The reaction given below $2 \mathrm{NH}_{3(\mathrm{~g})} \xrightarrow{\mathrm{Pt}}…

The reaction given below $2 \mathrm{NH}_{3(\mathrm{~g})} \xrightarrow{\mathrm{Pt}} \mathrm{N}_{2(\mathrm{~g})}+3 \mathrm{H}_{2(\mathrm{~g})}$ has rate of reaction $2.5 \times 10^{-6} \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{sec}^{-1}$ formation of $\mathrm{H}_{2(\mathrm{~g})}$ ?
  1. $2.5 \times 10^{-6} \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{sec}^{-1}$
  2. $3.75 \times 10^{-6} \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{sec}^{-1}$
  3. $5.0 \times 10^{-6} \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{sec}^{-1}$
  4. $7.5 \times 10^{-6} \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{sec}^{-1}$

Solution

$\begin{array}{ll} & \text { For the reaction, } 2 \mathrm{NH}_{3(\mathrm{~g})} \xrightarrow{\mathrm{Pt}} \mathrm{N}_{2(\mathrm{~g})}+3 \mathrm{H}_{2(\mathrm{~g})} \\ & \text { Rate of reaction }=-\frac{1}{2} \frac{\mathrm{~d}\left[\mathrm{NH}_3\right]}{\mathrm{dt}}=\frac{\mathrm{d}\left[\mathrm{N}_2\right]}{\mathrm{dt}}=\frac{1 \mathrm{~d}\left[\mathrm{H}_2\right]}{3} \\ \therefore \quad & 2.5 \times 10^{-6}=\frac{1}{3} \frac{\mathrm{~d}\left[\mathrm{H}_2\right]}{\mathrm{dt}} \\ \therefore \quad & \frac{\mathrm{d}\left[\mathrm{H}_2\right]}{\mathrm{dt}}=3 \times 2.5 \times 10^{-6}=7.5 \times 10^{-6} \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{sec}^{-1}\end{array}$

Asked in: MHT CET 2024 (15 May Shift 2)

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