The table given below gives kinetic data for the following reaction at 298 K:
`OCl^(-)+I^(-)rarrOI^(-)+Cl^(-)`

What is the rate law and what is the value of rate constant?

Explanation

Let the rate law be `=[OC1^(-)]^(x)[I^(-)]^(y)[OH^(-)]^(z)`
From expt. (1) `1.75xx10^(-4)=k[0.0017]^(x)[0.0017]^(y)[1.0]^(z).....(i)`
From expt. (2) `3.50xx10^(-4)=[00034]^(x)[0.0017]^(y)[1.01]^(z)....(ii)`
Dividing eq. (ii) by eq. (i) .
`(3.250xx10^(-4))/(1.75xx10^(-4))=([0.0034]^(x))/([0.0017]^(x))`
or `" " 2=2^(x)`
or `" " ` x=i.e., first order w.r.t. `OCI^(-)`
From expt. (1)
`1.75xx10^(-4)=k[0.0017]^(x)[0.0017]^(y)[1.0]^(z)....(i)`
From (3),
`3.50xx10^(4)=k[0.0017]^(x)[0.0017]^(y)[1.0]^(z)....(ii)`
Dividing eq. (ii) by eq. (i)
`(3.250xx10^(-4))/(1.75xx10^(-4))=([0.0034]^(y))/([0.0017]^(y))`
or `" " 2=2^(y)`
or y=i.e.., first order w.r.t. `I^(-)`
From (1), `1.75xx10^(-4)=k[0.0017]^(x)[0.0017]^(y)[1.0]^(z)....(i)`
From expt. (4),
`3.50xx10^(-4)k=[0.0017]^(x)[0.0017]^(y)[0.5]^(x)....(iv)`
Dividing eq. (i) by eq. (iv)
`(1.752xx10^(-4))/(3.50xx10^(-4))=([1.05]^(z))/([0.5]^(z))`
or `" " (1)/(2)=2^(z)`
or `" " 2^(-1)=2^(z)`
`z=-1i.e.,` first order w.r.t. `OH^(-) is -1`
Ratw law `=(k[OCI^(-)][I^(-)])/([OH^(-)])`
From expt. (1) `k=(1.75xx10^(-4)[OH^(-)])/([OCI^(-)][I^(-)])=(1.75xx10^(-4)xx1.0)/(0.0017xx0.0017)`
`=6.55s^(-1)`