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,

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

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

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

= 0.046 s