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I am reading Tamás Turányi & Alison S. Tomlin's Analysis of Kinetic Reaction Mechanisms (2015). On page 10-11, the authors state that for a mechanism with $N_R$ reactions involving $N_S$ species, where $\nu_{ij}^L$ denotes the stoichiometric coefficients on the left-hand side,

[...] the rates of elementary reactions can be calculated by assuming the rule of mass action kinetics. According to the chemical kinetic law of mass action $$ r_i = k_i \prod_j^{N_S} Y_j^{\nu_{ij}^L}, \tag{2.5} $$ where $r_i$ and $k_i$ are the rate and the rate coefficient, respectively, of reaction step $i$, and $Y_j$ is the molar concentration of species $j$.

[...]

The kinetic system of ordinary differential equations (ODEs) defines the relationship between the production rates of the species and rates of the reaction steps $r_i$: \begin{align}\label{eq26} \frac{\mathrm{d} Y_j}{\mathrm{d}t} &= \sum_i^{N_R} \nu_{ij}r_i;& j&=1,2,\dots,N_S. \tag{2.6} \end{align}

[...]

An analogous equation to Eq. (2.6) can be written when other concentration units are used, e.g. mass fractions or mole fractions [see, e.g. Warnatz et al. (2006)], but Eq. (2.5) is applicable only when the “amount of matter divided by volume” concentration units are used. The amount of matter can be defined, e.g., in moles or molecules, whilst volume is usually defined in $\pu{dm3}$ or $\pu{cm3}$ units.

Can anyone help me understand why the two bolded sentences are true? I looked in the book of Warnatz et al. (2006) and did not find the answer.

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  • $\begingroup$ Hello! Are you sure of how you've written Eq. (2.5)? Maybe $\nu_{ij}^L$ is the exponent, i.e., $Y_j^{\nu_{ij}^L}$? $\endgroup$ Aug 6, 2023 at 10:03
  • $\begingroup$ In addition, maybe Eq. (2.6) is a sum and not a product? $$ \frac{\mathrm{d}Y_j}{\mathrm{d}t} = \sum_{i}^{N_\mathrm{R}} \nu_{ij} r_i $$ $\endgroup$ Aug 6, 2023 at 10:10

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[AUTHOR] The kinetic system of ordinary differential equations (ODEs) defines the relationship between the production rates of the species and rates of the reaction steps $r_i$: $$ \frac{\mathrm{d}Y_j}{\mathrm{d}t} = \sum_i^{N_R} \nu_{ij}r_i \quad j=1,2,\dots,N_\mathrm{S} \tag{2.6} $$

This equation can be obtained by performing a mole balance of chemical species $j$ on an element of volume.

The rate of change of species $j$ takes place due to $N_\mathrm{R}$ reactions. Let me call this magnitude just $r_j$ $$ r_j := \sum_{i = 1}^{N_\mathrm{_R}} \nu_{i,j}r_{i} \tag{1} $$

The mole balance in this case is simplified, because in our reactor there are no species flowing into or out of it, so the first two brackets are zero \begin{align} \begin{bmatrix} \text{Flow} \\ \text{of species $j$} \\ \text{INTO the} \\ \text{system} \\ \text{[mol/s]} \end{bmatrix} + \begin{bmatrix} \text{Flow} \\ \text{of species $j$ } \\ \text{OUT OF the} \\ \text{system} \\ \text{[mol/s]} \end{bmatrix} + \begin{bmatrix} \text{Generation} \\ \text{of species $j$} \\ \text{within the} \\ \text{system} \\ \text{[mol/s]} \end{bmatrix} &= \begin{bmatrix} \text{Accumulation} \\ \text{of species $j$} \\ \text{within the} \\ \text{system} \\ \text{[mol/s]} \end{bmatrix} \\ \begin{bmatrix} \text{Generation} \\ \text{of species $j$} \\ \text{within the} \\ \text{system} \\ \text{[mol/s]} \end{bmatrix} &= \begin{bmatrix} \text{Accumulation} \\ \text{of species $j$} \\ \text{within the} \\ \text{system} \\ \text{[mol/s]} \end{bmatrix} \\ G_j &= \frac{\mathrm{d}N_j}{\mathrm{d}t} \tag{2} \end{align} where $G_j$ is the rate of generation and $N_j$ is the amount of species $j$. The rate of generation is a function of the space coordinates in the volume $V$. So, imagine that we break up this volume in a huge number of elements of volume $\Delta V_1$, $\Delta V_2$, and so on. These elements of volume are so small, that we assume that the rate is constant in those tiny volumes. The picture is the following:

