rate of reaction,order , molecularity, integrated rate law

1. REACTION KINETICS
Dr. Priyanka Jain

2. Important Terms
◦ Order
◦ Molecularity
◦ Rate Law
◦ Rate Constant
◦ Rate of reaction

3. Order of Reaction
◦ Defined as the power dependence of rate on the concentration of all reactants.
◦ It represents the number of species whose concentration directly affect the rate of reaction.
◦ It can be obtained by adding all the exponents of the concentration terms in the rate
expression.
◦ It does not depend on the stoichiometric coefficients corresponding to each species in the
balanced reaction.
◦ It is always defined with the help of reactant concentrations and not with product
concentrations.
◦ This value can be Integer or Fraction or even Zero
F2 (g) + 2ClO2 (g) → 2FClO2 (g)
rate = k [F2][ClO2]1

4. Molecularity
◦ The molecularity of a reaction refers to the number of atoms, molecules or
ions which must undergo a collision with each other in a elementary
reaction.
◦ It is the sum of stoichiometric coefficients of reactants in the elementary
reaction.
◦ It is always a whole number.
◦ It can be determined from the balanced chemical equation.
◦ Is only applicable in simple reactions.

5. Rate Law
◦ Rate laws or rate equations are mathematical expressions that describe the
relationship between the rate of a chemical reaction and the concentration
of its reactants.
◦ In general, a rate law or differential rate law takes this form:
rate = k[A]m[B]n[C]p…..
Where [A],[B] and [C] represents the molar concentrations of reactants and k is the rate constant

6. Rate Constant
◦ Rate constant is the proportionality factor in the rate law and is indicated by letter k.
For a general reaction:
aA + bB cC + dD
Rate = k[A]x[B]y
Where x+y = n = order of the reaction
Rate constant = rate/ [A]x[B]y
◦ The rate constant may be found experimentally using the molar concentrations of the reactants and the order of
reaction.
◦ Alternatively, it may be calculated using the Arrhenius equation.
◦ The units of rate constant depend on the order of reaction.
◦ Unit of rate constant k= (mole)1-n (litre)n-1 sec-1
◦ The rate constant is not a true constant since its value depends on temperature and other factors.

7. Rate of reaction
◦ The rate of reaction for a given chemical reaction is the measure of the change in
concentration of reactants or products per unit time.
◦ For the reaction A →B there are two ways of measuring rate:
(1) the speed at which the reactants disappear
Rate of reaction =
𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝐴
𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑐ℎ𝑎𝑛𝑔𝑒
= -
∆[𝑨]
∆𝒕
◦ As the reaction proceeds, concentration of the reactant decreases with time so
negative (-) sign is used in rate expression.

8. Rate of reaction
(2) the speed at which the products appear
rate of reaction =
𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑟𝑜𝑑𝑢𝑐𝑡
𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛 𝑓𝑜𝑟 𝑐ℎ𝑎𝑛𝑔𝑒
or average velocity
∆[𝐵]
∆𝑡
◦ Concentration of the product increases with time so positive (+) sign is used in
rate expression.
◦ For general reaction aA + bB → cC + dD
Rate = −
1
𝑎
Δ 𝐴
Δ𝑡
= −
1
𝑏
Δ 𝐵
Δ𝑡
=
1
𝐶
Δ 𝐶
Δ𝑡
=
1
𝑑
Δ 𝐷
Δ𝑡

9. Numerical
Question: Consider the reaction:
N2(g) + 3H2(g) → 2NH3(g)
The equality relationship between
− 𝑑 𝐻2
𝑑𝑡
𝑎𝑛𝑑
ⅆ 𝑁𝐻3
ⅆ𝑡
Solution: According to equation
1
2
ⅆ 𝑁𝐻3
ⅆ𝑡
= −
1
3
ⅆ 𝐻2
ⅆ𝑡
𝑑 𝑁𝐻3
𝑑𝑡
= −
2
3
𝑑 𝐻2
𝑑𝑡

10. Instantaneous rate and average rate
◦ The instantaneous rate of reaction is defined as the change in
concentration of an infinitely small time interval.
◦ An instantaneous rate is a rate at some instant in time.
◦ The average rate is the average of the instantaneous
reaction rate over a period of time during the reaction.
◦ We determine an instantaneous rate at time t by calculating the negative of the slope of
the tangent to the curve of concentration vs time.

