Question

The rate consant for the reaction
I+I+Ar---->I₂+Ar
is 0.59 x 10¹⁶ cm⁶mol^-2s^-1 at 293K. What is the half-life of I is [I]_o= 2 x 10^-5 mol L^-1 and [Ar]= 5 x 10^-3 mol L^-1?

Answer 1

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)

                 =( 0.59*10^16 cm⁶ mol^-2 s^-1)/(5 mol/cm^3)

                 =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

            =1/(0.118* 10^16 cm³ mol^-1 s^-1 )*( 2*10^-2 mol/cm³)

             =8.47 *10^14 s

               t(1/2)    =8.47 *10^14 s