For a first-order reaction, $20 \%$ of the initial concentration remains after $10 \text{ min}$. What is the rate constant of the reaction (in $\text{ min}^{-1}$)? (Given: $\log_{10}(5) = 0.6989$)

Hydrolysis of methyl acetate in aqueous solution has been studied by titrating the liberated acetic acid against sodium hydroxide. The concentration of the ester at different times is given below:

Time $(t)$ $(\min)$	$0$	$30$	$60$	$90$
Conc. of ester $(C)$ $(M)$	$0.850$	$0.800$	$0.754$	$0.710$

Show that it follows a pseudo first order reaction as the concentration of $H_{2}O$ remains nearly constant $(55 \ mol \ L^{-1})$ during the course of the reaction. What is the value of $k^{\prime}$ in this reaction?

The time taken for $10 \%$ completion of a first order reaction is $20$ minutes. The time required for the completion of $19 \%$ of the same reaction in minutes is

The half-life for the decomposition of gaseous compound $A$ is $240 \ s$ when the initial gaseous pressure is $500 \ Torr$. When the initial pressure is $250 \ Torr$,the half-life is found to be $4.0 \ min$. The order of the reaction is....... (Nearest integer)

For the first$-order$ reaction,$T_{av}$ (average life),$T_{50}$ and $T_{75}$ in the increasing order are:

Identify $True$ $(T)$ and $False$ $(F)$ statements of the following for a first order reaction $R \rightarrow P$.
$I. \log \frac{[R]}{[R]_0} = -\frac{kt}{2.303}$
$II. \log \frac{[R]}{[R]_0} = \frac{kt}{2.303}$