11. Numerical
◦ For the reaction A + B → C, it is found that doubling the concentration of A increases the rate by 4
times, and doubling the concentration of B doubles the reaction rate. What is the overall order of
the reaction?
◦ Solution: Let assume Rate = k[A]x[B]y (1)
If we double the concentration of A, rate increase by 4 times
4 Rate = k[2A]x[B]y (2)
From eq 1 and 2
4=2x , so x= 2
If we double the concentration of B, rate increase by 2 times
2Rate = k[A]x[2B]y (3)
From eq. 1 and 3
2=2y, so y = 1
So order of reaction is 2 + 1= 3

12. Factor affecting
rate of reaction
Nature of
reactant
Concentration
of the
reactant
temperature
catalyst
Surface area
of rectant
pH of the
medium

13. Factor affecting rate of reaction
◦ Nature of the reactant
if reactant are ionic than reaction proceed fast.
if reactant are covalent than reaction proceed slow.
◦ Concentration of the reactant
Increasing the concentration of reactant, increase the rate of reactions
The more particle present, the more often they collide.
◦ Temperature
Increasing the temperature, increase the rate of reaction
The higher the temperature, the higher the kinetic energy of particles, the more they collide.
◦ Catalyst
Catalyst: substance that increase the rate of reaction
Inhibitor: substance that decrease or inhibits reaction rates

14. Factor affecting rate of reaction
◦ pH of the medium
oxidation reactions of metal ions depend on hydrogen ion concentration.
Tl(III) oxidizes [Fe(CN)6]4- in acidic medium
Tl (I) oxidizes [Fe(CN)6]3- in basic medium not in acidic medium
◦ Nature of the solvent
Ionic and polar reactants react very fast in polar solvent
Non polar reactant react fast in non-polar solvent.
Properties of the solvent depend on the ionic strength and dielectric constant.
◦ Surface area of the reactant
The smaller the size of particle, the greater the surface area.
Increasing the surface area speeds up the rate of reaction by increasing the collision rate

15. Zero order reaction
Rate of reaction does not depend on the concentration of the reactant.
A
𝑘0
product
Differential rate equation :
− ⅆ 𝐴
ⅆ𝑡
= 𝑘0 𝐴 0
= k0CA
0
In terms of reactant
− ⅆ 𝐴
ⅆ𝑡
= 𝑘0
In terms of Product
ⅆ𝑥
ⅆ𝑡
= 𝑘0
rate of reaction = constant
ko is known as zero order rate constant

16. Integrated form of zero order reaction
Differential equation for zero order reaction is
ⅆ𝑥
ⅆ𝑡
= 𝑘0 (1)
or dx=k0dt (2)
Integrate equation (2)
𝑑𝑥 = 𝑘0 𝑑𝑡
x = k0t + C (3)
Where C is integration constant
When t = 0, x = 0, than equation (3) becomes
0 = k00 +C
C=0
Putting value of C in eq. (3)
x= k0t (4)
Eq(4) is the integrated form of zero order reaction

17. Graphical representation of zero order
reaction
◦ If we plot a graph between
concentration of product versus time,
a straight line with positive slope
passing through origin is obtained.
(Fig 1)
Slope= k0
◦ If we plot a graph between
concentration of reactant versus
time, a straight line with negative
slope is obtained.(Fig 2)
Slope = -k0
◦ Unit of rate constant k0 = mol L-1s-1
Fig 2

18. Example and numerical
◦ Example of zero order reaction
2NH3 → N2 +3H2 at gold or tungsten surface
2HI → H2 +I2 at gold surface
◦ Numerical
Question: A zero order reaction takes 160 minutes for 80% completion. Determine the
rate constant of the reaction.
Solution: given x = 0.8, t = 160 minute
Rate constant k0 =
𝑥
𝑡
=
0⋅8
160
= .005 mol lit-1s-1

19. First order reaction
Rate of a reaction is proportional to the first power of the concentration of a reactant.
A
𝑘1
product
Differential rate equation :
− ⅆ 𝐴
ⅆ𝑡
= 𝑘1 𝐴 1
= k1(CA)1
Let initial concentration of reactant is ‘a’, after time t x moles of A converted into
product. So CA at time t = a-x
Rate In terms of reactant
− ⅆ𝐶 𝐴
ⅆ𝑡
= 𝑘1 𝐶𝐴
−
𝑑 𝑎−𝑥
𝑑𝑡
= 𝑘1 𝑎 − 𝑥
In terms of Product
ⅆ𝒙
ⅆ𝒕
= 𝐤 𝟏(a-x)
k1 is known as first order rate constant.

