (x e x) a 2 (x e x) 2axx e dt x e k dx 12 dt (x e x) a (x e x) 2axx e x e 2
Integrating the above equation,
(x
k dx 12 dt xe e x) a (x e x) 2axx e 2
……… (3)
Using partial fraction method we get 1 A B 2 (x e x) a (x e x) 2axx e (x e x) a (x e x) 2axx e
……… (4)
1 A a 2 (x e x) 2axx e B(x e x)
……… (5)
2
For calculating the value of A, put (xe – x) = 0 or x = xe that is 1 A a 2 (x e x e ) 2ax e x e B(x e x e )
1 A(2a 2 x e 2ax e2 )
A
1 2ax e (a x e )
For calculating the value of B, consider equation (5), 1 A a 2 (x e x) 2axx e B(x e x) 1 Aa 2 x e Aa 2 x 2Aaxx e Bx e Bx
Comparing coefficients of x, we get 0 Aa 2 2Aax e B B A(a 2 2ax e )
Putting the value of A,
B
(a 2 2ax e ) 2ax e (a x e )
Putting the values of A and B in equation (4),
33
(a 2 2ax e ) 1 1 2 (x e x) a 2 (x e x) 2axx e 2ax e (a x e )(x e x) [2ax e (a x e )][a (x e x) 2axx e ] Hence equation (3) becomes
(a 2 2ax e )dx k1 1 dx dt 2ax e (a x e ) (x e x) a 2 x e a 2 x 2axx e x e2 k 1 ln(x e x) ln(a 2 x e a 2 x 2axx e ) 12 t c 2ax e (a x e ) xe
(a 2 x e a 2 x 2axx e ) k1 1 ln 2 tc 2ax e (a x e ) (x e x) xe
……… (6)
When t =0, x = 0, therefore the value of c comes out to be,
(a 2 x e a 2 0 2a0x e ) k1 1 ln 2 0c 2ax e (a x e ) (x e 0) xe
a 2xe 1 ln c 2ax e (a x e ) x e c
1 ln a 2 2ax e (a x e )
Putting the value of c in equation (6),
(a 2 x e a 2 x 2axx e ) k1 1 1 ln a 2 ln 2 t 2ax e (a x e ) (x e x) 2ax e (a x e ) xe (a 2 x e a 2 x 2axx e ) k 1 1 ln a 2 12 t ln 2ax e (a x e ) (x e x) xe 2ax e (a x e ) (a 2 x e a 2 x 2axx e ) k1 1 ln 2 t 2ax e (a x e ) (x e x)a 2 xe k1
a(ax e ax 2xx e ) x e2 ln 2atx e (a x e ) (x e x)a 2
k1
ax e ax 2xx e xe ln 2at(a x e ) (x e x)a
k1
x(a 2x e ) ax e xe ln 2at(a x e ) (x e x)a
……… (7)
This is the integrated rate equation for opposing reaction of the type second order opposed by second order.
34 4.10.2 CONSECUTIVE OR SEQUENTIAL REACTIONS The reactions which proceed from reactants to products in one or more steps through intermediates are known as consecutive reactions. Each step involved here has its own rate constant. Consider the simplest consecutive reaction as follows,
k1
k2
A
B
C
When t = 0
a
0
0
When t = t
[A]
[B]
[C]
According to law of mass action, rate of decomposition of A is given by,
d[A] k1[A] dt
Integrating this equation after separating variables,
d[A] k1 dt [A]
ln[A] k1t c
……… (1)
When t =0, [A] = a, therefore the value of c comes out to be,
ln a k1 0 c
ln a c Putting the value of c in equation (1),
ln[A] k1t ln a ln[A] ln a k1t ln
[A] k1 t a
Taking antilog on both sides, [A] e k1t a
[A] ae k1t
……… (2)
Now rate of formation of B, d[B] k1[A] k 2 [B] dt d[B] k 2 [B] k1[A] dt
d[B] k 2 [B] k1ae k1t dt
(By equation (2))
35 Multiplying both sides by e k 2 t d[B] k 2 [B] ek 2 t (k1ae k1t )e k 2 t dt
d[B] k 2 t e [B]k 2ek 2 t k1ae(k 2 k1 )t dt d [B]ek 2 t k1ae(k 2 k1 )t dt d [B]ek2 t k1ae(k 2 k1 )t dt
d [B]e k a e k2t
(k 2 k1 )t
1
[B]e
k2t
dt
e(k 2 k1 )t k1a D k 2 k1
……… (3)
