Question #19979

1 Answer
Oct 6, 2016

Answer:

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"#