6
$\begingroup$

Just to be clear, I do understand that the units of the rate constant k is selected to make the equation dimensionally consistent. That is not what I am asking.

It should be the case that k 'means' something physically, which is implied by its dimensions/units. And I would also assume that what k 'means' should not be dependent on the order of the reaction, since I cannot see any obvious reason why it should. But we know that the overall order of a reaction changes the units of the rate constant, so I must be wrong.

My question is as stated: Why do the dimensions of the rate constant change depending on different reaction orders, and what does that physically mean?

As an extension to this, I would also like to understand how the same thing works for the equilibrium constant, where the stoichiometric coefficients of the species or the number of products/reactants change the units of the constant. Again, I cannot see the reason why these would change what the constant means dimensionally.

Sorry if this an obvious question. I am not a chemist.

Edit: I wanted to clear up a few things. First of all, as I understand I was expecting there to be a deeper meaning to the dimensions of the rate constant even though there isn't always an obvious one, since it is an empirical equation which merely simplifies a process rather than describing the fundamental workings of it. But I still want to understand intuitively (if there is indeed an intuitive reason) why the units are different for each overall order. As in the examples Maurice gives in his answer, a reaction of order zero produces dimensions of amount per time whereas a first order reaction produces [ratio] per time and so on for other real number orders. It just seemed strange to me that a constant in the equation, which seems to be doing the same job regardless of order, express something in completely different dimensions depending on order.

After reading your answers, this is what I thought: In a zeroth order reaction where the rate is constant, amount per time seems to be the only reasonable way to describe the rate since ratio per time inevitably produces a curve. Similarly for a first order reaction, amount per time wouldn't work because this value changes constantly. So, the constant is expressed in terms of the units which can describe the reaction given its nature. This was stated/suggested in some replies but thinking of it over an example is what cleared it up (if it is correct) and thank you for your responses. Perhaps this was obvious but a constant with varying units was not something I was familiar with.

$\endgroup$
10
  • 2
    $\begingroup$ The units of the "equilibrium constants" $K_c$ or $K_p$ have no meaning. The One True Equilibrium Constant $K$ (note the lack of subscript) is defined in terms of thermodynamic activities and is dimensionless, see Which equilibrium constant is appropriate to use? or Unit of the equilibrium constant: contradiction of Bridgman's theorem? $\endgroup$ Jan 14, 2021 at 14:43
  • 3
    $\begingroup$ think he's talking about rate constants in kinetics and not in static equilibria $\endgroup$
    – Bertram
    Jan 14, 2021 at 16:02
  • $\begingroup$ @Bertram Thanks, but the second last paragraph certainly talks about equilibria. I don't pretend that my short comment is good enough to be an answer that addresses the complete question, I just wanted to post a couple of extra links that addressed that part. $\endgroup$ Jan 14, 2021 at 18:18
  • 1
    $\begingroup$ I seems the number of answers here rises with lack of clarity as usual. $\endgroup$
    – Mithoron
    Jan 14, 2021 at 20:33
  • 1
    $\begingroup$ @Mithoron :)) I think this is rooted in the way we, even the chemists, start learning. But I won't recommend teaching chemical potential and activities before of solutions and concentrations. $\endgroup$
    – Alchimista
    Jan 15, 2021 at 9:25

5 Answers 5

4
$\begingroup$

Maybe it would make more sense if rate constants would have different symbols depending on whether it is a zero, first or second order reaction.

As an analogy, we might be asking "how big" an object is depending on the length of it. Depending whether the object is, say, a square, a cube or a hypercube, the answer would be in terms of its area, volume or 4D-volume, and the units (and dimensions) would be different in each case.

For the equilibrium constant, we avoid the issue by switching from concentrations to either concentrations divided by a standard concentration (this is a simplification) or to activities (the former multiplied by an activity coefficient that accounts for non-ideality).

Why do the dimensions of the rate constant change depending on different reaction orders, and what does that physically mean?

They have to change with the chosen system of units so that the equations are consistent. We could choose instead to have a dimensionless measure of concentration when we set up the kinetic equations. This would describe the reality equally well, but the kinetic constants would always have the same dimensions (1/time).

According to this argument, there is no physical meaning in rate constants having distinct dimensions depending on the order of reaction.

$\endgroup$
4
$\begingroup$

The physical reason why the dimensions of k differ for different reactions is the different mechanism behind the reaction.

A process that only involves the internal vibration or rearrangement of one species differs from a process in which two or more equal or different species must first diffuse and collide*.

It seems intuitive to me that this should be reflected via power relationship between concentrations/activities and reaction rate,

or alternatively

as for the rate of reaction has dimensions of amount of substance over time, the dimensions of k forcedly must change for reactions of different order.

Finally, concerning the equilibrium constant K, consider that at equilibrium the rate of forward and back reactions are equal. This shows that the true equilibrium constant is a dimensionless quantity kf / kb.

Apparent dimensions occur because, operatively, concentration is used, but molar ratio units could be chosen instead.

*note that different mechanisms pose different upper limits to the reaction rate.

