# Mass action law and conservation of mass [closed]

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.

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

• That should be obvious from the reaction's equation... May 26, 2023 at 21:24
• I am a Math student, so I don't know too much about chemistry. Do you mean that the equations are wrong? May 26, 2023 at 21:44
• If you make one Y from two X than why would you expect to still have two? May 26, 2023 at 22:21
• 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. May 27, 2023 at 3:29
• Please MathJax the screenshots May 27, 2023 at 5:39

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.