When t =0, [B] = 0, therefore the value of D comes out to be,
[0]e
k2 0
e(k 2 k1 )0 k1a D k 2 k1
k1a D k 2 k1
Putting the value of D in equation (3),
[B]e
k2t
ka e(k 2 k1 )t k1a 1 k 2 k1 k 2 k1
[B]ek 2 t
k1a e(k2 k1 )t 1 k 2 k1
[B]ek 2 t
k1a ek2t ek1t 1 k 2 k1
[B]
k1a ek 2 t k1t 1 e k2t k 2 k1 e k 2 t e
[B]
k1a e k1t e k 2 t k 2 k1
The concentration of C can be found as, a [A] [B] [C]
[C] a [A] [B]
[C] a ae k1t
k1a e k1t e k 2 t k 2 k1
……… (4)
36
[C] a ae
k1t
(k1ae k1t k1ae k 2 t ) k 2 k1
[C]
a(k 2 k1 ) ae k1t (k 2 k1 ) k1ae k1t k1ae k 2 t k 2 k1
[C]
a(k 2 k1 ) k 2ae k1t k1ae k1t k1ae k1t k1ae k 2 t k 2 k1
[C]
a(k 2 k1 ) a(k 2e k1t k1e k 2 t ) k 2 k1
(k e k1t k1e k 2 t ) [C] a 1 2 k 2 k1
……… (5)
Radioactive decay is an example of such a process. Other examples include chemical processes such as polymerization, thermal cracking and chlorination of hydrocarbons. 4.10.3 SIMULTANEOUS OR PARALLEL OR SIDE REACTIONS The reactions in which reactants undergo two or more independent reactions simultaneously are called as parallel reactions. These are also known as side reactions. The reaction in which maximum yield of the product is obtained is called main or major reaction while the others are called side reactions. Consider a general parallel reaction,
k1 A a a-x
when t = 0 when t = t
k2
B 0 x
at t = 0 at t = t
C 0 x
at t = 0 at t = t
Rate equation for such a reaction is given by, dx k1 (a x) k 2 (a x) dt dx (k1 k 2 )(a x) dt
Integrating the above equation after separating variables,
dx
(a x) (k
1
k 2 ) dt
ln(a x) (k1 k 2 )t c
……… (1)
When t =0, x =0, therefore the value of c comes out to be,
ln(a 0) (k1 k 2 )0 c
37
ln a c Putting the value of c in equation (1),
ln(a x) (k1 k 2 )t ln a (k1 k 2 )t ln a ln(a x) 1 a (k1 k 2 ) ln t ax
……… (2)
The equation (2) is the rate constant expression for a parallel reaction. It is similar to the equation of a first order reaction, (k1 + k2) being the sum of the specific rates of the two simultaneous reactions. If k1 > k2 then A→B is main reaction and A→C is side reaction. Graphically this can be represented as shown in figure (4.1). For example, nitration of phenols gives ortho-nitrophenol and para-nitrophenol in two simultaneous side reactions Figure (4.2).
Fig(4.1): Graphical representation
Fig(4.2): Simultaneous side reactions
4.10.4 CHAIN REACTIONS The chemical reactions which take place in a series of successive processes involving the formation of free atoms and radicals are known as chain reactions. The kinetic laws for such reactions are considerably complex. A well-known example of these types of reactions is the hydrogen–bromine reaction. Chain reactions usually proceed very rapidly. Many explosive reactions occur by a chain mechanism; atomic fission and atomic fusion are of this type of reactions. 4.11
THERMAL REACTIONS A well known example of a thermal reaction is the hydrogen bromine reaction. The
stoichiometry for this reaction is,
H2
+
Br2
2HBr
