Question

For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained. Calculate the rate constant.
$\begin{array}{cc}t\left(s\right)& p\left(mmofHg\right)\\ 0& 35.0\\ 360& 54.0\\ 720& 63.0\end{array}$

Answer 1

$(C{H}_3{)}_{2}CHN=NCH(C{H}_3{)}_2\left(g\right)\to N_2\left(g\right)+C_6{H}_{14}\left(g\right)$
$\begin{array}{cccc}Initialpressure& p_i& 0& 0\\ Aftertimet& p_i-p& p& p\end{array}$
Total pressure after time t, i.e., $\left(p_t\right)=(p_i-p)+p+p=p_i+p$
or $p=p_t-p_i$
$a=p_i(a-x)=p_i-p$ on substituting the values of p;
$(a-x)=p_i-(p_t-p_i)$, i.e., $(a-x)=2p_i-p_t$
The decomposition reaction is of gaseous nature and the rate constant k can be calculated as:
$k=\frac{2.303}{t}log\frac{a}{a-x}$
substitution of value of a and (a - x) gives
$k=\frac{2.303}{t}log\left(\frac{p_i}{2p_i-p_t}\right)$
(i) Rate constant after 360s $=k_1$
$k_1=\frac{2.303}{\left(360s\right)}log\frac{35atm}{(70-54)atm}=\frac{2.303}{\left(360s\right)}log\frac{35}{16}=\frac{2.303}{\left(360s\right)}log2.1875=\frac{2.303\times 0.33995}{\left(360s\right)}=2.17\times {10}^{-3}{s}^{-1}$

(ii) Rate constant after 720 s $=k_2$
$k_2=\frac{2.303}{\left(720s\right)}log\frac{\left(35atm\right)}{(70-63)atm}=\frac{2.303}{\left(720s\right)}log5=\frac{2.303\times 0.6990}{\left(720s\right)}=2.24\times {10}^{-3}{s}^{-1}$
Average rate constant k $=\frac{(2.17+2.24)\times {10}^{-3}{s}^{-1}}{2}$
$k=2.21\times {10}^{-3}{s}^{-1}$