Question

show that the time required for the completation of 3/4 of the reaction of 1st order is twice the time required for the complete reaction?

Answer 1

For a first order reaction we have:

t=2.303klog[A]o[A]
where t is the time, k is the rate constant and [Ao] and [A] is the initial concentration of the reactant and concentration of the reactant at time t, respectively.

Then, at t = t12/ ; [A] = [A]o2/t12/=2.303klog[A]o[A]o2/t12/=2.303klog 2 −−−−(i)and for, t =t34/ ; [A] = [A]o−34[A]o = [A]o4/t34/=2.303klog[A]o[A]o4/t34/=2.303klog 4 −−−−(ii)Dividing equation (ii) by (i), we gett34/t12/=log 4log 2=2 log 2log2=2Hence, t34/ is twice the t12/.