The experimental data for decomposition of $N_2O_5$ in the gas phase at $318 \, K$ are given below:

$t/s$	$0$	$400$	$800$	$1200$	$1600$	$2000$	$2400$	$2800$	$3200$
$10^2 \times [N_2O_5] / mol \, L^{-1}$	$1.63$	$1.36$	$1.14$	$0.93$	$0.78$	$0.64$	$0.53$	$0.43$	$0.35$

$(i)$ Plot $[N_2O_5]$ against $t$.
$(ii)$ Find the half-life period for the reaction.
$(iii)$ Draw a graph between $\log[N_2O_5]$ 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 $(ii)$.

(N/A) $(i)$ Plot $[N_2O_5]$ vs $t$ shows a curve indicating a first-order reaction.
$(ii)$ Initial concentration $[N_2O_5]_0 = 1.63 \times 10^{-2} \, mol \, L^{-1}$. Half-life corresponds to the time when concentration becomes half,i.e.,$0.815 \times 10^{-2} \, mol \, L^{-1}$. From the graph,$t_{1/2} \approx 1450 \, s$.
$(iii)$ Plot $\log[N_2O_5]$ vs $t$ gives a straight line.
$(iv)$ Since the plot of $\log[N_2O_5]$ vs $t$ is a straight line,the reaction is of the first order. The rate law is: $\text{Rate} = k[N_2O_5]$.
$(v)$ Slope of $\log[N_2O_5]$ vs $t$ graph $= \frac{-k}{2.303}$.
Using points $(0, -1.79)$ and $(3200, -2.46)$:
$\text{Slope} = \frac{-2.46 - (-1.79)}{3200 - 0} = \frac{-0.67}{3200} = -2.09 \times 10^{-4} \, s^{-1}$.
$k = -\text{Slope} \times 2.303 = 2.09 \times 10^{-4} \times 2.303 \approx 4.82 \times 10^{-4} \, s^{-1}$.
$(vi)$ $t_{1/2} = \frac{0.693}{k} = \frac{0.693}{4.82 \times 10^{-4}} \approx 1438 \, s$. This is in close agreement with the value obtained in $(ii)$.