Question

At a particular temperature, K = 2.0 × 10-6 for the reaction

Answer 1

Answer:

y = 0.037 M = [O₂]

Explanation:

Let's write the equilibrium reaction:

2CO₂ <--------> 2CO + O₂      K = 2x10⁻⁶

The problem states that we have initially 15.95 mol of CO₂ in a 1.6 L  vessel. With this we can determine the concentration:

M = moles / V

[CO₂] = 15.95 / 1.6 = 9.97 M

Now, we need to write an ICE chart so we can have the equilibrium concentrations:

          2CO₂ <--------> 2CO + O₂      K = 2x10⁻⁶

I.         9.97                     0       0

C.         -2y                  +2y     +y

E.        9.97-2y              2y      y

K = [CO]² [O₂] / [CO₂]²            Replacing we have:

2x10⁻⁶ = (2y)² (y)² / (9.97 - 2y)²      But K is <10⁻⁴ so we can neglect the substraction of 9.97 - 2y, we have then:

2x10⁻⁶ = 4y² (y) / (9.97)²

2x10⁻⁶ * (9,97)² = 4y³

y = ∛2x10⁻⁶ * (9,97)² / 4

y = 0.037 M = [O₂]

Hope this helps