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

What percentage of O2 O 2 will react to form NO if 0.883 0.883 mol N2 N 2 and 0.883 0.883 mol O2 O 2 are added to a 0.605 0.605 L container and allowed to come to equilbrium at 1700 1700 °C?

Answer 1

K_C = [NO]²/([N₂][O₂] Let x be O₂ moles reacted in order to reach equilibrium

4.1 x 10^-4 = (2x/V)²/((.883-x)/V)² V cancels and we are left with:

4.1 x 10^-4 = 4x²/(.883-x)²

If we make the assumption that x << .883:

sqrt((4.1 x 10^-4)(.883)²/4) = x = .00894 (about 1%. You can solve quadratic to get better answer: .0885 or so)

Please consider a tutor. Take care.

Answer 2

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

0.883......0.833.............0...........Initial

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

0.883-x...0.833-x.........2x..........Equilibrium

Kc = 4.10x10^-4 = [NO]² / [N₂][O₂]

4.10x10^-4 = (2x)² / (0.883-x)(0.833-x) ... assume x is small relative to 0.833 and ignore it

4.10x10^-4 = (2x)² / (0.883)(0.833) = 4x² / 0.7797

4x2 = 3.2x10^-4

x = 9.94x10^-3 = moles O2 that reacted

% O₂ reacted = 9.94x10^-3 mols/0.883 mols (x100%) = 1.13% (which is small so above assumption was valid)