Question #19979

1 Answer
Ernest Z.
Oct 6, 2016

The partial pressure of "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.

color(white)(mmmmmm)"2NO" ⇌ "N"_2 + "O"_2
"I/atm:"color(white)(mmll)36.1color(white)(mmll)0color(white)(mm)0
"C/atm:"color(white)(mml)"-2"xcolor(white)(mml)"+"xcolor(white)(ml)"+"x
"E/atm:"color(white)(ml)"36.1 - 2"xcolor(white)(mll)xcolor(white)(mm)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_"O₂" = "18.0 atm"

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