For an elementary chemical reaction,$A_2 \underset{k_{-1}}{\overset{k_1}{\longleftrightarrow}} 2A$,the expression for $\frac{d[A]}{dt}$ is

The hypothetical reaction : $2A + B \to C + D$ is catalyzed by $E$ as indicated in the possible mechanism below -
Step-$1$ : $A + E \rightleftharpoons AE$ (fast)
Step-$2$ : $AE + A \to A_2 + E$ (slow)
Step-$3$ : $A_2 + B \to C + D$ (fast)
What rate law best agrees with this mechanism?

The rate for the reaction $2 A + B \rightarrow \text{product}$ is $6 \times 10^{-4} \ mol \ dm^{-3} \ s^{-1}$. Calculate the rate constant if the reaction is first order in $A$ and zeroth order in $B$,given $[A] = [B] = 0.3 \ M$.

For the first order decomposition reaction of $N_2O_5$,it is found that -
$(a)$ $2N_2O_5 \rightarrow 4NO_{2(g)} + O_{2(g)}$ ; $-\frac{d[N_2O_5]}{dt} = k[N_2O_5]$
$(b)$ $N_2O_5 \rightarrow 2NO_{2(g)} + 1/2 O_{2(g)}$ ; $-\frac{d[N_2O_5]}{dt} = k'[N_2O_5]$
Which of the following is true?

For a gaseous reaction between $X$ and $Y$,$X + 3Y \rightarrow XY_3$,the initial rate data is given below:
$[X] = 0.1 \ M, [Y] = 0.1 \ M, \text{Rate} = 0.002 \ Ms^{-1}$
$[X] = 0.2 \ M, [Y] = 0.1 \ M, \text{Rate} = 0.002 \ Ms^{-1}$
$[X] = 0.3 \ M, [Y] = 0.2 \ M, \text{Rate} = 0.008 \ Ms^{-1}$
$[X] = 0.4 \ M, [Y] = 0.3 \ M, \text{Rate} = 0.018 \ Ms^{-1}$
What is the rate law?

What is the order of reaction $A + B \to C$?

Observation	$[A] \ (mol \ L^{-1})$	$[B] \ (mol \ L^{-1})$	Rate $(mol \ L^{-1} \ sec^{-1})$
$1$	$0.1$	$0.1$	$2 \times 10^{-3}$
$2$	$0.4$	$0.1$	$3.2 \times 10^{-3}$
$3$	$0.1$	$0.2$	$8 \times 10^{-3}$