In the reaction $A \rightarrow B$,if the concentration of $A$ is doubled,then the reaction rate increases by $1.59$ times. What will be the order of the reaction?

The experimental data for the reaction $2A + B_2 \longrightarrow 2AB$ is given below:

Exp.	$[A] \ (mol \ L^{-1})$	$[B_2] \ (mol \ L^{-1})$	Rate $(mol \ L^{-1} \ S^{-1})$
$1$	$0.50$	$0.50$	$1.6 \times 10^{-4}$
$2$	$0.50$	$1.00$	$3.2 \times 10^{-4}$
$3$	$1.00$	$1.00$	$3.2 \times 10^{-4}$

Determine the rate law for the reaction.

For a second-order reaction,the rate constant is $8 \times 10^{-5} \, M^{-1} \, \text{min}^{-1}$. In how much time will a $1 \, M$ solution decrease to $0.5 \, M$?

The reaction $N_2O_5$ (in $CCl_4$ solution) $\to 2NO_2$ (solution) $+ \frac{1}{2}O_{2(g)}$ is of first order in $N_2O_5$ with rate constant $6.2 \times 10^{-1} \, s^{-1}$. What is the value of the rate of reaction when $[N_2O_5] = 1.25 \, mol \, L^{-1}$?

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 temperature coefficient of most reactions lies between $.......$.