The unit of the rate constant for a second-order reaction is ....

For the reaction,$2N_2O_5 \to 4NO_2 + O_2$,the rate equation can be expressed in two ways $-\frac{d[N_2O_5]}{dt} = k[N_2O_5]$ and $+\frac{d[NO_2]}{dt} = k'[N_2O_5]$. $k$ and $k'$ are related as:

The rate equation for the reaction $2A + B \to C$ is found to be: $\text{rate} = k[A][B]$. The correct statement in relation to this reaction is that the

For the reaction $2NOBr(g) \rightarrow 2NO(g) + Br_2(g)$, the rate law is $r = k[NOBr]^2$. If the rate constant $k = 1.62 \times 10^{-2} \text{ L mol}^{-1} \text{ s}^{-1}$ and the concentration of $[NOBr] = 2 \times 10^{-3} \text{ mol L}^{-1}$, what is the rate of 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:

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?