2. 16.1(b) Hess’s Law 1

3. 2 POINT > Recall enthalpies of reaction, formation and combustion POINT > Define Hess’s Law POINT > Use Hess’s law to determine enthalpy of reaction

4. 3 POINT > Recall enthalpy of reaction ∆H = H products - H reactants 2CO(g) + O 2 (g)  2CO 2 (g)  H = -566.0 kJ  H is determined experimentally. You have charts of Enthalpy of Formation and Combustion in your book (pages R66 and R60, respectively)

5. 4 POINT > Recall enthalpy of reaction 2CO(g) + O 2 (g)  2CO 2 (g)  H = -566.0 kJ Thermochemistry requires that you consider equation coefficients as moles (not molecules) So, if you wanted  H f for the reaction producing 1 mole of CO 2 you could (and will) do this: CO(g) + ½ O 2 (g)  CO 2 (g)  H = -283.0 kJ

6. 5 POINT > Recall enthalpy of formation For liquid water, H 2 O,  H f = -285.8 kJ You must generate the equation for this: 2H 2 (g) + O 2 (g)  2H 2 O(l) By definition,  H f is for the formation of 1 mole from its elements: H 2 (g) + ½ O 2 (g)  H 2 O(l)  H f = -285.8 kJ

7. 6 WB CHECK: What is the  H f for iron(III) oxide?  H f = -824.2kJ Generate the thermochemical equation that represents the formation of 1 mole of iron(III) oxide. 4Fe(s) + 3O 2 (g)  2Fe 2 O 3 (s) By definition,  H f is for the formation of 1 mole 2Fe(s) + 3/2O 2 (g)  Fe 2 O 3 (s)  H f = -824.2 kJ

8. 7 WB CHECK: What is the  H c for butane?  H c = -2877.6kJ Generate the thermochemical equation that represents the combustion of 1 mole of butane. 2C 4 H 10 (g) + 13O 2 (g)  10H 2 O(g) + 8CO 2 (g) By definition,  H c is for the combustion of 1 mole C 4 H 10 (g) + 13/2O 2 (g)  5H 2 O(g) + 4CO 2 (g)  H c = -2877.6 kJ

9. 8 POINT > Define Hess’s Law Hess’s Law: The overall enthalpy change  H) for any reaction is equal to the sum of enthalpy changes for each step in the reaction

10. POINT > Define Hess’s Law 9 Start Finish Enthalpy of reaction is a State function: it does not depend on the pathway Therefore, we can pick any pathway to calculate  H for a reaction

11. 10 POINT > Use Hess’s law to determine enthalpy of reaction Hess’s Law allows us to determine unknown  H of reactions by combining known  H reaction values (from the Enthalpy Tables in your Appendix)  H f for elements = 0

12. 11 WB CHECK: What is a “state function”? What is the enthalpy of formation  H f ) of Cl 2 ? Given: N 2 (g) + O 2 (g)  2NO(g)  H = 180.6 kJ What is the  H f for nitrogen monoxide?

13. 12 WB CHECK: Given: H 2 (g) + ½ O 2 (g)  H 2 O(g)  H = -241.8 kJ What is  H for: 2H 2 (g) + O 2 (g)  2H 2 O(g)? What is  H for: H 2 O(g)  H 2 (g) + ½ O 2 (g) ?

14. 13 POINT > Use Hess’s law to determine  H of reaction Given the following reaction, determine  H: 4NH 3 (g) + 5O 2 (g)  4NO(g) + 6H 2 O(g) The task is to piece together known reactions (from the Appendix) to be combined algebraically You need a known value for each reactant/product

15. 14 Determine enthalpy of reaction (  H) for the reaction: 4NH 3 (g) + 5O 2 (g)  4NO(g) + 6H 2 O(g) Using the following sets of known reactions: ½ N 2 (g) + ½ O 2 (g)  NO(g)*  H = 90.3 kJ ½ N 2 (g) + 3/2 H 2 (g)  NH 3 (g)  H = -45.9 kJ H 2 (g) + ½ O 2 (g)  H 2 O(g)  H = -241.8 kJ These three reactions must be algebraically manipulated to sum up to the desired reaction. and.. the  H values must be treated accordingly.

16. 15 4NH 3 (g) + 5O 2 (g)  4NO(g) + 6H 2 O(g) Using the following sets of known reactions: ½ N 2 (g) + ½ O 2 (g)  NO(g)  H = 90.3 kJ ½ N 2 (g) + 3/2 H 2 (g)  NH 3 (g)  H = - 45.9 kJ H 2 (g) + ½ O 2 (g)  H 2 O(g)  H = -241.8 kJ Goal: NH 3 : O 2 : NO: H 2 O: Reverse and x 4 4NH 3  2N 2 + 6H 2  H = +183.6 kJ  H = 0 x4 2N 2 + 2O 2  4NO  H = 361.2 kJ x6 6H 2 + 3O 2  6H 2 O  H = -1451.1 kJ

17. 16 4NH 3 (g) + 5O 2 (g)  4NO(g) + 6H 2 O(g) Goal: NH 3 : O 2 : NO: H 2 O: 4NH 3  2N 2 + 6H 2  H = + 183.6 kJ  H = 0 2N 2 + 2O 2  4NO  H = 361.2 kJ 6H 2 + 3O 2  6H 2 O  H = -1451.1 kJ Cancel terms and take sum. 4NH 3 + 5O 2  4NO + 6H 2 O  H = - 906.3 kJ Is the reaction endothermic or exothermic?

18. 17 Determine the enthalpy of reaction for the reaction: C 2 H 4 (g) + H 2 (g)  C 2 H 6 (g) Use the following reactions from the Appendix: C 2 H 4 (g) + 3O 2 (g)  2CO 2 (g) + 2H 2 O(l)  H = -1401 kJ C 2 H 6 (g) + 7/2O 2 (g)  2CO 2 (g) + 3H 2 O(l)  H = -1550 kJ H 2 (g) + 1/2O 2 (g)  H 2 O(l)  H = -286 kJ WB CHECK:

19. 18 Determine the heat of reaction for the reaction: Goal: C 2 H 4 (g) + H 2 (g)  C 2 H 6 (g)  H = ? C 2 H 4 (g) : C 2 H 4 (g) + 3O 2 (g)  2CO 2 (g) + 2H 2 O(l)  H = -1401 kJ H 2 (g): H 2 (g) + 1/2O 2 (g)  H 2 O(l)  H = -286 kJ C 2 H 6 (g) : 2CO 2 (g) + 3H 2 O(l)  C 2 H 6 (g) + 7/2O 2 (g)  H = +1550 kJ C 2 H 4 (g) + H 2 (g)  C 2 H 6 (g)  H = -137 kJ

20. 19 Homework: Finish reading 16.1 pp. 509-514 #6 page 514, #15a-c page 523

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