For the given first order reaction
$\mathrm{A} \rightarrow \mathrm{B}$
the half life of the reaction is $0.3010 \mathrm{~min}$. The ratio of the initial concentration of reactant to the concentration of reactant at time $2.0 \mathrm{~min}$ will be equal to ___________. (Nearest integer)
Answer (integer)
100
Solution
$\mathrm{t}_{1 / 2}=\frac{0.693}{\mathrm{~K}} \quad\quad \mathrm{t}_{1 / 2}$ given $=0.3010$
<br/><br/>
$$
\begin{aligned}
&K=\frac{0.693}{0.3010} \\\\
&K=2.30
\end{aligned}
$$
<br/><br/>
$\mathrm{K}=\frac{2.303}{\mathrm{t}} \log \frac{\left(\mathrm{A}_{0}\right)}{\left(\mathrm{A}_{\mathrm{t}}\right)}$
<br/><br/>
$A_{0} \rightarrow$ initial concentration of reactant
<br/><br/>
$A_{t} \rightarrow$ concentration of reactant at time $t$
<br/><br/>
$2.303=\frac{2.303}{2} \log \frac{\left(A_{0}\right)}{\left(A_{t}\right)}$
<br/><br/>
$2=\log \frac{\left(A_{0}\right)}{\left(A_{t}\right)}$
<br/><br/>
$100=\frac{A_{0}}{A_{t}}$
About this question
Subject: Chemistry · Chapter: Chemical Kinetics · Topic: Rate of Reaction
This question is part of PrepWiser's free JEE Main question bank. 96 more solved questions on Chemical Kinetics are available — start with the harder ones if your accuracy is >70%.