Chemical Kinetics : Order of the reaction & rate Constant

 

Order of Reaction

“The order of reaction is defined as the number of atoms or molecules whose concentration change during the chemical reaction.”

Or

“The total number of molecules or atoms whose concentration determine the rate of reaction is known as order of reaction.”

Order of reaction = Sum of exponents of the concentration terms in rate law

                        xA + yB → Products

By the rate law, Rate = [A]x[B]y, then the overall order of reaction. n = x + y, where x and y are the orders with respect to individual reactants.

If reaction is in the form of reaction mechanism then the order is determined by the slowest step of mechanism.

2A + 3B → A2B3

A + B → AB (fast)

                        AB + B2 → AB3 (slow) (Rate determining step)

                        AB3 + A → A2B3 (fast)

(Here, the overall order of reaction is equal to two.)

An order of a reaction may be zero, negative, positive or in fraction and greater than three. Infinite and imaginary values are not possible.

(1)     First order reaction : When the rate of reaction depends only on the one oncentration term of reactant.

            Examples :

A → Product

H2O2 → H2O + \frac { 1 }{ 2 } O2

All radioactive reactions are first order reaction.

Rate of growth of population if there is no change in the birth rate or death rate.

Rate of growth of bacterial culture until the nutrients are exhausted.

Exception : H2O, H+, OH– and excess quantities are not considered in the determining process of order.

Examples :        

            CH3COOC2H5 + H2O → CH3COOH + C2H5; Order = 1;

r = k [CH3COOC2H5]

2A (excess) + B → Product ; Order = 1; R = k [B]

2N2O5 → 4NO2 + O2 ; Order = 1; R = k [N2O5]

2Cl2O7 → 2Cl2 + 7O2 ; Order = 1 ; R = k [Cl2O7]

(CH3)3 – C – Cl + OH– → (CH3)3C – OH + Cl– ; Order = 1; 

R = k [(CH3)3C – Cl]

(i)      Velocity constant for first order reaction : Let us take the reaction

                                        A → Product

            Initially t = 0               a              0

            After time t = t          (a – x)                

Here, ‘a’ be the concentration of A at the starting and (a – x) is the concentration of A after time t i.e., x part has been changed in to products. So, the rate of reaction after time t is equal to

\frac { dx }{ dt }  ∝ (a-x) or  \frac { dx }{ dt } = (a-x) or  \frac { dx }{ (a-x) } = k.dt         …..(i)

  integrated rate constant is,

k = \frac { 2.303 }{ t }   log10   \frac { a }{ (a-x) }                             …..(ii)

t = \frac { 2.303 }{ k }   log10   \frac { a }{ (a-x) }                             …..(ii)

(ii)       Half life period of the first order reaction : when t = t1/2 ; x = \frac { a }{ 2 }   then eq. (ii) becomes

t1/2 = \frac { 2.303 }{ k }   log10   \frac { a }{ \left( a-\frac { a }{ 2 } \right) }   ;  t1/2 = \frac { 2.303 }{ k }   log10   \frac { a }{ a/2 }  

t1/2 = \frac { 2.303 }{ k }   log10 2  (∴ log 2 = 0.3010) ;  ∴ t1/2 = \frac { 2.303 }{ k }   ×  0.3010   t1/2 = \frac { 0.693 }{ k }

Half life period for first order reaction is independent from the concentration of reactant.

            Time for completion of nth fraction, t1/n = \frac { 2.303 }{ k }   log  \frac { 1 }{ \left( 1-\frac { 1 }{ n } \right) }

 

(iii)      Unit of rate constant of first order reaction :  k = (sec)–1

 

(2)     Second order reaction : Reaction whose rate is determined by change of two concentration terms is said to be a second order reaction. For example,

            CH3COOH + C2H5OH → CH3COOC2H5 + H2O

            S2O82– + 2I– → 2SO42– + I2

(i)      Calculation of rate constant : 2A → product   or   A + B → product

            When concentration of A and B are same.

