# Why are the "extents of reaction" the same for all reactants and products?

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.

• 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.) Feb 7 at 15:40
• 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: $$\left|a\right|x_A = \left|b\right|x_B = \left|c\right|x_C = \left|d\right|x_D \tag{3}$$ ? Feb 7 at 16:15
• How should some constituents of one reaction have reacted to a different extend than the others? That makes no sense.
– Karl
Feb 7 at 17:25
• 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. Feb 8 at 0:01
• 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. Feb 8 at 0:25

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.

• 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? Feb 8 at 12:37
• 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 Feb 8 at 13:07
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