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:

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 $\frac{r_2}{r_1}$?

For the reaction $2NO_{(g)} + O_{2_{(g)}} \rightarrow 2NO_{2_{(g)}}$,the volume of the reaction vessel is suddenly reduced to half of its original volume. If the reaction is first order with respect to $O_2$ and second order with respect to $NO$,the rate of the reaction will:

The order of a reaction can have:

The rate constant for the reaction $2N_2O_5 \rightarrow 4NO_2 + O_2$ is $3.0 \times 10^{-5} \text{ s}^{-1}$. If the rate of reaction is $2.4 \times 10^{-5} \text{ mol L}^{-1} \text{ s}^{-1}$, then the concentration of $N_2O_5$ in $\text{mol L}^{-1}$ is:

The mechanism of the reaction $2NO_2 + F_2 \to 2NO_2F$ is given by:
$(i)$ $NO_2 \xrightarrow{slow} NO + O$
$(ii)$ $F_2 + O + NO \xrightarrow{fast} NO_2F + F$
$(iii)$ $F + NO_2 \xrightarrow{fast} NO_2F$
Select the correct statement.