The rate of the reaction: $2N_2O_5 \rightarrow 4NO_2 + O_2$ can be written in three ways.
$-\frac{d[N_2O_5]}{dt} = k[N_2O_5]$
$\frac{d[NO_2]}{dt} = k'[N_2O_5]$ ; $\frac{d[O_2]}{dt} = k''[N_2O_5]$
The relationship between $k$ and $k'$ and between $k$ and $k''$ are:

Consider the following reaction $A + B \rightarrow C$.
The time taken for $A$ to become $1/4$ of its initial concentration is twice the time taken to become $1/2$ of the same. Also,when the change of concentration of $B$ is plotted against time,the resulting graph gives a straight line with a negative slope and a positive intercept on the concentration axis. The overall order of the reaction is . . . . .

The experimental data for the reaction $2A + B_2 \to 2AB$ is given below. Determine the rate equation for the reaction.

$Exp.$	$[A]_0$	$[B_2]_0$	$Rate \ (mol \ L^{-1} \ s^{-1})$
$(1)$	$0.50$	$0.50$	$1.6 \times 10^{-4}$
$(2)$	$0.50$	$1.00$	$3.2 \times 10^{-4}$
$(3)$	$1.00$	$1.00$	$3.2 \times 10^{-4}$

For a second-order reaction,the rate constant is $8 \times 10^{-5} \, M^{-1} \, \text{min}^{-1}$. In how much time will a $1 \, M$ solution decrease to $0.5 \, M$?

The order of a reaction with rate equals $k[C_A]^{3/2} [C_B]^{-1/2}$ is

For the reaction $aA + bB \to$ product,the rate of reaction is given by $Rate = k[A]^3 [B]^0$. If the concentration of $A$ is doubled and the concentration of $B$ is halved,the rate of reaction will be: (in $\times$)