- #1

Saladsamurai

- 3,020

- 7

## 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:

[itex]CO_2 \leftrightharpoons X_{CO}CO + X_{CO_2}CO_2 + X_{O_2}O_2[/itex]

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.

[tex]\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}}[/tex]Now I can see that what they have essentially written is

[tex]\frac{\text{No. oxygen atoms}}{\text{No. carbon atoms}}

=

\frac{\text{mole fractions of everything with carbon in it}}{\Sum\text{mole fractions of everything with oxygen in it}}[/tex]

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

[tex]

\frac{X_{CO}+X_{CO_2}}{X_{CO}+X_{CO_2}+X_{O_2}} =

\frac{N_{CO}/N_T+N_{CO_2}/N_T}{N_{CO}/N_T+N_{CO_2}/N_T+N_{O_2}/N_T} =

\frac{N_{CO}+N_{CO_2}}{N_{CO}+N_{CO_2}+N_{O_2}}

[/tex]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?

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