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For example, the rate of a chemical reaction can be expressed in $\mathrm{mol}/\mathrm{L}^{-1}/\mathrm{sec}^{-1}$. Why is it ‘−1’ and not, say, ‘−2’? Does it change the meaning if the minus is removed and we simply express the rate in $\mathrm{mol}/\mathrm{L}/\mathrm{sec}$?

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    $\begingroup$ The answers below are correct but none seem to mention that in math $x^{-1}$ equals $\dfrac{1}{x}$ for some variable $x$. The same thing applies here. $\endgroup$ – Calvin's Hobbies Oct 21 '14 at 21:51
  • $\begingroup$ @Calvin'sHobbies while my answer doesn't say that explicitly, it does say it implicitly with the depiction of the example as a fraction. $\endgroup$ – John Snow Oct 21 '14 at 22:09
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    $\begingroup$ Please note that a solidus (/) shall not be followed by a multiplication sign or a division sign on the same line unless parentheses are inserted to avoid any ambiguity. Besides, the unit symbol for ‘second’ is ‘s’ (not ‘sec’). $\endgroup$ – Loong Apr 9 '15 at 16:07
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The -1 means "per" unit. So your first example mol/L-1/s-1 is not correct - it would actually be written as mol L-1 s-1, OR mol/(L s). It is also sometimes written as mol/L/s, but the double division is ambiguous and should be avoided unless parentheses are used.

If it were mol L-1 s-2, this would mean moles per litre per second per second.

This is really just a question of notation, and is not chemistry-specific at all. Yes, all the minus/plus signs and the value of numbers are important. Good examples of units can include:

  • area, measured in m2, or metres squared
  • volume, measured in m3, or metres cubed
  • pressure, measured in N m-2, or Newtons per metre squared
  • velocity, measured in m s-1, or metres per second
  • acceleration, measured in m s-2, or metres per second per second
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The $^{-1}$ superscript can be thought of as saying "per" or as being the denominator of the fraction.

So in your example $\mathrm{mol \cdot L^{-1} sec^{-1}}$ can be thought of as saying moles per liter per second.

This is easier than writing $\mathrm{\frac{mol}{(L \cdot sec)}}$

Changing the super script from $1$ to $2$ or $3$ would change the meaning of the value.

Ex

$$1 \mathrm{cm^{3}\ is\ 1mL}$$ So, $\mathrm{cm}^{-1}$ is per centimeter, which would be a measurement of something per distance, but $\mathrm{cm^{-3}}$ would be talking about something in a given volume.

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  • $\begingroup$ Generally correct, but fails to mention that the unit abbreviation for the second is simply s, not sec. $\endgroup$ – Jan Nov 18 '17 at 2:31
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It may have its roots even earlier than that, but this was mainly due to people using typewriters to write scientific papers, etc.

Now we have the ability to format things like $\mathrm{\frac{mol}{L}}$, both on-screen and in print, but adjusting the carriage and line feed knob every time you had to type a complicated formula was tedious, so it was easier to type "mol-L-1" instead. Even when the -1s became superscripts, as John points out in his answer, it was still used in typesetting to keep formulas, etc. all on the same line in books.

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    $\begingroup$ Even if we do not use typewriters anymore, an inline fraction just looks plain awful and makes a manuscript very hard to read, since there will be different spacing between lines in a single paragraph. $\endgroup$ – Martin - マーチン Oct 22 '14 at 9:10
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First off: your suggestion $\require{cancel}\cancel{\mathrm{mol/L^{-1}/sec^{-1}}}$ is very wrong for three principal reasons:

  • the unit symbol for seconds is $\pu{s}$, not $\pu{sec}$ or anything else
  • you should never include two slashes for division. Does $\mathrm{mol/l/s}$ equal to $\mathrm{mol/(l/s)}$ or to $\mathrm{(mol/l)/s}$? This is ambiguous. One should always indicate with brackets which units are ‘per’ and which are not; in your example it should be $\pu{mol/(l\cdot s)}$.
  • your suggestion does not mean what you think it means; more on that below.

Mathematically, a negative exponent has the same effect placing the expression associated with it into the denominator.

$$\begin{align}x^{-1} &= \frac 1x\\[0.3em] 2^{-2} &= \frac1{2^2}\\[0.3em] e^{-i\phi} &= \frac1{e^{i\phi}}\end{align}$$

Units in the natural sciences are treated much like variables in general mathematics, i.e. they can be multiplied and thereby raised to powers (e.g. $\mathrm{m^2}$) or divided by each other (e.g. $\mathrm{m/s^2}$). Only if the unit is identical, two numeric values can be added or subtracted; so $\pu{2m}+\pu{3m}=\pu{5m}$ makes sense as does $2a + 3a=5a$, but $\pu{2m}+\pu{3s}$ cannot be added akin to $2a+3b$.

The combination of units usually means what common sense would read them as. So $\pu{1m^2}$ is equivalent to a square area with the side length being $\pu{1m}$. $\pu{1 N\cdot m}$ is equivalent to a force of one newton applied over the distance of 1 meter (with a lever). And $\pu{1m/s}$ means travelling one meter per second. While more complex expressions such as $\mathrm{kg \cdot m^2 / s^2}$ do not always immediately make intuitive sense, they can usually be broken down into fragments that would make intuitive sense.

After this excursion, it becomes clear that an expression such as $\pu{mol\cdot l^-1\cdot s^-1}$ is equivalent to a fractional unit of $\mathrm{\frac{mol}{l\cdot s}}$, meaning that the concentration is increased by $\pu{1 mol/l}$ in one second. This also means that:

  • it does not make sense to replace the exponent of $-1$ with e.g. $-2$ as that would result in a different unit (e.g: $\mathrm{kg\cdot m^2\cdot s^{-2}}$ is joule, the unit of energy, while $\mathrm{kg\cdot m^2\cdot s^{-3}}$ is watt, the unit of power).

  • it does not make sense to remove the negative sign from the exponent as that would result in a different unit (e.g. $\pu{10Hz} = \pu{10s-1}$ corresponds to a frequency — ten times per second — while $\pu{10s}$ obviously corresponds to a duration).

  • one has to choose between either the slash or the negative exponent as both would cancel each other out.

This last one is implied by the general laws of mathematics: $$\begin{align}\frac1{x^-1} &= \frac1{\frac1x}\\[0.5em] &= \left(\frac11\right) / \left(\frac1x\right)\\[0.5em] &= \left(\frac11\right) \times \left(\frac x1\right)\\[0.5em] &= x\end{align}$$ which is also the third wrong factor in your suggestion.

In general, I would give preference to the negative exponents ($\pu{mol l-1 s-1}$) except in cases where there is only a single unit raised to a power of $-1$ and no other powers exist; in these cases, e.g. $\pu{mol/l}$ usually integrates itself better into the flow of text.

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