first we have to find out the rate constant of the reaction i.e. K
So, for qa 1st order reaction K=(1/16)X2.303X log[a/(a-50a/100) where a is the initial substrate concentration.
i.e. K= (1/16) X 2.303 X log 2
i.e. K= (1/16) X 0.693
i.e. K= 0.043
Now, we have to find out the time required for 75% completion
t=(1/0.043)X2.303X log[a/(a-75a/100)
t=(1/0.043)X2.303X log 4
t=(1/0.043) X 2 X 2.303 X log 2
t=(1/0.043)X 2 X 0.693
t=32.23min or 32 min
Thanks
Souvik Mukherjee
M.Sc. Chemistry, IIT Hyderabad