એઝોઆઇસોપ્રોપેનનું હેઝેન અને નાઇટ્રોજનમાં વિઘટન $543$ $K$ તાપમાને કરવામાં આવે છે. માહિતી નીચે પ્રમાણે મળેલી છે :

$t$ $(sec)$	$P(m m \text { of } H g)$
$0$	$35.0$
$360$	$54.0$
$720$	$63.0$

વેગ અચળાંક ગણો.

The decomposition of azoisopropane to hexane and nitrogen at $543 K$ is represented by the following equation.

After time, $t,$ total pressure$P _{t}=\left( P _{0}-p\right)+p+p$

$\Rightarrow P _{t}= P _{0}+p$

$\Rightarrow p= P _{t}- P _{0}$

Therefore, $P_{0}-p=P_{0}-\left(P_{t}-P_{0}\right)$

$=2 P_{0}-P_{t}$

For a first order reaction,

$k=\frac{2.303}{t} \log \frac{ P _{0}}{ P _{0}-p}$

$=\frac{2.303}{t} \log \frac{P_{0}}{2 P_{0}-P_{t}}$

When $t=360 \,s$,

$=2.175 \times 10^{-3} \,s ^{-1}$

$k=\frac{2.303}{360 s } \log \frac{35.0}{2 \times 35.0-54.0}$

When $t=720\, s$

$=2.235 \times 10^{-3} \,s ^{-1}$

$k=\frac{2.303}{720 s } \log \frac{35.0}{2 \times 35.0-63.0}$

Hence, the average value of rate constant is

$k=\frac{\left(2.175 \times 10^{-3}\right)+\left(2.235 \times 10^{-3}\right)}{2} s ^{-1}$

$=2.21 \times 10^{-3} \,s ^{-1}$