2. Lecture 4: Hess’ Law Reading: Zumdahl 9.5 Outline –Definition of Hess’ Law –Using Hess’ Law (many examples)

3. Hess’ Law Defined From lecture 3: Enthalpy is a state function. As such,  H for going from some initial state to some final state is pathway independent. Hess’ Law:  H for a process involving the transformation of reactants into products is not dependent on pathway. Therefore, we can pick a single-step pathway to calculate  H for a reaction.

4. Hess’ Law: An Example

5. Using Hess’ Law When calculating  H for a chemical reaction as a single step, we can use combinations of reactions as “pathways” to determine  H for our “single step” reaction.

6. Example (cont.) Our reaction of interest is: N 2 (g) + 2O 2 (g) 2NO 2 (g)  H = 68 kJ This reaction can also be carried out in two steps: N 2 (g) + O 2 (g) 2NO(g)  H = 180 kJ 2NO (g) + O 2 (g) 2NO 2 (g)  H = -112 kJ

7. Example (cont.) If we take the previous two reactions and add them, we get the original reaction of interest: N 2 (g) + O 2 (g) 2NO(g)  H = 180 kJ 2NO (g) + O 2 (g) 2NO 2 (g)  H = -112 kJ N 2 (g) + 2O 2 (g) 2NO 2 (g)  H = 68 kJ

8. Example (cont.) Note the important things about this example, the sum of  H for the two reaction steps is equal to the  H for the reaction of interest. Big point: We can combine reactions of known  H to determine the  H for the “combined” reaction.

9. Hess’ Law: Details One can always reverse the direction of a reaction when making a combined reaction. When you do this, the sign of  H changes. N 2 (g) + 2O 2 (g) 2NO 2 (g)  H = 68 kJ 2NO 2 (g) N 2 (g) + 2O 2 (g)  H = -68 kJ

10. Details (cont.) The magnitude of  H is directly proportional to the quantities involved (it is an “extensive” quantity). As such, if the coefficients of a reaction are multiplied by a constant, the value of  H is also multiplied by the same integer. N 2 (g) + 2O 2 (g) 2NO 2 (g)  H = 68 kJ  N 2 (g) + 4O 2 (g) 4NO 2 (g)  H = 136 kJ

11. Using Hess’ Law When trying to combine reactions to form a reaction of interest, one usually works backwards from the reaction of interest. Example: What is  H for the following reaction? 3C (gr) + 4H 2 (g) C 3 H 8 (g)

12. Example (cont.) 3C (gr) + 4H 2 (g) C 3 H 8 (g)  H = ? You’re given the following reactions: C (gr) + O 2 (g) CO 2 (g)  H = -394 kJ C 3 H 8 (g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O (l)  H = -2220 kJ H 2 (g) + 1/2O 2 (g) H 2 O (l)  H = -286 kJ

13. Example (cont.) Step 1. Only reaction 1 has C (gr). Therefore, we will multiply by 3 to get the correct amount of C (gr) with respect to our final equation. C (gr) + O 2 (g) CO 2 (g)  H = -394 kJ 3C (gr) + 3O 2 (g) 3CO 2 (g)  H = -1182 kJ Initial: Final:

14. Example (cont.) Step 2. To get C 3 H 8 on the product side of the reaction, we need to reverse reaction 2. 3CO 2 (g) + 4H 2 O (l) C 3 H 8 (g) + 5O 2 (g)  H = +2220 kJ C 3 H 8 (g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O (l)  H = -2220 kJ Initial: Final:

15. Example (cont.) Step 3: Add two “new” reactions together to see what is left: 3C (gr) + 3O 2 (g) 3CO 2 (g)  H = -1182 kJ 3CO 2 (g) + 4H 2 O (l) C 3 H 8 (g) + 5O 2 (g)  H = +2220 kJ 2 3C (gr) + 4H 2 O (l) C 3 H 8 (g) + 2O 2  H = +1038 kJ

16. Example (cont.) Step 4: Compare previous reaction to final reaction, and determine how to reach final reaction: 3C (gr) + 4H 2 O (l) C 3 H 8 (g) + 2O 2  H = +1038 kJ H 2 (g) + 1/2O 2 (g) H 2 O (l)  H = -286 kJ 3C (gr) + 4H 2 (g) C 3 H 8 (g) Need to multiply second reaction by 4

17. Example (cont.) Step 4: Compare previous reaction to final reaction, and determine how to reach final reaction: 3C (gr) + 4H 2 O (l) C 3 H 8 (g) + 2O 2  H = +1038 kJ 4H 2 (g) + 2O 2 (g) 4H 2 O (l)  H = -1144 kJ 3C (gr) + 4H 2 (g) C 3 H 8 (g)

18. Example (cont.) Step 4 (cont.): 3C (gr) + 4H 2 O (l) C 3 H 8 (g) + 2O 2  H = +1038 kJ 4H 2 (g) + 2O 2 (g) 4H 2 O (l)  H = -1144 kJ 3C (gr) + 4H 2 (g) C 3 H 8 (g)  H = -106 kJ

19. Another Example Calculate  H for the following reaction: H 2 (g) + Cl 2 (g) 2HCl(g) Given the following: NH 3 (g) + HCl (g) NH 4 Cl(s)  H = -176 kJ N 2 (g) + 3H 2 (g) 2NH 3 (g)  H = -92 kJ N 2 (g) + 4H 2 (g) + Cl 2 (g) 2NH 4 Cl(s)  H = -629 kJ

20. Another Example (cont.) Step 1: Only the first reaction contains the product of interest (HCl). Therefore, reverse the reaction and multiply by 2 to get stoichiometry correct. NH 3 (g) + HCl (g) NH 4 Cl(s)  H = -176 kJ 2NH 4 Cl(s) 2NH 3 (g) + 2HCl (g)  H = 352 kJ

21. Another Example (cont.) Step 2. Need Cl 2 as a reactant, therefore, add reaction 3 to result from step 1 and see what is left. 2NH 4 Cl(s) 2NH 3 (g) + 2HCl (g)  H = 352 kJ N 2 (g) + 4H 2 (g) + Cl 2 (g) 2NH 4 Cl(s)  H = -629 kJ N 2 (g) + 4H 2 (g) + Cl 2 (g) 2NH 3 (g) + 2HCl(g)  H = -277 kJ

22. Another Example (cont.) Step 3. Use remaining known reaction in combination with the result from Step 2 to get final reaction. N 2 (g) + 4H 2 (g) + Cl 2 (g) 2NH 3 (g) + 2HCl(g)  H = -277 kJ ( N 2 (g) + 3H 2 (g) 2NH 3 (g)  H = -92 kJ) H 2 (g) + Cl 2 (g) 2HCl(g)  H = ? Need to take middle reaction and reverse it

23. Another Example (cont.) Step 3. Use remaining known reaction in combination with the result from Step 2 to get final reaction. N 2 (g) + 4H 2 (g) + Cl 2 (g) 2NH 3 (g) + 2HCl(g)  H = -277 kJ 2NH 3 (g) 3H 2 (g) + N 2 (g)  H = +92 kJ H 2 (g) + Cl 2 (g) 2HCl(g)  H = -185 kJ 1