enter image description here

The rate of generation $G_j$ can be calculated by summing all the rate of reactions evaluated in their respected tiny volumes, multiplied by this same tiny volume. If we take the limit and use the definition of a Riemannian integral \begin{align} G_j &= \lim_{M \to \infty} \sum_{k = 1}^M r_j(\Delta V_k)\Delta V_k \\ G_j &= \int_{0}^V r_j(V) \, \mathrm{d}V \tag{3} \end{align} combining Eqs. (2) and (3) and relating the amount with the concentration via $N_j = C_jV$ $$ \frac{\mathrm{d}(C_jV)}{\mathrm{d}t} = \int_{0}^V r_j(V) \, \mathrm{d}V \tag{4} $$ Eq. (4) is super general. Now comes the two most important assumptions:

  • In the case that the reactor is perfectly mixed, the rate of reaction may be taken out of the integral because it is not a function of position. This mathematical step is done in real life when you use a magnetic stirrer in the laboratory to ensure a homogeneous behavior of the variables in the beaker, e.g., during a titration. There is no mistery.
  • The volume $V$ is constant. This is an excellent approximation in liquid phase reactions, because it is generally assumed incompressible. Strictly, the molar volume of the mixture is going to change in time, because species are being formed and consumed during the reaction (pure $A$ will not have the same molar volume than pure $B$ in the academic reaction $A \to B$).

Thus, Eq. (4) becomes \begin{align} \require{cancel} \frac{\mathrm{d}(C_jV)}{\mathrm{d}t} &= \int_{0}^V r_j \, \mathrm{d}V \\ V\frac{\mathrm{d}C_j}{\mathrm{d}t} &= r_j \int_{0}^V \, \mathrm{d}V \\ \cancel{V}\frac{\mathrm{d}C_j}{\mathrm{d}t} &= r_j \cancel{V} \\ \frac{\mathrm{d}C_j}{\mathrm{d}t} &= r_j \qquad \text{(Use Eq. (1))} \rightarrow \boxed{\frac{\mathrm{d}C_j}{\mathrm{d}t} = \sum_{i = 1}^{N_\mathrm{_R}} \nu_{i,j}r_{i}} \tag{5} \end{align} Eq. (5) is the one that is presented in the book.

1. Eq. (5) in terms of the mass fraction The relation between the amount and the mass fraction of species $j$ is \begin{align} N_j &= \frac{m_j}{M_j} \\ N_j &= \frac{\omega_j m}{M_j} \\ \frac{\mathrm{d}N_j}{\mathrm{d}t} &= \left(\frac{m}{M_j}\right) \frac{\mathrm{d}\omega_j}{\mathrm{d}t} \tag{6} \end{align} since the mass $m$ and the molar mass $M_j$ are constant over time. Combining Eq. (4) and (6), under the assumption that $r_j$ is not a function of position, yields \begin{align} \frac{\mathrm{d}N_j}{\mathrm{d}t} &= r_jV \\ \left(\frac{m}{M_j}\right)\frac{\mathrm{d}\omega_j}{\mathrm{d}t} &= r_jV \\ \left(\frac{m}{M_j}\right)\frac{\mathrm{d}\omega_j}{\mathrm{d}t} &= \sum_{i = 1}^{N_\mathrm{R}} \nu_{i,j}r_{i}V \rightarrow \boxed{\frac{\mathrm{d}\omega_j}{\mathrm{d}t} = \left(\frac{M_jV}{m}\right)\sum_{i = 1}^{N_\mathrm{R}} \nu_{i,j}r_{i}} \tag{7} \end{align} The mass $m$, which will not change because we are dealing with chemical reactions, can be calculated by the initial conditions of the differential equations. Knowing the concentrations $C_1, C_2, ..., C_\mathrm{S}$ and the volume $V$, we can obtain the amounts $N_1, N_2, ..., N_\mathrm{S}$ and then with the molar mass of each species get the individual masses.

