I stumped the AP Chemistry and AP Calculus teacher: Where are the units in this calculation?

I’m the AP Chemistry teacher, btw. I spent 20 minutes with the Calc teacher, and we can’t find the units in this calculation. Where are they?

We discover that our chemical reaction is a first order reaction, so rate = k[A] (This happens before the math shown.) A is the reactant. k is the constant that makes the 2 sides actually equal. t is time after the reaction starts. Brackets shows that the value of A is the concentration of A in water, in moles/L, which is molarity, or M. The integrated rate law is ln[Ao]=ln[At]+kt. This lets us find molarity at a certain time if we know the concentration at another time.

My contention is that the brackets on the left side only indicate that we should find a M unit when we solve for it, not that it has M as its unit. It is a concept, not a measurement, and I should find a number with the correct unit on the right, to which the unknown is equal. A chem teacher I emailed about this seems to be saying that there is an M unit on the ln[At] value, but then I couldn’t add it to the other number on the right side, because those units have canceled out.

*I just now talked this through with my top math/science student (10th grade), and his idea is that taking ln of the molarity on the right side removes the unit, allowing me to add it to the unitless number, but then taking the inverse ln of the right side restores the unit. It’s the best idea I’ve heard on this, but if true, it’s news to me.


Use t = 31.28min, k = 0.0376/min ln[At] at 31.28min = -0.329
The actual chemical is N2O5

ln [N2O5]0 = ln [N2O5]t + k · t
ln [N2O5]0 = (–0.329) + (0.0376/min)(31.28min)
= 0.847
[N2O5]0 = e^0.847 = 2.33

Taking a logarithm of anything with units is… problematic. You can come up with a mathematical formalism to make it work, but why bother? It’s usually much more straightforward to re-arrange the equation so that you’re only ever taking logarithms of dimensionless ratios. Using your equation of ln[Ao]=ln[At]+kt, I’d re-arrange that to kt = ln[Ao/At]. Now, whatever units you use for A, as long as it’s the same for both, it cancels out and doesn’t matter.

This blog post is exactly on point: log, ln, units and activities.

The second to last equation on the page is exactly what you are talking about.

Also… your post would be a lot easier to read with proper use of sub and sup tags. e.g…

ln ([N[sub]2[/sub]O[sub]5[/sub]][sub]0[/sub] / 1 mol L[sup]-1[/sup]) = ln ([N[sub]2[/sub]O[sub]5[/sub]][sub]t[/sub] / 1 mol L[sup]-1[/sup]) + k · t

First Order:
rate = k[A]
M/t = k M

k units are: s[sup]-1[/sup] or min[sup]-1[/sup]

Second Order:
rate = k[A][sup]2[/sup]
M/t = k M[sup]2[/sup]

k units are: M[sup]-1[/sup]s[sup]-1[/sup] or M[sup]-1[/sup]min[sup]-1[/sup]

Yes, part of our problem is that the unit appears when it’s arranged that way, but not in the way it was presented in the textbook. I’ve never seen that. Should I just present it in the fraction form? It bothers me that this doesn’t work both ways, though.

k is a reaction rate constant, being unitlles once a reaction reaches a state of equilibrium is simply an artifact of the coordinate system chosen.

Just as your longitude becomes undefined at the north pole, the rate of your approach to equilibrium becomes undefined once you reach equilibrium.

No, the rate of approach to equilibrium is perfectly well defined at equilibrium. It’s zero.

The point being, no matter what units you use for your rate it doesn’t matter that the units become undefined.

The formula I posted above work fine for the approach, the required applied mathematics don’t require moving to another coordinate system that would preserve the units at k.

order increases by 1, we divide the previous k unit by M and if you see the progression.

0: M/s
1: 1/s
2: 1/sM
3: 1/s

Thus, for order = n:

k units = 1/s*M[sup]n-1[/sup]

Also consider what is being tracked with the ingratiated form:

0: [A][sub]0[/sub] - [A][sub]t[/sub] = kt
1: ln([A][sub]t[/sub]/[A][sub]0[/sub]) = - kt
2: (1/[A][sub]t[/sub]) - (1/[A][sub]0[/sub]) = kt

Which lets you find the linear plots:

0: [A] over time
1: ln([A]) over time
2: (1/[A]) over time

Which will get you to the same units I gave above by finding the slope.

Chemical Engineer here and have dealt with this problem all my career.

IMO - You unnecessarily complicate things when you deal with molarity instead of molality. It is the same with cooking and recipes when people express components in volumes rather than weight (mass).

The order of a reaction in strict terms depends on the activity of the component. The activity can be represented by concentration, partial pressure, etc. etc. The best way (IMO) is to use Moles of reactant / Moles of solvent - because volume changes with temperature.

I hope you see how easy the equation becomes once you express the concentration of N2O5 in Mols of N2O5 / Mols of water. Then the only units you are dealing with is time - which is the whole idea of learning reaction order.

I have a big big gripe about this. This whole concentration things Mols/liter, grams / liter, lbs/ gallon - leaves students doing unit conversion most of their valuable time and prevents them from learning the actual concepts. This actually turns off a lot of people away from chemistry (and also cooking btw).

Sorry - I messed up the above post.

The best way to represent concentration is Bu the mole fraction. That is the (mole of reactants)/ total Moles.

This makes all calculations easy and intuitive . Because fundamentally, concentration is a dimensionless number.

I didn’t understand one word of this thread, but I loved reading it because it demonstrates one of the things I love most about the SDMB: there are so many smart people here.

Carry on.

The rate constant does not change as the reaction proceeds

I was referencing the units issue once you reach the end of the reaction, obviously a ratio of K won’t change.

FWIW, here is a link that will go over the method I outlined above.

That page will have good syntax, uses the integration method to find the slope and reaches the same conclusions I did.

It is basic calculus to find the unit on a Slope.

Rate constants do not become unitless at equilibrium.

Well, one could make a case for instead measuring concentrations in mass per mass, given that masses are what one actually measures. Or, if it’s convenient to compare numbers of particles, you could just skip the middleman and keep track of numbers of particles, instead of number of particles divided by Avogadro’s number, and just accept that you’re going to be using scientific notation. But yeah, comparing things by volume is odd, since volume can change even without changes in pressure or temperature (as we’re reminded by the classic experiment of mixing water and alcohol).

I’ll spitball a summary from a safe distance.
The OP has moles and wants the same number as on the other side of the fence. The rat poster put in an order for more. Then the super secret science code started and I lost track of the moles.

The rate of a reaction has dimensions of concentration per unit time, and neither mass[sub]0[/sub] nor volume is conserved quantity.

As the equilibrium constant - K is a rational constant, as long as you make sure to use the same method of determining the concentration those units really just don’t matter at all as what you are measuring is the shape of reaction.

If I had to guess, Molarity is probably used because graduated cylinders are plentiful and cheap and the precision isn’t high enough during labs to care about variations due to ambient temperature. As the mass[sub]0[/sub] changes during all reactions, whether chemical or nuclear; the property that most people think is invariant would not be in this case too. That removes the advantage of typically using inertial mass in other domains IMHO.

That said, for the OP the correct answer is answered through calculus and not algebraic rules. The units of the function are what are important in this context.

Thank you for reading the thread. Allow me to simplify things.

A mole is like a dozen - a very large dozen. Just like it makes sense to count eggs by dozen, not their weight (you don’t buy 2 lbs of eggs) , not their volume (you don’t buy half a gallon of eggs) - it makes sense to count molecules by dozens too ( the dozen in this case is called a Mole or Avogadro number which is 6.023 x 10^23).

So with that in mind, lets look at a simple setup that you want to study :

You have 10 dozen chicken eggs incubating in a tub of 100 dozen marbles. Every hour a dozen chicken eggs hatch and the chicks walks out of the tub. You want to plot the eggs remaining in the tub on a graph. Concentration of eggs is some measure of how many eggs there are divided by some measure of how many marbles there are. Lets look at how Chemists will measure this situation :

  1. Concentration in Dozen Fraction (Mole Fraction) - This is very intuitive. At the start, the dozen fraction of eggs is 10/100 or 0.1 and every hour it reduces like 9/100, 8/100, 7/100, 6/100 …

2.Concentration in Dozenality (Molarity) :

Volume of one marble is about 2 cm3. (0.07 fl oz)

So the volume of marbles is 2 x 12 x100 = 2,400 cm3

So the concentration of eggs at the start is = 10 dozen / 2400 cm3 = 0.0042 dozen /cm3

So per hour the concentration of eggs (in dozen eggs per cm3 ) goes like 0.0042, 0.00378, 0.00336 …
There other ways of expressing this concentration too. Say like dozAnility (Molality) - where you divide the dozen of eggs by the weight of the marbles.

There are other ways too - like mass fraction = weight of eggs divided by weight of marbles. Or volume fraction = volume of eggs divided by volume of marbles.

**My point is that chemistry students are taught with all sorts of concentration measurement systems like shown above where they end up doing useless math when they should be counting chickens. **

A mole of gas at standard conditions measures 22.4 liters or 1 cu ft approx. This is the typical amazon box of 1 x 1 x 1 ft.

So consider the reaction N2O --> N2 + 1/2 O2.

This tells us that a amazon box full of Laughing gas (N2O) will give us 1 box of Nitrogen (N2) and half a box of oxygen (O2). - ** Intuitive **

In mass terms, it tells us that 44 grams (1.6 oz) of N2O will give us 28 grams (1 oz) of nitrogen and 16 grams (0.6 oz) of oxygen. ** Not Intuitive **

In the case of the OP, using moles can also tell you the rate of formation of products. Or in the case of hatching chickens, the dozens measurement can easily account for weight gain of the hatched chicken with time.

One can do that but again it is not intuitive. A mole of gas measures about 1 ft3 is intuitive and units and measurements have tried to be intuitive. The inch or cm can be related to body parts. People are used to measuring large items in dozens or bushels or the like.

I think the whole decimal number system is designed to be intuitive. So you count by 1s when it makes sense and by 1000s when it make sense.

Since you are a physics person, you know that in relativistic physics, it makes more sense to express velocities as a fraction of c like 0.2c because it is intuitive than in say m/s or ft/s.

Agree fully. It defeats the whole purpose of tracking chemical changes .