Hydrogen peroxide may decompose to form water and oxygen according to the equation below. In a particular experiment, 1.75 moles of H2O2 were placed in a 2.5-L vessel at 307oC. After equilibrium, 1.20 moles of H2O2 remained. What is Kc?

Kc = 0.0231

Explanation:

2H₂O₂(g) <> 2H₂O(g) + O₂(g)

Before the reaction,

Conc of H₂O₂ = 1.75/2.5 = 0.7M

Conc of H₂O = 0M

Conc of O₂ = 0M

At equilibrium

Conc of H₂O₂ remaining = 1.2/2.5 = 0.48M

Conc of H₂0

We will obtain this from the stoichiometric balance

2 moles of H₂O₂ gives 2 moles of H₂O

(1.75 - 1.2) moles of H₂O₂ had reacted at equilibrium and will give 0.55 moles of H₂O

Conc of H₂O at equilibrium = 0.55/2.5 = 0.22M

Conc of O₂ at equilibrium = (0.55/2)/2.5 = 0.11M (the 0.55/2 is from the stoichiometric balance)

Kc = [H₂0]²[O₂]/[H₂O₂]²

Kc = [0.22]²[0.11]/[0.48]² = 0.0231 (L/mol) or (/M)

Hope this Helps!!!

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