### +2 UNIT 6 PAGE- 2

MOLECULARITY AND MECHANISM OF REACTION
According to collision theory a chemical reaction takes place due to collision between the particles of the reactants. The number of reacting species(atoms, molecules or ions) which must collide simultaneously in order to bring about the chemical reaction is called the molecularity of the reaction. The molecularity of the reaction can be 1, 2 or 3 . For example, the decomposition of ammonium nitrite is a unimolecular reaction.
NH4NO2 * N2 + 2 H2O
Similarly, the reaction involving simultaneous collision between two species is a bimolecular reaction.
2 H I (g) * H2 (g)+ I2(g)
In the same way, the reaction between NO and O2 is trimolecular reaction.
2 NO + O2 * 2 NO2
In most of the reactions, the molecularity does not exceed three. It is because the chances of simultaneous collisions between three or more particles are rare. In general , for elementary reactions, i.e., single step reactions, the molecularity of the reaction can be obtained from balanced chemical equation. However, in many reactions molecularity of the reaction as obtained from the balanced chemical equation may come out to be more than three. For example, in the reaction between HBr and O2 , the molecularity is five as given by balanced equation.
4 HBr + O2 * 2 H2O + 2 Br2
Since molecularity greater than three is not possible, therefore , the above reaction does not involve the simultaneous collision of all the reacting species in a single step. In fact, such chemical reactions proceed through a sequence of steps. Each step is an elementary step and involves the simultaneous collision of two or three species only. Such chemical reactions proceed through more than one steps are termed as complex reactions.
The detailed description of various steps (the stepwise reactions are not necessarily observable; they are merely proposed pathways from reactants to products which logically fit into all the available experimental data.) of the chemical reaction is called mechanism of reaction. For example, the above reaction occurs by the following steps :
HBr + O2 * HOOBr
HOOBr + HBr * 2 HOBr
[HOBr + HBr * H2O + Br2 ] x 2
------------------------------------------
4 HBr + O2 * 2 H2O + 2 Br2
It is clear that from the above discussions that a complex reaction occurs in two or more steps. The term molecularity of overall reaction has no significance in this case.
RATE LAW OR RATE EQUATION
In complex reactions, the rate expression written on the basis of the overall balanced equation has no significance at all. The true rate expression for such complex reactions can be evaluated on the basis of the experimental data only. For example, in the reaction between NO2 and F2 to yield nitryl fluoride, NO2F,
2 NO2 (g) + F2 (g) * 2 NO2F(g)
the expected rate expression is :
Rate = k [NO2]2 [F2]
But it is experimentally found that the rate of this reaction is proportional to the product of single concentration term of NO2 and and F2. Thus the experimental rate of the reaction is given as :
Rate = k [NO2] [F2]
Such a mathematical expression that gives the true rate of a reaction in terms of concentration of the reactants , which actually influence the rate ,is called rate law.
In general, for any hypothetical reaction:
a A + b B * c C + d D
The rate law expression may be written as :
Rate = k [A]m [B]n
In the above rate law expression, the numerical value of m and n may or may not be same as a and b . The constant k in the rate law expression is called rate constant, velocity constant or specific reaction rate. Now, if the concentration of each of the reactants involved in the reaction is unity, i.e.,
[ A] = [B] = 1 mol L1 , then
rate = k x 1 x 1 = k
Thus, the rate constant of a reaction at given temperature may be defined as the rate of the reaction when the concentration of each of the reacting species is unity.

Characteristics of rate constants
i) The value of k is different for different reactions.
ii) It is a measure of intrinsic rate of reaction. This means larger the value of k, faster will be the reaction. Similarly, small value of k reflects slower reaction.
iii) At a fixed temperature, the value of k is a constant and characteristic of a reaction.
iv) For a particular reaction, k is independent of concentration but depends on temperature.
The important differences between the rate of the reaction and rate constant of the reaction are given in the following TABLE.
TABLE
Rate of reaction Rate constant of reaction
1. It is the speed at which the reactants are converted into the products.
2. It depends upon the concentration of reactant species at that moment.

3. It decreases with the
progress of reaction
generally. 1. It is a constant of proportionality in the rate law expression.

2. It refers to the rate of reaction at specific point when concentration of every reacting species is unity.
3. It is a constant and does
not depend on the
progress of the reaction.

