In the reaction, $\mathrm{N}_{2}+3 \mathrm{H}_{2} \longrightarrow 2 \mathrm{NH}_{3}$, the rate of…

In the reaction, $\mathrm{N}_{2}+3 \mathrm{H}_{2} \longrightarrow 2 \mathrm{NH}_{3}$, the rate of disappearance of $\mathrm{H}_{2}$ is $0 \cdot 02 \mathrm{M} / \mathrm{s}$. The rate of appearance of $\mathrm{NH}_{3}$ is
  1. $0 \cdot 0133 \mathrm{M} / \mathrm{s}$
  2. $0 \cdot 023 \mathrm{M} / \mathrm{s}$
  3. $0 \cdot 004 \mathrm{M} / \mathrm{s}$
  4. $0 \cdot 032 \mathrm{M} / \mathrm{s}$

Solution

$\mathrm{N}_{2}+3 \mathrm{H}_{2} \longrightarrow 2 \mathrm{NH}_{3}$ Rate of reaction $=-\frac{\mathrm{d}\left(\mathrm{N}_{2}\right)}{\mathrm{dt}}=-\frac{1}{3} \frac{\mathrm{d}\left(\mathrm{H}_{2}\right)}{\mathrm{dt}}=\frac{1}{2} \frac{\mathrm{d}\left(\mathrm{NH}_{3}\right)}{\mathrm{dt}}$ From rate expression, $\frac{\mathrm{d}\left(\mathrm{NH}_{2}\right)}{\mathrm{dt}}=\frac{2}{3} \frac{\mathrm{d}\left(\mathrm{H}_{2}\right)}{\mathrm{d} \mathrm{t}}=\frac{2}{3} \times 0.02=0.0133 \mathrm{M} / \mathrm{s}$

Asked in: MHT CET 2020 (12 Oct Shift 1)

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