                                        A       +       B       →      Product

            Initially     t = 0  a                 a                 0      

            After time t = t   (a-x)           x

            \frac { dx }{ dt } = k[A] [B] = k[A][B] = k[A][B]

            \frac { dx }{ dt } = k [a-x]2 ; Integrated equation is  k = \frac { 1 }{ t } .\frac { x }{ a(a-x) }   ;   \frac { 1 }{ t } .\frac { x }{ a(a-x) }

            When concentration of A and B are taken different

                                                A       +        B     →       Product

            Initially      t = 0          a                 b                     0

            After time t = t          (a – x)         (b – x)             x

            \frac { dx }{ dt }   k[a-x].[b-x] , Integrated equation is,

k = \frac { 2.303 }{ t(a-b) } log\frac { b(a-x) }{ a(b-x) }   ;  t = \frac { 2.303 }{ k(a-b) } log\frac { b(a-x) }{ a(b-x) }

 

(ii)     Half life period of the second order reaction : When t = t1/2 ; x = \frac { a }{ 2 } ;

t1/2 = \frac { 1 }{ k } \left( \frac { \frac { a }{ 2 } }{ a\times (a-\frac { a }{ 2 } ) } \right) =\frac { 1 }{ ka }

Half life of second order reaction depends upon the concentration of the reactants. t1/2 ∝  \frac { 1 }{ a }

 

(iii)   Unit of rate constant : k = mol1–Δn litΔn–1 sec–1 ; Δn = 2

          k = mol–1 lit. sec–1 (Where Δn = order of reaction)

 

(3)     Third order reaction : A reaction is said to be of third order if its rate is determined by the variation of three concentration terms. When the concentration of all the three reactants is same or three molecules of the same reactant are involved, the rate expression is given as

            3A → Products or A + B + C → products

(i)      Calculation of rate constant: \frac { dx }{ dt } = k(a-x)3 , Integrated equation is K = \frac { 1 }{ t } .\frac { x(2x-x) }{ 2a^{ 2 }(a-x)^{ 2 } }

(ii)       Half life period of the third order reaction : Half life period  = ; \frac { 3 }{ 2a^{ 2 }k } ;  t1/2 ∝  \frac { 1 }{ a^{ 2 } }  Thus, half life is inversely proportional to the square of initial concentration.

(iii)    Unit of rate constant :  k = \left( \frac { mol }{ litre } \right) ^{ -2 }  time-1 or k =litre2 mol-2 time-1 

 

(4)     Zero order reaction : Reaction whose rate is not affected by concentration or in which the concentration of reactant do not change with time are said to be of zero order reaction. For example,

           H2 + Cl2  \underrightarrow { \quad Sunlight\quad } 2HCl

            Dissociation of HI on gold surface.

            Reaction between acetone and bromine.

            The formation of gas at the surface of tungsten due to adsorption.

  (i)      Calculation of Rate Constant : Let us take the reaction

                                         A →  Product

            Initially     t = 0           a       0

              \frac { dx }{ dt }   = k[A]o , \frac { dx }{ dt } = k ; dx = k. dt 

Integrated rate equation, k = \frac { x }{ t } ; The rate of reaction is independent of the concentration of the reacting substance.

(ii)     Half life period of zero order reaction: When t = t1/2 ; x = \frac { a }{ 2 } ; t1/2 = \frac { a }{ 2k } or  t1/2 ∞ a ;  The half life period is directly proportional to the initial concentration of the reactants.

(iii)   Unit of Rate constant : k = \frac { mole }{ lit. sec } ;  Unit of rate of reaction = Unit of rate constant.

Note  : In general, the units of rate constant for the reaction of nth order are:

                                      Rate = k[A]n    

                                      \frac { mol\quad L^{ -1 } }{ s } = k)mol L-1)n or k  = (mol L-1)1-n s-1

Units of rate constants for gaseous reactions: In case of gaseous reactions, the concentrations are expressed in terms of pressure in the units of atmosphere. Therefore, the rate has the units of atm per second.

