Question 19979

Oct 6, 2016

The partial pressure of ${\text{O}}_{2}$ is 18.0 atm.

Explanation:

${K}_{P}$ is so large that the position of equilibrium will be almost entirely to the right.

Let's set up an ICE table to solve the problem.

$\textcolor{w h i t e}{m m m m m m} {\text{2NO" ⇌ "N"_2 + "O}}_{2}$
$\text{I/atm:} \textcolor{w h i t e}{m m l l} 36.1 \textcolor{w h i t e}{m m l l} 0 \textcolor{w h i t e}{m m} 0$
$\text{C/atm:"color(white)(mml)"-2"xcolor(white)(mml)"+"xcolor(white)(ml)"+} x$
$\text{E/atm:"color(white)(ml)"36.1 - 2} x \textcolor{w h i t e}{m l l} x \textcolor{w h i t e}{m m} x$

K_P = (P_"N₂"P_"O₂")/P_"NO"^2 = (x·x)/(36.1 -2x) = 2.40 × 10^3

x^2 = 2.40 × 10^3(36.1 -2x) = 8.664 × 10^4 - 4.80 × 10^3x

x^2 + 4.80 × 10^3x - 8.664 × 10^4 = 0#

$x = 18.0$

${P}_{\text{O₂" = "18.0 atm}}$