Given the heat formation of KCl(s) (-435.9) and KClO3(s) (-391.4). Calculate the standard molar enthalpy of reaction for 2KCl(s) + 3O2(g) à 2KClO3(s)

Question

When potassium chloride reacts with oxygen under the right conditions, potassium chloride is formed. Given the heat formation of KCl(s) (-435.9) and KClO3(s) (-391.4). Calculate the standard molar enthalpy of reaction for 2KCl(s) + 3O2(g) à 2KClO3(s)

WITH FORMULA:

Answer 1

2KCl(s) + 3O₂(g) ==> 2KClO₃(s)

Standard molar enthalpy of reaction (∆Hºrxn) = difference in heat of formation of products and reactants.

∆Hf_reactants = 2 x -435.9 + 0 = -871.8 (note; ∆Hf for O2 is zero)

∆Hf_products = 2 x -391.4 = -782.8

∆Hrxn = -782.8 - (-871.8) = +89 (units omitted since you didn't include them in the question. Should be kJ/mol)