20. Integrated form of first order reaction
Differential form of first order reaction is:
ⅆ𝑥
ⅆ𝑡
= k1(a-x) (1)
To integrate above equation use variable separation method
ⅆ𝑥
𝑎−𝑥
= 𝑘1 ⅆ𝑡 (2)
Now integrate eq (2)
ⅆ𝑥
𝑎 − 𝑥
= 𝑘1 ⅆ𝑡
-ln(a-x) = k1t +C (3)
where C is an integration constant.
When t=0 than x = 0, so equation 3 becomes
-ln a = C (4)

21. Integrated form of first order reaction
From equation 3 and 4
-ln(a-x) = k1t – ln a
-ln(a-x) + ln a = k1t
ln
𝑎
𝑎−𝜘
= 𝑘1 𝑡 (5)
𝑘1 𝑡 = 2 ⋅ 303 log
𝑎
𝑎−𝑥
[ ln x = 2.303 log x]
𝑘1 =
2⋅303
𝑡
log
𝑎
𝑎−𝑥
(6)
Equation (6) is the integrated rate equation for first order reaction.
Unit of k1 = s-1

22. Different forms of integrated rate
equation
◦ Wilhelmey`s equation
Ct = C0 exp(-k1t)
Where C0 = initial concentration of reactant
Ct = concentration at time t
◦ Interval Formula
𝚫𝒕 = 𝒕 𝟐 − 𝒕 𝟏 = 𝟐 ⋅ 𝟑𝟎𝟑 𝐥𝐨𝐠
𝑪 𝟏
𝑪 𝟐
Where C1 is the concentration of reactant at time t1
C2 is the concentration of reactant at time t2

23. Graphical representation of first order
reaction
𝑘1 =
2⋅303
𝑡
log
𝑎
𝑎−𝑥
(6)
Rearrange equation (6)
𝑘1 𝑡
2⋅303
= log
𝑎
𝑎−𝑥
(7)
This shows graph between log a/(a-x) v/s t
Will be straight line passing through origin.
k1 = slope × 2.303
log 𝑎 − log 𝑎 − 𝑥 =
𝑘1 𝑡
2⋅303
log 𝑎 − 𝑥 = −
𝑘1
2⋅303
𝑡 + log 𝑎 (8)
Eq 8 also linear eq of form y = -mx +c
Graph between log(a-x) v/s t will be linear line with negative slope
k1 = - slope × 2.303

24. Example of first order reaction
◦ Thermal decomposition of Azoisopropane-
(CH3)2CHN=NCH(CH3)2 → N2 + C6H14
◦ Inversion of cane-sugar
C12H22O11 + H2O → C6H12O6 + C6H12O6
◦ Radioactive disintegration
◦ Decomposition of Oxalic acid
(COOH)2 → CO2 + CO+ H2O

25. Numerical
Question :
A first order reaction takes 10 min for 90 % completion. Calculate rate constant.
Solution:
Given: t = 10 min, a =100, x=90
𝑘1 =
2⋅303
10
log
100
100−90
2⋅303
10
log
100
10
k1 = 0.2303 min-1

26. Numerical
Question
A first order reaction, takes 60 minutes for 90% completion. Calculate the time required for 40%
completion of the reaction.
Solution
Given: t = 60 min, a=100, x=90
Thus 𝑘1 =
2.303
60
log
100
100−90
=
2.303
60
log 10
𝑘1 =
2 ⋅ 303
60
Time taken for 40%
𝑡 =
2 ⋅ 303
2 ⋅ 303
× 60 log
100
100 − 40
t = 13.2 min

