Question

For the decomposition of azoisopropane to hexane and nitrogen at $543K,$ 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)

${\left(C{H}_3\right)}_{2}CHN=NCH{\left(C{H}_3\right)}_{2g}\to N_{2g}+C_6{H}_{14g}$
$\begin{array}{cccc}& Azoisopropane& & Hexane\\ Initialpressure& P_0& 0& 0\\ Pressureaftertimet& P_0-P& P& P\\ Totalpressureaftertimet\left(P_t\right)=\left(P_0-P\right)+P+P=P_0+P& & & \end{array}$
or $P=P_t-P_0$
$R_0\propto P_0$ and $R\propto P_0-P$
On substituting the value of $p,$
$R\propto P_0(P_t-P_0)i.eR\propto 2P_0-P_t$
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 $t=360sec,$
$k=\frac{2.303}{360}log\frac{35.0}{2\times 35.0-54.0}$
$=\frac{2.303}{360}log\frac{35.0}{16}=2.175\times 10^{-3}{s}^{-1}$
When $t=720$ sec,
$k=\frac{2.303}{720}log\frac{35.0}{2\times 35.0-63}=\frac{2.303}{720}log5$
$=2.235\times 10^{-3}{s}^{-1}$
$\therefore$ Average value of $k=\frac{2.175+2.235}{2}\times 10^{-3}{s}^{-1}$
$2.20\times 10^{-3}{s}^{-1}$