Question

On heating, arsine $\left(As{H}_3\right)$ decomposes as:
$2As{H}_3\left(g\right)⟶2As\left(s\right)+3H_2\left(g\right)$
The total pressure measured at constant temperature and constant volume varies with time as follows:

$t$ $\left(min\right)$	$0$	$5$	$7.5$	$10$
$p(mm$ $Hg)$	$760$	$836$	$866.4$	$896.8$

Calculate the rate constant assuming the reaction to follow the first order rate law.

Answer 1

For first order, the rate equation is
Initially after time $t$
$k=\frac{2.303}{t}\log \frac{{\left[A\right]}_0}{\left[A\right]}=\frac{2.303}{t}\log \frac{p_0}{p}$
given, $p_0=760$ $mm$ $Hg$.
The decomposition reaction is:
$\begin{array}{c}2As{H}_3\left(g\right)\\ p_0\\ p_0-2x\end{array}\begin{array}{c}⟶\\ \end{array}\begin{array}{c}As\left(s\right)\\ 0\end{array}\begin{array}{c}+\\ \end{array}\begin{array}{c}3H_2\left(g\right)\\ 0\\ 3x\end{array}$
Total pressure, $p_t=p_0-2x+3x=p_0+x$
$x=p_t-p_0$
$p_{As{H}_3}=(p_0-2x)=p_0-2p_t+2p_0=3p_0-2p_t$
After $5$ minutes, $p_{As{H}_3}=(3\times 760)-(2\times 836)$
$=608$ $mm$ $Hg$
$k=\frac{2.303}{5}{\log }_{10}\frac{760}{608}=0.0446{min}^{-1}$
After $7.5$ minutes, $p_{As{H}_3}=(3\times 760)-(2\times 866.4)$
$=547.2$ $mm$ $Hg$
$k=\frac{2.303}{7.5}{\log }_{10}\frac{760}{547.2}=0.0438{min}^{-1}$
After $10$ minutes, $p_{As{H}_3}=(3\times 760)-(2\times 896.8)$
$=486.4$ $mm$ $Hg$
$k=\frac{2.303}{10}{\log }_{10}\frac{760}{486.4}=0.0446$