Chemical Kinetics - Result Question 33
33. Consider the chemical reaction,
$ N _2(g)+3 H _2(g) \longrightarrow 2 NH _3(g) $
The rate of this reaction can be expressed in terms of time derivatives of concentration of $N _2(g), H _2(g)$ or $NH _3(g)$. Identify the correct relationship amongst the rate expressions
(2002, 3M)
(a) Rate $=-\dfrac{d\left[N _2\right]}{d t}=-\dfrac{1}{3} \dfrac{d\left[H _2\right]}{d t}=\dfrac{1}{2} \dfrac{d\left[NH _3\right]}{d t}$
(b) Rate $=-\dfrac{d\left[N _2\right]}{d t}=-3 \dfrac{d\left[H _2\right]}{d t}=2 \dfrac{d\left[NH _3\right]}{d t}$
(c) Rate $=\dfrac{d\left[N _2\right]}{d t}=\dfrac{1}{3} \dfrac{d\left[H _2\right]}{d t}=\dfrac{1}{2} \dfrac{d\left[NH _3\right]}{d t}$
(d) Rate $=-\dfrac{d\left[N _2\right]}{d t}=-\dfrac{d\left[H _2\right]}{d t}=\dfrac{d\left[NH _3\right]}{d t}$
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Answer:
Correct Answer: 33. (a)
Solution:
- For any general reaction,
$ \begin{aligned} & a A+b B \longrightarrow c C+d D \\ & \text { Rate }=-\dfrac{1}{a} \dfrac{d[A]}{d t}=-\dfrac{1}{b} \dfrac{d[B]}{d t} \\ &= \quad \dfrac{1}{c} \dfrac{d[C]}{d t}=\dfrac{1}{d} \dfrac{d[D]}{d t} \\ & \Rightarrow \quad \text { For } \quad N _2+3 H _2 \longrightarrow 2 NH _3 \\ & \text { Rate }=-\dfrac{d\left[N _2\right]}{d t}=-\dfrac{1}{3} \dfrac{d\left[H _2\right]}{d t}=\dfrac{1}{2} \dfrac{d\left[NH _3\right]}{d t} \end{aligned} $
BETA
