Question

The half-life period of a first-order reaction is 30 min. The percentage of the reactant remaining after 70 min will be:

• A. $80$
• B. $40$
• C. $20$
• D. $10$

Answer 1

The correct option is C $20$

For the first-order reaction, the relationship between the half-life period and the rate constant is

$k=\frac{0.693}{t_{1/2}}=\frac{0.693}{30}=0.0231mi{n}^{-1}.$

The rate law expression is $k=\frac{2.303}{t}\log \frac{a}{(a-x)}.$

Substitute values in the above expression.

$0.0231=\frac{2.303}{70}\log \frac{a}{(a-x)}$

$\log \frac{a}{(a-x)}=0.7021$

$\frac{a}{(a-x)}=5.036$

$\frac{a-x}{a}=\frac{1}{5.036}=0.2$

Thus, the percentage of the reactant remaining after $70$ min will be $20$%.