But it does not follow the simple rate expression
d[HBr] k[H 2 ][Br2 ] dt
38 It follows the following empirical rate equation given by Bodenstein and Lind in 1906:
d[HBr] k[Br2 ]1/2 [H 2 ] [HBr] dt 1 k [Br2 ] The complexity of this equation could not be explained until 1919 when Christiansen, Herzfeld and Polanyi independently and almost simultaneously solved the problem. They proposed a chain of reactions with the following steps, k1
(1)
Br2
(2)
Br + H2
(3)
H + Br2
(4)
H
(5)
2 Br
k2 k3
+ HBr
k5
Chain initiation
2 Br
k4
HBr + H
Chain propagation
HBr + Br
Chain propagation
H2
Chain inhibition
+ Br
Br2
Chain termination
According to this mechanism HBr is formed in reactions (2) and (3) and removed in reaction (4). Consequently, the rate of formation of HBr is given by
d[HBr] k 2 [Br ][H 2 ] k 3[H ][Br2 ] k 4[H ][HBr] dt d[HBr] k 2 [Br ][H 2 ] k 3[Br2 ] k 4 [HBr][H ] dt
……… (1)
The concentration of [Br ] and [H ] can be determined using steady state approximation. The steady state approximation is also occasionally called stationary state approximation. This approximation assumes that the concentration of reaction intermediates remains constant throughout the reaction after an initial buildup. This approximation can only be applied to short lived or very reactive species. Therefore mathematically we can write,
d[Br ] k1[Br2 ] k 2 [Br ][H 2 ] k 3[H ][Br2 ] k 4[H ][HBr] k 5[Br ]2 0 ……… (2) dt d[H ] k 2 [Br ][H 2 ] k 3[H ][Br2 ] k 4[H ][HBr] 0 dt Solving equation (3), k 2 [Br ][H2 ] k 3[H ][Br2 ] k 4 [H ][HBr] 0
……… (3)
39 k 2 [Br ][H2 ] k 3[Br2 ] k 4 [HBr][H ] 0
k3[Br2 ] k 4[HBr][H ] k 2[Br ][H2 ] [H ]
k 2 [Br ][H 2 ] k 3[Br2 ] k 4 [HBr]
……… (4) ……… (5)
Now solve equation (2), k1[Br2 ] k 2 [Br ][H2 ] k 3[H ][Br2 ] k 4[H ][HBr] k 5[Br ]2 0 k1[Br2 ] k 2 [Br ][H2 ] k 3[Br2 ] k 4[HBr][H ] k 5[Br ]2 0
k1[Br2 ] k 2 [Br ][H2 ] k 2 [Br ][H2 ] k 5[Br ]2 0 (By equation (4))
k1[Br2 ] k 5[Br ]2 0 k 5 [Br ]2 k1[Br2 ] 1/2
k [Br ] [Br ] 1 2 k5
……… (6)
Now putting the value of [Br ] from equation (6) in equation (5), 1/2
k [Br ] k 2 1 2 [H 2 ] k5 [H ] k 3[Br2 ] k 4 [HBr]
……… (7)
Putting the values of [Br ] and [H ] from Eq.(6) and Eq.(7) in Eq.(1) we get
d[HBr] k 2 [Br ][H 2 ] k 3[Br2 ] k 4 [HBr][H ] dt 1/2
k [Br ] k 2 1 2 [H 2 ] 1/2 k5 k [Br ] d[HBr] k 2 1 2 [H 2 ] k 3[Br2 ] k 4 [HBr] dt k 3[Br2 ] k 4 [HBr] k5 1/2
k [Br ] k [Br ] k 4 [HBr] d[HBr] k 2 1 2 [H 2 ] 1 3 2 dt k5 k 3[Br2 ] k 4 [HBr] 1/2
k [Br ] k [Br ] k 4 [HBr] k 3[Br2 ] k 4 [HBr] d[HBr] k 2 1 2 [H 2 ] 3 2 dt k 3[Br2 ] k 4 [HBr] k5
d[HBr] k 2 dt
k1 k5
[Br2 ]
1/2
[H 2 ]2k 3[Br2 ]
k 3[Br2 ] k 4 [HBr]
40
1/2
k1 1/2 d[HBr] 2k 2 k5 [Br2 ] [H 2 ] k [HBr] dt 1 4 k 3[Br2 ]
d[HBr] k[Br2 ]1/2 [H 2 ] [HBr] dt 1 k [Br2 ] Where k 2k 2
and k k1 k5
k4 k3
……… (8)
are constants. This equation (8) is in agreement with the rate
equation given by Bodenstein and Lind. The appearance of the term [HBr] in the denominator implies that the velocity of the reaction is decreased by the product HBr and this product acts an inhibitor of the reaction. This is an example of self-inhibition. 4.12
