Reaction : $2Br^{-} + H_2O_2 + 2H^{+} \to  Br_2 + 2H_2O$

take place in two steps :

$(a)$ $Br^{-} + H^{+} + H_2O_2 \xrightarrow{{slow}} HOBr + H_2O$

$(b)$ $HOBr + Br^{-} + H^{+} \xrightarrow{{fast}} H_2O + Br_2$

The order of the reaction is 

For the reaction $A + 2B \to C,$ rate is given by $R$ $ = [A]{[B]^2}$ then the order of the reaction is

If the surface area of the reactants increases, then order of the reaction

$A + 2B \to C$, the rate equation for this reaction is given as Rate $= K[A][B]$ . If the concentration of $A$ is kept the same but that of $B$ is doubled what will happen to the rate it self ?

The hypothetical reaction : $2A + B \to C + D$ is catalyzed by $E$ as indicated in the possible mechanism below -

Step$-1$ : ${\text{A  +  E }} \rightleftharpoons AE$ (fast)

Step$-2$ :${\text{AE  +  A }} \to {A_2} + E$ (slow)

Step$-3$ :${{\text{A}}_2}{\text{ +  B }} \to {\text{D}}$ (fast)

what rate law best agrees with this mechanism

$A $ gaseous hypothetical chemical equation $2A$ $ \rightleftharpoons  $ $4B + C$ is carried out in a closed vessel. The concentration of $ B$  is found to increase by $5 \times {10^{ - 3}}mol\,\,{l^{ - 1}}$ in $10 $ second. The rate of appearance of $B$  is