IIT JEE , AIEEE Forum - Chemistry , Physical Chemistry

gayathri

i have this basic doubt from the ncert chemistry part1.please do clear it up!!

Qn :hydrolysis of methy acetate in aqueous soln. has been studied by titrating the liberated acetic acid against NaOH. the concentrationof ester at different times is gn
below

t/min         0       30      60 90

C/mol/L   .8500 .8004 .7538 .7096

Show that it follows a pseudo first order reaction, as the concentration of water remains nearly constant(55mol/L), during the course of the reaction what is the value of k' in this equation?

Rate=k'[CH3COOCH3][H2O]

Solution:
         For pseudo first order reaction , the reaction should be first order with respect to ester when H2O is constant. the rate constant k for the pseudo first order reaction is

k=2.303/t *logCo/C

where k=k'[H2O]

t/min    C/mol/L             k/mol/L        ( i understand only till this part of
      0                   .8500                  -                  the solution)

     30                  .8004             2.004*10^ -3
     60                  .7538             2.002*10^ -3
     90                  .7096             2.005*10^ -3

it can be seen that k[H2O] is constant and equal to 2.004*10^ -3 per min
and hence, it is pseudo first order reaction. we can now determine k from
k[H2O]= 2.004*10^ -3
k[55 mol/L]= 2.004*10^ -3
k=3.64* 10^ -5 per mole litre pre min

i can't understand the soln. kindly sort it out.

krishna.gopal

Rate of reaction is k[h2o][ester]
let's write it asa k'[ester] where k'=k[h2o]
from conc time data find k' (as u understood) and divde it with[h2o] to get k

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