Question

The decomposition of $N_2{O}_5$ according to the equation, $2N_2{O}_5\left(g\right)⟶4N{O}_2\left(g\right)+O_2\left(g\right)$, is a first order reaction. After $30$ minutes from the start of the decomposition in a closed vessel, the total pressure developed is found to be $284.5$ $mm$ $Hg$ and on completion, the total pressure is $584.5$ $mm$ $Hg$. Calculate the rate constant of the reaction.

Answer 1

$2N_2{O}_5⟶4N{O}_2+O_2$
On decomposition of $2$ moles of $N_2{O}_5$, $4$ moles of $N{O}_2$ and $1$ mole of $O_2$ are produced. Thus, the total pressure after completion corresponds to $5$ moles and initial pressure to $2$ moles.
Initial pressure of $N_2{O}_5$, $p_0=\frac{2}{5}\times 584.5=233.8$ $mm$ $Hg$
After $30$ minutes, the total pressure $=284.5$ $mm$ $Hg$
$\begin{array}{c}2N_2{O}_5\\ p_0-2p\end{array}\begin{array}{c}⟶\end{array}\begin{array}{c}4N{O}_2\\ 4p\end{array}\begin{array}{c}+\end{array}\begin{array}{c}{O}_2\\ p\end{array}$
or $p_0+3p=284.5$
or $3p=284.5-233.8=50.7$ $mm$ $Hg$
or $p=\frac{50.7}{3}=16.9$ $mm$ $Hg$
Pressure of $N_2{O}_5$ after $30$ minutes $=233.8-(2\times 16.9)$
$=200$ $mm$ $Hg$
$k=\frac{2.303}{30}{\log }_{10}\frac{233.8}{200.0}$
$=5.2\times {10}^{-3}{min}^{-1}$