1. Section 7.5—Hess’s Law How can we find the enthalpy of a reaction using step-wise reactions?

2. Hess’s Law Hess’ Law – The sum of the energy changes during a series of reactions is equal to the sum of the reaction. In other words…if you go from Reactant A to Product Z all in one step, you will have the same total energy change as someone that went from A to Z in 7 step—the energy from each of their 7 steps would add up to your 1 step energy change.

3. Example 1 Label each step-wise equations with letters (“a”, “b”, “c”) if not already done for you. Calculate the enthalpy of the reaction 2N 2 (g) + 5O 2 (g)  2N 2 O 5 (g)  H rxn = ? Use: 2 H 2 (g) + O 2 (g)  2 H 2 O (l)  H rxn = -571.6 kJ N 2 O 5 (g) + H 2 O (l)  2HNO 3 (l)  H rxn = -76.6 kJ N 2 (g) + 3 O 2 (g) + H 2 (g)  2 HNO 3 (l)  H rxn = -74.1 kJ abcabc

4. For the first reactant in the overall reaction, find the step-wise reaction that has the same chemical and the same state of matter. It doesn’t have to be on the reactants side of the step-wise equation If it is on the correct “side” write it as is. Write it’s label beside it, too (“a”, “b”) If it’s on the wrong “side”, flip the equation. If you flip it, write it’s label as “-a” or “-b”. 2 Calculate the enthalpy of the reaction 2N 2 (g) + 5O 2 (g)  2N 2 O 5 (g)  H rxn = ? Use: 2 H 2 (g) + O 2 (g)  2 H 2 O (l)  H rxn = -571.6 kJ N 2 O 5 (g) + H 2 O (l)  2HNO 3 (l)  H rxn = -76.6 kJ N 2 (g) + 3 O 2 (g) + H 2 (g)  2 HNO 3 (l)  H rxn = -74.1 kJ abcabc c N 2 (g) + 3 O 2 (g) + H 2 (g)  2 HNO 3 (l) Example

5. Calculate the enthalpy of the reaction 2N 2 (g) + 5O 2 (g)  2N 2 O 5 (g)  H rxn = ? Use: 2 H 2 (g) + O 2 (g)  2 H 2 O (l)  H rxn = -571.6 kJ N 2 O 5 (g) + H 2 O (l)  2HNO 3 (l)  H rxn = -76.6 kJ N 2 (g) + 3 O 2 (g) + H 2 (g)  2 HNO 3 (l)  H rxn = -74.1 kJ abcabc c N 2 (g) + 3 O 2 (g) + H 2 (g)  2 HNO 3 (l) Repeat Step 2 for each reactant & product in the overall equation. If a reactant appears in more than one step-wise reaction, skip that reactant or product and move onto the next one. 3 Example

6. Calculate the enthalpy of the reaction 2N 2 (g) + 5O 2 (g)  2N 2 O 5 (g)  H rxn = ? Use: 2 H 2 (g) + O 2 (g)  2 H 2 O (l)  H rxn = -571.6 kJ N 2 O 5 (g) + H 2 O (l)  2HNO 3 (l)  H rxn = -76.6 kJ N 2 (g) + 3 O 2 (g) + H 2 (g)  2 HNO 3 (l)  H rxn = -74.1 kJ abcabc c N 2 (g) + 3 O 2 (g) + H 2 (g)  2 HNO 3 (l) Repeat Step 2 for each reactant & product in the overall equation. If a reactant appears in more than one step-wise reaction, skip that reactant or product and move onto the next one. 3 -b 2HNO 3 (l)  N 2 O 5 (g) + H 2 O (l) Example

7. -b 2 HNO 3 (l)  N 2 O 5 (g) + H 2 O (l) Calculate the enthalpy of the reaction 2N 2 (g) + 5O 2 (g)  2N 2 O 5 (g)  H rxn = ? Use: 2 H 2 (g) + O 2 (g)  2 H 2 O (l)  H rxn = -571.6 kJ N 2 O 5 (g) + H 2 O (l)  2 HNO 3 (l)  H rxn = -76.6 kJ N 2 (g) + 3 O 2 (g) + H 2 (g)  2 HNO 3 (l)  H rxn = -74.1 kJ abcabc c N 2 (g) + 3 O 2 (g) + H 2 (g)  2 HNO 3 (l) Use any un-used step-wise equations to get rid of unwanted things. Putting them on opposite sides will allow them to cancel. 4 -a 2 H 2 O (l)  2 H 2 (g) + O 2 (g) Example

