Consider the reaction N2(g)+O2(g)↽−−⇀2NO(g) N 2 ( g ) + O 2 ( g ) ↽ − − ⇀ 2 NO ( g ) that has an equilibrium constant, 𝐾c K c , fof 4.10×10−4 4.10 × 10 − 4 at 1700 1700 °C.

Question

Consider the reaction N2(g)+O2(g)↽−−⇀2NO(g) N 2 ( g ) + O 2 ( g ) ↽ − − ⇀ 2 NO ( g ) that has an equilibrium constant, 𝐾c K c , fof 4.10×10−4 4.10 × 10 − 4 at 1700 1700 °C.

What percentage of O2 will react to form NO if 0.603 mol N2 and 0.603 mol O2 are added to a 0.901 L container and allowed to come to equilbrium at 1700 °C?

Answer 1

N₂(g) + O₂(g) <==> 2NO(g) ... Kc = 4.10x10^-4

Initial concentrations are:

[N₂] = 0.603 mols / 0.901 L = 0.669 M

[O₂] = 0.603 mols / 0.901 L = 0.669 M

Set up an ICE table:

N₂(g) + O₂(g) <==> 2NO(g)

0.669....0.669..............0.............Initial

-x..........-x..................+2x...........Change

0.669-x.....0.669-x...........2x...........Equilibrium

Write the equilibrium expression:

Kc = 4.10x10^-4 = [NO]² / [N₂][O₂]

Plug in values and solve for x:

4.10x10^-4 = [2x]² / [0.669-x][0.669-x]

4.10x10^-4 = 4x² / (0.669-x)² ... Assume x is small relative to 0.669 and ignore it in denominator

4.10x10^-4 = 4x² / (0.669)²

4x² = 1.84x10^-4

x² = 4.59x10^-5

x = 6.67x10^-3 M = 0.00667 M

[O₂] @ equilibrium = 0.669 - 000667 = 0.662 M

% O₂ unreacted = 0.662 / 0.669 (x100%) =99.0%

% O₂ reacted = 100 - 99.0 = 1.00% reacted

Be sure to check the math.