Question

For a reaction which is of first order, the fraction of reactants remaining after 160 min if its half life is 40 min is

• A. $\frac{1}{4}$
• B. $\frac{1}{8}$
• C. $\frac{1}{16}$
• D. $\frac{1}{32}$

Answer 1

The correct option is C

$\frac{1}{16}$

$K=\frac{2.303}{t}log\left[\frac{a}{(a-x)}\right];\frac{0.693}{t_{\frac{1}{2}}}=\frac{2.303}{t}log\left[\frac{a}{(a-x)}\right]$

Let intital concentration 'a' = 1 then

fraction of the reactant left = (a-x)

$\frac{0.693}{40}=\frac{2.303}{160}log\left[\frac{1}{(a-x)}\right];(a-x)=1/16$