RATE CONTROLLING STEP
A complex reaction proceeds through more than one steps. Out of the various steps of the reaction, the slowest step will decide the rate of overall reaction because the reaction cannot take place faster than the slowest step. The slowest step of the complex reaction is called the rate controlling step or rate determining step.
The rate law expression of a complex reaction gives an indication about the slowest step in the mechanism of a reaction. For example, the decomposition of nitrous oxide:
2 N2O * 2 N2 + O2
the rate law is found to be :
Rate = k [N2O]
Since the rate of the reaction depends only upon the single power of N2O, it indicates that only one molecule of N2O is involved in the slow step. Thus, with this knowledge and chemical intuition a possible mechanism of the reaction may be proposed as :

The above postulated mechanism is consistent with the rate law expression.
Note: It must be noted that there is no way to verify the absolute validity of the proposed mechanism. All we can do is to postulate a mechanism that is consistent with experimental data. If further investigations reveal information that is not consistent with postulated mechanism, it is to be modified accordingly so as to explain all the observations.

Problems
10. For the reaction:
2 NO2 + F2 * 2 NO2F
The experimental rate law :
Rate = k [NO2] [F2]
Propose the mechanism of the reaction.
11. For the reaction :
2 NO + Br2 * 2 NOBr
if proceeds by the following mechanism:

Suggest a rate law consistent with the mechanism.
ORDER OF A REACTION
It is an important parameter for every chemical reaction. It is always determined experimentally and cannot be written from the balanced chemical equation. It refers to the number of reacting particles whose concentration terms determine the reaction rate.
The Order of a reaction may be defined as the sum of the powers to which concentration terms in the rate law are raised to express the observed rate of a reaction.
For example, for a hypothetical reaction :
a A + b B + c C * Products
if the rate law expression for this reaction is :
Rate = k [A]m [B]n [C]p
Then the order of the reaction is equal to ( m + n + p ). These powers or exponents i.e., m, n and p have no relation to the stoichiometric coefficients a, b and c. Order of the reaction with respect to A is m, with respect to B is n and with respect to C is p. If the sum of the powers is equal to one, the reaction is first order reaction. If the sum of the powers is two or three, the reaction is second order or third order reaction respectively. The order of the reaction can also be zero or fractional.
Some examples of reactions of different orders
a) Reactions of first order
i) Decomposition of Nitrogen pentoxide.
N2O5(g) * 2 NO2(g) + (1/2) O2(g)
Rate = k [N2O5] ; Order = 1
ii) Decomposition of Hydrogen peroxide.
H2O2 * H2O + (1/2)O2
Rate = k[H2O2] ; Order = 1
iii) Dehydrohalogenation of Chloroethane.
C2H5Cl * C2H4 + HCl
Rate = k [C2H5Cl] ; Order = 1
iv) Decomposition SO2Cl2.
SO2Cl2 * SO2 + Cl2
Rate = k [SO2Cl2] ; Order = 1
b) Reactions of second order
i) Decomposition of nitrogen peroxide :
2 NO2 * 2 NO + O2
Rate = k [NO2]2 ; Order = 2
ii) Reaction between hydrogen and iodine to give HI.
H2 + I2 * 2 HI : Rate = k [H2] [I2] ; Order = 2
c) Reactions of Third order :
i) The reaction between nitric oxide and oxygen .
2 NO(g) + O2(g) * 2 NO2(g)
Rate = k [NO]2 [O2] ; Order = 3
ii) Reaction between nitric oxide and chlorine :
2NO + Cl2 **2 NOCl Rate = k [NO]2 [Cl2] ; Order = 3
It may be noted that the order of a reaction may or may not be a whole number. It can have zero or fractional values also.
d. Reactions of fractional order
i) Combination of CO and Cl2.
CO + Cl2 * COCl2
Rate = k [CO]2 [Cl2]1/2 ; Order = 2.5
ii) Decomposition of carbonyl chloride.
COCl2 * CO + Cl2
Rate = k [COCl2]3/2 ; Order = 3/2
iii) Decomposition of acetaldehyde.
CH3CHO * CH4 + CO
Rate = k [CH3CHO] 3/2 ; Order = 3/2
e. Zero order reaction : A number of zero order reactions are known in which the rate of reaction is independent of the concentration of the reactants. For example, the decomposition of ammonia at the surface of metals like gold, platinum etc. is a zero order reaction.

