In a first order reaction,$87.5 \%$ of the reactant is converted into the product in $15 \ minutes$. The rate constant for the reaction is given by:

For a first-order reaction,the rate constant is $60 \ s^{-1}$. How much time will it take for the initial concentration of the reactant to be reduced to $1/10$th of its initial value?

The following results were obtained during kinetic studies of the reaction $2A + B \to$ products:

Experiment	$[A]$ $(mol \ L^{-1})$	$[B]$ $(mol \ L^{-1})$	Initial rate $(mol \ L^{-1} \ min^{-1})$
$I$	$0.10$	$0.20$	$6.93 \times 10^{-3}$
$II$	$0.10$	$0.25$	$6.93 \times 10^{-3}$
$III$	$0.20$	$0.30$	$1.386 \times 10^{-2}$

The time (in minutes) required to consume half of $A$ is:

When the value of rate constant $(k)$ is $2.0 \text{ min}^{-1}$, then what will be the half-life of the reaction $(t_{1/2})$ in seconds?

$A$ first order reaction has a rate constant $1.1 \times 10^{-3} \ s^{-1}$. How long will $6 \ g$ of this reactant take to reduce to $3 \ g$ (in $s$)?

The half-life for the first-order reaction $N_2O_5 \to 2NO_2 + \frac{1}{2} O_2$ is $2.4 \text{ hours}$. Starting with $10.8 \text{ g}$ of $N_2O_5$ at $STP$,how many liters of oxygen will be obtained after a period of $9.6 \text{ hours}$? (Given: $10^{0.2} = 1.584$)