Question

  Iodide ion is oxidized to hypoiodite ion, IO^- , by hypochlorite ion, ClO^-, in basic solution. The equation is

The following initial rate experiments were run and, for each, the initial rate of formation of IO^- was determined. Find the rate law and the value of the rate constant.

               Initial concentrations initial rate

please explain with steps.

Answer 1

We are given, initial concentration of I^- , ClO^- and OH^- and rate of reaction

First we need to calculate order with respect to each reactant

Table-

            I^-      ClO^-         OH^-         Rate

     0.010   0.020   0.010    12.2x10^-2

   0.020   0.010   0.010    12.2x10^-2

     0.010   0.010   0.010    6.1 x10^-2

     0.010   0.010   0.020    3.0 x10^-2

We assume the rate law for this reaction

Rate = k [I^-]^x [ClO^-]^y [OH^-]^z

The x, y and Z are the order with respect to I^- , ClO^- and OH^-

Rate3/ Rate2 = k [I^-]₃^x [ClO^-]₃^y [OH^-]₃^z / k [I^-]₂^x [ClO^-]₂^y [OH^-]₂^z

6.1x10^-2 / 12.2x10^-2 = (0.010)^x /(0.020)^x * (0.010)^y /(0.010)^y *(0.010)^z /(0.010)^z

0.5 = (0.5)^x

  x = 1

Now need to calculate y

Rate3/ Rate1 = k [I^-]₃^x [ClO^-]₃^y [OH^-]₃^z / k [I^-]₁^x [ClO^-]₁^y [OH^-]₁^z

6.1x10^-2 / 12.2x10^-2 = (0.010)^x /(0.010)^x * (0.010)^y /(0.020)^y *(0.010)^z /(0.010)^z

0.5 = (0.5)^y

  y = 1

now for z

Rate4 / Rate3 = k [I^-]₄^x [ClO^-]₄^y [OH^-]₄^z / k [I^-]₃^x [ClO^-]₃^y [OH^-]₃^z

3.0*10^-2 / 6.1*10^-2 = (0.010)^x /(0.010)^x * (0.010)^y /(0.010)^y *(0.010)^z /(0.020)^z

0.5 = (0.50)^z

z = 1

so, rate law