NEET UG 2006 — Chemistry Physical Chemistry

2006mcqmedium

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:

Official previous-year question

Held on 30 Apr 2006 · Verified 9 Jul 2026.

Options

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

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Solution

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

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Options

$\frac{d\left[\mathrm{NH}_3\right]}{d t}=-\frac{d\left[\mathrm{H}_2\right]}{d t}$

$\frac{d\left[\mathrm{NH}_3\right]}{d t}=-\frac{1}{3} \frac{d\left[\mathrm{H}_2\right]}{d t}$

$+\frac{d[\mathrm{NH}]_3}{d t}=-\frac{2}{3} \frac{d\left[\mathrm{H}_2\right]}{d t}$

$+\frac{d\left[\mathrm{NH}_3\right]}{d t}=-\frac{3}{2} \frac{d\left[\mathrm{H}_2\right]}{d t}$

Solution

NEET UG 2006 — Chemistry Physical Chemistry

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NEET UG 2006 — Chemistry Physical Chemistry

Options

Did you get this right?

Solution

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NEET UG Physical Chemistry in other years

More NEET UG Physical Chemistry questions

Other Chemistry topics for NEET UG

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry