Question

The decomposition of $N_2{O}_5$ according to the equation:
$2N_2{O}_5\left(g\right)\to 4N{O}_2\left(g\right)+O_2\left(g\right)$ 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 $284.5\text{mm}$ of Hg and on complete decomposition, the total pressure is $548.5$ mm of Hg.
$(\log 1.169=0.06)$

• A. $\text{Initial pressure of}{N}_2{O}_5=350.80\text{mm of Hg}$
• B. Pressure of $N_2{O}_5$ after $30$ min = $200$ mm of Hg
• C. Rate constant of the reaction is $10.2\times {10}^{-3}{\text{min}}^{-1}$
• D. Rate constant of the reaction is $4.6\times {10}^{-3}{\text{min}}^{-1}$

Answer 1

Answer: (B), (D)

Rate constant of the reaction is $4.6\times {10}^{-3}{\text{min}}^{-1}$

Given,
$2N_2{O}_5\left(g\right)\to 4N{O}_2\left(g\right)+O_2\left(g\right)$
$2$ mol of gaseous nitrogen pentoxide on complete decomposition gives $5$ mol of gaseous products.
$\text{Therefore initial pressure of}{N}_2{O}_5=584.5\times \frac{2}{5}=233.8\text{mm of Hg}$
Let, $x$ be the amount of $N_2{O}_5$ decomposed after $30$ min.
$\therefore$ After $30$ min:
Pressure due to $N_2{O}_5=(233.8-x)\text{mm of Hg}$
Pressure due to $N{O}_2=2x$
Pressure due to $O_2=\frac{x}{2}$
Total pressure after $30$ min
$233.8-x+2x+\frac{x}{2}$
Hence, $284.5=233.8+\frac{3x}{2}$
$\Rightarrow x=33.8\text{mm of Hg}$
Hence pressure of $N_2{O}_5$ after 30 min
$=233.8-33.8=200\text{mm of Hg}$
Again, it is a first order reaction.
So, rate constant $k=\frac{2.303}{t}\log \frac{a}{a-x}=\frac{2.303}{30}\log \frac{233.8}{200}=\frac{2.303}{30}\times \log 1.169=4.606\times {10}^{-3}{\text{min}}^{-1}$