3. Friday, Oct. 25 th : “A” Day Monday, Oct. 28 th : “B” Day Agenda  Homework questions/Quick review  Section 10.2 Quiz: “Using Enthalpy”  Section 10.3: “Changes in Enthalpy During Chemical Reactions” Calorimetry, calorimeter, adiabatic calorimetry, Hess’s Law, standard enthalpy of formation  Homework Pg. 15 practice worksheet (MUST show work) Sec. 10.3 review, pg. 357: #1-5 Concept Review: “Changes in Enthalpy During Chemical Reactions”

4. Homework Questions/Problems  Pg. 349: #1-8

5. Section 10.2 Quiz: “Using Enthalpy”  You can use both your book and your notes.  You’ll need both 10.1 AND 10.2 notes. May the FORCE be with you!  #4: M = molar mass  #8: Use the “25 J rule”

6. Changes in Enthalpy Accompany Reactions  Changes in enthalpy occur during chemical reactions.  A change in enthalpy during a reaction depends on many variables, but temperature is one of the most important variables.  To standardize enthalpies of reaction, data are presented for reactions in which both reactants and products have the standard thermodynamic temperature of 25˚C or 298.15 K.

7. Chemical Calorimetry  Calorimetry: the measurement of heat- related constants, such as specific heat or latent heat.  Calorimeter: a device used to measure the heat absorbed or released in a chemical or physical change.

8. Nutritionists Use Bomb Calorimetry  A bomb calorimeter is used to measure enthalpy changes caused by combustion reactions.

9. Adiabatic Calorimetry is Another Strategy  Instead of using a water bath to absorb the energy generated in a combustion reaction, adiabatic calorimetry uses an insulating vessel that doesn’t allow energy to pass through.  As a result, the temperature of the reaction mixture will change and can be recorded.  Adiabatic calorimetry is used for reactions that are not ignited, such as for reactions in aqueous solution.

10. Hess’s Law  Hess’s Law: the law that states that the amount of heat released or absorbed in a chemical reaction does not depend on the number of steps in the reaction.  The overall enthalpy change in a reaction is equal to the sum of the enthalpy changes for the individual steps in the process.

11. Standard Enthalpies of Formation  Standard enthalpy of formation: the enthalpy change in forming 1 mol of a substance from elements in their standard state.  By definition, the values of the standard enthalpies of formation for elements are zero.  Symbol: ΔH˚ f  Unit: kJ/mol

12. Calculating Enthalpy Change for a Chemical Reaction  Using a list of standard enthalpies of formation, the enthalpy change of any reaction for which there is data available can be calculated: ΔH reaction = ΔH f ° products - ΔH f ° reactants  ΔH reaction is in kJ or Joules (moles cancel out)

13. Table 2: Standard Enthalpies of Formation

14. Example  Calculate the enthalpy change for the following reaction. Use the standard enthalpies of formation listed in Table A-11 on pg 833-834. HCl(g) + NH 3 (g) NH 4 Cl(s) ΔH reaction = ΔH f 0 products - ΔH f 0 reactants ΔH f 0 product = (1 mol)(-314.4 kJ/mol) = -314/4 kJ ΔH f 0 reactants =[(1 mol)(-92.3 kJ/mol)+(1 mol)(-45.9 kJ/mol)] = -138.2 kJ ΔH reaction = (-314.4 kJ) – (-138.2 kJ) -176.2 kJ (exothermic reaction)

15. Additional Practice Calculate the enthalpy change for the following reaction. Use the standard enthalpies of formation listed in Table A-11 on pg 833-834. N 2 (g) + 3 H 2 (g) 2 NH 3 (g) State whether the reaction is exothermic or endothermic. ΔH reaction = ΔH f 0 products - ΔH f 0 reactants ΔH f 0 prod = [(2 mol)(-45.9 kJ/mol) = -91.8 kJ ΔH f 0 reactants = [(1 mol)(0 kJ/mol) + (3 mol)(0 kJ/mol)] = 0 kJ ΔH reaction = (-91.8 kJ) – (0 kJ) = -91.8 kJ *Reaction is exothermic because ΔH is negative.*

16. Calculating a Reaction’s Change in Enthalpy Sample Prob. E, pg.356  Calculate the change in enthalpy for the reaction below using data from Table 2 on pg 355. 2 H 2 (g) + 2 CO 2 (g) 2 H 2 O(g) + 2 CO(g) State whether the reaction is exothermic or endothermic. ΔH reaction = ΔH f 0 products - ΔH f 0 reactants ΔH f 0 prod = [(2 mol)(-241.8 kJ/mol) + (2 mol)(-110.5 kJ/mol)] = -704.6 kJ ΔH f 0 reactants = [(2 mol)(0 kJ/mol) + (2 mol)(-393.5 kJ/mol)] = -787 kJ ΔH reaction = (-704.6 kJ) – (-787 kJ) = 82.4 kJ *Reaction is endothermic because ΔH is positive.*

17. Homework  Pg. 15 practice workshet MUST show work!  Section 10.3 review, pg. 357: #1-5  Concept Review: “Changes in Enthalpy During Chemical Reactions”