Question

The rate law for the reaction H+(aq) + OH−(aq) ---> H2O(l) is Rate = k[H+ ][OH− ], with
 k = 1.3 x 10^11 mol-1 L s-1 at 25 deg C. In an experiment, 0.500 L of 0.020 mol L-1 KOH(aq) is rapidly mixed with an equal volume of 0.020 mol L-1 HBr(aq). Calculate the time, in seconds, required for [H+ ] to decrease to 1.0x 10^-7 mol L-1.


Answer 1

given, Rate=K[H+][OH-]

K=1.3*10^11 mol-1 L s-1

order of reaction=2

for second order reaction with equal initial concentration, [H+]=[OH-]

Initial Rate can be written as =k[H+]^2

t=1/K[1/[H+]t- 1/[H+]o]

[H+]o=initial conc =0.020 mol/L

[H+]t=conc after time=t=1.0*10^-7 mol/L

t=1/(1.3*10^11 mol-1 L s-1) [1/(1.0*10^-7 mol/L)-1/(0.020mol/L)]

   =(0.77*10^-11 mol L-1 s)[1.0*10^7 L/mol -50 L/mol]

   =0.77 *10^-4 s-38.5 *10^-11 s

    =0.77 *10^-4 s-38.5 *10^-7*10^-4 s

    =0.77*10^-4-0.38*10^-6 *10^-4

    =(0.77 * 10^-4 -0.38 *10^-4)(1-10^-6)s

     =(0.39*10^-4) (1)

     t=0.39*10^-4 s