Chemical Kinetics - Result Question 74
73. The decomposition of $N _2 O _5$ according to the equation,
$ 2 N _2 O _5(g) \longrightarrow 4 NO _2(g)+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 $284.5$ $ mm$ of $Hg$. On complete decomposition, the total pressure is $584.5$ $ mm$ of $Hg$. Calculate the rate constant of the reaction.
(1991, 6M)
Show Answer
Answer:
Correct Answer: 73. $(5.2 \times 10^{-3} $ $\min ^{-1})$
Solution:
- For the reaction :
$ 2 N _2 O _5 \longrightarrow 4 NO _2+O _2 $
If $p _0$ is the initial pressure, the total pressure after completion of reaction would be $\dfrac{5}{2} p _0$.
$ \Rightarrow \quad 584.5=\dfrac{5}{2} p _0 \Rightarrow p _0=233.8 $ $mm $
Let the pressure of $N _2 O _5$ decreases by ’ $p$ ’ amount after $30$ $ min$. Therefore,
$ \hspace{17mm} 2 N _2 O _5 \longrightarrow 4 NO _2+\underset{2}{O _2} $
At $ 30 $ $min$ : $ p _0 \quad - p \quad \quad \quad 2 p \quad \quad \dfrac{p}{2}$
Total pressure $=p _0+\dfrac{3}{2} p=284.5$
$\Rightarrow \quad p=\dfrac{2}{3}(284.5-233.8)=33.8$
Now, $\quad k t=\ln \dfrac{p _0}{p _0-p}$
$\Rightarrow \quad k=\dfrac{1}{30} \ln \dfrac{233.8}{233.8-33.8} \min ^{-1}=5.2 \times 10^{-3} $ $\min ^{-1}$
BETA
