Consider the following reaction,

$2 H _2( g )+2 NO ( g ) \rightarrow N _2( g )+2 H _2 O ( g )$

which following the mechanism given below:

$2 NO ( g ) \underset{ k _{-1}}{\stackrel{ k _1}{\rightleftharpoons}} N _2 O _2( g )$

$N _2 O _2( g )+ H _2( g ) \stackrel{ k _2}{\rightleftharpoons} N _2 O ( g )+ H _2 O ( g )$

$N _2 O ( g )+ H _2( g ) \stackrel{ k _3}{\rightleftharpoons} N _2( g )+ H _2 O ( g )$

(fast equilibrium)

(slow reaction)

(fast reaction)

The order of the reaction is

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

The variation of the rate of an enzyme catalyzed reaction with substrate concentration is correctly represented by graph

Decay of $_{92}{U^{235}}$is .....order reaction

The experimental data for decomposition of $N _{2} O _{5}$

$\left[2 N _{2} O _{5} \rightarrow 4 NO _{2}+ O _{2}\right]$

in gas phase at $318 \,K$ are given below:

$(i)$ Plot $\left[ N _{2} O _{5}\right]$ against $t$

$(ii)$ Find the half-life period for the reaction.

$(iii)$ Draw a graph between $\log \left[ N _{2} O _{5}\right]$ and $t$

$(iv)$ What is the rate law $?$

$(v)$ Calculate the rate constant.

$(vi)$ Calculate the half-life period from $k$ and compare it with $(ii)$.

For the reaction $A + B \to $ products, what will be the order of reaction with respect to $A$ and $B$ ?