Using the provided table and the equation below, determine the heat of formation for KClO₂. 2 KClO₃ (s) → 2 KClO₂ (s) + O₂ (g) ∆H° = 296.2 kJ/mol

Question

Substance ∆H_f° (kJ/mol)

KClO₃ (s) -391.2

O_{2 (g)} 0

Answer 1

2KClO₃(s) ===> 2KClO₂(s) + O₂(g)

To find ∆H_f for KClO₂, we use the fact that the ∆H_reaction = ∑n∆H_products - ∑n∆H_products ...(equation 1)

∑n∆Hproducts = 2x∆HKClO2 + 0 = 2x since we don't know ∆HKClO2 we call it x

∑n∆Hreactants = 2x-391.2 = -782.4

Using equation 1 and substituting:

296.2 = 2x - (-782.4)

296.2 = 2x + 782.4

2x = 296.2 - 782.4

2x = -486.2

x = -243.1 kJ/mol