In the Haber's process of ammonia manufacture,
$N_{2} (g) + 3 H_{2} (g) \to 2 N H_{3} (g)$
The rate of appearance of $N H_{3}$ is
$\frac{d [N H_{3}]}{d t} = 2 \times 1 0^{- 4} m o l L^{-} s e c^{- 1}$
The rate of disppearance of $N_{2}$ and $H_{2}$ will be

Question

Rate in terms of

H_{2}

(mol

L^{- 1} s e c^{- 1}

)

Rate in terms of

N_{2}

(mol

L^{- 1} s e c^{- 1}

)

TestHub · Accepted Answer

Rate = + 1 2 d [ N H 3 ] d t = d [ N 2 ] d t = - 1 3 d [ H 2 ] d t - d [ H 2 ] d t = 3 2 × d [ N H 3 ] d t = 3 2 × 2 × 1 0 - 4 = 3 × 1 0 - 4 m o l L - 1 s e c - 1 - d [ N 2 ] d t = 1 2 d [ N H 3 ] d t = 1 2 × 2 × 1 0 - 4 = 1 × 1 0 - 4 m o l L - 1 s e c - 1

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