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Understanding Hess's Law: Calculation of Enthalpy Changes in Chemical Reactions

Hess's Law states that the total enthalpy change of a reaction is the sum of the enthalpy changes for each individual step, regardless of the pathway taken. This foundational principle in thermodynamics allows chemists to calculate enthalpy changes for complex reactions through simpler steps. Examples include reactions like CH₄ + 2O₂ → CO₂ + 2H₂O, where the overall enthalpy change is the sum of the enthalpy changes for the individual steps. Practicing Hess's Law improves understanding of thermochemical equations and reaction energetics.

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Understanding Hess's Law: Calculation of Enthalpy Changes in Chemical Reactions

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  1. Ch. 5.6 Notes Hess’s Law

  2. Hess’s Law • If a reaction is carried out in a number of steps, H for the overall reaction is the sum of H for each individual step • Example CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) ∆H = -802 kJ 2H2O(g)  2H2O(l) ∆H = -88 kJ CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) ∆H = -890 kJ 5.6

  3. Note that: H1 = H2 + H3

  4. Example Given: Fe2O3 (s) + 3 CO (g)  2 Fe (s) + 3 CO2 (g) ΔH = -23 kJ 3 Fe2O3 (s) + CO (g)  2 Fe3O4 (s) + CO2 (g) ΔH = -39 kJ Fe3O4 (s) + CO (g)  3 FeO (s) + CO2 (g) ΔH =+18 kJ • What is ΔH for the following? FeO (s) + CO (g)  Fe (s) + CO2 (g) 5.6

  5. Example Given: 1/2 (Fe2O3(s) + 3 CO (g)  2 Fe (s) + 3 CO2 (g) ΔH = -23 kJ) -1/6 (3 Fe2O3 (s) + CO (g)  2 Fe3O4 (s) + CO2 (g) ΔH = -39 kJ) -1/3 (Fe3O4(s) + CO (g)  3 FeO (s) + CO2 (g) ΔH =+18 kJ) • What is ΔH for the following? FeO (s) + CO (g)  Fe (s) + CO2 (g) 5.6

  6. Example Given: 1/2 (Fe2O3(s) + 3 CO (g)  2 Fe (s) + 3 CO2 (g) ΔH = -23 kJ) -1/6 (3 Fe2O3 (s) + CO (g)  2 Fe3O4 (s) + CO2 (g) ΔH = -39 kJ) -1/3 (Fe3O4(s) + CO (g)  3 FeO (s) + CO2 (g) ΔH =+18 kJ) • What is ΔH for the following? FeO (s) + CO (g)  Fe (s) + CO2 (g) ∆H = -11 kJ 5.6

  7. Your turn! C6H4(OH)2 (aq)  C6H4O2 (aq) + H2 (g) ΔH = +177.4 kJ H2 (g) + O2 (g)  H2O2 (aq) ΔH = -191.2 kJ H2 (g) + ½ O2 (g)  H2O (g) ΔH = -241.8 kJ H2O (g)  H2O (l) ΔH = -43.8 kJ C6H4(OH)2 (aq) + H2O2 (aq) C6H4O2 (aq) + 2 H2O (l) ΔH = ??? 5.6

  8. Your turn! 1 (C6H4(OH)2(aq)  C6H4O2 (aq) + H2 (g) ΔH = +177.4 kJ) -1 (H2(g) + O2 (g)  H2O2 (aq) ΔH = -191.2 kJ) 2 (H2(g) + ½ O2 (g)  H2O (g) ΔH = -241.8 kJ) 2 (H2O (g)  H2O (l) ΔH = -43.8 kJ) C6H4(OH)2 (aq) + H2O2 (aq) C6H4O2 (aq) + 2 H2O (l) ΔH = -202.6 kJ 5.6

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