For a chemical reaction $A \to B$,it is found that the rate of reaction doubles when the concentration of $A$ is increased four times. The order of the reaction with respect to $A$ is:

For a reaction $2A + B \to \text{Products}$,doubling the initial concentration of both the reactants increases the rate by a factor of $8$,and doubling the concentration of $B$ alone doubles the rate. The rate law for the reaction is

The rate constant for the reaction,$2N_2O_5 \to 4NO_2 + O_2$ is $3.0 \times 10^{-4} \ s^{-1}$. If the reaction starts with $1.0 \ mol \ L^{-1}$ of $N_2O_5$,calculate the rate of formation of $NO_2$ at the moment when the concentration of $O_2$ is $0.1 \ mol \ L^{-1}$.

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:

Which among the following statements is $NOT$ true about rate constant?

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: