For the reaction, $\mathrm{N}_{2(\mathrm{~g})}+3 \mathrm{H}_{2(\mathrm{~g})} \longrightarrow 2…

For the reaction, $\mathrm{N}_{2(\mathrm{~g})}+3 \mathrm{H}_{2(\mathrm{~g})} \longrightarrow 2 \mathrm{NH}_{3(\mathrm{~g})}$ $\mathrm{NH}_3$ is formed at a rate of $0.088 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}$. Calculate consumption rate of $\mathrm{N}_{2(\mathrm{~g})}$.
  1. $0.011 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}$
  2. $0.022 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}$
  3. $0.033 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}$
  4. $0.044 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}$

Solution

$\begin{aligned} & \mathrm{N}_2+3 \mathrm{H}_2 \longrightarrow 2 \mathrm{NH}_3 \\ & \text { Rate }=-\frac{\mathrm{d}\left[\mathrm{~N}_2\right]}{\mathrm{dt}}=-\frac{1}{3} \frac{\mathrm{~d}\left[\mathrm{H}_2\right]}{\mathrm{dt}}=\frac{1}{2} \frac{\mathrm{~d}\left[\mathrm{NH}_3\right]}{\mathrm{dt}} \\ \therefore \quad & -\frac{\mathrm{d}\left[\mathrm{~N}_2\right]}{\mathrm{dt}}=\frac{1 \mathrm{~d}\left[\mathrm{NH}_3\right]}{2}=\frac{1}{\mathrm{dt}} \times 0.088 \end{aligned}$ $\therefore \quad$ Rate of consumption of $\mathrm{N}_2=0.044 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}$

Asked in: MHT CET 2024 (09 May Shift 2)

Practice more Chemical Kinetics questions on Aicharya