The conversion of $A \to B$ follows second order kinetics. Doubling the concentration of $A$ will increase the rate of formation of $B$ by a factor of:

In a reaction,$A + B \rightarrow$ product,the rate is doubled when the concentration of $B$ is doubled,and the rate increases by a factor of $8$ when the concentrations of both the reactants $(A$ and $B)$ are doubled. The rate law for the reaction can be written as:

The experimental data for the reaction $2A + B_2 \longrightarrow 2AB$ is given below:

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

Determine the rate law for the reaction.

The mechanism of the reaction $A + 2B \to D$ is given by:
$2B \xrightarrow{k} B_2$ [Slow]
$B_2 + A \to D$ [Fast]
The rate law expression,order with respect to $A$,order with respect to $B$,and overall order are respectively:

The unit of rate constant for a second-order reaction is usually expressed as:

The rate law for the reaction $A + 2B \rightarrow C + 2D$ is given by: