prove that for a first order reaction the time taken for 3/4 completion of a reaction is twice of half life Share with your friends Share 0 Vartika Jain answered this Dear User, For the first order reaction, t = 2.303klog[A]o[A]For 3/4 of a reaction to take place, t= t34, [A]= [A]o-34[A]o=14[A]oThus, t34= 2.303klog [A]o14 [A]o= 2.303klog 4....(i)Now, for half of a reaction to take place,t= t12, [A]= [A]o-12[A]o=12[A]oThus, t12= 2.303klog [A]o12 [A]o= 2.303klog 2.....(ii)Dividing eq (i) by eq (ii)t34t12=log 4log 2=2or, t34=2t12 1 View Full Answer Karthik Venkat answered this That's obvious let T1/2 be x min. and we have 100 g of a substancecase I so time taken to complete 50% of substance that is 50g is T1/2 i.e., x min. so the quantity remaining is 50g.case II so again it takes x min to disintegrate 50% of the remaining substance, so remaining quatity is 25g or 25%. so 75% of the initial value gets disintegrated in 2x min. i.e., 2 x T1/2. - hope that helps :P 1
