Question

For the first order reaction $A \rightarrow 2 B, 1$ mole of reactant $A$ gives $0.2$ moles of $B$ after $100$ minutes. The half life of the reaction is ...... min.
(Round off to the nearest integer).
[Use : $\ln 2=0.69, \ln 10=2.3$ Properties of logarithms : $\ln x^{y}=y \ln x; .\ln \left(\frac{x}{y}\right)=\ln x-\ln y$ ]
(Round off to the nearest integer)

Answer 1

Now, $t=\frac{t_{1 / 2}}{\ln 2} \times \frac{\left[A_{0}\right]}{\left[A_{t}\right]}$
$100=\frac{t_{1 / 2}}{\ln 2} \times \ln \frac{1}{0.9} $
$\Rightarrow t_{1 / 2}=690 \min .$
( taking $\ln 3=1.11$ )
Ans. 600 to $700$