equilibrium chem

Question

At a certain temperature, Ke = 8 for the reaction

2NO2(g) NO3(g) + NO(g)

If 5.0 mol of NO2(g) were added to a 1.0 L vessel, calculate the equilibrium concentrations of all the compounds.

Answer 1

Start with the balanced equilibrium reaction and set up an ICE chart. We'll use x to represent the amout of substances gained or lost and solve.

2 NO₂ ⇐⇒ NO₃ + NO

I 5.0 M 0 0

C - 2x + x + x

E 5.0 - 2x x x

_Keq = {[NO₃] [NO]}/ [NO₂]²

8 = (x) (x) / (5.0 - 2x)² since (x)(x) = x² we can simply take the square root of both sides then solve for x.

The square root of 8 is ∼ 2.8, so I'll use that value. 2.8 = x/(5.0 - 2x)

solving for x one gets x ≅ 2.1. Since K has only 1 sig fig, I'll use 2 for x.

Hence, the equilibrium concentrations will be: [NO₃] = [NO] = 2 M and [NO₂] = 5.0 - 2(2) = 1 M