It has been observed that the rate of reaction is independent of the concentration of ammonia, i.e.,
Rate = k [NH3]0
 Order of the reaction = zero.
Units of rate constants
The rate is the change in concentration with time. Therefore, the rate of reaction is expressed by concentration units divided by time units. If the concentrations are expressed in mol L1 and time in seconds, then the units for rate of a reaction are
mol L1 s1 as :

The units of rate constants of different orders are different.
For an nth order reaction:

If concentration is expressed in mol L1 and time in seconds,

For First order reaction , the unit of rate constant is s1 .
The unit of second order rate constant is L mol 1 s1
The unit of third order rate constant is L2 mol2 s1
For gaseous reactions of nth order , k has units of ;
(atm) 1 n s1
INTEGRATED RATE EXPRESSION
This is the most common method for studying the kinetics of chemical reactions. The instantaneous rate of a reaction is given by differential rate law equations. For example, for a general reaction :
a A + b B  Products
The differential form of the rate law is transformed to integrated form of rate law by simple mathematics (calculus). The integrated form is more convenient because it helps to understand the variation of concentrations of the reactants with time. The integrated rate equations for different orders can be derived.
Zero order reaction
A reaction is said to be of zero order, if its rate is independent of the concentration of the reactants. Consider a general reaction
A  Product
Let [A] be the concentration of the reactant A and ko is the rate constant for the zero order reaction. For the zero order reaction, the rate of reaction is independent of the concentration of A.

Thus,

This form of the rate law is known as differential rate equation. Rearranging the above equation,
 d [A] = k0 d t
Itegrating the above equation, we get,

 [A] = k0 t + I …………(2)
where I is the constant of integration. The value of I can be calculated from the initial concentration . For example, the initial concentration of A be [A]0 at t = 0 . Then equation becomes
 [A]o = k0 x 0 + I
I =  [A]0
Substituting the value of I in (2) we get,
 [A] = k0 t  [A]0
k0 t = [A]0  [A]

where [A]o is the initial concentration of A and [A] is the concentration at time t.
Alternately, if the initial concentration of A is ‘a’ moles per litre and x moles of reactants gets changed to products in time t. Then concentration of A left after t be (a  x).
A  Product
Initial concentration : a 0
Concentration at time t : (a  x) x
 [A]o = a , [A] = a  x so that
[A]o  [A] = a  (a  x) = x
Thus equation (3) becomes,

The above equation is general equation for zero order reaction. The amount of the substance reacted is proportional to time.
First order reaction
Consider the general first order reaction
A  Products.
The differential rate law equation is  d[A] = k [A]
dt
The differential form of rate law is transformed to integrated form of rate law by simple calculus. This equation is used to calculate the rate constant k.
Let us illustrate this method by applying it to a first order reaction. Consider a first order reaction. For the first order reaction, the rate of this reaction is directly proportional to the concentration of the reactant A. Thus :
Rate =  d[A] = k1 [A] ……….(5)
dt
This form of rate law is known as differential rate equation. Rearranging the above equation :
 d[A] = k1dt ……….(6)
[A]
Integrating the above equation we get:

 ln [A] = k1 t + I ……(7)
where I is the constant of integration. The value of I can be calculated from the initial concentrations. For example, the initial concentration of A can be [A]0 at t = 0. Then Equ.(7) becomes :
 ln [A]0 = k 1 x 0 + I
 I =  ln [A]0
Substituting the value of I in Eq.(7) , we get :
 ln [A] = k1 t  ln [A]0
Rearranging ,
ln [A]0  ln [A] = k 1 t
or ln [A]0 = k 1 t
[A]
or k 1 = 1 ln [A]0 ……..(8)
t [A]
Changing the above expression to log base 10
( ln x = 2.303 log x), we get:

The Equ .(9 ) is integrated rate expression for first order reaction.
The above equation may also be written in an alternate form. Let the initial concentration of A is 'a' moles per litre. Suppose at time t , x moles of reactants get changed into products. Then the concentration of A left after time t is (a  x). Therefore, [A]0= a and [A] = (a  x ).
Thus Equ. (9) becomes :

Significance of integrated Rate equation
The integrated Rate equation can be used in the following ways :
i) Order of reaction : All reactions of first order obey the following equation :

Starting with the known concentration of A, the concentration of the reactant [A], at different intervals of time (t) may be measured. The value of [A] at different time intervals are substituted in the above equation. If the value of k comes out to be a constant, then the reaction is of first order.
(ii) Calculation of Rate constant : The integrated rate equation can be used to calculate the rate constant of a reaction. If the reaction is known to be first order , then by substituting initial concentration of reactant [A]0 and concentrations at different times (t) in the above equation , the value of k can be calculated.
The rate constants can be calculated graphically . The Equation (9) may be written as :

This equation is comparable to the equation of a straight line ( y = m x + c ). When a graph is plotted between log[A] against corresponding time interval we get a straight line. This is shown in Fig.

Plot of t versus log[A] to calculate rate
constant for First Order Reaction.

The slope of this line is equal to :
Slope =  k 1
2.303
From this the rate constant can be calculated.
k 1 =  2.303 (slope)
EXPERIMENTAL DETERMINATION OF ORDER OF A REACTION
The following methods employed to determine the rate law , the rate constant and order of a reaction.
1. GRAPHICAL METHOD : This method is used to determine the rate law of the reaction which involves only one reactant species. The various steps involved are :
i) The concentrations of reacting substances are determined at different intervals by suitable method.
ii) A graph is plotted between concentration and time.
iii) From the plot of concentration vs time , the instantaneous rates corresponding to different concentrations determined by drawing tangents to the curve and subsequently calculating their slopes.
iv) The rate of reaction (calculated above step 3) is plotted versus concentration [A] or (concentration)2 , [A]2 and so on.
(a) If rate of reaction rate remains constant in the rate versus concentration graph, it means that the rate is independent of the concentration of the reactant, i.e.,
Rate = k [A]0 = k
Therefore, the reaction is of zero order.
(b) If a straight line is obtained in rate versus concentration graph, it means that the rate is directly proportional to the concentration of the reactant, i.e.,
Rate = k [A]
Therefore, the reaction is of first order.
(c) If a straight line is obtained in the rate versus (concentration)2 graph, it means that
Rate = k [A]2
Therefore, the order of the reaction is 2.
(d) Similarly , if we get straight line in the rate versus (concentration)3 graph, then
Rate = k [A]3
And the order of the reaction is 3.
In general, , if we get straight line by plotting graph of the rate versus (concentration)n, where n = 1,2,3… so on, then
Rate = k [A]n
And order of the reaction is n.

These graphs are shown below.

Illustration
Consider the decomposition of nitrogen pentoxide
2 N2O5(g)  4 NO2(g) + O2(g)
A convenient method to follow the reaction is to measure at different times , the increase in pressure accompanying the reaction. From the measured values it is possible to first calculate the partial pressure of N2O5 and then the concentration in moles per litre of N2O5. The molar concentrations so obtained for different times are plotted in the form of curve(Fig 5). The values of slopes(or reaction rate) progressively decreases as the concentration of N2O5 decreases. This shows that the reaction depends on its concentration. On plotting rate against concentration, i.e., the rate against [N2O5], a straight line is obtained (Fig 6).

It means that the rate of reaction is proportional to the concentration of N2O5 or :
Rate  [N2O5]
which in turn means that the rate law is :
Rate = k [N2O5]
However, we do not get straight line by plotting rate of a reaction against [N2O5]2. This means that the reaction is not of second order.
Thus the correct rate law for this reaction is :

and the order of the reaction is one.
Calculation of Rate constant
The rate constant can be calculated by substituting the values of rate and concentration of [N2O5] in the rate expression :

Note : From the above discussion, we notice that the coefficient of N2O5 in the balanced equation is 2 while the exponent in the rate law is 1. Thus the order of the reaction is not same as the coefficient in the balanced chemical equation.
Graphical method for integrated rate equation
Graphical method can also be applied for integrated rate equation. In this method, appropriate function of concentration is plotted against time. The resulting curve will be straight line only for the case in which the appropriate integrated equations has been used. The integrated rate equations for zero, first and second order reactions are given below :

As shown below, straight lines are obtained for a plot of [A] versus t for a zero order reaction, of log [A] versus t for a first order reaction and 1/[A] versus t for second order reaction as shown below.

The curves also help to calculate the value of k from the slope of the straight line.
2. INITIAL RATE METHOD
The graphical method cannot be applied for the reactions which involve more than one reactants. The rates of such reactions can be determined by initial rate method. This method involves the following steps:
(a) The initial rate of the reaction, i.e., the rate at the beginning of the reaction is measured. This may be taken as the rate over an initial time interval that is short enough so that concentrations of the reactants do not change appreciably from their initial values. This corresponds to slope of the tangent to the concentration versus time graph at t = 0.
(b) The initial concentration of only one reactant is changed(keeping other concentration constant) and the rate is determined again. From this the order with respect to that particular reactant is calculated.
(c) The procedure is repeated with respect to each reactant until the overall rate law is fully determined.
(d) The sum total of the individual orders with respect to each reactant gives the order of the reaction.
Illustration
The method is illustrated by taking hypothetical reaction :
2 A + 2 B  Products.
The experimental data for this reaction is given below:
Experiment Concentration
[A] [B] Rate of reaction
I 0.01 0.01 0.005
II 0.02 0.01 0.020
III 0.02 0.03 0.060
The general form of the rate law may be written as :
Rate = k [A]p[B]q
Then, the expression for initial rate is :
(Rate)0 = k [A]op[B]oq
where subscript zero denotes initial values. The problem involves the determination of p and q.
Consider the experiments I and II and substituting the values we get,
(Rate)1 = k (0.01)p(0.01)q = 0.005 …….(1)
(Rate)2 = k (0.02)p(0.01)q = 0.020 ……..(2)
Dividing Eq(2) by Eq.(1):

