Chemical Kinetics - Result Question 13
####13. Consider the reaction
$N_2(g)+3 H_2(g) \to 2 NH_3(g)$
[2006]
The equality relationship between $\frac{d[NH_3]}{d t}$ and $-\frac{d[H_2]}{d t}$ is
(a) $+\frac{d[NH_3]}{d t}=-\frac{2}{3} \frac{d[H_2]}{d t}$
(b) $+\frac{d[NH_3]}{d t}=-\frac{3}{2} \frac{d[H_2]}{d t}$
(c) $\frac{d[NH_3]}{d t}=-\frac{d[H_2]}{d t}$
(d) $\frac{d[NH_3]}{d t}=-\frac{1}{3} \frac{d[H_2]}{d t}$
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Solution:
- (a) If we write rate of reaction in terms of concentration of $NH_3$ and $H_2$, then
Rate of reaction $=\frac{1}{2} \frac{d[NH_3]}{d t}=-\frac{1}{3} \frac{d[H_2]}{d t}$
So, $\frac{d[NH_3]}{d t}=-\frac{2}{3} \frac{d[H_2]}{d t}$
