I+I+Ar→I2+Ar
Rate=k[I]^2 [Ar]
As Ar is constant ,so the rxn is independent of concentration of
Ar,[Ar]
Can be represented as I+I→I2
rate =k’ [I]^2
k’=k*[Ar]
It is second order with respect to I,
So integrated rate law,
t(1/2) =1/k[I]o
given, k’=rate constant=0.59*10^16 cm6
mol-2 s-1=k*[Ar]
k=k’/[Ar]=( 0.59*10^16 cm6 mol-2
s-1)/(5*10^-3 mol/L*1000cm^3/L)
               Â
=( 0.59*10^16 cm6 mol-2 s-1)/(5
mol/cm^3)
               Â
=0.118 *10 ^16 cm3 mol-1 s-1
[I]o=2*10^-5 mol/L=2*10^-5 mol/L*1000cm3/L=2*10^-2 mol/cm
t(1/2) =1/k[I]o
          Â
=1/(0.118* 10^16 cm3 mol-1 s-1 )*(
2*10^-2 mol/cm3)
           Â
=8.47 *10^14 s
  Â
           t(1/2)  Â
=8.47 *10^14 s
                                                                                         Â
