The possible mechanism for the reaction $2NO + Br_2 \to 2NOBr$ is:
$NO + Br_2 \rightleftharpoons NOBr_2$ (Fast)
$NOBr_2 + NO \to 2NOBr$ (Slow)
The rate law expression is:

If the half-life of a reaction is halved when the initial concentration of the reactant is doubled,what is the order of the reaction?

Consider the reaction:
$Cl_{2(aq)} + H_2S_{(aq)} \rightarrow S_{(s)} + 2H^{+}_{(aq)} + 2Cl^{-}_{(aq)}$
The rate equation for this reaction is:
$\text{rate} = k[Cl_2][H_2S]$
Which of these mechanisms is/are consistent with this rate equation?
$A.$ $Cl_2 + H_2S \rightarrow H^{+} + Cl^{-} + Cl^{+} + HS^{-}$ (slow)
$Cl^{+} + HS^{-} \rightarrow H^{+} + Cl^{-} + S$ (fast)
$B.$ $H_2S \rightleftharpoons H^{+} + HS^{-}$ (fast equilibrium)
$Cl_2 + HS^{-} \rightarrow 2Cl^{-} + H^{+} + S$ (slow)

The following results are obtained for the reaction $S + Nu \rightarrow \text{product}$. By which reaction mechanism does this reaction occur?

Experiment	$[S]$	$[Nu]$	Rate
$1$	$0.1$	$0.1$	$2.2 \times 10^{-3}$
$2$	$0.2$	$0.1$	$4.4 \times 10^{-3}$
$3$	$0.1$	$0.2$	$4.4 \times 10^{-3}$

The rate of a reaction is given by $r = K[x][y] / [OH^-]$. If the concentration of $[OH^-]$ is increased,the order of the reaction will be ........

The rate equation for the reaction $2A + B \longrightarrow \text{products}$ is $\text{rate} = k[A][B]^2$. If $k$ at $T \, K$ is $5.0 \times 10^{-6} \, mol^{-2} \, L^2 \, s^{-1}$,the initial rate of the reaction,when $[A] = 0.05 \, mol \, L^{-1}$ and $[B] = 0.1 \, mol \, L^{-1}$ is: