Let consider a steady state CSTR in which occurs the following reaction:

$$\ce{aA + bB \leftrightarrow cC}$$


$F_i$ is the molar flow of $i$, $r$ is the rate of reaction, $V$ is the volume, $X$ is the conversion at the end of the reactor, $\nu_i$ is the stoichiometric coefficient of $i$ and $\xi$ is the extent of reaction.

Mass balance:

$$\ce{IN + PROD - CONS = OUT + VAR}$$


$$ \begin{cases} F_{A,in} + 0 - arV = F_{A,out} \\ F_{B,in} + 0 - brV = F_{B,out} \\ F_{C,in} + crV - 0 = F_{C,out} \\ \end{cases} $$

As $$\xi = -rV = -X\frac{F_{i,in}}{\nu_i}$$ we have for $\ce{A}$:

$$\begin{alignat}{} F_{A,out} &= F_{A,in} - arV \\ & = F_{A,in} - a(-\xi) \\ & = F_{A,in} - a\left(X\frac{F_{A,in}}{a}\right) \\ & = \left(1 - X \right)F_{A,in} \end{alignat}$$

For $B$ with can find as well:

$$F_{B,out} = \left(1 - X \right)F_{B,in} = \frac{b}{a}\left(1 - X \right)F_{A,in}$$

My problem with C:

Now if I follow my calculus for $\ce{C}$, here is what I get:

$$\begin{alignat}{} F_{C,out} &= F_{C,in} + crV \\ & = F_{C,in} + c(-\xi) \\ & = F_{C,in} + c\left(X\frac{F_{C,in}}{c}\right) \\ & = \left(1 + X \right)F_{C,in} \end{alignat}$$

Which states that if there is no $C$ at the beginning, there is no $C$ produced at all! Which is wrong. Can you tell me where I am wrong?

NB: I have no clue, which tags fit best for this question.

  • $\begingroup$ Sure about the $F_{B, out}$ expression? $\endgroup$ – Eashaan Godbole Aug 14 '19 at 21:18
  • 1
    $\begingroup$ @EashaanGodbole well I guess I am not wrong but please tell me what you think is wrong? $$\begin{alignat}{} F_{B,out} &= F_{B,in} - brV \\ & = F_{B,in} - b(-\xi) \\ & = F_{B,in} - b\left(X\frac{F_{B,in}}{b}\right) \\ & = \left(1 - X \right)F_{B,in} \end{alignat}$$ As $$\frac{a}{b} = \frac{F_A}{F_B}$$ it gives the result:$$F_{B,out} = \left(1 - X \right)F_{B,in} = \frac{b}{a}\left(1 - X \right)F_{A,in}$$ $\endgroup$ – Eichhörnchen Aug 14 '19 at 21:25
  • $\begingroup$ Oh, then it's no issue. The $\dfrac{a}{b} = \dfrac{F_A}{F_B}$ condition isn't mentioned in the question. $\endgroup$ – Eashaan Godbole Aug 16 '19 at 6:38

I can't tell you precisely where your reasoning starts to be wrong, but here is another way to think about the problem.

Notice that for a specific reaction as you gave here, the term $rV$ is always the same independently of the molecule you choose to determine it. It does vary for sure over time though (don't read I said it was a constant).

As you wrote:

$$\begin{cases} F_{A,in} + 0 - arV = F_{A,out} \\ F_{B,in} + 0 - brV = F_{B,out} \\ F_{C,in} + crV - 0 = F_{C,out} \\ \end{cases}$$

Thus, $$rV = \frac{F_{A,in} - F_{A,out}}{a} = \frac{F_{B,in} - F_{B,out}}{b} = \frac{F_{C,in} - F_{C,out}}{-c}$$

You can write what happens for $C$ in terms of $A$: $$\frac{F_{A,in} - F_{A,out}}{a} = \frac{F_{C,in} - F_{C,out}}{-c} \Leftrightarrow F_{C,out} = F_{C,in} + \frac{c}{a} \left(F_{A,in} - F_{A,out}\right)$$

Which boils down to, $$F_{C,out} = F_{C,in} + \frac{c}{a} X F_{A,in}$$

And if you write equations in function of a particular molecule, $A$, as you did here, make sure that the conversion you wrote $X$, is to be more precise, the conversion of $A$ at the exit of the reactor. Otherwise, it doesn't make sense.


Note as well that your equation for $B$ is wrong as well, using the same reasoning as before you should find: $$F_{B,out} = F_{B,in} - \frac{b}{a} X F_{A,in}$$ which make sense because the flow of $B$ must be less at the end of the reactor if this reaction really occurs.


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