Question

For the following reaction, with `\sf{K_p=28.5" @ 25°C"}`; if `\tt{KO_2}` solid and `\tt{CO_2(g)}` are placed in an evacuated flask and equilibrium is established, what is `\tt{P_(O_2)}` at equilibrium...?

...if `\tt{P_(CO_2)}` at equilibrium is determined to be `\sf{"0.0721 atm"}`?

The reaction is
`\sf{4KO_2(s)+2CO_2(g)\harr2K_2CO_3(s)+3O_2(g)}`

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I tried to set up an ICE table, but... I can't really calculate when I have two variables for the `\sf{2CO_2}` column.
It's literally "`\tt{P_{CO_2,"init"}+2x=0.0721}`" right now.

Also, if I've asked this before, sorry in advance!

Answer 1

You're overthinking it. This is the product of getting into a habit with ICE tables and just doing what you think you should do, rather than writing down what variables you know.

Write the mass action expression:

`K_P = P_(O_2)^3/(P_(CO_2)^2) = 28.5` (implied units of `"atm"`)

ignoring the solids and putting them as `1`.

Since `"CO"_2(g)` was "placed" into the flask, `P_(CO_2,i) > 0`. But this is fine, we have enough info. You know `P_(CO_2)` at equilibrium.

Solve for `P_(O_2)` at equilibrium.

`color(blue)(P_(O_2)) = (K_P P_(CO_2)^2)^(1//3)`

`= ("28.5 atm" cdot ("0.0721 atm")^2)^(1//3)`

`=` `color(blue)("0.529 atm")`