$N _{2} O _{5}$ ના વાયુમય કલામાં $318$ $K$ તાપમાને વિઘટનની $\left[2 N _{2} O _{5} \rightarrow 4 NO _{2}+ O _{2}\right]$ પ્રાયોગિક માહિતી નીચે આપેલ છે :

$t/s$	$0$	$400$	$800$	$1200$	$1600$	$2000$	$2400$	$2800$	$3200$
${10^2} \times \left[ {{N_2}{O_5}} \right]/mol\,\,{L^{ - 1}}$	$1.63$	$1.36$	$1.14$	$0.93$	$0.78$	$0.64$	$0.53$	$0.43$	$0.35$

$(i)$ $\left[ N _{2} O _{5}\right]$ વિરુદ્ધ $t$ આલેખ દોરો.

$(ii)$ પ્રક્રિયા માટેનો અર્ધઆયુષ્ય સમય શોધો.

$(iii)$ $\log \left[ N _{2} O _{5}\right]$ અને $t$ વચ્ચેનો આલેખ દોરો.

$(iv)$ વેગ નિયમ શું હશે ?

$(v)$ વેગ અચળાંક ગણો.

$(vi)$ $k$ ઉપરથી અર્ધઆયુષ્ય સમય ગણો અને $(ii)$ સાથે સરખાવો.

$(i)$

$(ii)$ Time corresponding to the concentration, $\frac{1.630 \times 10^{2}}{2} \,mol\, L ^{-1}=81.5 \,mol\, L ^{-1}$ is the half life. From the graph, the half life is obtained as $1450$ $s$

$t(s)$	$10^{2} \times\left[ N _{2} O _{5}\right] / mol\, L ^{-1}$	$\log \left[ N _{2} O _{5}\right]$
$0$	$1.63$	$-1.79$
$400$	$1.36$	$-1.87$
$800$	$1.14$	$-1.94$
$1200$	$0.93$	$-2.03$
$1600$	$0.78$	$-2.11$
$2000$	$0.64$	$-2.19$
$2400$	$0.53$	$-2.28$
$2800$	$0.43$	$-2.73$
$3200$	$0.35$	$-2.46$

$(iv)$ The given reaction is of the first order as the plot, $\log \left[ N _{2} O _{3}\right]_{ v / s } t$ is a straight line. Therefore, the rate law of the reaction is

$(v)$ From the plot, $\log \left[ N _{2} O _{5}\right]$

Slope $=\frac{-2.46-(-1.79)}{3200-0}$

$=\frac{-0.67}{3200}$   $v / s t,$ we obtain

Again, slope of the line of the plot $\log \left[ N _{2} O _{5}\right]_{ v / s } t$ is given by

$-\frac{k}{2.303}$

Therefore, we obtain,

$-\frac{k}{2.303}=-\frac{0.67}{3200}$

$\Rightarrow k=4.82 \times 10^{-4} s ^{-1}$

$(vi)$ Half-life is given by,

$t_{1 / 2}=\frac{0.639}{k}$

$=\frac{0.693}{4.82 \times 10^{-4}} s$

$=1.438 \times 10^{3} \,s$

$=1438 \,s$