PHOTOCHEMICAL REACTIONS Photochemical reaction is a chemical reaction initiated by the absorption of
energy in the form of light. In photochemical reactions free radicals are produced which initiate the chain reactions. An example is the photosynthesis of HCl gas. All photochemical reactions are of zero order. These reactions are independent of concentration of reactants. Change in concentration of reactants has no effect on rate of photochemical reactions. a) Hydrogen–Bromine Reaction Consider the following photochemical reaction in which H2 and Br2 are going to produce HBr,
H2
+
h
Br2
2HBr
The mechanism of this reaction consists of following five elementary steps, (1)
Br2
(2)
Br + H2
(3)
H + Br2
(4)
H
(5)
2 Br
+
h
k1 k2 k3
+ HBr
k5
k4
2 Br
Chain initiation
HBr + H
Chain propagation
HBr + Br
Chain propagation
H2
Chain inhibition
+ Br
Br2
Chain termination
According to this mechanism HBr is formed in reactions (2) and (3) and removed in reaction (4). Consequently, the rate of formation of HBr is given by
41
d[HBr] k 2 [Br ][H 2 ] k 3[H ][Br2 ] k 4[H ][HBr] dt
d[HBr] k 2 [Br ][H 2 ] k 3[Br2 ] k 4 [HBr][H ] dt
……… (1)
The concentration of [Br ] and [H ] can be determined using steady state approximation. This approximation assumed that at steady state the rate of formation of reaction intermediates can be considered to be equal to their rate of disappearance. Mathematically we can write,
d[Br ] k1Ia k 2 [Br ][H 2 ] k 3[H ][Br2 ] k 4 [H ][HBr] k 5[Br ]2 0 dt
……… (2)
d[H ] k 2 [Br ][H 2 ] k 3[H ][Br2 ] k 4[H ][HBr] 0 dt
……… (3)
Solving equation (3), k 2 [Br ][H2 ] k 3[H ][Br2 ] k 4 [H ][HBr] 0 k 2 [Br ][H2 ] k 3[Br2 ] k 4 [HBr][H ] 0
k3[Br2 ] k 4[HBr][H ] k 2[Br ][H2 ] [H ]
k 2 [Br ][H 2 ] k 3[Br2 ] k 4 [HBr]
……… (4) ……… (5)
Now solve equation (2), k1Ia k 2 [Br ][H2 ] k 3[H ][Br2 ] k 4[H ][HBr] k 5[Br ]2 0 k1Ia k 2 [Br ][H2 ] k 3[Br2 ] k 4[HBr][H ] k 5[Br ]2 0
k1Ia k 2 [Br ][H2 ] k 2 [Br ][H2 ] k5[Br ]2 0
(By equation (4))
k1Ia k 5[Br ]2 0 k 5[Br ]2 k1Ia 1/2
k [Br ] 1 Ia k5
……… (6)
Now putting the value of [Br ] from equation (6) in equation (5), 1/2
k k 2 1 Ia [H 2 ] k5 [H ] k 3[Br2 ] k 4 [HBr]
……… (7)
Putting the values of [Br ] and [H ] from Eq.(6) and Eq.(7) in Eq.(1) we get
42
d[HBr] k 2 [Br ][H 2 ] k 3[Br2 ] k 4 [HBr][H ] dt 1/2
k k 2 1 Ia [H 2 ] 1/2 k k d[HBr] k 2 1 Ia [H 2 ] k 3[Br2 ] k 4 [HBr] 5 dt k 3[Br2 ] k 4 [HBr] k5 1/2
k k [Br ] k 4 [HBr] d[HBr] k 2 1 Ia [H 2 ] 1 3 2 dt k5 k 3[Br2 ] k 4 [HBr] 1/2
k k [Br ] k 4 [HBr] k 3[Br2 ] k 4 [HBr] d[HBr] k 2 1 Ia [H 2 ] 3 2 dt k 3[Br2 ] k 4 [HBr] k5
1/2
k1 d[HBr] k 2 k5 Ia [H 2 ]2k 3[Br2 ] dt k 3[Br2 ] k 4 [HBr]
1/2
k1 Ia1/2 [H 2 ] d[HBr] 2k 2 k5 k [HBr] dt 1 4 k 3[Br2 ]
kIa1/2 [H 2 ] d[HBr] [HBr] dt 1 k [Br2 ] Where k 2k 2
and k k1 k5
k4 k3
……… (8)
are constants. The equation (8) is the final expression.
