$$A+B \rightarrow C$$ $$rate = k[A] [B]$$ [A]= x mol of A/L [B]= y mol of B/L So why the unit of the rate constant is merely $$L/M$$ not having mol of A or mol of B?

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    $\begingroup$ Why the unit of weight is merely a kilogram, and not "kilogram of lead" when you weigh lead, or "kilogram of fluff" when you weigh fluff? $\endgroup$ Jan 11, 2018 at 12:53
  • $\begingroup$ The rate has units of concentration/time or $\ce{mol dm^{-3}s^{-1}}$. As this is an equation the left had side equals the right hand side both in value and units thus the rate constant $k$ has units 1/(concentration x time) or $\ce{dm^{3}mol^{-1} s^{-1}}$ $\endgroup$
    – porphyrin
    Jan 11, 2018 at 16:44

1 Answer 1


The heart of your problem is that you're assuming that there exist different types of the unit mole.

Consider the unit of length one meter. Take two meters of cotton thread, and two meters of jute thread. Place them together and you get a nice square of $1\times1 \text{ <unit of area>}$. Now what should the unit of area be? Should it be $=\text{cotton meter} \times \text{jute meter}$?

Well, it should NOT be $\text{cotton meter} \times \text{jute meter}$ because a meter, being a unit, is same for all substances being measured. A meter of jute is exactly the same length as a meter of cotton. Hence, the unit of area should be $\text{meter}^2$, which it always is.

Similarly, in your solution, you are assuming there exist two different types of the unit mole, one for $x$ and one for $y$. However, there is only one fundamental unit, and it measures the same amount of substance for both $x$ and $y$.

Hope it helps!


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