Q78 / 200 · NEET 2006 · Chemistry

Consider the reaction: $\mathrm{N}_{2(g)}+3 \mathrm{H}_{2(g)} \rightarrow$ $2 \mathrm{NH}_{3(g)}$ The equality relationship between $\frac{d\left[\mathrm{NH}_3\right]}{d t}$ and $-\frac{d\left[\mathrm{H}_2\right]}{d t}$ is:

  • A
    $\frac{d\left[\mathrm{NH}_3\right]}{d t}=-\frac{d\left[\mathrm{H}_2\right]}{d t}$
  • B
    $\frac{d\left[\mathrm{NH}_3\right]}{d t}=-\frac{1}{3} \frac{d\left[\mathrm{H}_2\right]}{d t}$
  • Correct answer
    C
    $+\frac{d[\mathrm{NH}]_3}{d t}=-\frac{2}{3} \frac{d\left[\mathrm{H}_2\right]}{d t}$
  • D
    $+\frac{d\left[\mathrm{NH}_3\right]}{d t}=-\frac{3}{2} \frac{d\left[\mathrm{H}_2\right]}{d t}$

Solution

For the reaction $$ \begin{aligned} \mathrm{N}_2(g)+3 \mathrm{H}_2(g) & \rightarrow 2 \mathrm{NH}_3(g) \\ \frac{-d\left[\mathrm{~N}_2\right]}{d t} & =-\frac{1}{3} \frac{d}{d t}\left[\mathrm{H}_2\right] \\ & =\frac{1}{2} \frac{d}{d t}\left(\left[\mathrm{NH}_3\right]\right) \\ \therefore \frac{d}{d t}\left(\left[\mathrm{NH}_3\right]\right) & =-\frac{2}{3} \frac{d}{d t}\left[\mathrm{H}_2\right] \end{aligned} $$

Hence, the correct answer is option: 'C'.

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