b) Hydrogen–Chlorine Reaction The chemical reaction between hydrogen and chlorine in the presence of light produces hydrogen chloride with explosion. In 1930 Bodenstein and Hanger investigated the kinetics of this photochemical reaction. The rate law expression for the reaction on the basis of their experimental observations is given below,
H2
+
Cl2
h
2HCl
d[HCl] kIa [H 2 ] dt Where k is constant and Ia is the intensity of absorbed light (expressed in Einstein dm-3 sec-1). They proposed the following mechanism for the reaction, (1)
Cl2
+
h
k1
2 Cl
Initiation step
43
(2)
Cl + H2
(3)
H + Cl2
(4)
Cl
k4
k2 k3
HCl + H
Propagation step
HCl + Cl
Propagation step Termination step
1/2 Cl2
HCl is more stable than HBr. Bond between HCl is strong so [H ] radical do not react with HCl molecule as in case of HBr. The net rate of reaction in terms of rate of formation of HCl according to above mechanism can be written as,
d[HCl] k 2 [Cl ][H 2 ] k 3[H ][Cl2 ] dt
……… (1)
The concentration of [Cl ] and [H ] can be determined applying steady state approximation. This approximation assumed that at steady state the net change in concentration of reaction intermediates is zero. Mathematically,
d[Cl ] k1Ia k 2 [Cl ][H 2 ] k 3[H ][Cl2 ] k 4 [Cl ] 0 dt
……… (2)
d[H ] k 2 [Cl ][H 2 ] k 3[H ][Cl2 ] 0 dt
……… (3)
Solving equation (3), k 2 [Cl ][H2 ] k 3[H ][Cl2 ] 0 k 2 [Cl ][H2 ] k 3[H ][Cl2 ]
……… (4)
k 2 [Cl ][H 2 ] k 3[Cl2 ]
……… (5)
[H ]
Now solve equation (2), k1Ia k 2 [Cl ][H2 ] k 3[H ][Cl2 ] k 4[Cl ] 0 k1Ia k3[H ][Cl2 ] k 3[H ][Cl2 ] k 4 [Cl ] 0
(By equation (4))
k1Ia k 4 [Cl ] 0
[Cl ]
k 1I a k4
……… (6)
Now putting the value of [Cl ] from equation (6) in equation (5), [H ]
k 2 k1Ia [H 2 ] k 3k 4 [Cl2 ]
……… (7)
44 Putting the values of [Cl ] and [H ] from Eq.(6) and Eq.(7) in Eq.(1) we get d[HCl] k 2 [Cl ][H 2 ] k 3[H ][Cl2 ] dt
kI k k I [H ] d[HCl] k 2 1 a [H 2 ] k 3 2 1 a 2 [Cl2 ] dt k4 k 3k 4 [Cl2 ] k k I [H ] d[HCl] k 2 k1Ia [H 2 ] 2 1 a 2 dt k4 k4 d[HCl] 2k 2 k1 Ia [H 2 ] dt k4
d[HCl] kIa [H 2 ] dt
……… (8)
Where k 2k 2 k1 / k 4 is a constant. The expression (8) is same as was given by Bodenstein and Hanger. 4.13
EFFECT OF TEMPERATURE ON RATE OF REACTION
Arrhenius Equation (1889) It is well established that the rate of reaction increases to an appreciable extent with the rise in temperature. Experimentally, it has been found that for a 10 °C rise in temperature, the velocity of the reaction is doubled or trebled. The first empirical equation to show the variation of rate constant k of a reaction with temperature T was suggested by Hood. His equation is, log k A
B T
……… (1)
Where A and B are positive empirical constants and T is the absolute temperature. The verification of this equation lies in the fact that a plot of log k versus 1/T is nearly a straight line with negative slope for most of the reactions. This equation was theoretically justified by Van’t Hoff in 1884. His arguments were based on the variation of equilibrium constant with temperature. Svante Arrhenius extended his idea and suggested a similar equation to show the variation of rate constant with temperature in the year 1889. The differential form of Arrhenius equation is given by, E d ln k a2 dT RT
……… (2)
Where, Ea = Energy of activation, T = Absolute temperature and R = Gas constant. Integrating equation (2) we get,
45
d ln k
E a dT R T2
Ea ln A(constant) RT
ln k
ln k ln A
Ea RT
ln k ln A ln
……… (3)
Ea RT
E k a A RT
Taking antilog on both sides, E
a k e RT A
k Ae
Ea RT
……… (4)
The constants A and Ea are related to A and B in equation (1). The factor A is known as Arrhenius pre-exponential factor, also called frequency factor, have same units as that of specific rate constant. The frequency factor is related to collision frequency and steric factor. The steric (or orientation) effect, also called probability factor, is the fraction of collisions in which the molecules have proper orientation favourable for reaction. Equation (3) and equation (4) are the alternate forms of the Arrhenius equation. Arrhenius equation shows that the value of k for a reaction is directly related to A. The value of A increases if collision frequency or steric factor increases. The second factor in Arrhenius equation Ea/RT depends on the value of Ea. This term usually has a positive value because Ea is usually positive. Since this term is subtracted from lnA in equation (3), we see that the value of k decreases as the value of Ea/RT increases. In other words, an increase in the value of Ea decreases the rate of the reaction. Conversely, an increase in temperature (which is in the denominator) reduces the value of the term Ea/RT and thus increases the value of k. This increase is consistent with the observation that the rates of almost all reactions increase as the temperature rises at the same concentration. Rewriting equation (3) as follows, log k
Ea log A 2.303RT
……… (5)
This equation is identical with the empirical equation (1). To test the validity of this equation; a plot of log k against 1/T should be a straight line having slope Ea/2.303R and intercept logA
46 as shown in figure. Knowing the slope, the value of constant ‘Ea’ which is the characteristic of the reaction can be calculated.
