For the reaction, $2 \mathrm{NO}_{(\mathrm{g})}+\mathrm{O}_{2(\mathrm{~g})} \rightarrow 2…

For the reaction, $2 \mathrm{NO}_{(\mathrm{g})}+\mathrm{O}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{NO}_{2(\mathrm{~g})} \text { If } \frac{\mathrm{d}\left[\mathrm{NO}_2\right]}{\mathrm{dt}}=0.052 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}$ Calculate rate of consumption of $\mathrm{NO}_{(\mathrm{g})}$.
  1. $0.114 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}$
  2. $0.078 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}$
  3. $0.026 \mathrm{~mol}^{-3} \mathrm{~s}^{-1}$
  4. $0.052 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}$

Solution

In the given reaction, the rate of consumption of $\mathrm{NO}=$ rate of formation of $\mathrm{NO}_2$ $\mathrm{So}, \frac{-\mathrm{d}[\mathrm{NO}]}{\mathrm{dt}}=\frac{+\mathrm{d}\left[\mathrm{NO}_2\right]}{\mathrm{dt}}$ $\begin{aligned} & \frac{-1}{2} \mathrm{~d} \frac{[\mathrm{NO}]}{\mathrm{dt}}=\frac{1}{2} \frac{\mathrm{~d}\left[\mathrm{NO}_2\right]}{\mathrm{dt}} \\ & \frac{-\mathrm{d}[\mathrm{NO}]}{\mathrm{dt}}=\frac{2}{2} \frac{\mathrm{~d}\left[\mathrm{NO}_2\right]}{\mathrm{dt}} \\ & \frac{-\mathrm{d}[\mathrm{NO}]}{\mathrm{dt}}=0.052 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}\end{aligned}$

Asked in: MHT CET 2022 (05 Aug Shift 2)

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