What is the rate of formation of $\mathrm{O}_2$ for the reaction stated below? $\begin{aligned} & 2…

What is the rate of formation of $\mathrm{O}_2$ for the reaction stated below? $\begin{aligned} & 2 \mathrm{~N}_2 \mathrm{O}_{5(\mathrm{~g})} \longrightarrow 4 \mathrm{NO}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})} \\ & {\left[\frac{\mathrm{d}\left[\mathrm{~N}_2 \mathrm{O}_5\right]}{\mathrm{dt}}=0.02 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}\right]} \end{aligned}$
  1. $0.01 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}$
  2. $0.02 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}$
  3. $0.03 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}$
  4. $0.04 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}$

Solution

$\begin{array}{ll} & \text { Rate of reaction }=\frac{1}{2} \frac{\mathrm{~d}\left[\mathrm{~N}_2 \mathrm{O}_5\right]}{\mathrm{dt}}=\frac{\mathrm{d}\left[\mathrm{O}_2\right]}{\mathrm{dt}} \\ \therefore \quad & \frac{\mathrm{d}\left[\mathrm{O}_2\right]}{\mathrm{dt}}=\frac{0.02}{2}=0.01 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}\end{array}$

Asked in: MHT CET 2024 (02 May Shift 1)

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