Question

For the decomposition of azoisopropane to hexane and nitrogen at $543 K, $ the following data are obtained

f( sec)	P(mm of Hg)
0	35.0
360	54.0
720	63.0

The rate constant is

• A. $ 3.2 \times 10^{-2} s^{-1} $
• B. $2.3 \times 10^{-3} s^{-1} $
• C. $4 . 3 \times 10^{-2} s ^{-1} $
• D. $7.0 \times 10^{-5} s^{-1} $

Answer 1

Answer: (B)

$\left(CH_{3}\right)_{2} CHN =NCH \left(CH_{3}\right)_{2g} \rightarrow N_{2g} + C_{6} H_{14g} $
$ \begin{matrix}&Azoisopropane&&Hexane\\ Initial pressure&P_{0}&0&0\\ Pressure after time t&P_{0}-P&P&P\\ Total pressure after time t \left(P_{t}\right)=\left(P_{0}-P\right)+ P +P =P_{0} +P&&&\end{matrix}$
or $ P=P_t - P_0 $
$ R_0 \propto P_0 $ and $ R \propto P_0 -P $
On substituting the value of $p,$
$ R \propto P_0 (P_t -P_0) i.e R \propto 2P_0 -P_t $
As decomposition of azoisopropane is a first order reaction,
$\therefore k=\frac{2.303}{t} log \frac{\left[R\right]_{0}}{\left[R\right]} = \frac{2.303}{t} log \frac{p_{0}}{2P_{0} -P_{10}} $
When $t = 360 sec,$
$ k=\frac{2.303}{360} \,log\, \frac{35.0}{2\times35.0-54.0} $
$ =\frac{2.303}{360} \,log\, \frac{35.0}{16} = 2.175\times10^{-3}s^{-1} $
When $ t = 720$ sec,
$ k=\frac{2.303}{720 } \,log\, \frac{35.0}{2\times35.0-63}=\frac{2.303}{720} log5 $
$=2.235\times10^{-3} s^{-1} $
$\therefore $ Average value of $k =\frac{ 2.175 + 2.235}{2} \times10^{-3} s^{-1} $
$2.20 \times10^{-3} s^{-1}$