Thus, the unit of different rate constants would be:

(i)      Zero order reaction : atm s–1                          

(ii)     First order reaction : s–1

(iii)    Second order reaction: atm–1 s–1 

(iv)    Third order reaction: atm–2 s–1

In general, for the gaseous reaction of nth order, the units of rate constant are (atm)1–n s–1

 

Modified expressions for rate constants of some common reactions of first order

Reaction Expression for rate constant
N­2O5 → 2NO2 + \frac { 1 }{ 2 }   O2

 

 k = \frac { 2.303 }{ t }   log  \frac { V_{ \infty } }{ V_{ \infty }-V_{ t } }

Here Vt = volume of O2 after time t and V∞ = volume of O2 after infinite time.
NH4NO2(aq) → 2H2O + N2 Same as above, here Vt and V∞ are volumes of N2 at time t and at infinite time respectively.
H2O2 → H2O + \frac { 1 }{ 2 } O2

 

 k = \frac { 2.303 }{ t }   log \frac { V_{ 0 } }{ V_{ t } }

Here V0 and Vt are the volumes of KMnO4 solution used for titration of same volume of reaction mixture at zero time (initially) and after time t.
CH3COOC2 + H2O \underrightarrow { \quad H^{ + }\quad }  CH3COOH + C2H5OH k = \frac { 2.303 }{ t }   log \frac { V_{ \infty }-{ V }_{ 0 } }{ V_{ \infty }-{ V }_{ t } }

Here V0, Vt and V∞ are the volumes of NaOH solution used for titration of same volume of reaction mixture after time, 0, t and infinite time respectively.
\underset { d-sucrose }{ C_{ 12 }H_{ 22 }O_{ 11 } } +H_{ 2 }O\underrightarrow { \quad H^{ + }\quad } \underset { d-glucose }{ C_{ 6 }H_{ 12 }O_{ 6 } } +\underset { 1-fructose }{ C_{ 6 }H_{ 12 }O_{ 6 } }

 

(After the reaction is complete the equimolar mixture of glucose and fructose obtained is laevorotatory)

 k = \frac { 2.303 }{ t } log \frac { r_{ 0 }-{ r }_{ \infty } }{ r_{ t }-{ r }_{ \infty } }

Here, r0, r1 and r∞ are the polarimetric readings after time 0, t and infinity respectively.

   

Examples of reactions having different orders

Examples Rate Law Order
First order reaction
2H2O2 → 2H2O + O2 r = k[H2O2] 1
C2H5Cl → C2H4 + HCl r = k [C2H5Cl] 1
2N2O5 → 4NO2 + O2 r = k [N2O5] 1
SO2Cl2 → SO2 + Cl2 r = k [SO2Cl2] 1
CH3COOC2H5 + H2O → CH3COOH + C2H5OH r = k [ester][H2O]0 1
C12H22O11 + H2O → C6H12O6 + C6H12O6 r = k [Sugar][H2O]0 1
All radioactive decay r = k [radioactive species] 1
Second order reactions
NO + O3 → NO2 + O2 r = k [NO] [O3] 2
2NO2 → 2NO + O2 r = k [NO2]2 2
H2 + I2 → 2HI r = k [H2][I2] 2
CH3COOC2H5 + OH– → CH3COO– + C2H5OH r = k [CH3CO2C2H5][OH–] 2
C2H4 + H2 → C2H6 r = k [C2H4][H2] 2
2N2O → 2N2 + O2 r = k [N2O]2– 2
2CH3CHO → 2CH4 + 2CO r = k [CH3CHO]2 2
Third order reactions
2NO + O2 → 2NO2 r = k [NO]2[O2] 3
2NO + Br2 → 2NOBr r = k [NO]2[Br2] 3
2NO + Cl2 → 2NOCl r = k [NO]2[Cl2] 3
Fe2+ + 2I– → FeI2 r = k [Fe2+][I–]2 3
Zero order reactions
H2 + Cl2 → 2HCl (over water) r = k [H2]0[Cl2] 0
2NH3  N2 + 3H2 r = k [NH3]0 0
Fractional order reactions
Para H2 → ortho H2 r = k [Para H2]1.5 1.5
CO + Cl2 → COCl2 r = k [CO]2[Cl2]1/2 2.5
COCl2 → CO + Cl2 r = k [COCl2]3/2 1.5
CH2CHO → CH4 + CO r = k [CH3CHO]3/2 1.5
Negative order  reaction
2O3 → 3O2 r = k[O3]2[O2]–1 –1 with respect to O2. Overall order = 1