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!
...if
The reaction is
I tried to set up an ICE table, but... I can't really calculate when I have two variables for the
It's literally "
Also, if I've asked this before, sorry in advance!
1 Answer
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
Solve for
#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")#