2. Eq. (5) in terms of the mole fraction With Eq. (4) under the assumption that $r_j$ is not a function of position, and that $N_j = x_jN$ \begin{align} \frac{\mathrm{d}N_j}{\mathrm{d}t} &= r_jV \tag{8} \\ \frac{\mathrm{d}(x_jN)}{\mathrm{d}t} &= r_jV \\ x_j\frac{\mathrm{d}N}{\mathrm{d}t} + N\frac{\mathrm{d}x_j}{\mathrm{d}t} &= r_jV \\ \frac{\mathrm{d}x_j}{\mathrm{d}t} &= \frac{1}{N}\left(r_jV - x_j\frac{\mathrm{d}N}{\mathrm{d}t}\right) \tag{9} \end{align} to obtain $\mathrm{d}N/\mathrm{d}t$ we use its definition, the total amount is equal to the sum of all the amounts (the presence of inert species won't change the final result, because the time derivative will kill them) \begin{align} N &= \sum_{j = 1}^{N_\mathrm{S}} N_j \\ \frac{\mathrm{d}N}{\mathrm{d}t} &= \sum_{j = 1}^{N_\mathrm{S}} \frac{\mathrm{d}N_j}{\mathrm{d}t} \qquad \text{(Use Eq. (8))} \\ \frac{\mathrm{d}N}{\mathrm{d}t} &= \sum_{j = 1}^{N_\mathrm{S}}r_jV \\ \frac{\mathrm{d}N}{\mathrm{d}t} &= \sum_{j = 1}^{N_\mathrm{S}} \sum_{i = 1}^{N_\mathrm{R}}\nu_{i,j}r_iV \\ \frac{\mathrm{d}N}{\mathrm{d}t} &= \sum_{i = 1}^{N_\mathrm{R}} \sum_{j = 1}^{N_\mathrm{S}} \nu_{i,j}r_iV \\ \frac{\mathrm{d}N}{\mathrm{d}t} &= \sum_{i = 1}^{N_\mathrm{R}} r_iV \sum_{j = 1}^{N_\mathrm{S}} \nu_{i,j} \\ \frac{\mathrm{d}N}{\mathrm{d}t} &= \sum_{i = 1}^{N_\mathrm{R}}\nu_ir_iV \tag{10} \\ \end{align} where we defined $\nu_i := \sum_{j = 1}^{N_\mathrm{S}} \nu_{i,j}$. This number is the sum of the stoichiometric coefficients of reaction $i$. For example, for the reaction scheme \begin{align} \ce{A + B &-> C + 2D} \qquad \nu_1 = (1 + 2) - (1 + 1) = 1 \\ \ce{B + 2C &-> 4C + 5E} \qquad \nu_2 = (4 + 5) - (1 + 2) = 6 \end{align} or you could use matrix form if you like. Note that in the case of a zero change of amounts, then $\nu_i = 0$, as expected.

Combining Eqs. (9) and (10) yields the final result \begin{align} \frac{\mathrm{d}x_j}{\mathrm{d}t} &= \frac{1}{N}\left(\sum_{i = 1}^{N_\mathrm{R}}\nu_{i,j}r_iV - x_j\sum_{i = 1}^{N_\mathrm{R}}\nu_ir_iV\right) \\ \frac{\mathrm{d}x_j}{\mathrm{d}t} &= \frac{V}{N}\left(\sum_{i = 1}^{N_\mathrm{R}}\nu_{i,j}r_i - x_j\sum_{i = 1}^{N_\mathrm{R}}\nu_ir_i\right) \rightarrow \boxed{\frac{\mathrm{d}x_j}{\mathrm{d}t} = \frac{V}{N}\sum_{i = 1}^{N_\mathrm{R}}(\nu_{i,j} - x_j\nu_i)r_i} \tag{9} \end{align}

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  • $\begingroup$ So in your explanation could C_j be the mole fraction or the mass fraction of species j? @MetalStorm $\endgroup$ Aug 9, 2023 at 1:53
  • $\begingroup$ @math_lover0105 There I added the results when you want the ODE in terms of the mass fraction or the mole fraction. Tell me if it is clear afterwards. $\endgroup$ Aug 9, 2023 at 10:20

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