2. Objectives: Today I will be able to: Correctly manipulate thermochemical equations to predict the enthalpy of reaction (Hess’s Law) Informal assessment – monitoring student interactions and questions as they complete the practice problems Formal assessment – analyzing student responses to the practice and exit ticket Common Core Connection Make sense of problems and persevere in solving them Build strong content knowledge Reason abstractly and quantitatively

3. Lesson Sequence Evaluate: Warm Up Explain: Thermochemical Equations Elaborate: Thermochemical Equations Practice Explain: Hess’s Law Elaborate: Hess’s Law Practice Evaluate: Exit Ticket

4. Warm Up 350 J are released as ice ( Specific Heat = 2.1 J / (g o C) ) cools from - 5.0 o C to -32 o C. What is the mass of ice?

5. Objectives Today I will be able to: Correctly manipulate thermochemical equations to predict the enthalpy of reaction (Hess’s Law)

6. Homework Finish practice problems

7. Agenda Warm Up Thermochemical Equations Notes Thermochemical Equations Practice Hess’s Law Notes Hess’s Law Practice Exit Ticket

9. Thermochemical Equations Balanced Equation with two additions: Enthalpy (heat) accompanying the reaction Coefficients represent moles – it is possible to have fractions – Example: ½ means half a mole of the substance State of matter is specified

10. Laws of Thermochemistry Δ H is directly proportional to the amount of substance produced or reacting in a reaction H 2(g) + ½ O 2(g)  H 2 O (l) Δ H = -285.8 kJ 2 H 2(g) + O 2(g)  2 H 2 O (l) Δ H = -571.6 kJ

11. Laws of Thermochemistry Δ H for a reaction is equal in magnitude but opposite in sign from the reverse reaction HgO (s)  Hg (l) + ½ O 2(g) Δ H = 90.7 kJ Hg (l) + ½ O 2(g)  HgO (s) Δ H = -90.7 kJ

14. Hess’s Law Δ H is independent of the number of steps involved If a reaction occurs in several steps, the sum of the enthalpy changes must equal Δ H for the overall reaction

15. Hess’s Law Example Determine ∆H of the reaction Sn (s) + 2 Cl 2(g)  SnCl 4(l) using the information provided below. Sn (s) + Cl 2(g)  SnCl 2(s) ∆H= -83.6 kJ SnCl 2(s) + Cl 2(g)  SnCl 4(l) ∆H = -46.7 kJ

16. Answer ∆H = -130.3 kJ

17. Hess’s Law Example 2

18. Answer + 15.3 kJ

20. Exit Ticket Which question on the Hess’s Law practice did you find the most challenging? We will start by reviewing this question tomorrow.