First of all, I would like to make clear that I don't know very much about chemistry, and I'm not very sure this is the right place on Stack Exchange for this question, so please let me know if it needs migration.

I was reading a PDF about differential equations that mentioned equilibrium reactions and where the law of mass action was used. I looked it up on Wikipedia then looked up reaction rate because I was unfamiliar with the concept, and went to its formal definition.

How does the derivative of a concentration with respect to time even make sense? I seems to me that the concentration of a substance is discontinuous in time and increases or decreases in steps.

Imagine a reaction

$$\ce{aA + bB <=> cC}$$

where $A,B,C$ are reactants and products and $a,b,c$ are stoichiometric coefficients. Now, we see that:

$$\ce{\frac{a}{N_A}A + \frac{b}{N_A}B <=> \frac{c}{N_A}C}$$

where $N_A$ is Avogadro's Number. These are the smallest stoichiometric coefficients for this equation. This reduced equation happens $N_A$ times in the above equation, considering it has an efficiency ($\eta$) of $100\%$. And this reduced equation takes some short time $\Delta t$. Before this reaction, the concentration of $A$ is $[A]_0 = \frac{\nu_A}{V}$ and after it, $[A]_{\Delta t} = \frac{\nu_A'}{V} = \frac{\nu_A - a/N_A}{V}$. I see no in-between that could link these two steps. How could we even define the concentration during the reaction? The only way I see it makes sense it to define the concentration on the interval $(0, \Delta t)$ as $[A]_t = [A]_0$. There is clearly a jump at $\Delta t$. And this doesn't happen only once. The graph of concentration vs time looks like a very compressed and scaled down floor function.

What am I missing?

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    $\begingroup$ You start with molecular dynamics, then approximate with stochastic PDEs, then average and approximate with PDEs. Eventually, you assume spatial homogeneity and approximate using ODEs. Which is why you may have read that the reaction is taking place while under magnetic stirring. If the number of molecules involved is small, then you use stochastic models, not ODEs. $\endgroup$ – Rodrigo de Azevedo Apr 15 '20 at 9:56
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    $\begingroup$ You seem to be assuming that, since chemistry happens one molecular interaction at a time, there cannot be a continuous equation describing the rate. But chemists work on large quantities of molecules at a time (1 mole is ~10^26 molecules) this means that most reactions are very well approximated by a continuous mathematical function (unless you can measure rates to 26 decimal places). $\endgroup$ – matt_black Apr 15 '20 at 10:04
  • $\begingroup$ So, everything is just approximated? Do chemists consider the graph of concentration vs time as the best linear fit? $\endgroup$ – Stefan Octavian Apr 15 '20 at 10:16
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    $\begingroup$ Regarding concentrations as continous is an extremely good approximation, unless perhaps if you try to model biochemical processes on a single cell level. (or are interested in the molecular/atomistic mechanism of the reaction obviously) $\endgroup$ – Karl Apr 15 '20 at 11:11
  • $\begingroup$ Additional Information There is another aspect to consider. For an elementary reaction the forward rate, $r_f$, would be given by the equation: $$r_f = k\ce{[A]^a[B]^b}$$ where $k$ is a constant. However in reality most reactions are not elementary and the coefficients $a$ and $b$ must be determined experimentally as well as the rate constant $k$. $\endgroup$ – MaxW Apr 15 '20 at 16:04

There are two ways to look at this. In the first, we note that all thermodynamic and kinetic analyses rely on the assumption of very large populations of molecules. The curve of concentration vs time is made of discrete points, but the large population ensures that the change in concentration resulting from addition or subtraction of one molecule is so small that it can be treated as infinitesimal. Thus, we can create a smooth curve by interpolating between points, and the result will be continuous and differentiable. From a practical standpoint, any experiment to determine the concentration will result in discrete points separated by much much more than the change of one molecule, so we are implicitly assuming that the behavior between points can be treated as a well-behaved continuous curve.

The second, and perhaps more mathematically rigorous, view is that we never differentiate the actual concentration v time "curve". Instead, we are claiming that the change in concentration over time can be represented by a function which is continuous and differentiable. The only difference between this and the first view is semantic. Does $[A](t)$ refer to the actual concentration of A or to a function that is a proxy for the concentration? It doesn't matter.

Either way, the key point is that traditional kinetic and thermodynamic analyses assume a population large enough that the discrete behavior can reasonably be treated as equivalent to a continuous function.

[On a side note, there are those who would extend your argument to the $dt$ part of the derivative, as there is no proof that time is infinitely divisible, but that's a discussion for another time and place]


I will try to present an answer, to the question, 'How does the derivative of concentration make sense?' simply from the perspective of advanced mathematics, as the question directly relates to the concept of 'derivative'.

Now, yes, delta, a measure of a finite change in concentration with time does NOT exactly equal the first derivative of concentration (relating to instantaneous change). Clearly, an approximation, but here, courtesy of the Calculus of Finite Differences, is the mathematical link between the first difference (delta) in an interval of change 'h' (not necessarily approaching zero) which can be derived, by the way, from the so-called Taylor series:

$\ce{f(x + h) - f(x) = hf'(x) + {h^2}/2!f''(x) + {h^3}/3!f'''(x) + ...}$

Source: An online available ebook, Page 13 at Calculus of Finite Differences - Charles Jordon.

Note, if one divides the equation through by 'h' and takes the limit as h approaches zero, we have the calculus definition of the first derivative as other terms involving h disappear.

So, the incremental change in the concentration can be approximated by a series expansion involving terms of higher derivatives, and as a very rough approximation, one could just use the first term only.

The appropriateness of this approximation is arguable depending on reaction particulars.


Let us assume that we can view molecules as point particles. We uniformly discretize space into "small" boxes of volume $\Delta x \Delta y \Delta z$ and assume that this volume is "sufficiently small", though not "too small" — otherwise we can no longer view molecules as point particles.

Take a snapshot. We have many adjacent boxes and many (though fewer) particles. Most boxes will be empty. Some will have $1$ particle. Fewer will have $2$ particles. Even fewer will have $3$ particles. Take the total mass of the particles in each box, divide by $\Delta x \Delta y \Delta z$ and call it concentration. Note that the concentration at most boxes will be zero.

Unfreeze time. Let a "tiny" amount of time elapse and take another snapshot. The concentration at most boxes will still be zero, but since the particles are moving, different boxes will be empty. Hence, the concentration at a box does indeed undergo discrete changes over time. It jumps up and down. The set of admissible concentrations should be discrete, as each particle has an integer number of protons, neutrons and electrons.

However, over a "small" — yet "sufficiently long" — period of time, each box will be "visited" by "many" particles. Hence, though the concentration at each box undergoes discrete changes, averaging over a period of time, the concentration will appear to be continuous, somewhat smooth and slowly-varying. This concentration can be differentiated with respect to time. Differentiating before averaging is problematic, however.


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