For a gaseous reaction,the rate is given by $Rate = k [A] [B]$. If the volume of the container is reduced to $1/4$ of its initial volume,the rate of the reaction will become how many times the initial rate?

The following results have been obtained during the kinetic studies of the reaction: $2 \ NO + 2 \ H_2 \longrightarrow N_2 + 2 \ H_2O$

Expt	$\frac{-d[NO]}{dt} \ (mol \ L^{-1} \ s^{-1})$	$[NO] \ (mol \ L^{-1})$	$[H_2] \ (mol \ L^{-1})$
$1$	$4.8 \times 10^{-5}$	$1 \times 10^{-2}$	$1 \times 10^{-3}$
$2$	$43.2 \times 10^{-5}$	$3 \times 10^{-2}$	$1 \times 10^{-3}$
$3$	$86.4 \times 10^{-5}$	$3 \times 10^{-2}$	$2 \times 10^{-3}$

The half-life period of a $second$ order reaction is:

Which function of $[X]$ plotted against time will give a straight line for a second order reaction? $X \to \text{Product}$

The rate law for the reaction $A + B \rightarrow \text{product}$ is given by $\text{rate} = k[A][B]$. Calculate $[A]$ if the rate of reaction and rate constant are $0.25 \ mol \ dm^{-3} \ s^{-1}$ and $6.25 \ mol^{-1} \ dm^3 \ s^{-1}$ respectively,and $[B] = 0.25 \ mol \ dm^{-3}$.

Rate of reaction is $r_1 = k[A]^a[B]^b$. If the concentration of $A$ is doubled and $B$ is halved,the new rate is $r_2$. What is the value of $\frac{r_2}{r_1}$?