# Mass transfert in process chemistry engineering, two film theory

I'm having troubles to understand a part of my lesson. I have no idea where to ask my question, but it is consider as chemistry so here we go.

Let me introduce some notations :

$$x$$ and $$y$$ are the mole fractions of the same compound but respectively in liquid phase and in gas phase.

$$k_{c_l}$$ and $$k_{c_g}$$ are the coefficient of exchange of matter in $$\ce{m.s^{-1}}$$

$$N_{1n}$$ is the molar flux density of the compound $$1$$ in the normal position

Let now me introduce (translating from french to english) you the part of the lessons in which I have troubles :

No phenomenas of production or consommation of the compound $$1$$ occur in the interface $$I$$, and we are in stationnary state then the molar flux density is the same in both parts and we have $$N_{1n}=c_lk_{c_l}\left(x_1^{I}-x_1\right)=c_gk_{c_g}\left(y_1-y_1^{I}\right)$$

The fluids at the interface are supposed to be in thermodynamical equilibrum then we have $$y_1^I=K_1^{eq}x_1^I$$ We can for example define a fictive gas phase which will be in equilibrium with the liquid phase (or the contrary). The mole fraction of the compound $$1$$ is by definition $$y_1^*=K_1^{eq}x_1$$

The total driving force is then \begin{align} (y_1-y_1^*)&=(y_1-y_1^I)+(y_1^I-y_1^*)\\ &=(y_1-y_1^I)+ K_1^{eq}\left(x_1^I-x_1\right)\end{align} if we suppose to have a unique average equilibrum constant in the liquid film.

Now I'm having troubles their :

We can then define a global transfer constant in the ralation to the driving force in gas phase : $$\frac{N_{1n}}{K_{c_g}c_g}=\frac{N_{1n}}{k_{c_g}c_g}+\frac{N_{1n}K_1^{eq}}{k_{c_l}c_l}$$

I really don't understand this last relation, if someone can help me, I would be glad. I apologize in advance for the troubles you can have too, to understand what I tried to translate here.

Thank you

PS : I have really no idea for the tags, feel free to edit them

## 1 Answer

You already derived this relation (eq 1),
\begin{align} (y_1-y_1^*)&=(y_1-y_1^I)+(y_1^I-y_1^*)\\ &=(y_1-y_1^I)+ K_1^{eq}\left(x_1^I-x_1\right)\end{align}

Now, to keep the molar flux density between the gas phase concentration $y_1$ and fictitious conc. on the other side $y_1^*$ we assumed a fictitious overall transfer coefficient $K_{cg}$. So, now we have another relation for molar flux density
$$N_{1n}=K_{cg}c_g\times(y_1-y_1^*)$$ that follows, $$y_1-y_1^*=\frac{N_{1n}}{K_{cg}c_g}$$ Similarly, $$y_1-y_1^I=\frac{N_{1n}}{k_{cI}c_I}$$ and $$x_1^I-x_1=\frac{N_{1n}}{k_{cg}c_g}$$ Plug these above three relations into eq 1 and you will get your last equation.

• Thank you a lot. I was on this for more than 4 hours... studies ... bam – ParaH2 Aug 23 '16 at 20:09