Answer:
[N₂] = 0.098M
[O₂] = 0.098M
[NO] = 0.004M
Step-by-step explanation:
The reaction is:
N₂(g) + O₂(g) ⇄ 2 NO(g)
Where K of equilibrium is:
4.1x10⁻⁴ = [NO]² / [N₂] [O₂]
Concentrations of each species are concentrations in equilibrium.
If some N₂ and O₂ gases reacts producing NO, the equilibrium concentrations are:
[N₂] = 0.100M - X
[O₂] = 0.100M - X
[NO] = 2X
Only 1 mole of N₂ and O₂ reacts producing 2 moles of NO
Replacing:
4.1x10⁻⁴ = [NO]² / [N₂] [O₂]
4.1x10⁻⁴ = [X]² / [0.100-X] [0.100-X]
4.1x10⁻⁴ = [X]² / [X²-0.2X + 0.01]
4.1x10⁻⁴X² - 8.2x10⁻⁵X + 4.1x10⁻⁶ = [X]²
-0.99959X² - 8.2x10⁻⁵ X+ 4.1x10⁻⁶ = 0
Solving for X:
X = -0.002. False solution. There is no negative concentrations.
X= 0.002. Right solution.
[N₂] = 0.100M - 0.002M = 0.098M
[O₂] = 0.100M - 0.002M = 0.098M
[NO] = 2*0.002 = 0.004M
[N₂] = 0.098M
[O₂] = 0.098M
[NO] = 0.004M