Alternatively, Ea can also be determined if the rate constants are known at two different temperatures. For this purpose Arrhenius equation (2) can be integrated between the limits k1 (the rate constant at T1) and k2 (the rate constant at T2), k2
d ln k
k1
ln k k 2 k
1
Ea R
T2
dT
T
2
T1
E a 1 T2 R T T1
ln k 2 ln k1
Ea R
1 1 T2 T1
ln
k 2 E a T2 T1 k1 R T1T2
ln
E a T2 T1 k2 k1 2.303R T1T2
……… (6)
Equation (6) is another form of Arrhenius equation that is useful for calculating the value of activation energy of a reaction when the value of rate constant at two different temperatures is known for that reaction. The Arrhenius equation correlates rate constant with temperature. It gives the idea of energy of activation. Energy of Activation and Activated Complex The concept of energy of activation was developed by Arrhenius in 1888 which constitute the backbone of all the modern theories. According to Arrhenius, the molecules must acquire a discrete minimum amount of energy before the end products are formed. These activated molecules will then collide and lead to the reaction. Collisions between molecules which are not activated will be of no use and no reaction will take place. Thus the reactants must pass through an energy rich or activated state before they can react. The
47 minimum amount of energy required by the molecules to overcome the activated state or energy barrier before the reaction takes place is known as energy of activation. It follows from the concept of activation that the reactants are not directly converted into products. The molecules first acquire the energy to form an activated complex and this activated complex is then decomposed into products. Activated Complex
Reactants
Products
In other words, there exists an energy barrier between the reactants and the products. If the reactant molecules can cross this energy barrier, they will be converted into products. The energy of activation is usually expressed in J mol-1 and kJ mol-1 and is denoted by the symbol Ea. The relationship between enthalpy of reaction and energy of activation for an exothermic reaction and an endothermic reaction is shown in figure (4.3) and (4.4).
Fig (4.3): Exothermic Reaction
Fig (4.4): Endothermic Reaction
In an exothermic reaction products are at lower energy level than the reactants and in an endothermic reaction the products are at higher energy level than the reactants. With both types of reactions the activation energy is an energy barrier which must be overcome before products are formed. Surmounting this barrier is similar to carrying a ball to the top of a hill and then rolling down the either side. But if the ball cannot be carried to the top, it will roll back. Similarly if Ea is not supplied, the reaction will not start and the reactants will never get converted into the products. According to the arguments given above, a general reaction of the type A2 + B2
2AB
takes place as follows;
It must be kept in mind that the energy of activation is always positive whether the reaction is exothermic or endothermic.
48 4.14
THEORIES OF REACTION RATES
There are two main theories of reaction rates, 1) Collision theory 2) Transition state theory or activated complex theory 4.14.1 Collision Theory This theory is based upon the kinetic theory of gases according to which the molecules of a gas are continuously moving and hence colliding with each other. Since for a collision to takes place, at least two molecules must be involved, therefore the simple case to consider is that of bimolecular reactions. Hence in discussion that follows, we shall first discuss the collision theory for bimolecular reactions and then for unimolecular reactions. 1) Collision Theory for Bimolecular Reactions Bimolecular collision theory for the reactions consisting of two reactant molecules was proposed independently by Max Trautz in 1916 and William Lewis in 1918. This theory is based on following assumptions: 1. For a chemical reaction to takes place reactants must collide with each other. 2. Not all the collisions occurring lead to reaction but only those which form activated molecules. 3. The molecules must be suitably oriented at the time of collisions. Consider a gas–phase bimolecular elementary reaction,
A+A
Products
e.g.,
2HI
A+B
Products
e.g.,
H2 + I2
H2 + I2 2HI
From reaction kinetics, the rate of reaction is given by, Rate k[A]2 kn A2
……… (1)
Rate k[A][B] kn A n B
……… (2)
Where nA and nB are the number of molecules per dm3 of the reactants. The rate of reaction can also be written according to collision theory as follows, Rate Z q
……… (3)
Where Z is number of binary collisions per second per dm3 of the reaction mixture and q is fraction of activated molecules. From kinetic theory of gases, number of collisions per second per dm3 between like molecules of a gas in the reaction mixture is given by,
ZAA
1 A2 n 2A u 2
……… (4)
49 Where σA is the collision diameter of the reactant molecules, nA is the number of molecules per dm3 of reactants, u is the average velocity of molecules. Now the value of u is equal to,
