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Consider the follwoing reaction:

$$\ce{a A + b B -> c C + d D}\tag{1}$$

where $\ce A$ and $\ce B$ are the reactants, $\ce C$ and $\ce D$ are the products, and $a (<0)$, $b(<0)$, $c(>0)$, and $d(>0)$ their respective stoichiometric coefficients. The extents of reaction are respectively defined for these reactants and products as:

$$\begin{align} \xi_A(t) &= \frac{n_A(t) - n_A(0)}{a}\tag{2.1}\\ \xi_B(t) &= \frac{n_B(t) - n_B(0)}{b}\tag{2.2}\\ \xi_C(t) &= \frac{n_C(t) - n_C(0)}{c}\tag{2.3}\\ \xi_D(t) &= \frac{n_D(t) - n_C(0)}{d}\tag{2.4} \end{align}$$

However, I have found in many references (e.g. source) the extent of reaction defined for the whole reaction as:

$$\xi(t) = \frac{n_A(t) - n_A(0)}{a} = \frac{n_B(t) - n_B(0)}{b} = \frac{n_C(t) - n_C(0)}{c} = \frac{n_D(t) - n_D(0)}{d}\tag{3}$$

I do not understand how you can justify that the extent of reaction of the different reactants and products are equal.

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  • $\begingroup$ It has to be, because of the stoichiometry of the reaction: if your reaction consumes $x_\ce{A}$ moles of $\ce{A}$, then it must also consume $x_\ce{B} = x_\ce{A} \cdot b / a$ moles of $\ce{B}$. Now note that (by definition) $x_\ce{A} = n_\ce{A}(t) - n_\ce{A}(0)$, and $x_\ce{B} = n_\ce{B}(t) - n_\ce{B}(0)$. (I might have been a bit careless with the signs, but it doesn't affect the argument.) $\endgroup$
    – orthocresol
    Feb 7 at 15:40
  • $\begingroup$ So, you seem to start your reasoning from $\left|a\right|x_A = \left|b\right|x_B$. But, form where this expression is coming from? You refer to the stoichiometry of the reaction, but I was thinking that the stoichiometry (only) says: $$\begin{align} ax_A + bx_B + cx_C + dx_D &= 0 \tag{1} \\ \\ -\left|a\right|x_A -\left|b\right|x_B + \left|c\right|x_C + \left|d\right|x_D &= 0 \tag{2} \end{align}$$ I have the impression I am still missing something... Did you actually mean that: $$ \left|a\right|x_A = \left|b\right|x_B = \left|c\right|x_C = \left|d\right|x_D \tag{3} $$ ? $\endgroup$
    – Patrice
    Feb 7 at 16:15
  • $\begingroup$ How should some constituents of one reaction have reacted to a different extend than the others? That makes no sense. $\endgroup$
    – Karl
    Feb 7 at 17:25
  • $\begingroup$ Re. eq. (3): yes, that is dictated, enforced, by the stoichiometry of the reaction. I am sorry, but I have no idea how to explain that any better. It is a very fundamental concept in chemistry, as you can judge from Karl's comment. You cannot consume B without also consuming a proportionate amount of A (the proportionality factor being $a/b$); you cannot produce C or D without also consuming a proportionate amount of A. $\endgroup$
    – orthocresol
    Feb 8 at 0:01
  • $\begingroup$ There is a slight error, though, in eq. (3) of your comment: the $x$'s should be divided by the stoichiometric coefficients, not multiplied. In eq. (3) of your question it is correct. $\endgroup$
    – orthocresol
    Feb 8 at 0:25
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Let's take a numerical example: $\ce{3 A + 1 B -> 8 C + 2 D}$

Here : $a = -3, b = -1, c = 8, d = 2$.

Let's start from $100$ $mol$ of $\ce{A}$, $120$ $mol$ of $\ce{B}$, $1$ $mol$ $\ce{C}$, and no $\ce{D}$. Suppose that $10$ $mol$ $\ce{B}$ have reacted with $30$ $mol$ $\ce{A}$, and producing $80$ $mol$ $\ce{C}$ and $20$ $mol$ $\ce{D}$. At the end of the reaction, the remaining amounts are :

For $\ce{A}$ : $100$ $mol$ - $30$ $mol$ = $70$ $mol$ $\ce{A}$.

For $\ce{B}$ : $\ce{120 $mol$ - $10$ $mol$ = $110$ $mol$}$ $\ce{B}$

For $\ce{C}$ : $\ce{1 $mol$ + 80 $mol$ = 81 $mol$}$ $\ce{C}$

$$\begin{align} \xi_A(t) &= \frac{n_A(t) - n_A(0)}{a} =\frac{70 - 100}{-3} = 10\tag{2.1}\\ \xi_B(t) &= \frac{n_B(t) - n_B(0)}{b} =\frac{110-120}{-1} = 10\tag{2.2}\\ \xi_C(t) &= \frac{n_C(t) - n_C(0)}{c} =\frac{81 - 1}{8} = 10\tag{2.3}\\ \xi_D(t) &= \frac{n_D(t) - n_D(0)}{d} =\frac{20 - 0}{2} = 10\tag{2.4} \end{align}$$ The extent of reaction are the same for all products.

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  • $\begingroup$ Many thanks! When you wrote "suppose that $10$ $mol$ of $\ce{B}$ have reacted with $30$ $mol$ of $\ce{A}$, and producing $80$ $mol$ of $\ce{C}$ and $20$ $mol$ of $\ce{D}$", you are actually using the (formal) expression: $$ \frac{\Delta N_A(t)}{a} = \frac{\Delta N_B(t)}{b} = \frac{\Delta N_C(t)}{c} = \frac{\Delta N_D(t)}{d} $$ where $a$, $b$, $c$ and $d$ are the stoichiometric coefficients/numbers (with the sign convention). Is that correct? $\endgroup$
    – Patrice
    Feb 8 at 12:37
  • $\begingroup$ I actually found it in "Physical Chemistry from a Different Angle", Eq. (1.15). Maybe I was just missing the "different angle" from my other sources ;-) | link.springer.com/book/10.1007%2F978-3-319-15666-8 $\endgroup$
    – Patrice
    Feb 8 at 13:07

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