$A$ chemical reaction proceeds through the following steps:
Step-$I$: $2A \rightleftharpoons X$ (fast)
Step-$II$: $X + B \rightarrow Y$ (slow)
Step-$III$: $Y + B \rightarrow \text{Product}$ (fast)
The rate law for the overall reaction is:

The rate law for the reaction $RCl + NaOH_{(aq)} \to ROH + NaCl$ is given by $\text{Rate} = K_1[RCl]$. The rate of the reaction will be

Which of the following options correctly represents the relationship between $t_{7/8}$ and $t_{1/2}$ where $t_{7/8}$ represents the time required for the concentration to become $1/8$ of the original for a reaction of order $n$?

For the reaction,$2A + B \to \text{products}$,when the concentrations of $A$ and $B$ both were doubled,the rate of the reaction increased from $0.3 \ mol \ L^{-1} \ s^{-1}$ to $2.4 \ mol \ L^{-1} \ s^{-1}$. When the concentration of $A$ alone is doubled,the rate increased from $0.3 \ mol \ L^{-1} \ s^{-1}$ to $0.6 \ mol \ L^{-1} \ s^{-1}$. Which one of the following statements is correct?

The reaction $2NO_{(g)} + O_{2(g)} \rightleftharpoons 2NO_{2(g)}$ is of first order. If the volume of the reaction vessel is reduced to $1/3$ of its original volume,the rate of reaction would be .......... times.

Consider the following reaction,
$2H_{2(g)} + 2NO_{(g)} \rightarrow N_{2(g)} + 2H_2O_{(g)}$
which follows the mechanism given below:
$2NO_{(g)} \underset{k_{-1}}{\stackrel{k_1}{\rightleftharpoons}} N_2O_{2(g)}$ (fast equilibrium)
$N_2O_{2(g)} + H_{2(g)} \stackrel{k_2}{\rightarrow} N_2O_{(g)} + H_2O_{(g)}$ (slow reaction)
$N_2O_{(g)} + H_{2(g)} \stackrel{k_3}{\rightarrow} N_{2(g)} + H_2O_{(g)}$ (fast reaction)
The order of the reaction is