Answer:
[N₂] = 0.098M
[O₂] = 0.098M
[NO] = 0.004M
Explanation:
<em>The reaction is:</em>
N₂(g) + O₂(g) ⇄ 2 NO(g)
<em>Where K of equilibrium is:</em>
4.1x10⁻⁴ = [NO]² / [N₂] [O₂]
<em>Concentrations of each species are concentrations in equilibrium.</em>
<em />
If some N₂ and O₂ gases reacts producing NO, the equilibrium concentrations are:
[N₂] = 0.100M - X
[O₂] = 0.100M - X
[NO] = 2X
<em>Only 1 mole of N₂ and O₂ reacts producing 2 moles of NO</em>
<em />
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
<em />
<em>Solving for X:</em>
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
<h3 /><h3>[N₂] = 0.098M</h3><h3>[O₂] = 0.098M</h3><h3>[NO] = 0.004M</h3>
