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A somewhat infamous law (to me at least), that states:

$$a.A_{eq}=b.B_{eq}=c.C_{eq}=d.D_{eq}$$

for any and every balanced chemical equation -

$$a.A+b.B = c.C+d.D$$

where $i$ denotes the stoichiometric coefficient(/no. of moles) of the chemical species $I$ and $I_{eq}$ denotes the number of equivalents in 1 mole(/formula mass...?) of the chemical species $I$.

Am I wrong about any of the facts I've presented above? Please mention any errors you spot.

Assuming that I did fine, my simple question is,

What's the mathematical proof for this law?

I tried surfing online, but all I came across (with my not-so-deep searches) were verbal explanations of the law.

Is there no mathematical proof to this law whatsoever? I'll be satisfied to just find a proof, even if I don't understand the complicated math going into it (for I'm an eleventh grader).

Edit to provide an example:

$$\ce{2NaOH + H2SO4 -> Na2SO4 + 2H2O}$$

For A, $a \cdot A_{eq} = 2 \ (moles) × 1 \ (equiv.\ per\ mole) = 2$

For B, $b \cdot B_{eq} = 1 \ (moles) × 2 \ (equiv.\ per\ mole) = 2$

And this follows for other species, hence showing $a.A_{eq}=b.B_{eq}=c.C_{eq}=d.D_{eq}$

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  • $\begingroup$ You are missing something. Where is i and where is I? $\endgroup$ – M. Farooq May 21 at 15:33
  • $\begingroup$ @M.Farooq I believe he means the $i$=lower case symbols and $I$=upper case symbols. $\endgroup$ – Tyberius May 21 at 15:41
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    $\begingroup$ It's not so much that this relationship has a mathematical proof, but more that it proceeds from the law of conservation of mass/matter. $\endgroup$ – hBy2Py May 21 at 15:58
  • $\begingroup$ @BaiduryaMathaddict Could you please write an example with a real reaction? $\endgroup$ – M. Farooq May 21 at 16:37
  • $\begingroup$ @BaiduryaMathaddict I will when I get the chance, but I can't guarantee how soon that will be. $\endgroup$ – hBy2Py May 21 at 18:42
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An example

Here is a balanced equation:

$$\ce{2 H2 + O2 -> 2 H2O}$$

Both sides of the equation show four atoms of hydrogen and two atoms of oxygen, so it is balanced. The coefficients are $\nu_{\ce{H2}} = 2$, $\nu_{\ce{O2}} = 1$ and $\nu_{\ce{H2O}} = 2$.

If 20 moles of $\ce{H2}$ and 10 moles of $\ce{O2}$ are used up in a reaction forming water (you can figure out 20 moles are formed), I can write:

$$\frac{n_{\ce{H2}}}{\nu_{\ce{H2}}} = \frac{n_{\ce{O2}}}{\nu_{\ce{O2}}},$$ where $n$ is the amount that is used up in the reaction.

I can not write:

$$n_{\ce{H2}} \cdot \nu_{\ce{H2}} = n_{\ce{H2}} \cdot \nu_{\ce{H2}},$$

as you will find when substituting the values from the example.

I should also not write:

$$2 \cdot \ce{H2} = 1 \cdot \ce{O2} $$

Hydrogen atoms are not oxygen atoms, no matter what we multiply it (nuclear reactions aside)

What about $I$ and $i$?

If you call the stoichiometric coefficients $i$ instead of $\nu$, and you want the equation below to be true:

$$i_{\ce{H2}} \cdot I_{\ce{H2}} = i_{\ce{O2}} \cdot I_{\ce{O2}},$$

you have to make sure that $I$ is proportional to $\frac{1}{i}$, i.e. define them by

$$I_{\ce{H2}} = \frac{ \mathrm{const} }{ i_{\ce{H2}} } $$

$$I_{\ce{O2}} = \frac{ \mathrm{const} }{ i_{\ce{O2}} }$$

If you look at definitions of "number of equivalents" (a concept that is outdated), you will find that this is the case. Once that is in place, it is a matter of showing that for any non-zero $a$ or $b$,

$$\mathrm{const} \cdot \frac{a}{a} = \mathrm{const} \cdot \frac{b}{b} $$

enter image description here

Source: https://www.askiitians.com/iit-jee-chemistry/physical-chemistry/stoichiometry-and-redox-reactions/law-of-equivalents.aspx

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  • $\begingroup$ @BaiduryaMathaddict Thanks for the edits (balance vs balanced). The edits to the mhchem code were unnecessary, so I rejected the edit and fixed the mistake (less work for me). If what follows an underscore is another backslash command, you don't need the braces around it, i.e. $Z_\delta$ is fine and $Z_{\delta}$ is harder to read. You do need it in $Z_delta$ vs $Z_{delta}$. $\endgroup$ – Karsten Theis May 22 at 15:38
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    $\begingroup$ @BaiduryaMathaddict Sorry, I don't see the problem. Maybe my rendering engine is different (phone vs computer sometimes makes a difference), or some other subtlety. For me, your edits and the original looked identical. $\endgroup$ – Karsten Theis May 22 at 15:44
  • $\begingroup$ @BaiduryaMathaddict I tried to edit it accordingly. It is hard to test without seeing the problem, but I think I got them all (and found another formatting problem). I learned something today! $\endgroup$ – Karsten Theis May 22 at 15:55
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There are a couple of problems with the main gist of the question. First of all we do not need to invoke the concept of moles here. Equivalents originated in 1700, survived until the 1960s and vanished slowly. Since the govt. did not bother to update chemistry syllabi, this concept is still lingering in some areas. It is creating havoc because students have to deal with moles and then equivalents and then they try to find relations between them. Gram equivalents do not need the crutches of moles, because they pre-date moles by centuries.