(2)p = 4 or (2)p = 22  p = 2
Similarly, comparing experiments II and III,
(Rate)2 = k ( 0.02)p(0.01)q = 0.020 ………(3)
(Rate)3 = k (0.02)p (0.03)q = 0.060 ………(4)
Dividing equation (4) by (3), we get :

3q = 3 or q = 1
Therefore , the order with respect to A is 2 and the order with respect to B is 1.
Thus, the rate law may be written as :
Rate = k [A]2 [B]
3. Ostwald Isolation method
This method was introduced by Ostwald in 1902 and is used to find the order of a reaction with respect to one reactant at a time. The total order of the reaction is equal to the sum of the orders of reaction for individual reactants. This method is based on the principle that if the concentrations of all but one reactant are taken in excess, then during the course of the reaction, the concentration of those reactants taken in excess will remain almost constant and hence variation in rate will correspond to the concentration of that reactant whose concentration is small. This process is repeated with other reactants and order with respect to each reactant is determined. The overall order will be the sum of all these orders. For example, consider the general reaction :
a A + b B + c C  Products
Suppose we isolate A by taking B and C in large excess and get the order of the reaction with respect to A ( say p) . Similarly, we isolate B by taking A and C in excess and isolate C by taking A and B in excess and get order with respect to B and (say q) and C (say r) respectively.
Overall order of the reaction n = p + q + r
Problems
12. Three experimental runs were carried out for the reaction between Cl2 and NO.
Cl2 + 2 NO(g)  2 NOCl(g).
The following rate law are determined.
Run Initial concentration (mol/L)
[Cl2] [NO] Initial rate
mol / L / s
1 0.020 0.010 2.4 x 104
2 0.020 0.030 2.16 x 103
3 0.040 0.030 4.32 x 103
Determine :
i) The order with respect to Cl2 and NO.
ii) The rate law. (iii) The rate constant.
13. The decomposition of N2O5 in carbon tetrachloride solution has been investigated.
N2O5(solution) 2 NO2(solution) + (1/2)O2(g)
The reaction has been found to be first order with
first order rate constant 6.2 x 104 s1. Calculate the
rate of reaction when :
(a) [N2O5] = 1.25 mol L-1 and (b) [N2O5] = 0.25 mol L1
(c) What concentration of N2O5 would give a rate 2.4 x 103 s1 ?
14. With the help of the following rate expressions of the reaction,
find out the overall order of the reactions and order with
respect to each reaction ?
2 NO(g) + O2(g)  2 NO2(g) ; Rate = k[NO]2[O2]
2 N2O(g)  2 N2 + O2(g) ; Rate = k [N2O]
2 NO(g) + 2 H2(g)  N2(g)+2H2O(g) ; Rate = k[NO]2[H2]
15. Identify the reaction order from the following rate constants ;
i) k = 2.3 x 105 L mol1 s1
ii) k = 2.3 x 101 s1
16. What is the rate of reaction and order of the reaction , if the mechanism is :
2 NO + H2  N2 + H2O2 (slow)
H2O2 + H2  2 H2O (fast)
17. The experimental data for the reaction :
2 A + B2  2 AB is :
Experiment [A] [B2] Rate
(mol L1 s1)
1 0.50 0. 50 1.6 x 104
2 0.50 1.00 3.2 x 104
3 1.00 1.00 3.2 x 104
Write the most probable rate equation for the reaction
18. For the following reactions, state the order with respect to each reactant and overall order :
a) 3 NO(g)  N2O(g) + NO2(g)
Rate = k [NO]2
b) H2O2 + 3 I + 2 H+  2 H2O + I3(aq)
Rate = k [H2O2][ I]
c) CH3CHO(g)  CH4(g) + CO(g)
Rate = k [CH3CHO]3/ 2
d) CHCl3(g) + Cl2(g)  CCl4(g) + HCl(g)
Rate = k [CHCl3][ Cl2]1/2
What are the dimensions of the rate constants in each case.
19. For the following reaction : 2 A + B + C  A2B + C
the rate law has been determined to be :
Rate = k[A] [B]2 : k = 2.0 x 106 mol2Ls1
(a) For the reaction, determine the initial rate of reaction with [A] = 0.1 mol L1;
[B] = 0.2 mol 1; [C] = 0.8 mol L1.
(b) Determine the rate after 0.04 mol L1 of A has reacted.
20. The rates of reaction starting with initial concentrations 2 x 103 M and 1 x 103 M are equal to 2.4 x 104 M s1 and 0.60 x 104 M s1 respectively. Calculate the order of the reaction with respect to the reactant and also the rate constant.
21. The initial rate of the reaction : A + 5 B + 6 C  3 L + 3 M
has been determined by measuring the rate of disapperance of A under the following conditions :
Expt No [A]o M [B]o M [C]o M Initial rate M min1
1 0.02 0.02 0.02 2.08 x 103
2 0.01 0.02 0.02 1.04 x 103
3 0.02 0.04 -.02 4.16 x 103
4 0.02 0.02 0.04 8.32 x 103
Determine the order of the reaction with respect to each reactant and overall order of the reaction. What is the rate constant ? Calculate the initial rate of the reaction when the concentration of all reactants is 0.01 M. Calculate the initial rate of change in concentration of B and L.
22. The pressure of a gas decomposing at the surface of a solid catalyst has been measured at different times and results are given below;
t (s) 0 100 200 300
P (Pa) 4.00 x 103 3.50 x 103 3.00 x 103 2.5 x 103
Determine the order of the reaction, its rate constant and half life period.
23. The rate of decomposition of N2O5 in CCl4 solution has been studied at 318 K and the following results are obtained :
t min 0 135 339 683 1680
C M 2.08 1.91 1.68 1,35 0. 57
Find the order of the reaction and calculate its rate constant. What is the half-life period ?
4. Half Life Period Method
The order of a reaction can also be determined by another method known as half life period method. This is discussed below.
HALF LIFE PERIOD OF A REACTION
Half life of a reaction is defined as the time during which the concentration of the reactants is reduced to half of the initial concentration or it is the time required for the completion of half of the reaction. It is denoted by t1/2 . Let us calculate the half life of a First Order reaction.
For a First Order reaction, we know that :

