For the reaction $2N_2O_5 \to 4NO_2 + O_2$,the rate of reaction and rate constant are $1.02 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$ and $3.4 \times 10^{-5} \ s^{-1}$ respectively. The concentration of $N_2O_5$ at that time will be:

Consider the reaction:
$Cl_{2(aq)} + H_2S_{(aq)} \rightarrow S_{(s)} + 2H^{+}_{(aq)} + 2Cl^{-}_{(aq)}$
The rate equation for this reaction is:
$\text{rate} = k[Cl_2][H_2S]$
Which of these mechanisms is/are consistent with this rate equation?
$A.$ $Cl_2 + H_2S \rightarrow H^{+} + Cl^{-} + Cl^{+} + HS^{-}$ (slow)
$Cl^{+} + HS^{-} \rightarrow H^{+} + Cl^{-} + S$ (fast)
$B.$ $H_2S \rightleftharpoons H^{+} + HS^{-}$ (fast equilibrium)
$Cl_2 + HS^{-} \rightarrow 2Cl^{-} + H^{+} + S$ (slow)

Give the relation between half-life $(t_{1/2})$ and initial concentration of reactant $([R]_0)$ for an $(n-1)^{th}$ order reaction.

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

$Exp.$	$[A] \ (mol \ L^{-1})$	$[B] \ (mol \ L^{-1})$	Initial rate $(mol \ L^{-1} \ s^{-1})$
$1$	$2.5 \times 10^{-4}$	$3 \times 10^{-5}$	$5 \times 10^{-4}$
$2$	$5 \times 10^{-4}$	$6 \times 10^{-5}$	$4 \times 10^{-3}$
$3$	$1 \times 10^{-3}$	$6 \times 10^{-5}$	$1.6 \times 10^{-2}$

If the half-life of a reaction is halved when the initial concentration of the reactant is doubled,what is the order of the reaction?

Differential form of the rate equation is $\frac{dx}{dt} = k[P][Q]^{0.5}[R]^{0.5}$. Which statement about the above equation is wrong?