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For the reaction $N_{2} (g) + 3 H_{2} (g) \to 2 {NH}_{3} (g)$ under certain conditions of temperature and partial pressure of the reactants, the rate of formation of ${NH}_{3}$ is 0.001 Kg $h^{- 1}$ . The rate of conversion of $H_{2}$ under the same conditions is

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$1 .82 \times 10^{- 4} kg / hr$

$0 .0015 kg / hr$

$1 .52 \times 10^{4} kg / hr$

$1 .82 \times 10^{- 14} kg / hr$

answer is B.

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Detailed Solution

$\begin{array}{l} \frac{- {dN}_{2}}{dt} = \frac{- 1}{3} \frac{{dH}_{2}}{dt} = \frac{1}{2} \frac{{dNH}_{3}}{dt} \\ \frac{{dH}_{2}}{dt} = \frac{3}{2} \times 0 .001 = 0 .0015 {kg hr}^{- 1} . \end{array}$

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