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Suppose I have a reaction (Say it is the Haber process):

$$\ce{N2 + 3H2 <=> 2NH3}$$

If we have to calculate the average rate of reaction of the above we follow the reasoning the rate of consumption of $\ce{N2}$ is one- third that of $\ce{H2}$ and half that of production of $\ce{NH3}$. We go ahead and divide the change in concentration of $\ce{N2}$ or $\ce{H2}$ or $\ce{NH3}$ by their stoichiometric coefficient. My doubt is why rate of reaction is taken as the rate of consumption of $\ce{N2}$? Why can't it be the rate of production of $\ce{NH3}$? Why can't we say that the rate of reaction is the rate of production of $\ce{NH3}$ or half the rate of consumption of $\ce{N2}$ or one and half times that of rate of consumption of $\ce{H2}$?

Mathew Mahindaratne
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Sharad1
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    See wikipedia's article on reaction rate – MaxW Feb 08 '20 at 20:23
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    If you added a space after each punctuation, and perhaps a line break somewhere, this would become a lot more readable. Btw. if you put chemical formulas into curly braces with backslash "ce" in front, they look a lot better. – Karl Feb 08 '20 at 20:37
  • The Haber process is reversible (corrected). The last phrase of question is "...one and half times that of rate of consumption of $\ce{H2}$" (not $\ce{NH3}$, whch is also corrected). If that's not what you have intended, you may change the corrections back. – Mathew Mahindaratne Feb 08 '20 at 20:41
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    The rate of this reaction is the same as the rate of N2 consumption, because N2 happens to have a stochiometric factor of 1. – Karl Feb 08 '20 at 20:44
  • See also https://chemistry.stackexchange.com/questions/38167/do-coefficients-matter-in-rate-law and https://chemistry.stackexchange.com/questions/32304/law-of-mass-action – Curt F. Feb 08 '20 at 20:51

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My doubt is why rate of reaction is taken as the rate of consumption of N2?

It's done that way because rates have units, and most people don't pay enough attention to the units. Usually in chemistry, rates are units of moles per unit time (per unit volume or per unit catalyst mass or per unit surface area or something). But you need to know moles of what?

By writing the reaction as $\ce{N2 + 3 H2 <=> 2 NH3}$, you are implicitly defining that a mole of "reaction" corresponds to one mole of $\ce{N2}$, two moles of $\ce{NH3}$, and three moles of $\ce{H2}$. If you wrote the reaction as $\ce{\frac{1}{2}N2 + \frac{3}{2}H2 <=> NH3}$, then one mole of this new reation would correspond to one mole of $\ce{NH3}$, but only half a mole of $\ce{N2}$, etc.

Why can't it be the rate of production of NH3? Why can't we say that the rate of reaction is the rate of production of NH3 or half the rate of consumption of N2 or one and half times that of rate of consumption of H2?

It could be any of those things. The choice is up to you. You just have to be consistent. You might this related chem.SE question and answer helpful.

When you say "the rate", you have to specify what rate you are referring to. And if you talk about rates "of the reaction", the agreed upon reference is that you are counting moles of species that have stoichiometric coefficients of 1.

Curt F.
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    "The choice is up to you" -1 // "The rate" has a definition just as assigning coefficients to the chemical equation does. The equation $\ce{\frac{1}{2}N2 + \frac{3}{2}H2 <=> NH3}$ has not assigned the coefficients "properly." – MaxW Feb 08 '20 at 21:21
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    What I try to do when I answer questions is to explain things logically and to dispel confusion. Your strategy might be more focused on what you feel is "proper". Multiple strategies for answering could be effective, so I urge you to consider writing your own. – Curt F. Feb 08 '20 at 21:29
  • It depends how we define the rate,so we have different values of $k$ – Adnan AL-Amleh Feb 09 '20 at 02:20