# Concentration of products in tank reactor

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!

• You are interested in steady state conditions, that don't depend on start conditions. Nov 28, 2022 at 11:40
• @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? Nov 28, 2022 at 12:01
• 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) Nov 28, 2022 at 12:59
• @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$_{C1}$? Nov 28, 2022 at 13:49
• There are 3 steady concentrations and 3 equations for them balancing their positive and negative rates. Nov 28, 2022 at 14:08

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 $$A$$

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

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

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.

• How do you get those equations from the reaction stoichiometry? Nov 28, 2022 at 14:46
• 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. Nov 28, 2022 at 14:54
• @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. Nov 28, 2022 at 14:59
• @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. Nov 28, 2022 at 15:08
• 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. Nov 28, 2022 at 15:19