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I want to calculate the concentrations of all components in the outlet stream of the reaction: C$_2$H$_6$ -> C$_2$H$_4$ + H$_2$ (A -> B + C)

I have a tank reactor (isothermic and isobar): V= $305$ dm$^3$, C$_{A0}$ = $53.46$ mol/m$^3$, P = $2$ bar, E$_a$ = $130$ kJ/mol, A = $10^{13}$ s$^{-1}$. The in flow is $10$ Nm$^3$/h and has the same temperature as the reactor.

First of all, what is $10$ Nm$^3$/h (some places it says that it is normal cubic meter)?

I began by using the gas law to calculate the temperature, which I got to be $450$ K. Then by the use of the Arrhenius equation I got that the rate constant $k$ = $0.0081$ s$^{-1}$.

I then calculated the outlet concentration of A by using the material balance:

C$_{A0}$ $v$ - $k$ C$_{A1}$ V = C$_{A1}$ $v$

C$_{A1}$ = (C$_{A0}$ $v$) / ($v$ + $k$ V) = C$_{A0}$ / ($1$ + $k$ τ)

where τ = $0.305$ m$^3$ / $10$ Nm$^3$/h = $109.8$ s (assuming it is normal cubic meter)

So C$_{A1}$ = $28.3$ mol/m$^3$

But what I am having trouble with is how to calculate the concentration of B and C. How can I calculate B and C without having their start concentrations?

All help is appreciated!

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  • $\begingroup$ You are interested in steady state conditions, that don't depend on start conditions. $\endgroup$
    – Poutnik
    Nov 28, 2022 at 11:40
  • $\begingroup$ @Poutnik I don't really understand what that means. How could I calculate B and C then? Also is the calculations I've done correct or do I have to convert the inlet flow by using STP? I am also uncertain now if I should calculate the temperature or assume that it is STP? $\endgroup$
    – katara
    Nov 28, 2022 at 12:01
  • $\begingroup$ It was meant as a hint, was not going to write a full answer. If you open a water tap above an unplugged bath tube, there will get established such a water level, where outlet flow equals the inlet flow. This level ( a steady state) does not depend on what was the water level at the beginning. That applies to all A, B and C. Rate of in-flow + creation of X = rate of out-flow + consumption of X. (I.e. all A, B, C have steady concentrations) $\endgroup$
    – Poutnik
    Nov 28, 2022 at 12:59
  • $\begingroup$ @Poutnik I'm not expecting a full answer. I just don't really understand how to use the fact that A, B, C have steady state concentrations. I thought about making a material balance like this: C$_{A0}$ $v$ + k (C$_{B1}$+C$_{C1}$) V = (C$_{B1}$+C$_{C1}$)V, but then I would have two unknowns, unless because of the stoichometry I can say that C$_{B1}$=C$_{C1}$? $\endgroup$
    – katara
    Nov 28, 2022 at 13:49
  • $\begingroup$ There are 3 steady concentrations and 3 equations for them balancing their positive and negative rates. $\endgroup$
    – Poutnik
    Nov 28, 2022 at 14:08

1 Answer 1

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Since you have a flow reactor with constant $P$ and $T$, you have to consider the generation of volume.

Stoichiometry

The reaction stoichiometry gives

$$ C_A = \frac{C_{A0} (1-X)}{1+\varepsilon X} $$

$$ C_B = \frac{C_{A0} (\Theta_B+X)}{1+\varepsilon X} $$

$$ C_C = \frac{C_{A0} (\Theta_C+X)}{1+\varepsilon X} $$

Where $\Theta_B = \dfrac{C_{B0}}{C_{A0}}$ and $\Theta_C = \dfrac{C_{C0}}{C_{A0}}$ are the ratios of initial concentration of species. $X$ is the conversion and $\varepsilon$ is the fractional change in volume flowrate:

$$ v = v_0 (1+\varepsilon X) $$

To calculate $\varepsilon$, we start with

$$ \varepsilon = y_{A0} \delta$$

Where $$y_{A0} = \frac{C_{A0}}{C_{A0}+C_{B0}+C_{C0}}$$

and $\delta$ is the change in stoichiometric coefficients,

$$ \delta = \frac{\gamma_B + \gamma_C - \gamma_A}{\gamma_A} = \frac{1 + 1 - 1}{1} = 1$$

Where $\gamma$ is the stoichiometric coefficeint of each species in the reaction considered.

Mole balance of $A$

$$\text{(In) - (Out) - (Consumed) = (Accumulation)}$$

At steady-state, accumulation is zero.

$$ F_{A0} - F_{A} - r_A V = 0 $$

Dividing by $F_{A0}$:

$$ \frac{F_{A0} - F_{A}}{F_{A0}} - \frac{r_A V}{F_{A0}} = 0 $$

Note $\dfrac{F_{A0} - F_{A}}{F_{A0}}$ is the conversion $X$,

$$ X - \frac{r_A V}{F_{A0}} = 0 $$

Considering the rate law $r_A = k C_A$,

$$ V = \frac{F_{A0} X}{k C_A} $$

Using the stoichiometry,

$$ V = \frac{F_{A0} X (1+\varepsilon X)}{k C_{A0} (1-X) } $$

Note that $F_{A0}/C_{A0} = v_0$,

$$ V = \frac{v_0 X (1+\varepsilon X)}{k(1-X) } $$

You can solve for $X$ using this last equation. After that, you can calculate whichever concentration - $C_A, C_B, C_C$ - using the stoichiometric relations.

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  • $\begingroup$ How do you get those equations from the reaction stoichiometry? $\endgroup$
    – katara
    Nov 28, 2022 at 14:46
  • $\begingroup$ Start from the definiton of conversion (of A): $F_A=F_{A0}(1−X)$ (it's important to know that conversion defined for absolute molar quantities, not concentration). When you divide by $v_0$ to get concentrations, the RHS $F_{A0}$ becomes $C_{A0}$, but the LHS will have a $v/v_0$ factor, which leads to the $1+\varepsilon X$ factor of volume generation. $\endgroup$ Nov 28, 2022 at 14:54
  • $\begingroup$ @katara For the other species $B$ and $C$, you can derive the equations by considering the meaning of stoichiometry, that is $dF_A = -dF_B$. Integration, using the initial conditions and incorporating the volume factor to convert to concentrations will get you those equations. $\endgroup$ Nov 28, 2022 at 14:59
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    $\begingroup$ @katara I'm trying to be brief and thus not very rigorous as to keep a short comment but please check these infos on your preferred Reaction Engineering textbook. $\endgroup$ Nov 28, 2022 at 15:08
  • $\begingroup$ I am don't quite understand the $dF_A$=−$dF_B$ part. There isn't anything like this in my textbook hence why I am asking on here. I have never come across a question like this where the products are calculated without some sort of information about them. $\endgroup$
    – katara
    Nov 28, 2022 at 15:19

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