Atomic Ratio and Ratio of Mole Fractions

[Saladsamurai]

Homework Statement

Hi folks!

This is just a conceptual question that has arisen during some reading. At one point the author states that for the reaction:

$CO_2 \leftrightharpoons X_{CO}CO + X_{CO_2}CO_2 + X_{O_2}O_2$

where X is the mole fraction of each component at equilibrium, that we can relate the ratio of oxygen atoms:carbon atoms to the mile fractions by the following.

$$\frac{\text{No. carbon atoms}}{\text{No. oxygen atoms}}=\frac{1}{2}=\frac{X_{CO}+X_{CO_2}}{X_{CO}+X_{CO_2}+X_{O_2}}$$Now I can see that what they have essentially written is

$$\frac{\text{No. oxygen atoms}}{\text{No. carbon atoms}}<br /> =<br /> \frac{\text{mole fractions of everything with carbon in it}}{\Sum\text{mole fractions of everything with oxygen in it}}$$

Now intuitively this makes sense to me and I can dig it! BUT, I would like to make the math work to prove it to myself, but I cannot seem to figure it out

The Attempt at a Solution

This is what I did to try to "prove" it. Since X_i = N_i / N_T where N_T is the total number of moles in the mixture at equilibrium, I can write

$$\frac{X_{CO}+X_{CO_2}}{X_{CO}+X_{CO_2}+X_{O_2}} = <br /> \frac{N_{CO}/N_T+N_{CO_2}/N_T}{N_{CO}/N_T+N_{CO_2}/N_T+N_{O_2}/N_T} =<br /> \frac{N_{CO}+N_{CO_2}}{N_{CO}+N_{CO_2}+N_{O_2}}$$I am just not sure where to go from here? I thought about writing each N_i as something like: N_CO = (N_C + N_O), but did not get too far.

Any thoughts?

 

[Borek]

$$\frac{\text{No. oxygen atoms}}{\text{No. carbon atoms}}=\frac{1}{2}$$

And not

$$\frac{\text{No. oxygen atoms}}{\text{No. carbon atoms}}=\frac{2}{1}$$?

After all that's the ratio in CO₂ which is present initially.

I will try to get back to the problem, have to do something else at the moment. Seems like your approach should lead to the correct conclusion.

 

[Borek]

How many atoms of everything in 1 mole of CO₂?

And you probably need to use stoichiometry of the reaction:

2CO₂ <-> 2CO + O₂

 

[Saladsamurai]

And not

$$\frac{\text{No. oxygen atoms}}{\text{No. carbon atoms}}=\frac{2}{1}$$?

After all that's the ratio in CO₂ which is present initially.

I will try to get back to the problem, have to do something else at the moment. Seems like your approach should lead to the correct conclusion.

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Yes Borek; my mistake. I have edited to reflect the correct problem.

How many atoms of everything in 1 mole of CO₂?

And you probably need to use stoichiometry of the reaction:

2CO₂ <-> 2CO + O₂

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So if me let a_i be the number of atoms in the i^th species and N_Ais Avagadros number, we have for 1 mole of CO₂:

$$a_{\text{1mol}_{CO_2}} = \text{1mol}_C*N_A\frac{\text{atoms}} + 2*\text{1mol}_O*N_A\frac{\text{atoms}}{\text{mol}} = 3N_A \, \text{atoms}$$

I am can presumably do this for all remaining species and find that the ration is indeed 1/2. Now that I think about it, if I really wanted to *prove* it in general I should have let the subscripts be dummy variables as well as the species names...but that is for another time. When I get the time I will do this on paper since I am still not sure where I would use the stoichiometry of the rxn.

 

[Borek]

I am still not sure where I would use the stoichiometry of the rxn.

For example - to combine amounts of CO and O₂ present. If you assume initially there was CO₂ only, you can also calculate X_CO2 knowing X_CO (or X_O2)