$\endgroup$
2
  • 2
    $\begingroup$ Hmm, I don't agree re. eqm constant $K$. Both the forward and backward rate constants $k_\mathrm{f}$ and $k_\mathrm{b}$ have units which include concentration and don't necessary cancel out; the units of $k_\mathrm{f}/k_\mathrm{b}$ are analogous to $K_c$, which is in general not dimensionless. $\endgroup$ Jan 14, 2021 at 15:37
  • 1
    $\begingroup$ @orthocresol isn't the same, you can express kinetic laws in terms of mole fraction as well. Actually what is true is alway the activity, both expressing equilibrium as well as the kinetics. Physically there is no reason to use [M]. $\endgroup$
    – Alchimista
    Jan 14, 2021 at 17:25
3
$\begingroup$

Maybe a numerical example may help you compare the different orders. Let's take first an example of a reaction of zeroth order : the combustion in air of a candle containing $n$ $mol$ wax. At every second, the same amount of wax is burned. The rate of the reaction is $$r = -\frac {dn}{dt} = k_o·n^0 = k_o$$ If the candle contains $1.44$ mole wax and burns in $2$ hours ($7200$ s), the rate constant is $$k_o = r = \frac{1.44 mol}{7200 s} = 2·10^{-4} mol/s$$ Here $k_o$ is the number of moles burning per second. This number is constant during the whole chemical reaction. Its unit is $mol/s$. If, instead of studying a candle, we are working in solution, $k_o$ is defined in $mol·L^{-1} s^{-1}$

Let's compare this result with a first order reaction, namely the decay of a radioactive isotope . The rate law is : $$r = - \frac{dn}{dt} = k_1·n^1 = k_1· n$$ This can be written so $$- \frac{dn}{n ·dt} = k_1 $$ You should observe that $k_1$ is not the number of moles reacting per second. It is the proportion or the fraction of the initial number of moles that decays per second. $k_1$ has not the same dimension as $k_o$. The unit of $k_1$ is $s^{-1}$.

$\endgroup$
5
  • 1
    $\begingroup$ This answer comes close to what I was trying to address. Is there an intuitive reasoning to 'why' the constant expresses something different (e.g. amount per time, ratio per time, whatever per time etc.) for different orders, and why should it not be the same? Does it simply have to do with different reaction mechanisms for different orders? I'm aware my wording isn't very clear but thank you for the answer. $\endgroup$
    – Aerovolo
    Jan 14, 2021 at 22:59
  • $\begingroup$ @Aerovolo are you comfortable with the fact that rate has always dimensions of amounts of substance over time, because it is what you are looking for? At least this would fix one halve of your question. $\endgroup$
    – Alchimista
    Jan 15, 2021 at 8:28
  • $\begingroup$ @Alchimista Even though the units always have a term relating to amount (or concentration directly) over time, its power changes surely? (e.g. mol^0 and mol^-1 as given) So wouldn't it express something slightly different for different powers (ratio for 0, amount itself for -1, and something else for other real numbers which I'm not entirely sure)? I was wondering what the powers or the fact that they change for different orders signify; if it does at all. $\endgroup$
    – Aerovolo
    Jan 15, 2021 at 9:03
  • $\begingroup$ I don't understand fully the example you give, but say, if something requires two things colliding, it can even be a billiard pool, the number of a red and a white ball collisions change by adding just balls of one colour or both. If the balls have an internal mechanism that causes them to broken in pieces, you can change nothing. The number of balls keeping crashing (as compared to the total of course) will keep its pace. Note that all this is due to the use of concentration as Mol. $\endgroup$
    – Alchimista
    Jan 15, 2021 at 9:21
  • $\begingroup$ @Aerovolo rereading your comment... You must fix Rate not k, while thinking on this. Rate is surely amount of substance over mole, it cannot be anything else. For k, refer to my answer and comments. And also M. Farooq answer that puts my billiard pool in a formal context. A being related to the speed of the balls. The rest mechanism dictated. $\endgroup$
    – Alchimista
    Jan 16, 2021 at 13:27
2
$\begingroup$

The simplest way to approach this is that the rate constant has whatever units that make the rate have the correct units.

Generally, the rate is in units of molar per second, so based on how you're multiplying concentrations (in molar) in the rate law, you should be able to figure out the appropriate units for the rate constant.

$\endgroup$
1
  • 3
    $\begingroup$ I'll add to this response that different rate laws have different differential equations they are associated with, as the physics of the reactions are distinct. Naturally the units of the rate constant would depend on which differential equation modeled the scenario. $\endgroup$
    – Bertram
    Jan 14, 2021 at 16:23
2
$\begingroup$

Well, it may not be appropriate to say that the rate constant, k, has no physical meaning. It is the only factor in the rate equation which is temperature dependent, so larger k means faster rate of reaction. Another way of looking at the units of k is in terms of Arrhenius equation.

$$ k(T)=A e^{-E_{\mathrm{a}} / R T} $$

The factor $A$ is called the frequency factor and it has the same dimensions as the rate constant, because the exponential term is dimensionless. You can see why $A$ is called the frequency factor for any order.

For order $(m + n)$, the rate constant has units of mol$^{1−(m+n)}·L^{(m+n)−1}·s^{−1}$, (see Wikipedia on Reaction rate constant)

so you can associate the units of $k$ as in terms of the units of the frequency factor- roughly the frequency of collisions. It is called frequency because of the second$^{-1}$ factor.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.