27. Second order reaction
Rate of reaction depends on two molecules of one reactant or one molecule each of two reactant.
Case 1
When there is only one reactant taking part in the reaction or two different reactants have same initial concentration
A + A
𝑘2
products
A + B
𝑘2
products
Where initial concentration of A and B is same ‘a’ moles per litre.
Differential rate equation :
− ⅆ𝐶 𝐴
ⅆ𝑡
= 𝐾2 𝐶𝐴
2
(1)
In terms of product
ⅆ𝑥
ⅆ𝑡
= 𝑘2 𝐶𝐴 ⋅ 𝐶𝐴 = 𝑘2 𝐶𝐴 ⋅ 𝐶 𝐵 (2)
Let CA = a-x
ⅆ𝒙
ⅆ𝒕
= 𝒌 𝟐 𝒂 − 𝒙 𝟐
(3)
This eq 3 is known as differential equation for second order

28. Integrated second order reaction
Differential form of second order reaction is:
𝑑𝑥
𝑑𝑡
= 𝑘2 𝑎 − 𝑥 2
To integrate above equation use variable separation method
ⅆ𝑥
𝑎−𝑥 2 = 𝑘2 ⅆ𝑡 (4)
Integrating above equation
ⅆ𝑥
𝑎−𝑥 2 = 𝑘2 ⅆ𝑡 (5)
Or
1
𝑎−𝑥
= 𝑘2 𝑡 + 𝑐 (6)
Where C is integration constant
When t=0 than x = 0, so equation 6 becomes
1
𝑎
= 𝑐 (7)

29. Integrated second order reaction
From equation 6 and 7
1
𝑎−𝑥
= 𝑘2 𝑡 +
1
𝑎
(8)
𝑘2 =
1
𝑡
1
𝑎−𝑥
−
1
𝑎
(9)
Or 𝑘2 =
1
𝑡
𝑥
𝑎 𝑎−𝑥
(10)
This equation is known as integrated rate equation for second order reaction.
Unit of k2 = mol-1 lit1 s-1

30. Graphical representation of second
order reaction
𝑘2 =
1
𝑡
𝑥
𝑎 𝑎−𝑥
(10)
Rearrange above eq
𝑥
𝑎 𝑎−𝑥
= 𝑘2 𝑡 (11)
Eq 11 is linear eq of form y = mx
Graph between x/a(a-x) v/s t will be
Linear line
k2 = slope
We also rearrange eq 11 as
𝑘2 =
1
𝑡
1
𝑎−𝑥
−
1
𝑎
(9)
1
𝑎−𝑥
= 𝑘2 𝑡 +
1
𝑎
(12)
Graph between 1/(a-x) v/s t will be
Linear line of form y = mx + c
k2 = slope

31. Second order reaction
Case -2
If there are two reactant and their initial concentration are not same
A + B
𝒌 𝟐
products
Differential rate equation :
In terms of reactant: −
𝑑 𝐴
𝑑𝑡
= −
𝑑 𝐵
𝑑𝑡
= 𝑘2 𝐶𝐴 𝐶 𝐵 (1)
In terms of product:
𝑑𝑥
𝑑𝑡
= 𝑘2 𝐶𝐴 𝐶 𝐵 (2)
Let CA = (a-x) at time t
CB = (b-x) at time t
Thus
ⅆ𝒙
ⅆ𝒕
= 𝒌 𝟐 𝒂 − 𝒙 𝒃 − 𝒙 (3)
This equation known as second order differential equation

32. Integrated form
Differential form
ⅆ𝑥
ⅆ𝑡
= 𝑘2 𝑎 − 𝑥 𝑏 − 𝑥 (3)
𝑑𝑥
𝑎−𝑥 𝑏−𝑥
= 𝑘2 𝑑𝑡 (4)
Now integrate eq 4
ⅆ𝑥
𝑎−𝜘 𝑏−𝑥
= 𝑘2 ⅆ𝑡 (5)
To integrate above equation use partial fraction method
1
𝑎−𝑥 𝑏−𝑥
=
1
𝑎−𝑏
1
𝑏−𝑥
−
1
𝑎−𝑥
(6)

33. Integrated form
From equation 5 and 6
1
𝑎−𝑏
ⅆ𝑥
𝑏−𝑥
−
ⅆ𝑥
𝑎−𝑥
= 𝑘2 ⅆ𝑡 (7)
1
𝑎−𝑏
− ln 𝑏 − 𝑥 + ln 𝑎 − 𝑥 = 𝑘2 𝑡 + 𝑐 (8)
C is integration constant
When t = 0, x = 0 thus equation 8 becomes
1
𝑎 − 𝑏
− ln 𝑏 + ln 𝑎 = 𝑐
Or −
1
𝑎−𝑏
ln
𝑏
𝑎
= 𝑐 (9)

