# Kinetics of Sabatier reaction

I am modelling a methanation process in MATLAB, where I have a rate equation for methanation

$$\ce{CO2 + 4 H2 → CH4 + 2 H2O}$$

in which I need partial pressure of $$\ce{CO2},$$ $$\ce{H2},$$ $$\ce{CH4},$$ $$\ce{H2O}.$$

I am struggling to understand the link between the mass flow rate of $$\ce{H2}$$, $$\ce{CO2}$$ and partial pressure of all other species. The required amount of catalyst and the volume for the fixed bed reactors in accordance with the chosen stimulation parameters.

If suppose my operational pressure is $$\pu{7 atm}$$, $$\pu{16 mol s^-1}$$ of $$\ce{H2},$$ $$\pu{4 mol s^-1}$$ of $$\ce{CO2},$$ what will be the partial pressures of all those species and required amount of catalyst and reactor volume.

The paper in which I referred for rate reaction is this.

• As a mechanical engineer you should keep in mind that "sec" refers to secant, whereas "s" refers to second. Jul 5, 2020 at 11:54
• If the gas is a stoichiometric mixture of $\ce{CO_2}$ and $\ce{H_2}$, and if the total pressure is 7 atm, the partial pressure of $\ce{CO_2}$ before reaction is 1.4 atm, and it is 5.6 atm for $\ce{H_2}$. Now to get the corresponding values out of the reactor, some more information is needed, like the equilibrium constant, and the kinetics of the reaction. Jul 5, 2020 at 12:24
• The amount of catalyst required, and the volume of the reactor, will be dependent on the conversion of CO2 and H2 that is desirable. For example, for a stiochiometric amount of CO2 and H2, if you'd like to convert at least 95% of these reactants to CH4 and H2O, this will require more catalyst than if you were trying to convert 85%. The reaction system will be dynamic throughout the length of the Reactor. Take a look at the rate equation: as you convert more reactant to product, the partial pressures of the species important for driving the reaction forward (CO2 and H2) will be lowered.
– samp
Jul 5, 2020 at 17:32
• Not if water is removed. Remember, the reaction is under pressure. Mar 19, 2023 at 22:43

So you need three things:

1. Partial pressures of species.
2. Amount of catalyst.
3. Reactor volume.

We will design a PBR in order to calculate the mass of catalyst $$W$$ needed to achieve a certain conversion $$X$$.

Partial Pressures You have a heterogeneous reaction, since it takes place with a catalyst, but all the chemical species of the rxn are in the gas phase. It seems that you have no limiting reactant, as the stoichiometric ratio $$\ce{H2}/\ce{CO2} = 4/1$$ is the same as the ratio of the entering molar flow rates. I will choose $$\ce{CO2}$$ as the basis component, so the conversion refers to $$\ce{CO2}$$. Say the reaction is $$A + 4B \rightarrow C + 2D$$. The molar flow rate of any species is $$F_j = F_{A0} \bigg(\dfrac{F_{j0}}{F_{A0}} + \nu_j X\bigg)$$ The molar flow rates of all components, and the total molar flow rate, are \begin{align} F_A &= F_{A0} (1 - X) \\ F_B &= 4F_{A0} (1 - X) \\ F_C &= F_{A0}X \\ F_C &= 2F_{A0}X \\ F &= \sum_j F_j = F_{A0} (5 - 2X) \end{align}

Thus, with $$P_j = y_j P = (F_j/F) P$$, we have the partial pressures as a function of the conversion \begin{align} P_A &= \left(\dfrac{1 - X}{5 - 2X}\right)P \tag{1} \\ P_B &= \left(\dfrac{4 - 4X}{5 - 2X}\right)P \tag{2} \\ P_C &= \left(\dfrac{X}{5 - 2X}\right)P \tag{3} \\ P_C &= \left(\dfrac{2X}{5 - 2X}\right)P \tag{4} \end{align} Now, these equations are strictly true at the entrance of the reactor, i.e., where the conversion is $$X =0$$. Once we are in the reactor, the gas phase will suffer pressure drop.

