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Hess's Law

Hess's Law. Hess’s Law. Start. Finish. Path independent. Both lines accomplished the same result, they went from start to finish. Net result = same. USING ENTHALPY. Consider the formation of water H 2 (g) + 1/2 O 2 (g) --> H 2 O(g) + 241.8 kJ.

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Hess's Law

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  1. Hess's Law

  2. Hess’s Law Start Finish Path independent Both lines accomplished the same result, they went from start to finish. Net result = same.

  3. USING ENTHALPY Consider the formation of water H2(g) + 1/2 O2(g) --> H2O(g) + 241.8 kJ Exothermic reaction — heat is a “product” and ∆H = – 241.8 kJ

  4. USING ENTHALPY Making liquid H2O from H2 + O2 involves twoexothermic steps. H2 + O2 gas H2O vapor Liquid H2O

  5. The standard enthalpy of reaction (DH0 ) is the enthalpy of a reaction carried out at 1 atm. rxn Standard enthalpy of formation (DH0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm. The standard enthalpy of formation of any element in its most stable form is zero. f USING ENTHALPY ∆H0f Making H2O from H2 and O2 involves two steps. H2(g) + 1/2 O2(g) ---> H2O(g) + 242 kJ H2O(g) ---> H2O(liq) + 44 kJ ----------------------------------------------------------------- H2(g) + 1/2 O2(g) --> H2O(liq) + 286 kJ

  6. Hess’s Law & Energy Level Diagrams Forming H2O can occur in a single step or in a two steps. ∆Htotal is the same no matter which path is followed.

  7. ∆H along one path = ∆H along another path

  8. aA + bB cC + dD - [ + ] [ + ] = - S S = DH0 DH0 rxn rxn dDH0 (D) nDH0 (products) mDH0 (reactants) aDH0 (A) cDH0 (C) bDH0 (B) f f f f f f Hess’s Law: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps. (Enthalpy is a state function. It doesn’t matter how you get there, only where you start and end.)

  9. Enthalpy Values Depend on how the reaction is written and on phases of reactants and products H2(g) + 1/2 O2(g) --> H2O(g) ∆H˚ = -242 kJ 2 H2(g) + O2(g) --> 2 H2O(g) ∆H˚ = -484 kJ H2O(g) ---> H2(g) + 1/2 O2(g) ∆H˚ = +242 kJ H2(g) + 1/2 O2(g) --> H2O(liquid) ∆H˚ = -286 kJ

  10. 6.5

  11. Determine the heat of reaction for the reaction: 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g) Using the following sets of reactions: (1) N2(g) + O2(g)  2NO(g) H = 180.6 kJ (2) N2(g) + 3H2(g)  2NH3(g) H = -91.8 kJ (3) 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.

  12. Goal: 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g) Using the following sets of reactions: (1) N2(g) + O2(g)  2NO(g) H = 180.6 kJ (2) N2(g) + 3H2(g)  2NH3(g) H = -91.8 kJ (3) 2H2(g) + O2(g)  2H2O(g) H = -483.7 kJ 4NH3 2N2 + 6H2 H =+183.6 kJ NH3: (2)(Reverse and x 2) Found in more than one place, SKIP IT (its hard). O2 : NO: (1) (Same x2) 2N2 + 2O24NO H = 361.2 kJ H2O: (3)(Same x3) 6H2 + 3O26H2O H = -1451.1 kJ

  13. Goal: 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g) 4NH3  2N2 + 6H2 H =+183.6 kJ NH3: Reverse and x2 Found in more than one place, SKIP IT. O2 : NO: x2 2N2 + 2O24NO H = 361.2 kJ 6H2 + 3O26H2O H = -1451.1 kJ H2O: x3 Cancel terms and take sum. + 5O2 + 6H2O H = -906.3 kJ  4NH3 4NO Is the reaction endothermic or exothermic?

