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Consider the following reaction $N_{2} + 3 H_{2} \to 2 {NH}_{3}$ . The rate of the reaction in terms of N₂ at T (k) is $- \frac{d [N_{2}]}{dt}$ =0.02 mole lit^-1sec^-1. What is the value of $- \frac{d [H_{2}]}{dt}$ (in mole. lit^-1.sec^-1) at the same temperature ?

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0.02

0.06

0.04

answer is C.

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Detailed Solution

${N_2} + 3{H_2}xrightarrow{{}}2N{H_3}$

$Rate = - frac{{dleft[ {{N_2}} right]}}{{dt}} = - frac{1}{3}frac{{dleft[ {{H_2}} right]}}{{dt}} = + frac{1}{2}frac{{dleft[ {N{H_3}} right]}}{{dt}}$

$given - frac{{dleft[ {{N_2}} right]}}{{dt}} = 0.02M{S^{ - 1}}$

$- frac{{dleft[ {{H_2}} right]}}{{dt}} = ?$

$- frac{{dleft[ {{N_2}} right]}}{{dt}} = - frac{1}{3}frac{{dleft[ {{H_2}} right]}}{{dt}}$

$0.02 = - frac{1}{3}frac{{dleft[ {{H_2}} right]}}{{dt}}$

$- frac{{dleft[ {{H_2}} right]}}{{dt}} = 0.06;MS^{-1}$

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