Question

For the decomposition of azo-isopropane to hexane and nitrogen at $543K$, the following data are obtained.

$t\left(sec\right)$	$P$(mm of $Hg$)
$0$	$35.0$
$360$	$54.0$
$720$	$63.0$

Calculate the rate constant.

Answer 1

Let $p_i$ be the initial pressure of azoisopropane and at time $t,p$ atm of azoisopropane decomposes.

$(C{H}_3{)}_{2}CHNH=NHCH(C{H}_3{)}_2⟶N_2+C_6{H}_{14}$

At time, $t=0$

$P_{azoisopropane}=P_i{P}_{N_2}=0P_{C_6{H}_{14}}=0$

At time, $t$

$P_{azoisopropane}=P_i-p{P}_{N_2}=p{P}_{C_6{H}_{14}}=p$

(The pressure is in atm.)

Total pressure, $P_t=(P_i-p)+p+p$

$p=P_t-P_i$

$P_{azoisopropane}=2P_i-P_t$

$k=\frac{2.303}{t}log\frac{[A{]}_0}{\left[A\right]}$

$k-\frac{2.303}{t}log\frac{P_i}{2P_i-P_t}$

When time is $360$ s,

$k=\frac{2.303}{360}log\frac{35.0}{2\times 35.0-54}$

$k=2.17\times {10}^{-3}/s$

When time is $720$ s,

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

$k=2.24\times {10}^{-3}/s$

Average value of k $=2.20\times {10}^{-3}/s$