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PSYCHO15rus [73]
3 years ago
7

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

ibrium concentrations of all species
Chemistry
1 answer:
agasfer [191]3 years ago
3 0

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>
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