This is why I was continuously asking you to improve the notation because there is no need of moles in the notation of your query.

$$\ce{aA + bB -> cC +dD}$$

$$\ce{2NaOH + H2SO4 -> Na2SO4 + 2H2O}$$

You can now parse this reaction as follows:

1 equivalent weight of NaOH will react with 1 equivalent weight of $\ce{H2SO4}$ to produce 1 equivalent weight of $\ce{Na2SO4}$ and 1 equivalent weight of $\ce{H2O}$

Now, let us see what equivalent weights we are talking about (I assume you know how to calculate equivalent weights from the chemical equation)

equivalent weight of NaOH = 40

equivalent weight of sulfuric acid = 49

equivalent weight of sodium sulfate = 71

equivalent weight of water = 18.

Does it make sense if I say that 40 g of NaOH will react with 49 g of sulfuric acid to produce 71 g of sodium sulfate and 18 g of water?

Now, equivalents and equivalent weight are two different things. Equivalents is similar to calculating "moles" and equivalent weight is analogous to "molecular weight".

So, equivalents = weight in gram/ equivalent weight

So let us say, someone tells you that we have 0.1 equivalents of NaOH, how many equivalents of sulfuric acid are needed?

The answer is 0.1.

In terms of grams, you would need 0.1 x 40 = 4 g NaOH and, 0.1 x 49 = 4.9 g sulfuric acid.

Using your notation and formulas, I will arrive at wrong numbers!

[Post Edit]. Now that you have clarified the notation using my example, and their units it makes perfect sense. Earlier I told you that equivalents do not need the mole concept. They are a parallel system.

Now the formula $$a.A_{eq}=b.B_{eq}=c.C_{eq}=d.D_{eq}$$ connects the mole with the concept of equivalent weight (not equivalent).

It originates from the observational fact that molecular weight or the formula weight of any substance is an integer multiple of equivalent weight, i.e.,

$$\mathrm{Formula\,weight} = \mathrm{n\,x\,equivalent\,weight}$$.

How many moles are there in 1 formula weight? 1, right? The value of $\mathrm{n}$ is derived independently based on the chemical properties of the substance.

$$\ce{2NaOH + H2SO4 -> Na2SO4 + 2H2O}$$

You can now read this reaction as follows:

Fact no. 1: 1 equivalent weight of NaOH will react with 1 equivalent weight of $\ce{H2SO4}$ to produce 1 equivalent weight of $\ce{Na2SO4}$ and 1 equivalent weight of $\ce{H2O}$

Fact no. 2: 2 mole of NaOH will react with 1 mole of $\ce{H2SO4}$ to produce 1 mole of $\ce{Na2SO4}$ and 2 mole of $\ce{H2O}$

How would we connect the two facts:

We can say, in 1 formula weight of NaOH there is only 1 equivalent weight of NaOH,

In 1 formula weight of $\ce{H2SO4}$ there are 2 equivalent weights of sulfuric acid and so on. So your formula works and produces 2 in each case this reaction.

Will this formula work or fail in disproportionation or comproportionation reaction, I don't know. Most likely, it fails there.

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    $\begingroup$ "My question has more to do with why 1 equivalent of a species reacts with exactly 1 of another" Now this is a different question. It has nothing to do with the original post. Please post this as a separate question (with you background understanding and search). Also mention that you don't wish to invoke moles and it is a hitorical query rather. Your current question will be closed soon. Only one more person has to vote for its closure as being unclear out of (4/5 close votes). You see if I use your (corrected) formula, I get wrong numbers. So your formula is not correct. $\endgroup$ – M. Farooq May 23 at 5:25
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    $\begingroup$ If you can edit the question with this particular example using the numbers in this answer, it would be helpful to understand your formula. I am not a moderator, but I can see the reason for closure votes. $\endgroup$ – M. Farooq May 23 at 6:51
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    $\begingroup$ Please also edit the moles issue. You don't need to bring it in moles. That is adding to confusion. $\endgroup$ – M. Farooq May 23 at 6:55
  • $\begingroup$ Your answer befits my question. Of course, it uses the fact that 1 equiv. of a species reacts with 1 equiv. of another. That is one proposition that you're needing to compulsorily accept (like a postulate...?). If I want to interrogate that postulate, it'll be in the form of a new question. But thanks, my doubts are sated. $\endgroup$ – BaiduryaMathaddict May 23 at 11:19
  • $\begingroup$ On a second thought, no. Not satisfied. Because law of equivalence states that "x equiv.s of a substance reacts with and produces x equiv.s of reactants and products respectively." That the law holds for x=1 is obvious, because 1 is just another real number (x). And I'm not willing to assume that law of equivs. for x=1 holds, or in other words, that a specific case of the law holds, before the law is proved in general (or even in specific). However, just to be clear, it is not that I'm willing to disprove the law. Then again, if somebody does it, I won't mind :) $\endgroup$ – BaiduryaMathaddict May 23 at 11:34

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