For a reaction, $2 \mathrm{~N}_2 \mathrm{O}_{5(\mathrm{~g})} \longrightarrow 4…

For a reaction, $2 \mathrm{~N}_2 \mathrm{O}_{5(\mathrm{~g})} \longrightarrow 4 \mathrm{NO}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})}$ $\mathrm{N}_2 \mathrm{O}_5$ disappears at a rate of $0.06 \mathrm{moldm}^{-3} \mathrm{~s}^{-1}$. What is rate of $\mathrm{NO}_{2(\mathrm{~g})}$ formation?
  1. $0.06 \mathrm{moldm}^{-3} \mathrm{~s}^{-1}$
  2. $0.12 \mathrm{moldm}^{-3} \mathrm{~s}^{-1}$
  3. $\quad 0 \cdot 18 \mathrm{moldm}^{-3} \cdot \mathrm{~s}^{-1}$
  4. $0.24 \mathrm{moldm}^{-3} \mathrm{~s}^{-1}$

Solution

$2 \mathrm{~N}_2 \mathrm{O}_5 \longrightarrow 4 \mathrm{NO}_2+\mathrm{O}_2$
Overall rate of reaction can be expressed as: $-\frac{1}{2} \frac{\mathrm{~d}\left[\mathrm{~N}_2 \mathrm{O}_5\right]}{\mathrm{dt}}=\frac{1}{4} \frac{\mathrm{~d}\left[\mathrm{NO}_2\right]}{\mathrm{dt}}=\frac{\mathrm{d}\left[\mathrm{O}_2\right]}{\mathrm{dt}}$ Rate of formation of $\mathrm{NO}_2$ : $\frac{\mathrm{d}\left[\mathrm{NO}_2\right]}{\mathrm{dt}}=-\frac{4}{2} \frac{\mathrm{~d}\left[\mathrm{~N}_2 \mathrm{O}_5\right]}{\mathrm{dt}}=\frac{4 \times 0.06}{2}=0.12 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}$

Asked in: MHT CET 2024 (03 May Shift 1)

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