In combustion, for a kinetic mechanism involving $m$ reactions and $n$ chemical species, of the form \begin{align} \begin{cases} \nu_{11}' \mathcal{S}_1 + \nu_{12}' \mathcal{S}_2 + \dots + \nu_{1n}' \mathcal{S}_{n} &\ce{->[k_1]} \hspace{0.75cm} \nu_{11}'' \mathcal{S}_1 + \nu_{12}'' \mathcal{S}_2 + \dots + \nu_{1n}'' \mathcal{S}_{n} \\ \nu_{21}' \mathcal{S}_1 + \nu_{22}' \mathcal{S}_2 + \dots + \nu_{2n}' \mathcal{S}_{n} &\ce{->[k_2]} \hspace{0.75cm} \nu_{21}'' \mathcal{S}_1 + \nu_{22}'' \mathcal{S}_2 + \dots + \nu_{2n}'' \mathcal{S}_{n} \\ \hspace{2.63cm} \vdots & \hspace{0.45cm} \vdots \hspace{3.85cm} \vdots \\ \nu_{m1}' \mathcal{S}_1 + \nu_{m2}' \mathcal{S}_2 + \dots + \nu_{mn}' \mathcal{S}_{n} &\ce{->[k_{m}]} \hspace{0.75cm} \nu_{m1}'' \mathcal{S}_1 + \nu_{m2}'' \mathcal{S}_2 + \dots + \nu_{mn}'' \mathcal{S}_{n} \end{cases}, \end{align} we have the rate of the $i$-th reaction given by \begin{equation}\label{eq1} q_i = k_i[\mathcal{S}_1]^{\nu_{i1}'}[\mathcal{S}_2]^{\nu_{i2}'}\dots [\mathcal{S}_n]^{\nu_{in}'}, \end{equation} where $[\mathcal{S}_j]$ corresponds to the molar concentration of the specie $\mathcal{S}_j$, and $k_i$ corresponds to the Arrhenius equation \begin{equation*} %ARRHENIUS MODIFICADA k_i = A_i T^{\beta} e^{-E_a / (RT)}. \end{equation*} The change in the concentrations of all species with time $t$ is, then, given by the system of differential equations \begin{equation}\label{eq2} \frac{\mathrm{d}[\mathcal{S}_j]}{\mathrm{d}t} = \sum_{i=1}^{m} (\nu_{ij}''-\nu_{ij}') q_{i}, \hspace{3cm} j=1,\dots,n. \end{equation}

I received a computational code from a combustion researcher, in which the molar concentrations of the species, $[\mathcal{S}_j]$, in the equations above are replaced by the mass fraction of the species, $Y_j$. Thus, the code solves the system \begin{equation} \frac{\mathrm{d} Y_j}{\mathrm{d}t} = \sum_{i=1}^{m} (\nu_{ij}''-\nu_{ij}') q_{i}, \hspace{3cm} j=1,\dots,n; \end{equation} with \begin{equation} q_i = k_iY_1^{\nu_{i1}'}Y_2^{\nu_{i2}'}\dots Y_n^{\nu_{in}'}. \end{equation} Can this substitution be made so that the equations still make sense?

  • 1
    $\begingroup$ With mass fractions, calculated rates for particular reactions would be strongly affected by presence of components that do not take part in the reaction at all.. $\endgroup$
    – Poutnik
    Apr 22, 2023 at 21:21

1 Answer 1


The relation between molar concentration and mass fraction is \begin{align} [S_j] = \frac{N_j}{V} = \frac{m_j}{M_jV} = \frac{Y_jm}{M_jV} \rightarrow [S_j] = \left(\frac{\rho}{M_j}\right)Y_j \tag{1} \end{align} where $M_j$ is the molar mass of chemical species $ j $ and $\rho$ is the mass density.

It seems that your hope of writing the rate law by analogy, i.e. changing $[S_j]$ for $Y_j$, is true under very restrictive conditions: (1) irreversible elementary reaction $A \rightarrow B $, and (2) constant volume process. We prove this by using Eq. (1) in a batch reactor, like your equations show, where there is no inlet or outlet of any chemical species \begin{align} \require{cancel} \text{[Rate of acumulation of A within the system]} =& \text{[Rate of flow of A into the system = 0]} \\ -& \text{[Rate of flow of A out of the system = 0]} \\ +& \text{[Rate of generation of A within the system]} \\ \frac{dN_A}{dt} &= \nu_A rV \\ \frac{d([S_A]V)}{dt} &= -k[S_A]V \\ \bcancel{V}\cancel{\left(\frac{\rho}{M_A}\right)}\frac{dY_A}{dt} &= -k\cancel{\left(\frac{\rho}{M_A}\right)}Y_A \bcancel{V} \\ \frac{dY_A}{dt} &= -kY_A \therefore r' = k Y_A \end{align} The condition of constant volume, in your system, is satisfied by the way you have written the differential equations. The only way you can arrive to that form, is by taking out $V$ from the time derivative, and cancelling it on both sides. I will continue with this assumption, and since the mass is constant, then the mass density $\rho$ also is.

