Question

For the first order reaction $A\to 2B$, 1$\text{mole}$ of reactant A gives 0.2 moles of B after 100 minutes. The half life of the reaction in min (nearest integer) is :

Take
$\ln 2=0.693,\ln 3=1.1\ln 10=2.3$


<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> JEE MAINS 2021

• A. 658
• B. 658.00
• C. 658.0

Answer 1

Answer: (A), (B), (C)

$A\to 2B$
$t=010$
$t=100min1-x2x$
Given:
$2x=0.2\Rightarrow x=0.1$

$k=\frac{1}{t}\text{ln}\frac{[A{]}_0}{\left[A\right]}$
$\left[A_0\right]$ is the initial concentration
$\left[A\right]$ is the final concentration at time t


$k=\frac{\ln 2}{t_{\frac{1}{2}}}$

$\frac{\ln 2}{t_{\frac{1}{2}}}=\frac{1}{100}\ln \frac{1}{0.9}$

$t_{\frac{1}{2}}=\frac{\ln 2\times 100}{ln10-\ln 9}=657.88\approx 658min$