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)
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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 $\frac{5}{2} p _0$.
$$ \Rightarrow \quad 584.5=\frac{5}{2} p _0 \Rightarrow p _0=233.8 mm $$
Let the pressure of $N _2 O _5$ decreases by ’ $p$ ’ amount after $30 min$. Therefore,
$$ \begin{array}{lrlc} & 2 N _2 O _5 & \longrightarrow & 4 NO _2+\underset{2}{O _2} \ \text { At } 30 min: & p _0-p & 2 p & \frac{p}{2} \end{array} $$
Total pressure $=p _0+\frac{3}{2} p=284.5$
$\Rightarrow \quad p=\frac{2}{3}(284.5-233.8)=33.8$
Now, $\quad k t=\ln \frac{p _0}{p _0-p}$
$\Rightarrow \quad k=\frac{1}{30} \ln \frac{233.8}{233.8-33.8} \min ^{-1}=5.2 \times 10^{-3} \min ^{-1}$
