I am starting to learn about the use of mass action law in chemistry. I am doing some exercises to practice. In particular, I am considering the reaction $$2 X \underset{3}{\stackrel{4}{\rightleftharpoons}} Y.$$ By mass action law I have obtained the following system of ODE's $$ \left\{ \begin{array}{ l } \frac{d}{dt}x(t)=-8x^2+6y\\ \frac{d}{dt}y(t)=4x^2-3y, \end{array}\label{SEDO} \right.$$

where I have used this result. enter image description here

I am wondering if it is usual that the amount $x(t)+y(t)$ is not conservated.

  • 1
    $\begingroup$ That should be obvious from the reaction's equation... $\endgroup$
    – Mithoron
    May 26, 2023 at 21:24
  • $\begingroup$ I am a Math student, so I don't know too much about chemistry. Do you mean that the equations are wrong? $\endgroup$
    – mejopa
    May 26, 2023 at 21:44
  • 1
    $\begingroup$ If you make one Y from two X than why would you expect to still have two? $\endgroup$
    – Mithoron
    May 26, 2023 at 22:21
  • 2
    $\begingroup$ There is law of mass conservation, there is no law of molar amount conservation. 1 mol of Y can be formed from 2 mol of X with the same mass. $\endgroup$
    – Poutnik
    May 27, 2023 at 3:29
  • $\begingroup$ Please MathJax the screenshots $\endgroup$ May 27, 2023 at 5:39

1 Answer 1


The rate laws that you show don't give an identity to $x$ or $y$. I am going to assume they are moles, even though it is uncommon to show rate laws as a function of moles. The time evolution of the amount of substance of both species can be written as \begin{align} \frac{\mathrm{d}x}{\mathrm{d}t} &= -8x^2 + 6y = (-2) (4x^2 - 3y) \tag{1} = -2r_1\\ \frac{\mathrm{d}y}{\mathrm{d}t} &= 4x^2 - 3y = (+1) (4x^2 - 3y) = r_1 \tag{2} \end{align} where we defined $r_1 := 4x^2 - 3y$. Thus:

  1. It is clear that $x(t) + y(t)$ is not a conserved quantity. This is because for every $2$ moles disappeared of $x$, you get only $1$ mol of $y$. But the quantity $x(t)+2y(t)$ is conserved, by using Eqs. (1-2) \begin{align} x + 2y &= \text{constant} \\ \frac{\mathrm{d}x}{\mathrm{d}t} + 2\frac{\mathrm{d}x}{\mathrm{d}t} &= 0 \\ (-2r_1) + 2(r_1) &= 0 \rightarrow 0=0 \tag{3} \end{align}
  2. The mass for a chemical reaction is conserved, but this quantity is not $x(t) + y(t)$ rather it is $M_x x(t) + M_yy(t)$, where $M$'s are the molar masses of the compounds. Again, by Eqs. (1-2) \begin{align} \require{cancel} M_xx + M_yy &= \text{constant} \\ M_x\frac{\mathrm{d}x}{\mathrm{d}t} + M_y\frac{\mathrm{d}x}{\mathrm{d}t} &= 0 \\ (-2M_x)r_1 + (M_y)r_1 &= 0 \rightarrow -2M_x + M_y = 0 \tag{4} \\ \end{align} Eq. (4) is always satisfied. To give an example that is similar to your chemical reaction, we can illustrate the decomposition of nitrogen dioxide in the gas phase $$ \ce{2NO2(g) \rightleftharpoons N2O4(g)} \tag{5} $$ And this chemical equation satisfies \begin{align} -2M_\ce{NO2} + M_\ce{N2O4} &= 0 \\ (-2) \left(\pu{46.01 \frac{g}{mol}}\right) + \left(\pu{92.02 \frac{g}{mol}}\right) &= 0 \\ (-92.02 + 92.02) \; \pu{\frac{g}{mol}} &= 0 \rightarrow 0=0 \tag{6} \end{align}

To generalize, for a general chemical reaction that comprises $m$ species, we always have $$ \boxed{\sum_{j = 1}^m \nu_j M_{j} = 0} \tag{7} $$ So, if you solve a system that has at time $t$ an amount of moles of $x_1(t)$, $x_2(t)$, ..., $x_m(t)$, then $M_1x_1(t) + M_2x_2(t) + \dots + M_m x_m(t)$ is constant. It is a good check to see if the code is satisfying the mass balance by calculating this quantity and viewing that it does not deviate, apart from the absolute/relative errors that you specify.

  • $\begingroup$ Thank you! Now is completely clear for me! $\endgroup$
    – mejopa
    May 27, 2023 at 22:53

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