Concentration of species based on what basis is used

[gfd43tg]

Homework Statement

For the gas phase reaction with an equimolar feed of ##N_{2}## and ##H_{2}##

## \frac {1}{2} N_{2} + \frac {3}{2} H_{2} \rightarrow NH_{3}##

If you took ##N_{2}## as your basis of calculation, could 60% conversion of ##N_{2}## be achieved?

Homework Equations

The Attempt at a Solution

First, I will assume this is isothermal and isobaric.

Since I use nitrogen as my basis, I will divide all species by 1/2 and use letters to differentiate the species,

##A + 3B \rightarrow 2C##

And I set up a stoichiometric table (that LaTeX is not approving of for some reason... Edit by Borek - there was a parentheses mismatch, classic typo)

[tex]
\begin{pmatrix}
Species & Initial & Change & Final \\
A & C_{A0} & -C_{A0}X & C_{A0}(1-X) \\
B & C_{A0} \theta_{B} & -3C_{A0}X & C_{A0}(\theta_{B} - 3X) \\
C & 0 & +2C_{A0}X & 2C_{A0}X
\end{pmatrix}[/tex]

I calculate ##\delta = \sum_{i} \nu_{i} = 2 - 3 - 1 = -2##. Given that this is an equimolar mixture in the feed, that means ##y_{A0} = 0.5##. So I calculate ##\epsilon \equiv y_{A0} \delta = 0.5(-2) = -1##. Now I find the concentration of species A as a function of conversion with the equation

[tex] C_{i} = C_{A0} \frac{\theta_{i} + \nu_{i}X}{1 + \epsilon X} [/tex]

Since ##\theta_{A} = 1## and ##\nu_{A} = -1## and ##\epsilon = -1##, this simplifies to

[tex]C_{A} = C_{A0} \frac{1 - X}{1 - X} [/tex]

So obviously the answer is no, this is confirmed by the fact that I did the same with other species and get a negative concentration, which is not possible. However, when I saw ##C_{A} = C_{A0}##, and that the terms involving conversion cancel out, I was very curious to the implications of this.

Is this interpreted to mean that no reaction occurs? I mean, the value of conversion has no baring if it cancels out anyways. This means the concentration of nitrogen is the same no matter what the conversion is? Can't I manipulate variables to work out in such a way that concentration as a function of conversion would give something like this where those terms cancel out?

 

[epenguin]

Homework Statement

For the gas phase reaction with an equimolar feed of ##N_{2}## and ##H_{2}##

## \frac {1}{2} N_{2} + \frac {3}{2} H_{2} \rightarrow NH_{3}##

If you took ##N_{2}## as your basis of calculation, could 60% conversion of ##N_{2}## be achieved?

Homework Equations

The Attempt at a Solution

First, I will assume this is isothermal and isobaric.

Since I use nitrogen as my basis, I will divide all species by 1/2 and use letters to differentiate the species,

##A + 3B \rightarrow 2C##

And I set up a stoichiometric table (that LaTeX is not approving of for some reason...)

[tex]
\begin{bmatrix}
Species & Initial & Change & Final \\
A & C_{A0} & -C_{A0}X & C_{A0)(1-X) \\
B & C_{A0} \theta_{B} & -3C_{A0}X & C_{A0}(\theta_{B} - 3X) \\
C & 0 & +2C_{A0}X & 2C_{A0}X
\end{bmatrix}[/tex]

I calculate ##\delta = \sum_{i} \nu_{i} = 2 - 3 - 1 = -2##. Given that this is an equimolar mixture in the feed, that means ##y_{A0} = 0.5##. So I calculate ##\epsilon \equiv y_{A0} \delta = 0.5(-2) = -1##. Now I find the concentration of species A as a function of conversion with the equation

[tex] C_{i} = C_{A0} \frac{\theta_{i} + \nu_{i}X}{1 + \epsilon X} [/tex]

Since ##\theta_{A} = 1## and ##\nu_{A} = -1## and ##\epsilon = -1##, this simplifies to

[tex]C_{A} = C_{A0} \frac{1 - X}{1 - X} [/tex]

So obviously the answer is no, this is confirmed by the fact that I did the same with other species and get a negative concentration, which is not possible. However, when I saw ##C_{A} = C_{A0}##, and that the terms involving conversion cancel out, I was very curious to the implications of this.

Is this interpreted to mean that no reaction occurs? I mean, the value of conversion has no baring if it cancels out anyways. This means the concentration of nitrogen is the same no matter what the conversion is? Can't I manipulate variables to work out in such a way that concentration as a function of conversion would give something like this where those terms cancel out?

I cannot follow your math with its undefined terms, but I am not motivated to as your result is, as you recognise, impossible. I also do not see how you can solve this problem without knowing an equilibrium constant for the reaction. Congratulations for making the effort, but one of the things you need to develop a sense for is knowing whether you have enough information for a solution to be possible.

It seems to me if you can get pressures as high has you like an equilibrium equation tell you you can get as close as you like to 100% conversion - but I think near that is way above the practical. However the industrial process is not done in a single passage but the ammonia is removed and the reactants recycled. If you do that enough times without introducing new material it seems to me you can get to whatever percentage you like - well not 100 or above.

So I don't think you are given - or have relayed - enough information.

If you remove the phrase "if you took N2 as basis of the calculation", does the sentence still correspond to the question?

I don't imagine anyone worries about the N2 conversion, since it is cheap and H2 is expensive, but we are into understanding of principle.

 

[gfd43tg]

Here is the attachment of the problem statement. I was doing part (c), so I didn't want to clutter up the page.

I don't think the equilibrium concentration is necessary. The definition of is ##\theta_{i} = \frac {y_{i0}}{y_{A0}}##.

 

[Borek]

We still have no idea what yi0 and yA0 are.

(I am not saying I will try to solve the problem, all I am saying is, epenguin's statement about not being able to follow your math still holds).

Could be not being able to reach some "level of conversion" is just a side effect of the way it is defined (which doesn't have to follow logic nor intuition; as we know set can be both open and closed at the same time ;) ).

 

[gfd43tg]

I thought that y is standard notation for the mole fraction of a gaseous species, so that's why I didn't mention it.

 

[epenguin]

Now you have quoted the original I am confirmed in thinking the phrase "if you took N₂ as basis of the calculation" meaningless and if it were not there the question would be clearer.

As this math suggests you are supposing profundities which are not there let me put the question another way.
H₂ and N₂ are present in equal molarities, but they react in molar ratio 3:1
If all the hydrogen reacted how much of the nitrogen reacts, and is that more than 60% or not?

Simples.