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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”
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Friday, Oct. 25th: “A” DayMonday, Oct. 28th: “B” DayAgenda • 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”
Homework Questions/Problems • Pg. 349: #1-8
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”
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 temperatureof 25˚C or 298.15 K.
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.
Nutritionists Use Bomb Calorimetry • A bomb calorimeter is used to measure enthalpy changes caused by combustion reactions.
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.
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.
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
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: ΔHreaction = ΔHf°products - ΔHf°reactants • ΔHreaction is in kJ or Joules (moles cancel out)
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) + NH3(g) NH4Cl(s) ΔHreaction = ΔHf0products - ΔHf0reactants ΔHf0product = (1 mol)(-314.4 kJ/mol) = -314/4 kJ ΔHf0reactants=[(1 mol)(-92.3 kJ/mol)+(1 mol)(-45.9 kJ/mol)] = -138.2 kJ ΔHreaction = (-314.4 kJ) – (-138.2 kJ) -176.2 kJ (exothermic reaction)
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. N2(g) + 3 H2(g) 2 NH3(g) State whether the reaction is exothermic or endothermic. ΔHreaction = ΔH f0products - ΔH f0reactants ΔHf0prod = [(2 mol)(-45.9 kJ/mol) = -91.8 kJ ΔHf0reactants = [(1 mol)(0 kJ/mol) + (3 mol)(0 kJ/mol)] = 0 kJ ΔHreaction = (-91.8 kJ) – (0 kJ) = -91.8 kJ *Reaction is exothermic because ΔH is negative.*
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 H2(g) + 2 CO2(g) 2 H2O(g) + 2 CO(g) State whether the reaction is exothermic or endothermic. ΔHreaction = ΔH f0products - ΔH f0reactants ΔHf0prod = [(2 mol)(-241.8 kJ/mol) + (2 mol)(-110.5 kJ/mol)] = -704.6 kJ ΔHf0reactants = [(2 mol)(0 kJ/mol) + (2 mol)(-393.5 kJ/mol)] = -787 kJ ΔHreaction= (-704.6 kJ) – (-787 kJ) = 82.4 kJ *Reaction is endothermic because ΔH is positive.*
Homework • Pg. 15 practice workshet • MUST show work! • Section 10.3 review, pg. 357: #1-5 • Concept Review: “Changes in Enthalpy During Chemical Reactions”