It is unusual to carry reactions, specially gas-phase reactions like combustions, in this manner. I will leave at the end the mathematical restriction that it imposes.

Nevertheless, lets try to obtain an expression according to your case.

1. Rate law in terms of $Y_j$ Combined with Eq. (1) it gives \begin{align} q_i &= k_i \prod_{k = 1}^n [S_k]^{\nu_{ik}'} \\ q_i &= k_i \prod_{k = 1}^n \left(\frac{\rho Y_k}{M_k}\right)^{\nu_{ik}'} \\ q_i &= k_i \rho^{\sum_{k = 1}^n \nu_{ik}'} \prod_{k = 1}^n \left(\frac{Y_k}{M_k}\right)^{\nu_{ik}'} \\ q_i &= k_i \rho^{\nu_{i}'} \prod_{k = 1}^n \left(\frac{Y_k}{M_k}\right)^{\nu_{ik}'} \tag{3} \\ \end{align} where I have defined $\nu_i' := \sum_{k = 1}^n \nu_{ik}'$. If we go back to the the reaction scheme, and stare at row $i$, this guy is the sum of all the stoichiometric coefficients of the reactants for that row.

2. Mole balance in terms of $Y_j$ The rate of change of the concentration of species $j$, using Eqs. (1) and (3), yields \begin{align} \dfrac{d[S_j]}{dt} &= \sum_{i = 1}^m (\nu_{ij}'' - \nu_{ij}')q_i \\ \left(\frac{\rho}{M_j}\right)\frac{dY_j}{dt} &= \sum_{i = 1}^m (\nu_{ij}'' - \nu_{ij}') k_i \rho^{\nu_{i}'} \prod_{k = 1}^n \left(\frac{Y_k}{M_k}\right)^{\nu_{ik}'} \\ \end{align} $$ \boxed{\frac{dY_j}{dt} = M_j \sum_{i = 1}^m (\nu_{ij}'' - \nu_{ij}') k_i \rho^{\nu_{i}' - 1} \prod_{k = 1}^n \left(\frac{Y_k}{M_k}\right)^{\nu_{ik}'}} \tag{4} $$

Eq. (4) is as far as we can get, but is the desired expression, as we have the ODE in terms of the mass fraction of species $j$.

3. Constant volume process In the simplest of cases, the mixture of gases will obey the ideal gas law \begin{align} \rho &= \frac{P}{MRT} \\ \ln\rho &= \ln(P) - \ln(MRT) \\ \frac{d\ln\rho}{dt} &= \left(\frac{1}{P}\right)\frac{dP}{dt} - \left(\frac{1}{\cancel{MR}T}\right) (\cancel{MR})\frac{dT}{dt} - \frac{1}{M\cancel{RT}}(\cancel{RT}) \frac{dM}{dt} \\ \frac{d\ln\rho}{dt} &= \left(\frac{1}{P}\right)\frac{dP}{dt} - \left(\frac{1}{T}\right)\frac{dT}{dt} - \left(\frac{1}{M}\right)\frac{dM}{dt} \tag{5} \\ \end{align}

Where $M$ is the molar mass of the mixture. At any instant in time, it has a value of $ M = \sum_{k = 1}^n M_ky_k $, and continuing with Eq. (5) \begin{align} \frac{d\ln\rho}{dt} &= \left(\frac{1}{P}\right)\frac{dP}{dt} - \left(\frac{1}{T}\right)\frac{dT}{dt} - \frac{1}{\sum_{k = 1}^n M_ky_k} \frac{d}{dt}\left(\sum_{k = 1}^n M_ky_k\right) \\ \frac{d\ln\rho}{dt} &= \left(\frac{1}{P}\right)\frac{dP}{dt} - \left(\frac{1}{T}\right)\frac{dT}{dt} - \frac{1}{\sum_{k = 1}^n M_ky_k} \sum_{k = 1}^n M_k\frac{dy_k}{dt} \tag{6} \\ \end{align} In consequence, if the mass density is to remain constant, by Eq. (6) $$ \boxed{\left(\frac{1}{P}\right)\frac{dP}{dt} = \left(\frac{1}{T}\right)\frac{dT}{dt} + \frac{1}{\sum_{k = 1}^n M_ky_k} \sum_{k = 1}^n M_k\frac{dy_k}{dt}} \tag{7} $$ Eq. (7) must be satisfied at any instant of time while solving the system of differential equations. It requires the need of the energy balance, in order to obtain an expression of $dT/dt$. Of course, you can always sweep it under the carpet, and the pressure will "adjust" itself to meet the demands of the process. Pretty much like a normal force in a body diagram. However, if you can couple Eq. (7) to the system and solve it, then perhaps high values of pressure will be obtained (e.g. if the system of reactions is excessively exothermic). This is an indication that the isochoric operation may be unfeasable in reality.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.