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Hess's law
Hess's law
Hess's law
Hess's law
Hess's law
Hess's law
Hess's law
Hess's law
Hess's law
Hess's law
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Hess's law

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  • 1. 1 Hess’s Law Start Finish A State Function: Path independent. Both lines accomplished the same result, they went from start to finish. Net result = same.
  • 2. 2 Determine the heat of reaction for the reaction: 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g) Using the following sets of reactions: N2(g) + O2(g)  2NO(g) ∆H = 180.6 kJ N2(g) + 3H2(g)  2NH3(g) ∆H = -91.8 kJ 2H2(g) + O2(g)  2H2O(g) ∆H = -483.7 kJ Hint: The three reactions must be algebraically manipulated to sum up to the desired reaction. and.. the ∆H values must be treated accordingly.
  • 3. 3 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g) Using the following sets of reactions: N2(g) + O2(g)  2NO(g) ∆H = 180.6 kJ N2(g) + 3H2(g)  2NH3(g) ∆H = -91.8 kJ 2H2(g) + O2(g)  2H2O(g) ∆H = -483.7 kJ Goal: NH3: O2 : NO: H2O: Reverse and x 2 4NH3  2N2 + 6H2 ∆H = +183.6 kJ Found in more than one place, SKIP IT (its hard). x2 2N2 + 2O2  4NO ∆H = 361.2 kJ x3 6H2 + 3O2  6H2O ∆H = -1451.1 kJ
  • 4. 4 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g)Goal: NH3: O2 : NO: H2O: Reverse and x2 4NH3  2N2 + 6H2 ∆H = +183.6 kJ Found in more than one place, SKIP IT. x2 2N2 + 2O2  4NO ∆H = 361.2 kJ x3 6H2 + 3O2  6H2O ∆H = -1451.1 kJ Cancel terms and take sum. 4NH3 + 5O2  4NO + 6H2O ∆H = -906.3 kJ Is the reaction endothermic or exothermic?
  • 5. 5 Determine the heat of reaction for the reaction: C2H4(g) + H2(g)  C2H6(g) Use the following reactions: C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(l) ∆H = -1401 kJ C2H6(g) + 7/2O2(g)  2CO2(g) + 3H2O(l) ∆H = -1550 kJ H2(g) + 1/2O2(g)  H2O(l) ∆H = -286 kJ Consult your neighbor if necessary.
  • 6. 6 Determine the heat of reaction for the reaction: Goal: C2H4(g) + H2(g)  C2H6(g) ∆H = ? Use the following reactions: C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(l) ∆H = -1401 kJ C2H6(g) + 7/2O2(g)  2CO2(g) + 3H2O(l) ∆H = -1550 kJ H2(g) + 1/2O2(g)  H2O(l) ∆H = -286 kJ C2H4(g) :use 1 as is C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(l) ∆H = -1401 kJ H2(g) :# 3 as is H2(g) + 1/2O2(g)  H2O(l) ∆H = -286 kJ C2H6(g) : rev #2 2CO2(g) + 3H2O(l)  C2H6(g) + 7/2O2(g) ∆H = +1550 kJ C2H4(g) + H2(g)  C2H6(g) ∆H = -137 kJ
  • 7. 7 Summary: enthalpy is a state function and is path independent.
  • 8. 8
  • 9. 9 Standard Enthalpies of formation:
  • 10. 10 Thermodynamic Quantities of Selected Substances @ 298.15 K

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