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)}#

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!

1 Answer
Truong-Son N.
Aug 6, 2018

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")#

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