Q.

For a reaction $N_{2} + 3 H_{2} \to 2 {NH}_{3}$ , the rate of formation of ammonia was found to be $2.0 x 10^{- 4} mol {dm}^{- 3} s^{- 1}$ . The rate of consumption of $H_{2}$ will be

Moderate

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$1.0 \times 10^{- 4} mol {dm}^{- 3} s^{- 1}$

$2.0 \times 10^{- 4} mol {dm}^{- 3} s^{- 1}$

$3.0 \times 10^{- 4} mol {dm}^{- 3} s^{- 1}$

$4.0 \times 10^{- 4} mol {dm}^{- 3} s^{- 1}$

answer is C.

(Detailed Solution Below)

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Detailed Solution

We will have $- \frac{d [N_{2}]}{d t} = - \frac{1}{3} \frac{d [H_{2}]}{d t} = \frac{1}{2} \frac{d [{NH}_{3}]}{d t}$

It is given that $\frac{d [{NH}_{3}]}{d t} = 2 \times 10^{- 4} mol {dm}^{- 3} s^{- 1}$ . Hence $- \frac{d [H_{2}]}{d t} = \frac{3}{2} \frac{d [{NH}_{3}]}{d t} = 3 \times 10^{- 4} mol {dm}^{- 3} s^{- 1}$

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