Question

In a first-order reaction, 10% of the reactant is consumed in 25 minutes  Calculate: 

(i) The half-life of the reaction.
(ii) The time required for completing 17% of the reaction.

Answer 1

(i) For 1^st  order reaction, 

          `K = 2.303/t  log  a/(a-x)`

            `= 2.303/25 log  100/(100-10)`

             = 0.0042145 min^-1 

∴   Half life t_1/2 = `0.693/K= 0.693/0.0042145`

                          t_1/2 = 164.43 minutes.  

(ii) Again, for the given reaction,

      K = `2.303/t  log  a/(a-x)`

     `t = 2.303/0.0042145  log  100/(100-17)`

     t =  44.21 minutes 

Thus, time required for completion of 17% of the reaction is 44.21 minutes.