The basic physical definition of speed is "distance/time", and the rate of a reaction can be defined as "concentration/time".

However, how is distance (d) mathematically transformed into concentration (C) to go from the first definition to the second (reaction rate)?


I've been messing around with variables and might have found the answer. If speed (units: m/s) is multiplied by area (m^2), then multiplied by density (g/m^3), then divided by molar mass (g/mol), then divided by volume (m^3) we get to reaction rate (mol/L*s). However, I'm not sure if this is what was expected of me by my professor.

  • 4
    $\begingroup$ I'm not really sure what you are looking for here. Speed doesn't really have anything to do with reaction rate. $\endgroup$
    – bon
    Commented Jan 14, 2018 at 17:42
  • $\begingroup$ According to my Chemical Equilibrium Professor, it is possible to start from the first definition, do math, and obtain the second. He said it has to do with Fugacity and Activity. $\endgroup$
    – Sam202
    Commented Jan 14, 2018 at 18:10
  • 1
    $\begingroup$ You either grossly misunderstood you teacher, or he is fooling you. If he said what you say he did, i suspect the latter. Or you (not the teacher) are just months behind on the curriculum. Catch up! $\endgroup$
    – Karl
    Commented Jan 14, 2018 at 18:49

1 Answer 1


The speed (or rate) of a chemical reaction is by definition how frequently it happens. Since it is happening everywhere in our reaction container, we describe it using how many times it happens per unit volume per unit time. Since the number of times it happens to molecules is enormous, we use the unit mole to do the counting. So if a chemical reaction is happening at $1\,\mathrm{mol/(L\cdot min)}$, it means in $1$ minute, $6.022\times 10^{23}$ reactions happened in every liter of reaction space.

For example, if we have a reaction

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

when every $\ce{N2}$ molecule combines with $3$ $\ce{H2}$ molecules to form $2$ $\ce{NH3}$ molecules, we write down one such reaction equation. Then if we find ourselves writing down $1\,\mathrm{mol}$ of such chemical reaction equations for all reactions that happend in $1$ liter of volume and $1$ minute of time, we say the speed of the reaction is $1\,\mathrm{mol/(L\cdot min)}$. In a fixed volume this means the concentration of $\ce{N2}$ is reducing at the speed of $1\,\mathrm{mol/(L\cdot min)}$, the concentration of $\ce{H2}$ is reducing at the speed $3\,\mathrm{mol/(L\cdot min)}$, and the concentration of $\ce{NH3}$ is increasing at the speed of $2\,\mathrm{mol/(L\cdot min)}$. If the volume is not fixed, or if there is extra source of these materials, then the change of concentration is not only due to reactions, so the speed of reaction is not measured by how quickly the concentrations change, but is still measured by how many reactions happen per unit volume per unit time.

Another point to make is that the overall coefficients matter. For example, if the above reaction is happening at $1\,\mathrm{mol/(L\cdot min)}$, then the reaction below,

$$\ce{\frac{1}{2}N2 + \frac{3}{2}H2 <=> NH3},$$

is happening at twice the speed, i.e., $2\,\mathrm{mol/(L\cdot min)}$. This is because $1$ reaction above is equivalent to $2$ reactions below. The same goes for thermochemical equations.

  • $\begingroup$ My Chemical Equilibrium Professor says there's a way to start with the first definition (d/t), do math, and end up with the second definition (C/t). He also mentioned it is related with Fugacity and Activity. $\endgroup$
    – Sam202
    Commented Jan 14, 2018 at 18:09
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    $\begingroup$ @Sam202, then perhaps he's trying to calculate how many successful collisions there are based on thermodynamic properties of the gases and statistical physics, to link the "speed" in physics (how quickly molecules move) to the "speed" in chemistry (how frequently reactions happen). But these are different concepts with different definitions. The former determines the latter given some model. $\endgroup$
    – Zhuoran He
    Commented Jan 14, 2018 at 18:21
  • $\begingroup$ It very well could be something along those lines; however, this question was given as the first assignment of the course, and he said it wasn't too complicated. I attempted to use dimensional analysis to try to find factors of units that would lead from meters/seconds to moles/(Liters * seconds), but that would require a factor of (moles/(Liters*meters)), which doesn't translate to other variables (or at least none of the ones I can think about). $\endgroup$
    – Sam202
    Commented Jan 14, 2018 at 18:43

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