Calculate the change of enthalpy for: 4S + 6O2= 4SO3 Delta H= ___ kJ

Solutions

Expert Solution

_fH⁰ S (monoclinic) = 0.3 KJ/mol

_fH⁰ O_2(g) = 0 KJ/mol

_fH⁰ SO₃_(g) = -395.7 KJ/mol

H⁰ = _fH⁰ (Product) - _fH⁰ (Reactant)

H⁰ = [(4_fH⁰ SO₃] - [(4_fH⁰ S) + (6_fH⁰ O₂)]

= [(4 (-395.7)) ] - [4(0.3) + (6 (0)]

= [-1582.8] - [1.2]

H⁰ = -1584 KJ/mol

H⁰ for reaction = -1584 KJ/mol

queen_honey_blossom answered 9 months ago

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