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Ch. 5.6 Notes

Ch. 5.6 Notes. Hess’s Law. 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 CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g) ∆H = -802 kJ

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Ch. 5.6 Notes

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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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