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!




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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