Question

Question #19979

Answer 1

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"`