  14. C(graphite) + O2(g) CO2(g)DH0 = -393.5 kJ rxn S(rhombic) + O2(g) SO2(g)DH0 = -296.1 kJ rxn CS2(l) + 3O2(g) CO2(g) + 2SO2(g)DH0 = -1072 kJ rxn 2S(rhombic) + 2O2(g) 2SO2(g)DH0 = -296.1x2 kJ C(graphite) + 2S(rhombic) CS2 (l) C(graphite) + 2S(rhombic) CS2 (l) rxn rxn C(graphite) + O2(g) CO2(g)DH0 = -393.5 kJ + CO2(g) + 2SO2(g) CS2(l) + 3O2(g)DH0 = +1072 kJ rxn DH0 = -393.5 + (2x-296.1) + 1072 = 86.3 kJ rxn Calculate the standard enthalpy of formation of CS2 (l) given that: 1. Write the enthalpy of formation reaction for CS2 2. Add the given rxns so that the result is the desired rxn. 6.5

  15. Determine the heat of reaction for the reaction: TARGET  C2H4(g) + H2(g)  C2H6(g) Use the following reactions: (1) C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(l) H = -1401 kJ (2) C2H6(g) + 7/2O2(g)  2CO2(g) + 3H2O(l) H = -1550 kJ (3) H2(g) + 1/2O2(g)  H2O(l) H = -286 kJ Consult your neighbor if necessary.

  16. Determine the heat of reaction for the reaction: TARGET  C2H4(g) + H2(g)  C2H6(g) Use the following reactions: (1) C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(l) H = -1401 kJ (2) C2H6(g) + 7/2O2(g)  2CO2(g) + 3H2O(l) H = -1550 kJ (3) H2(g) + 1/2O2(g)  H2O(l) H = -286 kJ Consult your neighbor if necessary.

  17. 2C6H6(l) + 15O2(g) 12CO2(g) + 6H2O (l) - S S = DH0 DH0 DH0 - [ ] [ + ] = rxn rxn rxn [ 12x–393.5 + 6x–285.8 ] – [ 2x49.04 ] = -6534.88 kJ = 12DH0 (CO2) 2DH0 (C6H6) f f = - 3267.44 kJ/mol C6H6 6DH0 (H2O) -6534.88 kJ f 2 mol nDH0 (reactants) nDH0 (products) f f Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is 49.04 kJ/mol. 6.5

  18. Use the values of the heat of formation given to determine the molar heat Of combustion for acetylene (ethyne). ∆H0f (C2H2(g)) = +227 kJ/mol ∆H0f (CO2(g)) = - 394 kJ/mol ∆H0f (H2O(g)) = - 242 kJ/mol ANS: -1257 kJ/mol

  19. Hess’s Law Example Problem CH4 + 2O2 CO2 + 2H2O Calculate H for the combustion of methane, CH4

  20. Summary: Used for reactions that cannot be determined experimentally Summary: H is independent of the path taken

  21. Estimating Heats of Reaction Using BOND ENERGIES Extra

  22. Chemical bonding is actually a continuum, from very ionic (i.e., complete electron transfer) to pure covalent (i.e., equal electron sharing): • Very ionic----------------------Pure Covalent Incr. bond decreasing bond "strength“ --------------- "strength“ • Bond "strength" can thus be thought of in terms of the extent of electron transfer . • Another way to think about bond "strength" is how much energy is needed to break a bond apart (or how much energy is released when a bond forms).

  23. Bond ENERGY • These "bond energies" can be experimentally determined for simple molecules and estimated for others ASSUMPTIONS: • Bond energies are determined (measured and/or calculated) for gases only. • most bond energies are averages.

  24. WHY WE CARE ….? • b/c the changes in bond energy during a reaction are responsible for the heat energy associated with a chemical process. • Old bonds being broken • New bonds being formed

  25. Why DO WE CARE…. • Bond energies can be used to estimate ∆H for a reaction without doing an experiment or when an experiment is not practical. H2 + F2 2 HF (all gases) Is this reaction endothermic or exothermic?

  26. H2 + F2 2 HF (all gases) • So: • to break H-H bond = 436.4 kJ/mol • to break F-F bond = 156.9 kJ/mol • to make 2 H-F bonds = 2(568.2 kJ/mol) But which energy is going in and which is coming out? • The first two steps are endothermic; energy is required; we give them + signs.  breaking • The last step is exothermic; energy is released; we give it a - sign.  making

  27. Example: • Use average bond energies to estimate the enthalpy of reaction for the combustion of methane.

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