In general for an nth order reaction :

Thus it is clear from the above expression , that the half-life period or half-change time for first order reaction does not depend upon the initial concentration of the reactants. Similarly, the time required to reduce the concentration of reactant to any fraction of initial concentration for this type of reactions is also independent of the initial concentration.
In addition to calculation of t1/2, the time required to complete different fractions of a first order reaction can also be calculated.
For example, the time required to complete 1/3 rd of the reaction will be given as :

In general the time required to complete any fraction of a first order reaction is given as :

where n = 2 to 9.

Half life period of zero order and second order reactions
For zero order reaction,
k t = [A]o  [A]
For half life period, t½ ,

Similarly for second order reactions.

These are summed up below:

These results give very interesting observation that only the half lives of first order reactions are independent of the concentrations. The half-lives versus concentration [A]o or 1/[A]o plots are shown below.

DEPENDENCE ON REACTION RATE ON TEMPERATURE
The rate of a chemical reaction is significantly affected by a change in temperature. For most of the chemical reactions, the rate increases with rise in temperature. The rate usually becomes doubled to trebled for each 10 rise in temperature.

Temperature coefficient
The increase in the rate of reaction with rise in temperature is usually expressed in terms of a quantity known as temperature coefficient. It is defined as the ratio of rate constant of a reaction at two different temperatures separated by 10C. The two temperatures generally taken are 35C (308 K) and 25C (298 K).

where k toC is the rate constant for the reaction at tC and k t + 10OC is the rate constant for the same reaction at t + 10C. For most homogeneous reactions the value of temperature constant lies between 2 and 3. This means that the rate constant and hence rate of reaction increases two- to- three fold for every 10 rise in temperature. For example, temperature coefficient of the reaction 2 HI  H2 + I2 is 1.7. In some cases the value of temperature coefficient is found even greater than three. For example temperature coefficient for the decomposition of nitrogen pentoxide (N2O5) is found to be equal to 3.8.

## QUESTIONS

Atoms and Molecules
1.