# Are the units of rate constants affected if we have more than one substance in the rate law [closed]

$$A+B \rightarrow C$$ $$rate = k[A] [B]$$ [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?

• 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? Jan 11, 2018 at 12:53
• 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 $\ce{dm^{3}mol^{-1} s^{-1}}$ Jan 11, 2018 at 16:44

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 $y$.