At a particular temperature, K = 2.0 × 10-6 for the reaction 2CO2(g) ⇌ 2CO(g) + O2(g)
In an experiment 15.95 mol of CO2 is initially placed in a 1.6 L vessel. Calculate the equilibrium concentration of O2.
Give your answer to 4 decimal places.

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

Hope this helps

> b takes place in the gas phase in an isothermal tubular reactor. the feed is one mole of a per one mole of c, an inert. the entering temperature and pressure are 427°c and 10 atm, respectively. the gas constant r = 0.08206 atm · l/mol · k.

b. is the answer hoped i