1. Hess’s Law 5.3.1

2. Review  Q - What is the first Law of Thermodynamics?

3. The Laws of Thermodynamics  The First Law The total energy of the universe (which hates you) is constant. ○ This is similar to the law of conservation of energy. It can be written as ○ ΔE universe = ΔE system + ΔE surroundings = 0 Energy can only be transferred ○ The flow of heat is considered one such transfer

4. Review  Q - What kind of calorimeter is this?  Q - What is held constant in this type of calorimeter?

5. Calorimeters Cup Calorimeter * Constant P Bomb Calorimeter * Constant V

6. Enthalpies  Calorimetry under constant pressure is so common that the heat values (q p ) that are calculated using this method have a special name, Enthalpy (ΔH). If ΔH is > 0 the reaction is endothermic If ΔH is < 0 the reaction is exothermic

7. Enthalpies  ΔH is also known as ΔH reaction because it is often given for a particular reaction. Example: 2H 2 (g) + O 2 (g)  2H 2 O(g) ○ ΔH reaction = -483.6 kJ Q - Is this reaction endothermic or exothermic?

8. Enthalpies  ΔH is dependent on the state of matter so be very careful to note which state you are talking about when working problems. Making liquid water does not release the same amount of energy as the formation of steam… The same is true for nearly all variables in thermo chemistry (c, S, etc.)

9. Hess’s Law  You have already measured enthalpy using a calorimeter.  You can also calculate it using Hess’s Law which states: The heat transferred, ΔH, in a reaction is the same regardless whether the reaction occurs in a single step or in several steps. If a series of reactions are added together, the net change in the heat of the reaction is the sum of the enthalpy changes for each step. 1940 - Germain Henri Hess

10. Hess’s Law  Rules for using Hess's Law 1. If the reaction is multiplied (or divided) by some factor, H must also be multiplied (or divided) by that same factor. 2. If the reaction is reversed (flipped), the sign of H must also be reversed.

11. Hess’s Law  Easy Example Problem Nitrogen and oxygen gas combine to form nitrogen dioxide according to the following reaction: ○ N 2 (g) + 2O 2 (g)  2NO 2 (g) Calculate the change in enthalpy for the above overall reaction, given: ○ N 2 (g) + O 2 (g)  2NO (g) ΔH = +181 kJ ○ 2NO(g) + O 2 (g)  2NO 2 (g) ΔH = -131 kJ

12. Hess’s Law  Easy Example Problem - ANSWER Nitrogen and oxygen gas combine to form nitrogen dioxide according to the following reaction: ○ N 2 (g) + 2O 2 (g)  2NO 2 (g) Calculate the change in enthalpy for the above overall reaction, given: ○ N 2 (g) + O 2 (g)  2NO (g) ΔH = +181 kJ ○ 2NO(g) + O 2 (g)  2NO 2 (g) ΔH = -131 kJ + 50kJ

13. Hess’s Law  Another Easy Example Problem (# 2) Calculate the value of ΔH ⁰ for the reaction: ○ C (graphite) (s)  C (diamond) (s) Using the following equations: 1. C (graphite) (s) + O 2 (g)  CO 2 (g) ΔH ⁰ = ─ 394 kJ 2. C (diamond) (s) + O 2 (g)  CO 2 (g) ΔH ⁰ = ─ 396 kJ

14. Hess’s Law  ANSWER to Example Problem (# 2) Using the following equations: 1. C (graphite) (s) + O 2 (g)  CO 2 (g) ΔH ⁰ = ─ 394 kJ Use eqn #1 as is… 2. CO 2 (g)  C (diamond) (s) + O 2 (g) ΔH ⁰ = + 396 kJ need to reverse eqn #2 ----------------------------------------------------------------------------- C (graphite) (s)  C (diamond) (s) ΔH ⁰ = + 2 kJ

15. Hess’s Law  Harder Example Problem Calculate the value of ΔH for the reaction: ○ 2S(g) + 2OF 2 (g)  SO 2 (g) + SF 4 (g)

16. Hess’s Law  Harder Example Problem Calculate the value of ΔH for the reaction: ○ 2S(g) + 2OF 2 (g)  SO 2 (g) + SF 4 (g) If we look at the final reaction, we see that we need 2 S atoms on the reactants side. ○ Q - Which step reaction has just S? The only reaction with S atoms is the third reaction, and in order to get 2 S atoms, we need to multiply the whole reaction by a factor of 2. (put on board) The next reactant in the final reaction is 2 OF molecules. ○ Q - Which step reaction has OF? The only reaction with an OF molecule is the first reaction, and in order to get 2 OF molecules, we need to multiply the whole reaction by a factor of 2. (p.o.b.)

17. Hess’s Law  Harder Example Problem Calculate the value of ΔH for the reaction: ○ 2S(g) + 2OF 2 (g)  SO 2 (g) + SF 4 (g)  On the products side of the final reaction, there is 1 SF 4 molecule, and the only possible source of the SF 4 molecule is the second reaction.  However, the SF 4 molecule is on the reactants side, which is not the side we need it on.  So we'll have to FLIP the second reaction to get the SF 4 molecule where we need it.  Now if we total up the reactions, we should end up with the given overall reaction:

19. Hess’s Law  Another Example Problem (# 3) Calculate the value of ΔH ⁰ for the reaction: ○ 2N 2 (g) + 5O 2 (g)  2N 2 O 5 (g) Using the following equations: 1. H 2 (g) + ½ O 2 (g)  H 2 O(l)ΔH ⁰ = ─285.5 kJ 2. N 2 O 5 (g) + H 2 O(l)  2 HNO 3 (l) ΔH ⁰ = ─ 76.6 kJ 3. ½ N 2 (g) + 3/2 O 2 (g) + ½ H 2 (g)  HNO 3 (l) ΔH ⁰ = ─ 174.1 kJ

20.  ANSWER to Example Problem (# 3) Using the following equations:  Multiply equation # 3 by 4 2 N 2 (g) + 6 O 2 (g) + 2 H 2 (g)  4 HNO 3 (l) ΔH ⁰ = ─ 696.4 kJ  Reverse equation # 2 and multiply by 2 4 HNO 3 (l)  2 N 2 O 5 (g) + 2 H 2 O(l) ΔH ⁰ = + 153.2 kJ  Reverse equation # 1 and multiply by 2 2 H 2 O(l)  2 H 2 (g) + O 2 (g) ΔH ⁰ = + 571 kJ  Cancel any substances and add eqn together -------------------------------------------------------------------------- 2N 2 (g) + 5O 2 (g)  2N 2 O 5 (g) ΔH ⁰ = + 27.8 kJ

21. HW (Due Wednesday)  Read Section 5.3 – if you haven’t already…  Do Exercise 5.3 – Due on Thursday 5/2 P.143 ○ #1-5