Chemistry - Prove Law of Chemical Equivalence

Solution 1:

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If you look at definitions of "number of equivalents" (a concept that is outdated), you will find that this is the case.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 $b$,

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

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

Solution 2:

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.