Question

The time taken for 10% completion of a first order reaction is $20\text{min}$. How much time does it take for 19% completion?

• A. $40\text{min}$
• B. $60\text{min}$
• C. $30\text{min}$
• D. $50\text{min}$

Answer 1

The correct option is A $40\text{min}$

$t=\frac{2.303}{k}log\frac{a}{a-x}$
$\therefore 20=\frac{2.303}{k}log\frac{a}{a-0.10a}$
$\text{or},20=\frac{2.303}{k}log\frac{100}{90}$
$\text{or},20=\frac{2.303}{k}log\frac{10}{9}\to \left(i\right)$
In 2nd Case,
$t=\frac{2.303}{k}log\frac{a}{a-0.19a}$
$=\frac{2.303}{k}log\frac{100}{81}=\frac{2.303}{k}log{\left(\frac{10}{9}\right)}^2\to \left(ii\right)$
Eqn (ii) / Eqn (i) gives$\frac{t}{20}=2\text{or}t=40\text{min}$