The following graph shows the relationship between $(a-x)^{-1}$ and time $t$ for a second-order reaction. If $\theta = \tan^{-1}(1/2)$ and $OA = 2 \ L \ mol^{-1}$,then the rate at the start of the reaction will be (in $mol \ L^{-1} \ min^{-1}$):

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:

If the concentration of a first-order reaction is increased by $x$ times,then the rate constant $(k)$ becomes

For the chemical reaction $2A + 2B + C \rightarrow \text{Product}$,the rate law is given by $r \propto [A] [B]^2$. What is the order of the reaction?

The following mechanism has been proposed for the reaction of $NO$ with $Br_2$ to form $NOBr$:
$NO_{(g)} + Br_{2(g)} \rightleftharpoons NOBr_{2(g)}$
$NOBr_{2(g)} + NO_{(g)} \longrightarrow 2NOBr_{(g)}$
If the second step is the rate-determining step,the order of the reaction with respect to $NO_{(g)}$ is:

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