what is the half-life of first order reaction for active substance, if 75% of any given amount of the substance disintegrates in 60 minutes?

75% in 60 min means

for first half that is 50% completion it requires t min for other half i.e for 75% completion it requires t min mores 

therefore 2 * t= 60 min

hence t=30min half life 

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If we begin with 100 gms then 25 gms are left after 60 mins.

So a=100, a-x=25, t = 60 mins

Now use the formula

kt = 2.303 log (a/a-x)

sol :   we know that 

                                 here t=60 , x= 75%

                             k= (2.303/t) log { a/ (a-x)}

                  substituting the values we get k=( 2.303/t) log { a /(a-0.75a)}        

                                                                    =(2.303 / 60) log {a / 0.25a}

                                                                    =(2.303 / 60) 0.60

                                                                 k= 0.0231 min^-1

                                                t 1/2= o.693 / k

                                                       =0.693 / 0.0231

                                                        =29.98

                                                         =30 min