Find the rate of formation of $\mathrm{NO}_{2(\mathrm{~g})}$ in the following reaction. $\begin{aligned} & 2…

Find the rate of formation of $\mathrm{NO}_{2(\mathrm{~g})}$ in the following reaction. $\begin{aligned} & 2 \mathrm{~N}_2 \mathrm{O}_{5(\mathrm{~g})} \rightarrow 4 \mathrm{NO}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})} \\ & {\left[\frac{-\mathrm{d}\left[\mathrm{N}_2 \mathrm{O}_5\right]}{\mathrm{dt}}=0.02 \mathrm{~mol} \mathrm{dm}^{-3}\right]} \end{aligned}$
  1. $0.01 \mathrm{~mol} \mathrm{dm}^{-3}$
  2. $0.02 \mathrm{~mol} \mathrm{dm}^{-3}$
  3. $0.03 \mathrm{~mol} \mathrm{dm}^{-3}$
  4. $0.04 \mathrm{~mol} \mathrm{dm}^{-3}$

Solution

$\begin{aligned} \text { Rate of reaction } & =-\frac{1}{2} \frac{\mathrm{d}\left[\mathrm{N}_2 \mathrm{O}_5\right]}{\mathrm{dt}} \\ & =+\frac{1}{4} \frac{\mathrm{d}\left[\mathrm{NO}_2\right]}{\mathrm{dt}}=\frac{\mathrm{d}\left[\mathrm{O}_2\right]}{\mathrm{dt}}\end{aligned}$ Rate of formation of $\begin{aligned} \mathrm{NO}_2 & =\frac{\mathrm{d}\left[\mathrm{NO}_2\right]}{\mathrm{dt}} \\ & =-\frac{4}{2} \frac{\mathrm{d}\left[\mathrm{N}_2 \mathrm{O}_5\right]}{\mathrm{dt}} \\ & =-2 \frac{\mathrm{d}\left[\mathrm{N}_2 \mathrm{O}_5\right]}{\mathrm{dt}} \\ & =2 \times 0.02 \\ & =0.04 \mathrm{~mol} \mathrm{dm}^{-3}\end{aligned}$

Asked in: MHT CET 2023 (13 May Shift 1)

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