The rate of reaction between $A$ and $B$ increases by a factor of $100$ when the concentration of $A$ is increased $10$ folds. The order of reaction with respect to $A$ is

The rate constant for a second order reaction is $8 \times 10^{-5} \ M^{-1} \ min^{-1}$. How long will it take a $1 \ M$ solution to be reduced to $0.5 \ M$?

$[A]_0 / \text{mol } L^{-1}$	$t_{1/2} / \text{min}$
$0.100$	$200$
$0.025$	$100$

For a given reaction $R \rightarrow P$,$t_{1/2}$ is related to $[A]_0$ as given in the table:
Given: $\log 2 = 0.30$
Which of the following is true?
$A.$ The order of the reaction is $1/2$.
$B.$ If $[A]_0$ is $1 \text{ M}$,then $t_{1/2}$ is $200 \sqrt{10} \text{ min}$.
$C.$ The order of the reaction changes to $1$ if the concentration of reactant changes from $0.100 \text{ M}$ to $0.500 \text{ M}$.
$D.$ $t_{1/2}$ is $800 \text{ min}$ for $[A]_0 = 1.6 \text{ M}$.
Choose the correct answer from the options given below:

For a certain reaction,the rate constant $K = 2.37 \times 10^2 \ L^2 \ mol^{-2} \ s^{-1}$. What is the order of the reaction?

If the reaction between $A$ and $B$ to give $C$ shows first-order kinetics in $A$ and second-order in $B$,the rate equation can be written as:

For the gaseous reaction,$N_2O_5 \rightarrow 2NO_2 + \frac{1}{2}O_2$,the rate can be expressed as:
$-\frac{d[N_2O_5]}{dt} = K_1[N_2O_5]$
$+\frac{d[NO_2]}{dt} = K_2[N_2O_5]$
$+\frac{d[O_2]}{dt} = K_3[N_2O_5]$
The correct relation between $K_1, K_2$ and $K_3$ is: