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Given chemical equations for these reactions s(s) + o2(g)  so2(g) ∆h˚ –296.8 kj•mol–1 h2(g) + ½ o2(g)  h2o(l) ∆h˚ –285.8 kj•mol–1 h2(g) + s(s)  h2s(g) ∆h˚ –20.6 kj•mol–1 what is the value of ∆h for the reaction below? 2 h2s(g) + 3 o2(g)  2 h2o(l) + 2 so2(g) (a) –603.2 kj•mol–1 (b) –562.0 kj•mol–1

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Answer is: enthalpy is -1124 kJ/mol, if we divide reaction with two enthalpy is -562.0 kJ/mol.

Reaction 1: S(s) + O₂(g) → SO₂(g) ΔrH₁ = -296.8 kJ/mol.
Reaction 2: H₂(g) + ½ O₂(g) → H₂O(l) ΔrH₂ = -285.8 kJ/mol.
Reaction 3: H₂(g) + S(s) → H₂S(g) ∆rH₃ = -20.6 kJ/mol.
Reaction 4: 2H₂S(g) + 3O₂(g) → 2H₂O(l) + 2SO₂(g) ΔrH₄ = ?
Using Hess's law reaction number 4 is sum of reaction number 1 multiply with two, reaction number 2 multiply with two and reaction 3 reversed and multiply with two:
ΔrH₄ = 2 · (-296.8 kJ/mol) + 2 · (-285.8 kJ/mol) + 2 · 20.6 kJ/mol.
ΔrH₄ = -1124 kJ/mol.

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