Consider the reaction $\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) ightarrow 2…

Consider the reaction
$\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) ightarrow 2 \mathrm{NH}_{3}(\mathrm{~g})$
The equality relationship between $\frac{\mathrm{d}\left[\mathrm{NH}_{3}ight]}{\mathrm{dt}}$ and $-\frac{\mathrm{d}\left[\mathrm{H}_{2}ight]}{\mathrm{dt}}$ is
  1. $+\frac{\mathrm{d}\left[\mathrm{NH}_{3}ight]}{\mathrm{dt}}=-\frac{2}{3} \frac{\mathrm{d}\left[\mathrm{H}_{2}ight]}{\mathrm{dt}}$
  2. $+\frac{\mathrm{d}\left[\mathrm{NH}_{3}ight]}{\mathrm{dt}}=-\frac{3}{2} \frac{\mathrm{d}\left[\mathrm{H}_{2}ight]}{\mathrm{dt}}$
  3. $\frac{\mathrm{d}\left[\mathrm{NH}_{3}ight]}{\mathrm{dt}}=-\frac{\mathrm{d}\left[\mathrm{H}_{2}ight]}{\mathrm{dt}}$
  4. $\frac{\mathrm{d}\left[\mathrm{NH}_{3}ight]}{\mathrm{dt}}=-\frac{1}{3} \frac{\mathrm{d}\left[\mathrm{H}_{2}ight]}{\mathrm{dt}}$

Solution

If we write rate of reaction in terms of concentration of $\mathrm{NH}_{3}$ and $\mathrm{H}_{2}$, then
Rate of reaction $=\frac{1}{2} \frac{\mathrm{d}\left[\mathrm{NH}_{3}ight]}{\mathrm{dt}}=-\frac{1}{3} \frac{\mathrm{d}\left[\mathrm{H}_{2}ight]}{\mathrm{dt}}$
So, $\frac{\mathrm{d}\left[\mathrm{NH}_{3}ight]}{\mathrm{dt}}=-\frac{2}{3} \frac{\mathrm{d}\left[\mathrm{H}_{2}ight]}{\mathrm{dt}}$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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