Pressure drop I will assume that the gas flow is, at every coordinate, turbulent. The Ergun equation simplifies to $$\dfrac{dP}{dz} = -\dfrac{7}{4}\dfrac{(1 - \phi)G^2}{\phi^3 \rho D_P} \tag{5}$$ where $$G = \dot{m}/S$$ is the mass flux (constant along the reactor), $$D_P$$ is the catalyst particle diameter, and $$\phi$$ the void fraction. We need to relate the density as a function of things we know, by using the ideal gas law $$\dfrac{dP}{dz} = -\dfrac{7}{4}\dfrac{(1 - \phi)G^2}{\phi^3D_P} \dfrac{MRT}{P} \tag{6}$$ Now we need to relate the $$z$$ coordinate to the $$W$$ coordinate \begin{align} W &= (1 - \phi) S z \rho_c \\ dW &= (1 - \phi) S dz \rho_c \\ dz &= \dfrac{dW}{(1 - \phi) S \rho_c} \tag{7} \end{align} where $$\rho_c$$ is the mass density of the catalyst. Combining Eqs. (6) and (7) \begin{align} (1 - \phi) S \rho_c \dfrac{dP}{dW} = -\dfrac{7}{4} \bigg(\dfrac{(1 - \phi)G^2}{\phi^3D_P}\bigg)\dfrac{MRT}{P} \\ \dfrac{dP}{dW} = -\dfrac{7}{4} \left(\dfrac{G^2}{S \rho_c \phi^3D_P}\right) \dfrac{MRT}{P} \tag{8} \\ \end{align} If you are obsessive like me, the molar mass of the gas will change along the reactor, so you should also take into account $$M = \sum_j y_j M_j \tag{9}$$

Amount of catalyst This is obtained by a mole balance along the $$W$$ coordinate for species $$A$$ \begin{align} F_A\vert_W - F_A\vert_{W+\Delta W} + {r_A}' \Delta W = 0 \\ \dfrac{F_A\vert_W - F_A\vert_{W+\Delta W}}{\Delta W} + {r_A}' = 0 \\ -\dfrac{dF_A}{dW} + {r_A}' = 0 \tag{10} \end{align} with $$F_A = F_{A0}(1 - X)$$ or $$dF_A = -F_{A0}dX$$, Eq. (10) yields \begin{align} F_{A0}\dfrac{dX}{dW} + {r_A}' = 0 \\ \dfrac{dX}{dW} = \dfrac{-{r_A}'}{F_{A0}} \tag{11} \end{align} where $$-{r_A}'$$ is the rate of consumption of species $$A$$ in $$\text{mol A}/\text{kg cat} \cdot \text{s}$$.

Energy Balance As a starting point, before studying an adiabatic reactor, assume that the reactor is isothermic. This translates the energy balance in the amount of heat $$Q(W)$$ that you must remove at every mass coordinate in order for the temperature to remain constant. Hence, we disregard the energy balance for the first analysis, and $$T = T_0$$.

Thus, to know the mass of catalyst in the reactor, you need to numerically solve the following ODE's \begin{align} \dfrac{dX}{dW} &= \dfrac{-{r_A}'}{F_{A0}} \hspace{1 cm} X(W = 0) = 0 \tag{12} \\ \dfrac{dP}{dW} &= -\dfrac{7}{4} \left(\dfrac{G^2}{S \rho_c \phi^3D_P}\right)\dfrac{MRT}{P} \hspace{1 cm} P(W = 0) = P_0 \tag{13} \\ \end{align}

Other data we need to know

1. To evaluate the rate of reaction you need a kinetic law. To my surprise, there is one in the paper, and has the type of a homogenous kinetic rate law of the type $$-{r_A}' = f(P_A, P_B, P_C, P_D) = k'\bigg(P_A P_B^4 - \dfrac{P_C P_D^2}{K}\bigg)$$ at constant temperature, where $$k' = k(T=T_0)$$ and $$K = K(T=T_0)$$ are fixed. In the partial pressures you plug Eqs. (1-4), this is why we had to do all that math.
2. You stop the integration depending on your objective. If equilibrium allows you, you should choose the highest conversion possible, $$X = 0.95X_{eq}(T)$$ is a starting point.
3. Other things like the catalyst mass density $$\rho_c$$, the void fraction of the PBR $$\phi$$, the area $$S$$, and the particle diameter of the catalyst $$D_P$$.

Reactor Volume Once you solve the ODE this is just one single calculation, you just use the Eq. (7) but not differentiated, since the volume is $$V = Sz$$ $$V = \dfrac{W}{(1 - \phi)\rho_c}$$

This is an interesting problem, once I have more time, I will do some code to solve these balances. I think there is some other guy that asked for something similar.