u
8RT M
Where R is the general gas constant, T is absolute temperature and M is the molar mass. Putting this value in equation (4),
ZAA
1 8RT A2 n A2 M 2
ZAA 22A n 2A
RT M
……… (5)
When the two molecules are not alike then collision frequency (collision number) is given by, 8RT
ZAB n A n B2AB
……… (6)
MA MB B Where AB A is the is the mean collision diameter of the molecules; 2 MA MB
reduced mass; nA and nB are the number of molecules per dm3 of the reactants; MA and MB are the molecular weights of two gaseous reactants. The fraction of molecules ‘q’ having the necessary energy of activation is given by Boltzmann law as follows,
n q e Ea /RT n
……… (7)
Where n is the number of activated molecules having energy equal to or greater than the energy of activation Ea, and n is the total number of molecules per dm3. Thus the expression for the rate of reaction involving identical molecules is given by,
Rate ZAA q 2A2 n A2
RT Ea /RT e M
……… (8)
(By Eq. (3), (5) & (7))
……… (9)
(By Eq. (3), (6) & (7))
And for different reactant molecules 2 Rate ZAB q n A n BAB
8RT Ea /RT e
Hence the rate constant expressions for bimolecular reactions comes out to be, For like molecules, from equation (1) and (8)
kn 2A 22A n 2A
RT Ea /RT e M
50
RT Ea /RT e M
k 22A
……… (10)
For unlike molecules, from equation (2) and (9) 2 kn A n B n A n BAB
8RT Ea /RT e
8RT Ea /RT e
k 2AB
……… (11)
Now the rate constant estimated by Arrhenius equation is equal to,
k Ae
Ea RT
……… (12)
Comparing Eq.(10) and Eq.(11) with Eq.(12) we find that the frequency factor A in Arrhenius equation is proportional to ZAA or ZAB and depends on the square root of temperature. Hence the expression for the rate constant becomes,
k Ze
Ea RT
Where Z 22A RT / M
……… (13) (for like molecules) and Z 2AB 8RT /
(for unlike
molecules). It has been found experimentally that for reactions between simple molecules Z agrees well with the Arrhenius factor A. Therefore, collision theory successfully accounts for the rates of simple reactions but in many cases the rates differ considerably. For this reason the equation is further modified as
k PZe
Ea RT
……… (14)
Where, P is called the probability factor or steric factor. The factor P is a measure of the geometrical requirements that must be met when activated colliding molecules are interacting. This factor, the steric factor, limits the successful collisions to the ones in which molecules are favourably oriented. In other words, P is the fraction of collisions in which the molecules have an orientation favourable for reaction. Equation (14) is the equation of the collision theory. Limitations of Collision theory The collision theory in the form of equation (14) is not applicable to the following cases, 1. When the complex molecules are involved in the chemical reactions. 2. When the chain mechanism is to be obeyed by the chemical reactions. 3. When the reaction is surface catalyzed. It is applicable only to gas phase reactions and cannot be applied to kinetic studies in solutions. It does not give any explanation for abnormally high rates of reactions.
51 2) Collision Theory for Unimolecular Reactions (Lindemann’s Mechanism) In unimolecular reactions, only one molecule takes part in the reaction so consequently the question arises: How do molecules in unimolecular reactions attain their energy of activation? If the energy of activation is attained in a bimolecular collision, then the reaction should be of second order. But the fact is that the unimolecular reactions are first order except in gas phase at low pressure. Explanation of how a bimolecular collision can give rise to first order kinetics was given by F. A. Lindemann in 1922. Lindemann pointed out that the reacting molecules acquire activation energy through collisions with other molecules and as a result some molecules are activated. Further he postulated that the activated molecules do not decompose immediately, but remain in the activated form for a definite period. Meanwhile the added excess energy may increase the amplitude of vibration and thus can lead to rupture the bond, or may be robbed by a less energetic molecule in the ensuing collision. These possibilities can be represented by a mechanism consisting of following series of elementary steps, (1)
A + A
(2)
A* + A
A*
(3)
k1 k2
k3
A + A*
(Activation)
A + A
(Dectivation)
Products
(Decomposition)
Where, A represents a normal molecule and A* an activated molecule. k1, k2 and k3 are the rate constants of three steps. According to this mechanism rate of formation of product is given by, dx k 3[A* ] dt
……… (1)
The concentration of A*, the reaction intermediate, can be determined by using steady state approximation. That is
d[A* ] k1[A]2 k 2 [A* ][A] k 3[A* ] 0 dt k 2 [A* ][A] k 3[A* ] k1[A]2 [A* ](k 2 [A] k 3 ) k1[A]2
k1[A]2 [A ] k 2 [A] k 3 *
……… (2)
52 Purring the value of A* from equation (2) in equation (1), we get
