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The decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) according to the equation,
$$
2 \mathrm{~N}_{2} \mathrm{O}_{5}(g) \rightarrow 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)
$$
is a first order reaction. After 30 min from the start of the decomposition in a closed vessel, the total pressure developed is found to be \(284.5 \mathrm{~mm}\) of \(\mathrm{Hg}\). On complete decomposition, the total pressure is \(584.5 \mathrm{~mm}\) of \(\mathrm{Hg}\). Calculate the rate constant of the reaction.

3 Answers

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$$
\begin{array}{cccl}
2 \mathrm{~N}_{2} \mathrm{O}_{5}(g) & \longrightarrow & 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) \\
1 & 0 & 0 & \text { at } t=0 \\
(a-x) & 2 x & x / 2 & \text { at } t=30 \\
0 & 2 a & a / 2 & \text { at } t=\infty
\end{array}
$$
We know that
No. of moles \(\propto\) pressure developed
at \(\quad t=0, a \propto p_{0}\)
at \(t=30 \mathrm{~min}, \quad a+\frac{3 x}{2} \propto 284.5\)
at \(t=\infty, \frac{5 a}{2} \propto 584.5\)
Thus at \(t=0\),
\(a \propto \frac{2 \times 584.5}{5} \propto 233.8\) and at \(t=30 \mathrm{~min}, x \propto 33.8\) \(\therefore \quad a-x \propto 200.0\) Now, \(\quad k=\frac{2.303}{t} \log \frac{a}{(a-x)}\) \(\therefore \quad k=\frac{2.303}{30} \log \frac{233.8}{200}\) \(=5.206 \times 10^{-3} \mathrm{~min}^{-1}\)
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