Chemical Kinetics Question 8
Question 8 - 2024 (30 Jan Shift 2)
$\mathrm{NO}{2}$ required for a reaction is produced by decomposition of $\mathrm{N}{2} \mathrm{O}{5}$ in $\mathrm{CCl}{4}$ as by equation
$2 \mathrm{~N}{2} \mathrm{O}{5}(\mathrm{~g}) \rightarrow 4 \mathrm{NO}{2}(\mathrm{~g})+\mathrm{O}{2}(\mathrm{~g})$
The initial concentration of $\mathrm{N}{2} \mathrm{O}{5}$ is $3 \mathrm{~mol} \mathrm{~L}^{-1}$ and it is $2.75 \mathrm{~mol} \mathrm{~L}^{-1}$ after 30 minutes.
The rate of formation of $\mathrm{NO}_{2}$ is $\mathrm{x} \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}$, value of $\mathrm{x}$ is _________
Show Answer
Answer (17)
Solution
Rate of reaction (ROR)
$=-\frac{1}{2} \frac{\Delta\left[\mathrm{N}{2} \mathrm{O}{5}\right]}{\Delta \mathrm{t}}=\frac{1}{4} \frac{\left[\mathrm{NO}{2}\right]}{\Delta \mathrm{t}}=\frac{\Delta\left[\mathrm{O}{2}\right]}{\Delta \mathrm{t}}$
$\mathrm{ROR}=-\frac{1}{2} \frac{\Delta\left[\mathrm{N}{2} \mathrm{O}{5}\right]}{\Delta \mathrm{t}}=-\frac{1}{2} \frac{(2.75-3)}{30} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}$
$\mathrm{ROR}=-\frac{1}{2} \frac{(-0.25)}{30} \mathrm{molL}^{-1} \min ^{-1}$
$\mathrm{ROR}=\frac{1}{240} \mathrm{molL}^{-1} \min ^{-1}$
Rate of formation of $\mathrm{NO}{2}=\frac{\Delta\left[\mathrm{NO}{2}\right]}{\Delta \mathrm{t}}=4 \times \mathrm{ROR}$
$=\frac{4}{240}=16.66 \times 10^{-3} \mathrm{molL}^{-1} \mathrm{~min}^{-1} \simeq 17 \times 10^{-3}$.
