Question

The rate constant of a first order reaction is 60 s^-1. How much time will it take to reduce the initial concentration of the reactant to its 1/16^th value?

Answer 1

Rate constant of reaction, k = 60 s^-1

t_15/16 = ?

Rate constant of first order reaction is given as,

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

or, t = \(\frac{2.303}{k}log\frac{a}{a-x}\)

When \(\frac{15}{16}th\) of reaction is over,

if a = 1 M, then a - x = \(\frac{1}{16}\,M\)

Hence, \(t_\frac{15}{16} = \)\(\frac{2.303}{60\,s^{-1}}log\frac{1}{1-\frac{15}{16}}\)

= \(\frac{2.303}{60\,s^{-1}}log\,16\)

= \(\frac{2.303}{60\,s^{-1}}\times\,1.2041\,s\)

= 0.046 s.