# 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 $\setminus \texttt{{P}_{C {O}_{2}}}$ at equilibrium is determined to be $\setminus \textsf{\text{0.0721 atm}}$? The reaction is $\setminus \textsf{4 K {O}_{2} \left(s\right) + 2 C {O}_{2} \left(g\right) \setminus \leftrightarrow 2 {K}_{2} C {O}_{3} \left(s\right) + 3 {O}_{2} \left(g\right)}$ I tried to set up an ICE table, but... I can't really calculate when I have two variables for the $\setminus \textsf{2 C {O}_{2}}$ column. It's literally "$\setminus \texttt{{P}_{C {O}_{2} , \text{init}} + 2 x = 0.0721}$" right now. Also, if I've asked this before, sorry in advance!

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:

${K}_{P} = {P}_{{O}_{2}}^{3} / \left({P}_{C {O}_{2}}^{2}\right) = 28.5$ (implied units of $\text{atm}$)

ignoring the solids and putting them as $1$.

Since ${\text{CO}}_{2} \left(g\right)$ was "placed" into the flask, ${P}_{C {O}_{2} , i} > 0$. But this is fine, we have enough info. You know ${P}_{C {O}_{2}}$ at equilibrium.

Solve for ${P}_{{O}_{2}}$ at equilibrium.

$\textcolor{b l u e}{{P}_{{O}_{2}}} = {\left({K}_{P} {P}_{C {O}_{2}}^{2}\right)}^{1 / 3}$

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

$=$ $\textcolor{b l u e}{\text{0.529 atm}}$