Question

$N_2{O}_5$ decomposes according to the equation, $N_2{O}_5\left(g\right)\to 2N{O}_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)$
This is a first order reaction. A closed container has $N_2{O}_5$ and $Ne$ gases at a total pressure of 333.8 mm of Hg. Thirty minutes after $N_2{O}_5$ starts decomposing, the total pressure is found to be 384.5 mm Hg. On complete decomposition of $N_2{O}_5,$ the total pressure is 684.5 mm Hg. What is rate constant of decomposition of $N_2{O}_5$?
(Take ln(1.169) = 0.156)

• A. $2.6\times {10}^{-1}{s}^{-1}$
• B. $0.312s^{-1}$
• C. $5.2\times {10}^{-3}{s}^{-1}$
• D. $8.67\times {10}^{-5}{s}^{-1}$

Answer 1

The correct option is D $8.67\times {10}^{-5}{s}^{-1}$

$N_2{O}_5\left(g\right)\to 2N{O}_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)$

$\begin{array}{cccccccc}{}^0& & & 0& & 0& & initialpressure\\ p^0-p& & & 2p& & \frac{1}{2}p& & at30min\\ 0& & & 2p^0& & \frac{p^0}{2}& & atlongtime\end{array}$

Let $p_1$ be the pressure of Ne gas.
Then, $p^0+p_1=333.8mm$ of Hg .....(1)
$p^0+\frac{3}{2}p+p_1=384.5mm$ of Hg .....(2) and
$2p^0+\frac{1}{2}{p}^0+p_1=684.5mm$ of Hg (3)
Solving equation (1), (2) and (3)
$p^0=233.8mm$ of Hg and $p=33.8mm$ of Hg
So, $k=\frac{1}{30}ln\frac{p^0}{p^0-p}=\frac{1}{30}ln\frac{233.8}{200}$ $=\frac{1}{30}ln1.169=\frac{0.156}{30}=5.2\times {10}^{-3}mi{n}^{-1}$
$=8.67\times {10}^{-5}{s}^{-1}$