Q12 / 50 · Manipal (MET) 2022 · Chemistry
For the reaction, $\mathrm{N}_2 \mathrm{O}_5(g) \longrightarrow 2 \mathrm{NO}_2(g)+\frac{1}{2} \mathrm{O}_2(g)$ the value of the rate of disappearance of $\mathrm{N}_2 \mathrm{O}_5$ is given as $5.15 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$. The rate of formation of $\mathrm{NO}_2$ is..
Correct answer: 0.0103
Solution
$\left(1.03 \times 10^{-2}\right)$ For the given reaction,
$\begin{aligned}
\mathrm{N}_2 \mathrm{O}_5(g) \longrightarrow & 2 \mathrm{NO}_2(g)+\frac{1}{2} \mathrm{O}_2(g) \\
-\frac{d\left[\mathrm{~N}_2 \mathrm{O}_5\right]}{d t} & =\frac{1}{2} \frac{d\left[\mathrm{NO}_2\right]}{d t}=\frac{2 d\left[\mathrm{O}_2\right]}{d t} \\
\therefore \quad \frac{d\left[\mathrm{NO}_2\right]}{d t} & =-2 \frac{d\left[\mathrm{~N}_2 \mathrm{O}_5\right]}{d t}=2 \times 5.15 \times 10^{-3} \\
& =1.03 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} .
\end{aligned}$
Hence, the correct answer is 0.0103.