8. -b 2 HNO 3 (l)  N 2 O 5 (g) + H 2 O (l) Calculate the enthalpy of the reaction 2N 2 (g) + 5O 2 (g)  2N 2 O 5 (g)  H rxn = ? Use: 2 H 2 (g) + O 2 (g)  2 H 2 O (l)  H rxn = -571.6 kJ N 2 O 5 (g) + H 2 O (l)  2 HNO 3 (l)  H rxn = -76.6 kJ N 2 (g) + 3 O 2 (g) + H 2 (g)  2 HNO 3 (l)  H rxn = -74.1 kJ abcabc c N 2 (g) + 3 O 2 (g) + H 2 (g)  2 HNO 3 (l) -a 2 H 2 O (l)  2 H 2 (g) + O 2 (g) Begin to cancel things out that appear on both the reactants and products side. Your goal is to add up all the step-wise equations to equal the overall equation. Multiply equations by a whole number if you need more of something to match the overall reaction or to fully cancel something out that you don’t want in the overall equation. 5 2 2 6 2 4 Example

9. -b 2 HNO 3 (l)  N 2 O 5 (g) + H 2 O (l) Calculate the enthalpy of the reaction 2N 2 (g) + 5O 2 (g)  2N 2 O 5 (g)  H rxn = ? Use: 2 H 2 (g) + O 2 (g)  2 H 2 O (l)  H rxn = -571.6 kJ N 2 O 5 (g) + H 2 O (l)  2 HNO 3 (l)  H rxn = -76.6 kJ N 2 (g) + 3 O 2 (g) + H 2 (g)  2 HNO 3 (l)  H rxn = -74.1 kJ abcabc c N 2 (g) + 3 O 2 (g) + H 2 (g)  2 HNO 3 (l) -a 2 H 2 O (l)  2 H 2 (g) + O 2 (g) Begin to cancel things out that appear on both the reactants and products side. Your goal is to add up all the step-wise equations to equal the overall equation. Multiply equations by a whole number if you need more of something to match the overall reaction or to fully cancel something out that you don’t want in the overall equation. 5 2 2 6 2 4 2 4 22 Example

10. -b 2 HNO 3 (l)  N 2 O 5 (g) + H 2 O (l) Calculate the enthalpy of the reaction 2N 2 (g) + 5O 2 (g)  2N 2 O 5 (g)  H rxn = ? Use: 2 H 2 (g) + O 2 (g)  2 H 2 O (l)  H rxn = -571.6 kJ N 2 O 5 (g) + H 2 O (l)  2 HNO 3 (l)  H rxn = -76.6 kJ N 2 (g) + 3 O 2 (g) + H 2 (g)  2 HNO 3 (l)  H rxn = -74.1 kJ abcabc c N 2 (g) + 3 O 2 (g) + H 2 (g)  2 HNO 3 (l) -a 2 H 2 O (l)  2 H 2 (g) + O 2 (g) Begin to cancel things out that appear on both the reactants and products side. Your goal is to add up all the step-wise equations to equal the overall equation. Multiply equations by a whole number if you need more of something to match the overall reaction or to fully cancel something out that you don’t want in the overall equation. 5 2 2 6 2 4 2 4 22 5 2 N 2 (g) + 5 O 2 (g)  2 N 2 O 5 (g) Example

11. -b 2 HNO 3 (l)  N 2 O 5 (g) + H 2 O (l) Calculate the enthalpy of the reaction 2N 2 (g) + 5O 2 (g)  2N 2 O 5 (g)  H rxn = ? Use: 2 H 2 (g) + O 2 (g)  2 H 2 O (l)  H rxn = -571.6 kJ N 2 O 5 (g) + H 2 O (l)  2 HNO 3 (l)  H rxn = -76.6 kJ N 2 (g) + 3 O 2 (g) + H 2 (g)  2 HNO 3 (l)  H rxn = -74.1 kJ abcabc c N 2 (g) + 3 O 2 (g) + H 2 (g)  2 HNO 3 (l) -a 2 H 2 O (l)  2 H 2 (g) + O 2 (g) 2 2 6 2 4 2 4 22 5 2 N 2 (g) + 5 O 2 (g)  2 N 2 O 5 (g) Use the step-wise “labels” as a math expression for solving for  H rxn. 6 2 × (-74.1 kJ) 2 × -(76.6 kJ) 1 × -(-571.6 kJ) 270.2 kJ Example

12. What did you learn about hot/cold packs?

13. Hot/Cold Packs Transfer of energy Use System & Surroundings between Materials ability to absorb energy without noticeable temperature change Effect on temperature depends on Physical change Is determined with Calorimetry Can be done in Chemical change