The following reaction establishes equilibrium at 2000 K: If the reaction began with 0.110 of and 0.110 of , what were the equilibrium concentrations of all species

[N₂] = 0.098M

[O₂] = 0.098M

[NO] = 0.004M

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

i think   the answer is c hope i'm right