Some time ago, I was going through some exercises and I came across a odd question, for which my atomic species degrees of freedom did not match my degrees of freedom using molecular species. I showed my lecturer and he couldn't see why this was the case. The (summarized) question is as follows: 2 reactions occur in a reactor:

\begin{align} \text{C}_2\text{H}_6\text{O} + \text{H}_2\text{O} &\rightarrow 2\text{CH}_3\text{OH} \\ 2\text{CH}_3\text{OH} &\rightarrow \text{C}_2\text{H}_4 + 2\text{H}_2\text{O} \end{align}

The input feed of water and $\text{C}_2\text{H}_6\text{O}$ is specified and the selectivity of the first equation. Reactions do not go to completion.

So, there are 4 unknowns ($\text{H}_2\text{O}$, $\text{C}_2\text{H}_4$, $2\text{CH}_3\text{OH}$, and $\text{C}_2\text{H}_6\text{O}$).

Doing the atomic analysis: $$\text{DoF} = 4 - 3(\text{C},\text{H},\text{O}) - 1 (\text{Selectivity}) = 0$$

Doing the molecular species: $$\text{DoF} = 4 + 2 - 4(\text{H}_2\text{O}, \text{C}_2\text{H}_4, 2\text{CH}_3\text{OH}, \text{C}_2\text{H}_6\text{O}) - 1 (\text{Selectivity}) = 1$$

Do you see where the problem is?

  • $\begingroup$ Its been a while; where do the two additional unknowns come from in the molecular species dof analysis? $\endgroup$ – nluigi Apr 29 '16 at 15:53
  • 1
    $\begingroup$ You add an unknown per independent reaction to account for the unknown extent of reaction. $\endgroup$ – Ender Delat Apr 29 '16 at 15:59
  • $\begingroup$ Selectivity is the ratio of CH3OH to C2H4 produced. So we shouldn't talk about the selectivity of the first equation. $\endgroup$ – Salomon Turgman May 26 '16 at 16:50

The answer to your questions is that you calculate the DOF for a reactor with parallel reactions the same way you would any other process. Your problem is unrelated to the fact that you have parallel reactions.

The reason you have a different number of DOF for molecular species vs atomic species is because you have one less equation than you think for the atomic species balance. Remember that the equations resulting from your three atomic balances should be independent. If they are not, you have to count one less equation. The way to check for this (short of trying to solve the problem and failing) is to write the three atomic balance equations and check whether you can get one of them by algebraic manipulations of the other two. Let us label $n_1$ and $n_2$ as the input molar flow rate of diethyl ether (DE) and water (W), respectively; and $n_3$, $n_4$, $n_5$, and $n_6$ as the output molar flow rate of DE, W, ethylene (E), and methanol (M), respectively. Then the atomic balances read: $$ \begin{align} \text{(1) Oxygen: } & n_1+n_2=n_3+n_4+n_6 \\ \text{(2) Hydrogen: } & 6n_1+2n_2=6n_3+2n_2+4n_5+4n_6 \\ \text{(3) Carbon: } & 2n_1=2n_3+2n_5+n_6 \end{align} $$ If you take two times equation (1) and add two times equation (3) you get equation (2) (verify this). This means that the set of equations is not independent and you can only count two atomic species in your balance instead of three. Both approaches then give you 1 DOF.

On a side note, the selectivity should be defined as: $$ \frac{\text{moles of desired product}}{\text{moles of undesired product}} $$ which mean that it should not be defined with respect to the first reaction but with respect to the entire reactor.


The difference is due to you missing an unknown for the molecular degree of freedom analysis. You need two selectivity unknowns as you have two equations, which will progress at different rates.

i.e. The selectivity for the first equation will tell you how much methanol is present, but you need another selectivity variable to tell you how much of that methanol has been converted to ethene and water.

  • $\begingroup$ So if I added a unknown to the mol. species DoF would I not then get 5+2-4-1=2 Should not the Atomic species analysis give the same number of DoF as the other analysis? $\endgroup$ – Ender Delat Apr 29 '16 at 16:00
  • $\begingroup$ @EnderDelat No the number of unknowns don't change but the number of equations increased by 1 so you end up with a dof=0 $\endgroup$ – nluigi Apr 29 '16 at 16:04
  • $\begingroup$ @nluigi I understand that if a second selectivity/extent was specified then DoF=0 however in the question there wasn't any other data present. Surely, introducing a new selectivity equation would have that selectivity as a unknown in it so there would be no net change. $\endgroup$ – Ender Delat Apr 29 '16 at 19:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.