k k [A]2 dx 3 1 dt k 2 [A] k 3
……… (3)
Equation (3) predicts two limiting possibilities for gaseous reactions. Case (1): At high pressure the concentration of reacting molecules is relatively high. The activated molecules will have a greater chance of colliding with inactive molecules and thus become deactivated. Under such conditions, the rate of bimolecular deactivation will be much faster than that of unimolecular decomposition then k2 [A] >> k3 and equation (3) reduces to
dx k 3k1[A]2 k 3k1[A] k[A] dt k 2 [A] k2 Where, k k 3k1 / k 2 . The equation predicts a first order reaction. Case (2): At low pressure the concentration of A decreases and hence the number of collisions decreases. Consequently, the rate of decomposition will be much faster than its rate of deactivation. Under such conditions k3 >> k2 [A] and equation (3) reduces to
dx k 3k1[A]2 k1[A]2 dt k3 i.e., the reaction should be second order. Therefore it is the pressure which governs the order of reaction. This conclusion was experimentally verified by Hinshelwood. Examples: Such changes from first order kinetics in gaseous reactions at higher pressures to second order at lower pressures actually have been observed in many gaseous reactions. For example, the thermal decomposition of nitrogen pentoxide and of azomethane at high pressure are the first order reactions, N2O5
N2O4 + 1/2 O2
(CH3)2N2
C2H6 + N2
whereas at very low pressures these reactions become second order. Such observations confirm the Lindemann’s mechanism. 4.14.2 Transition State Theory This theory was first formulated in 1935 by Henry Eyring who was awarded Nobel prize for this valuable work. This theory is also known as absolute rate theory because with the help of it is possible to get the absolute value of the rate constant. The main postulates of this theory are:
53 1. All reactions proceed through an activated or transition state which is higher in energy than both reactants and products. 2. In the transition state the reactants are combined in a species called the activated complex. 3. In activated complex the bonds between the atoms are in the process of being formed and broken. 4. A true thermodynamic equilibrium exists between the reactant molecules and the activated complex species even though the overall chemical reaction is irreversible. 5. The rate of the reaction is equal to the concentration of activated complex species times the frequency at which the complex intermediates dissociate into products. In the light of above mentioned postulates, the reaction between two reactants A and B can be represented as follows
[AB]*
A + B
k
Products
Where [AB]* is the activated complex and k is the rate constant. On the basis of these ideas, Eyring derived an equation for the rate constant k of any reaction, given by k
RT * K NA h
……… (1)
In this equation, R is the gas constant, NA is Avogadro’s number, h is Plank’s constant, T is Kelvin temperature and K* is equilibrium constant for the formation of activated complex from the reactants. Now from thermodynamics equilibrium constant for the first step reversible process is related with the change in Gibb’s free energy according to following expression, G* RT ln K*
ln K*
K e *
G* RT
G* RT
……… (2)
(Taking antilog on both sides)
Where ΔG* is the change in free energy of activation and is obtained by subtracting the free energy of the activated complex from that of reactants. Putting this value of K* from equation (2) in equation (1) we get G RT RT k e NA h
*
……… (3)
54 Also from fundamental relations of thermodynamics we know that G* H* TS*
Where ΔG*, ΔH* and ΔS* represents the free energy, enthalpy and entropy of activation respectively and T is the absolute temperature. Putting this value of ΔG* equation (3) becomes, TS ) RT ( H RT k e NA h *
*
H RT RS RT k e e NA h *
*
RT RS RTa k e e NA h *
E
……… (4)
(∵ ΔH* = Ea)
Equation (4) is the equation of transition state theory and is known as Eyring equation. This equation gives valuable information about the reaction. A comparison of equation (4) with Arrhenius equation shows that the pre-exponential factor A is related to the entropy of activation by the expression,
RT RS A e NA h
*
The entropy of activation introduced by the transition state theory is analogous to the steric factor from collision theory. Transition state theory is applicable to all types of reactions. This is due to the presence of ΔS* factor in the theory. It has been found that the value of ΔS* is negative. The negative value of ΔS* indicates that the formation of activated complex (which is more ordered than the molecules of the reactants) from the reactants is accompanied by a large decrease of entropy. Larger the complexity of the reacting molecules, greater will be the decrease in the value of entropy. This results in the smaller value of the quantity eS /R and *
since RT / NA h is same for all reactions at same temperature, therefore, PZ value in collision theory would be much less than the interpreted value.
