Given the following equilibrium constants: Na2O(s) ⇌ 2 Na(l) + 1/2 O2(g) K1= 2 x 10^–25

NaO(g) ⇌ Na(l) + 1/2 O2(g) K2= 2 x 10^–5

Na2O2(s) ⇌ 2 Na(l) + O2(g) K3= 5 x 10^–29

NaO2(s) ⇌ Na(l) + O2(g) K4= 3 x 10^–14

Determine the value for the equilibrium constants for the following reaction:

2 NaO(g) ⇌ Na2O2(s)

K = 8x10¹⁸

Explanation:

When you sum a reaction, the result of this sum has a K equal to the multiplication of the K's of the reactions involved in the sum

The sum of two times the reaction:

NaO(g) ⇌ Na(l) + 1/2 O₂(g) K₂ = 2x10⁻⁵

2 NaO(g) ⇌ 2 Na(l) + O₂(g) K = K₂ₓK₂ = (2x10⁻⁵)² = 4x10⁻¹⁰

The result of the inverse reaction:

Na₂O₂(s) ⇌ 2 Na(l) + O₂(g) K₃= 5x10⁻²⁹

2 Na(l) + O₂(g) ⇌ Na₂O₂(s) K = 1/K₃ = 2x10²⁸

And the sum of the two bolded reactions:

2 NaO(g) + 2 Na(l) + O₂(g) ⇌ 2 Na(l) + O₂(g) + Na₂O₂(s)

2 NaO(g) ⇌ Na₂O₂(s) K = 4x10⁻¹⁰× 2x10²⁸

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the answer is true.