Chemistry › Chemical Kinetics › Rate of reaction
Find the average rate of formation of $\mathrm{NO}_{2(\mathrm{~g})}$, in following reaction. $2…
Find the average rate of formation of $\mathrm{NO}_{2(\mathrm{~g})}$, in following reaction.
$2 \mathrm{~N}_2 \mathrm{O}_{5(\mathrm{~g})} \rightarrow 4 \mathrm{NO}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})}$
$\left[-\frac{\Delta\left[\mathrm{N}_2 \mathrm{O}_5\right]}{\Delta \mathrm{t}}\right]=x \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}$
$x \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}$ $\frac{x}{2} \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}$ $2 x \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}$ $4 x \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}$ Solution
$\begin{aligned}
& \frac{-\Delta\left[\mathrm{N}_2 \mathrm{O}_5\right]}{\Delta \mathrm{t}}=x \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1} \\
& 2 \mathrm{~N}_2 \mathrm{O}_{5(\mathrm{~g})} \rightarrow 4 \mathrm{NO}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})}
\end{aligned}$
Average rate of reaction
$\begin{gathered}
=-\frac{1}{2} \frac{\Delta\left[\mathrm{N}_2 \mathrm{O}_5\right]}{\Delta \mathrm{t}}=\frac{1}{4} \frac{\Delta\left[\mathrm{NO}_2\right]}{\Delta \mathrm{t}}=\frac{\Delta\left[\mathrm{O}_2\right]}{\Delta \mathrm{t}} \\
\therefore \quad \frac{\Delta\left[\mathrm{NO}_2\right]}{\Delta \mathrm{t}}=-\frac{4}{2} \frac{\Delta\left[\mathrm{N}_2 \mathrm{O}_5\right]}{\Delta \mathrm{t}}=\frac{4}{2} \times x \\
=2 x \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}
\end{gathered}$
$\therefore \quad$ Average rate of formation of $\mathrm{NO}_{2(\mathrm{~g})}$ $=2 x \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}$
[Note: In the question, $\frac{\Delta\left[\mathrm{N}_2 \mathrm{O}_5\right]}{\Delta \mathrm{t}}$ is changed to $\frac{-\Delta\left[\mathrm{N}_2 \mathrm{O}_5\right]}{\Delta \mathrm{t}}$ to apply appropriate textual concepts.]
Asked in: MHT CET 2023 (10 May Shift 2)
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