## Second order reaction

source: SlidePlayer

## Second order reaction-

### Ex.1) Hydrolysis of ester by alkali-

CH3COOC2H5 + NaOH ———> CH3COONa + C2H5OH

### Ex.2) Decomposition of NO2–

2 NO2 ———–> 2 NO + O2

2O3 ———> 3 O2

### Ex.4) Thermal decomposition of Chlorine monoxide-

2 Cl2O ———> 2 Cl2 + O2

### Ex.5) Reaction of Potassium persulphate with KI-

K2S2O8 + 2 KI ———-> 2 K2SO4 + I2

## A) When concentration of both reactants are equal or two molecules of same reactant are involved-

A + B ——–>product

or

2 A ———> product

dx/dt ∝  (a- x ) 2

dx/dt = K (a- x )2

dx / (a – x )2 = K.dt

Taking integration of both sides,

∫ dx / (a – x )2 = K∫ dt

1 / (a – x) = Kt + C          ——— eq.1

when t = 0 then  x = 0

1 / a – 0 = K x 0 + C

C = 1/ a

Putting the value of ‘C’ in eq. 1

1 / (a – x) = Kt + 1 / a

K = 1/ t [ (1 / a- x) – (1/a)]

K = 1 / t [x / a (a – x)]

a = initial concentration

a – x = conc. of reactant after time ‘t’

## B) When concentration of both reactants are different –

A         +        B ——–>product

initial conc.                      a                    b

conc. after ‘t’ time        ( a- x)             ( b- x)

dx/dt ∝  (a- x ) ( b- x)

dx/dt  =  K (a- x ) ( b- x)

By solving ,

## Characteristics of second order reaction –

#### 1) Unit of ‘K’-

K = 1 / t [x / a (a – x)]

unit of ‘K’ = 1 / time ( mole  litre-1 / mole  litre-1  x mole litre-1)

#### unit of ‘K’ = litre mole -1 time-1.

2) The unit of velocity constant  depends upon  the units of concentration because,

unit of ‘K’ = litre mole -1 time-1

3) The time taken to complete a  half reaction( half life period )  is inversely proportional to the initial concentration of the reactant.

K = 1 / t [x / a (a – x)]

If t = t1/2 then x = a/2

K =  1/ t1/2  [ (a/2) / a(a- a/2)]

t 1/2 =  1 / K  [ (a/2) / a( a/2)]