34. Integrated form
From eq 8 and 9
1
𝑎 − 𝑏
− ln 𝑏 − 𝑥 + ln 𝑎 − 𝑥 = 𝑘2 𝑡 −
1
𝑎 − 𝑏
ln
𝑏
𝑎
𝑘2 𝑡 =
1
𝑎 − 𝑏
ln
𝑏
𝑎
+ ln
𝑎 − 𝑥
𝑏 − 𝑥
𝑘2 𝑡 =
1
𝑎 − 𝑏
ln
𝑏 𝑎 − 𝑥
𝑎 𝑏 − 𝑥
𝒌 𝟐 =
𝟐.𝟑𝟎𝟑
𝒕 𝒂−𝒃
𝐥𝐨𝐠
𝒃 𝒂−𝒙
𝒂 𝒃−𝒙
(10)
Above equation 10 is second order integrated rate equation where initial concentration of the
reactant A and B are different

35. Graphical representation
𝑘2 =
2.303
𝑡 𝑎−𝑏
log
𝑏 𝑎−𝑥
𝑎 𝑏−𝑥
(10)
On rearrange eq
log
𝑏 𝑎−𝑥
𝑎 𝑏−𝑥
=
𝑘2 𝑎−𝑏
2⋅303
𝑡 (11)
This equation is form of linear equation y = mx
Plot a graph between 𝒍𝒐𝒈
𝒃 𝒂−𝒙
𝒂 𝒃−𝒙
v/s t
Gives linear line passing through origin
𝒌 𝟐 =
𝟐⋅𝟑𝟎𝟑×𝒔𝒍𝒐𝒑𝒆
𝒂−𝒃

36. Numerical
A second order reaction where a=b, takes 400 seconds for 25 % completion. How much time
will it take for 60% completions.
Solution: For second order reaction 𝑘2 =
1
𝑡
𝑥
𝑎 𝑎−𝑥
Given t = 400 sec, x= 25, a = 100
𝑘2 =
1
400
⋅
25
100 75
=
1
120000
For 60 % completion of reaction
X= 60, a = 100, a-x = 40
𝑡 =
120000
1
⋅
60
100 × 40
= 800 sec

37. Half life
◦ Is a time period in which, half of the reactant is converted into products.
◦ Denoted by t½ or t0.5
◦ Concentration of reactant at half life = a/2
Half life for zero order reaction:
Rate equation 𝑘 = 𝑥
𝑡
For t½ , x = a/2
t½ = 𝒂
𝟐𝒌 𝟎
Half life is directly proportional to initial concentration of reactant.

38. Half life for first order reaction
Rate law : 𝑘1 =
2⋅303
𝑡
log
𝑎
𝑎−𝑥
For t½ , x= a/2
t½ =
2⋅303
𝑘1
log
𝑎
𝑎 2
=
2.303
𝑘1
log 2
=
Half life of first order reaction does not depend on initial concentration.𝟎 ⋅ 𝟔𝟗𝟑
𝒌 𝟏

39. Half life of second order reaction
Rate law : 𝑘2 =
1
𝑡
𝑥
𝑎 𝑎−𝑥
For half life t½ , x= a/2
t½ =
𝑎 2
𝑘2⋅𝑎 𝑎−
𝑎
2
=
𝟏
𝒌 𝟐
⋅
𝟏
𝒂
Half life period is inversely proportional to the initial concentration of reactant.

40. Numerical
Question:
In a second order reaction, concentration of reactant changes from 0.08 to 0.04 M within 10 minutes.
Calculate the time to change the concentration up to 0.01 M.
Solution:
For second order reaction: t½ =
1
𝑘2
⋅
1
𝑎
Given t½ = 10 min, a = 0.8
𝑘2 =
1
10×0.08
Integrated rate eq is k2 =
1
𝑡
𝑥
𝑎 𝑎−𝑥
X= 0.07, a = 0.08
𝑡 =
10×0.08×0.07
0.08×0.01
= 70 min