For the reaction $2HI \rightleftharpoons H_2 + I_2$,the rate of the reaction is proportional to $[HI]^2$. This means that the reaction is

The rate law for the reaction between substances $A$ and $B$ is given by $\text{Rate} = k[A]^n[B]^m$. If the concentration of $A$ is doubled and the concentration of $B$ is halved,what is the ratio of the new rate to the initial rate?

Write the differential rate expression for the following reaction and determine its order of reaction:
$H_{2}O_{2} + I^{-} \rightarrow H_{2}O + IO^{-}$
$H_{2}O_{2} + IO^{-} \rightarrow H_{2}O + I^{-} + O_{2}$

For the reaction $A_2 + B_2 \to 2AB$,the experimental data is given below. Determine the order of the reaction.

Experiment No.	$[A_2] \text{ (M)}$	$[B_2] \text{ (M)}$	Rate $(M \cdot s^{-1})$
$1$	$0.1$	$0.1$	$1.6 \times 10^{-4}$
$2$	$0.1$	$0.2$	$3.2 \times 10^{-4}$
$3$	$0.2$	$0.1$	$3.2 \times 10^{-4}$

The rate constant for the reaction, $2 \,N_2O_{5(g)} \rightarrow 2 \,N_2O_{4(g)} + O_{2(g)}$ is $4.98 \times 10^{-4} \,s^{-1}$. What is the order of the reaction?

Consider the following two reactions:
$A \to \text{Product}; -\frac{d[A]}{dt} = k_1[A]^0$
$B \to \text{Product}; -\frac{d[B]}{dt} = k_2[B]$
The units of $k_1$ and $k_2$ are expressed in terms of molarity $(M)$ and time $(sec^{-1})$ as: