[Gradient]

Hi! I am taking an online CLEP chemistry course to prepare for the CLEP exam. The exam covers material you would expect from a first-year chemistry course in college. This link has more info on the exam from the course website.

I just reached this chapter which introduces the idea of the reaction quotient (which I'll label as [itex]Q_c[/itex] for the purpose of this post). I understand how to solve basic problems with reaction quotients, but I am seriously confused by my book's definition. I'll try to explain how the book defines [itex]Q_c[/itex] in clear terms, and then I'll explain what confuses me.


To define [itex]Q_c[/itex], my book gives an example of a reversible reaction:

[tex]m\text{A} + n\text{B} \rightleftarrows x\text{C} + y\text{D}.[/tex]

Each chemical in the formula ([itex]\text{A}[/itex], [itex]\text{B}[/itex], [itex]\text{C}[/itex], and [itex]\text{D}[/itex]) has a certain molar concentration. The book's convention is to write the molar concentration of a chemical in square brackets. For example, the molar concentration of [itex]\text{A}[/itex] would be [itex][\text{A}][/itex].

Now, the book defines the reaction quotient [itex]Q_c[/itex] as a dimensionless quantity. This quantity is calculated by taking the product-to-reactant ratio of the molar concentrations raised to their coefficients' powers:

[tex]Q_c = \frac{ [\text{C}]^x [\text{D}]^y }{ [\text{A}]^m [\text{B}]^n }.[/tex]

The book puts serious emphasis on the fact that [itex]Q_c[/itex] is a dimensionless quantity.


Here's what confuses me. If my understanding is correct, [itex][A][/itex], [itex][B ][/itex], [itex][C][/itex], and [itex][D][/itex] are not dimensionless quantities! They are measured in units of amount per volume (like [itex]\frac{\text{mol}}{\text{L}}[/itex], for example). So, the dimension of [itex]Q_c[/itex] could be something like [itex]\frac{\text{mol}^2}{\text{L}^2}[/itex] or [itex]\frac{\text{mol}^3}{\text{L}^3}[/itex] or even [itex]\frac{\text{L}}{\text{mol}}[/itex]. For example, given the equation

[tex]2\text{NO} + \text{Cl}_2 \rightleftarrows 2\text{NOCl},[/tex]

let's assume we have the molar concentrations shown below:

[tex][\text{NO}] = 0.0500 \frac{\text{mol}}{\text{L}},[/tex]
[tex][\text{Cl}_2] = 0.0155 \frac{\text{mol}}{\text{L}},[/tex]
[tex][\text{NOCl}] = 0.500 \frac{\text{mol}}{\text{L}}.[/tex]

Now, going purely by my book's definition, the reaction quotient for this scenario is

[tex]Q_c = \frac{ \left(0.500 \frac{\text{mol}}{\text{L}}\right)^2 }{ \left(\left(0.0500 \frac{\text{mol}}{\text{L}}\right)^2 \left(0.0155 \frac{\text{mol}}{\text{L}}\right)\right) },[/tex]

which I calculated as [itex]6450 \frac{\text{L}}{\text{mol}}[/itex]. Compare with my book's answer of

[tex]Q_c \approx 6450.[/tex]


Why is it acceptable to toss out units in this way? What if I were given the exact same molar concentrations, but they were written in this form instead?

[tex][\text{NO}] = 50 \frac{\text{mol}}{\text{m}^3},[/tex]
[tex][\text{Cl}_2] = 15.5 \frac{\text{mol}}{\text{m}^3},[/tex]
[tex][\text{NOCl}] = 500 \frac{\text{mol}}{\text{m}^3}.[/tex]

That would make the reaction quotient [itex]6.45 \frac{\text{m}^3}{\text{mol}}[/itex], which is the same as what I got before. Clearly, it wouldn't be okay to simply chop off the units and write

[tex]Q_c \approx 6.45[/tex]

because that's a totally different constant!

So, what's going on here? What makes [itex]\frac{\text{mol}}{\text{L}}[/itex] such a special unit, and why do I only get right answers on the homework problem when I use measurements in [itex]\frac{\text{mol}}{\text{L}}[/itex]?

[Borek]

This question was asked many times here.

Common answer is that Q is actually defined not by using concentrations, but activities, which say how the substance behaves as compared to its standard state, which is defined as a 1 M (mol/L) concentration. So in fact you have not

[tex]Q_c = \frac{ [\text{C}]^x [\text{D}]^y }{ [\text{A}]^m [\text{B}]^n }[/tex]

(note: this is a pretty standard convention of marking concentrations and naming reaction quotient).

but

[tex]Q_c = \frac{ \left(\frac{[\text{C}]}{1~M}\right)^x \left(\frac{[\text{D}]}{1~M}\right)^y }{\left(\frac{ [\text{A}]}{1~M}\right)^m \left(\frac{[\text{B}]}{1~M}\right)^n }[/tex]


which is unitless.

Reaction quotient is present in many formulas in thermodynamics, like ΔG⁰ = -RTlnK - K is equilibrium constant, or Q at equilibrium, it must be unitless for the logarithm to be taken.

Note that defining it using activities (ie comparing it with some standard state) has several additional advantages - it will allow use of solids and gases in equations (solids have activity of just 1, for gases standard state is 1 bar).

And then there are activity coefficients, which help deal with nonlinearity.

Can of worms if you ask me

So, what's going on here? What makes [itex]\frac{\text{mol}}{\text{L}}[/itex] such a special unit, and why do I only get right answers on the homework problem when I use measurements in [itex]\frac{\text{mol}}{\text{L}}[/itex]?

Because we have chosen (completely arbitrary) 1 M to be the standard state.

[Gradient]

Thank you! This makes perfect sense. I'm baffled by the fact that [itex]Q_c = \frac{ [\text{C}]^x [\text{D}]^y }{ [\text{A}]^m [\text{B}]^n }[/itex] is technically inaccurate, but as you mention, it is a standard naming convention! It's printed in my textbook, and I'm sure it's printed in high school and college textbooks across the world.

That's probably why the question has been asked so many times on this website: the teachers and textbooks explain this formula inaccurately, creating a worldwide misconception.

Thanks again for your help, and I'll be sure to check the forum before I ask a different question.

[Borek]

Thank you! This makes perfect sense. I'm baffled by the fact that [itex]Q_c = \frac{ [\text{C}]^x [\text{D}]^y }{ [\text{A}]^m [\text{B}]^n }[/itex] is technically inaccurate

For diluted solutions activity and concentration are typically identical, so it is quite easy to do calculations using just concentrations. Students often have enough problems grasping the concept of Q and equilibrium, confusing them even further by explaining that these concentrations are actually not concentrations, but numbers which are something different but have exactly the same value - won't help.

[Gradient]

Students often have enough problems grasping the concept of Q and equilibrium, confusing them even further by explaining that these concentrations are actually not concentrations, but numbers which are something different but have exactly the same value - won't help.

I've gotta disagree with you on that point. I understand that the activity and concentration (in mol/L) are typically identical. However, we can't deny that this formula,

[tex]Q_c = \frac{ [\text{C}]^x [\text{D}]^y }{ [\text{A}]^m [\text{B}]^n },[/tex]

where [itex][\text{A}][/itex], [itex][\text{B}][/itex] and so on represent concentrations, is incorrect as written. I think that if teachers and textbooks actually started off with the correct formula,

[tex]Q_c = \frac{ \left(\frac{[\text{C}]}{1~M}\right)^x \left(\frac{[\text{D}]}{1~M}\right)^y }{\left(\frac{ [\text{A}]}{1~M}\right)^m \left(\frac{[\text{B}]}{1~M}\right)^n },[/tex]

then we could simply apply exponent rules to get

[tex]Q_c = \frac{ \frac{[\text{C}]^x}{(1~M)^x} \frac{[\text{D}]^y}{(1~M)^y} }{ \frac{ [\text{A}]^m}{(1~M)^m} \frac{[\text{B}]^n}{(1~M)^n} }.[/tex]

Then, we could multiply fractions in the numerator and denominator to get

[tex]Q_c = \frac{ \left(\frac{[\text{C}]^x[\text{D}]^y}{(1~M)^{x+y}}\right) }{ \left(\frac{[\text{A}]^m[\text{B}]^n}{(1~M)^{m+n}}\right) },[/tex]

and then rewrite as a product of two fractions,

[tex]Q_c = \frac{ [\text{C}]^x[\text{D}]^y }{ [\text{A}]^m[\text{B}]^n } \cdot \frac{ (1~M)^{m+n} }{ (1~M)^{x+y} }.[/tex]

Finally, we could use exponent rules for fractions to write

[tex]Q_c = \frac{ [\text{C}]^x[\text{D}]^y }{ [\text{A}]^m[\text{B}]^n } \cdot (1~M)^{m+n-x-y}.[/tex]

Now, if teachers and textbooks were to print THIS correct formula instead of the old, incorrect one, then students would be able to calculate [itex]Q_c[/itex] in the same general way. It would even work if the concentrations were expressed in a crazy unit like [itex]\frac{\text{mmol}}{\text{ft^3}}[/itex]. However, students would have to remember to multiply by the factor [itex](1~M)^{m+n-x-y}[/itex] to obtain the correct dimensionless value for [itex]Q_c[/itex].

I argue that this would NOT confuse students any further. In fact, it would show them exactly why [itex]Q_c[/itex] is a dimensionless quantity, and I think it would clear up a lot of confusion on the subject.

[Borek]

I am speaking from experience. Most students (especially most HS students first introduced to the these concepts) won't understand a single word of what you wrote.

Actually they would often not understand why you have any problem with this strange thing you call "units". Yes, it is sad, but that's the reality.