Lesson Objectives Classify reactions as endothermic or exothermic ( 11C )

Lesson Objectives • Classify reactions as endothermic or exothermic (11C) • Complete calculations using thermochemicalequations (11C) • Complete calculations involving specific heat, temperature change, mass, and heat (11D) • Thermochemical Equations

Endothermic and Exothermic Reactions • Energy can be absorbed or released • Endothermic– energy is absorbed by a reaction • Exothermic– energy is released from a reaction

Energy is released when bonds form • Energy is absorbed to break bonds • Releasing and Absorbing Energy Ex) 2H2+ O2 2H2O Energy absorbed Energy released Reactants Products

Exothermic ⇒ less E is absorbed to break bonds than is released when bonds form Bond Energy reactant < Bond Energy product • Endothermic ⇒ more E is absorbed to break bonds than is released when bonds form Bond Energy reactant > Bond Energy product • Classifying Reactions

Enthalpy change – (∆H); amount of energy released or absorbed as heat by a system when the pressure is constant • ∆Hrxn= Hproducts–Hreactants • Exothermic⇒ ∆H = negative values • Endothermic⇒ ∆H = positive values • Enthalpy Exothermic Endothermic Enthalpy Enthalpy Reaction Progress Reaction Progress Hreactants> Hproducts Hreactants< Hproducts

Thermochemical equation–chemical equation that includes the enthalpy change • Coefficients represent the number of moles • ∆H is directly proportional to the number of moles • Include physical states • Thermochemical Equations Ex) 4Fe(s) + 3O2(g)2Fe2O3(s) + 1625 kJ or 4Fe(s) + 3O2(g)2Fe2O3(s) ∆H= –1625 kJ

Standard enthalpy values for specific chemicals can be found in reference tables • Phase of the chemical will affect the enthalpy value • Elements in their natural state will have ∆H0f of zero • Enthalpy Values

Enthalpy of reaction equals the sum of the standard enthalpies of formation of the products minus the sum of the standard enthalpies of formation of the reactants • Calculating ∆Hrxn from ∆H0f Ex) CH4(g) + 2Cl2(g)  CCl4(l ) + 2H2(g) (–74.6 + (2×0)) ∆Hrxn= (–139 + (2×0)) – ∆Hrxn= ∆Hf0 (products) - ∆Hf0 (reactants) ∆Hrxn= –64.4 kJ/mol

Problem Use Table C-13 in your book to answer the following: Calculate the ∆Hrxn for the following: 2NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(g)

Specific Heat – quantity of energy needed to increase the temperature of one gram of a substance by one degree Celsius • Every substance has a unique specific heat • Specific heat of liquid water is 4.18 J/g∙∘C • Specific Heat • Specific heat =

Units • Heat in cal or J • Mass in g • Temperature in ∘C or K • Specific heat in J/g∙∘C Equivalent amounts • 1 cal = 4.18 J • 1000 cal = 1 kcal (Cal ) • Use as conversion factors • 0 ∘C = 273 K Temperature conversions Ex) 25 ∘C + 273 = 298 K Ex) 290 K – 273 = 17 ∘C • Specific Heat

Specific Heat • Write unknown and givens cp= ? J/g∙∘C Q = 750. J m = 50. g Tinitial = 25 ∘C Tfinal = 100. ∘C • Identify the formula and rearrange, if needed Q = m cp∆T cp= Ex) An unknown substance, with a mass of 50. g, is heated from 25 ∘C to 100. ∘C. During the heating process the substance absorbs 750. J of energy. What is the specific heat of the unknown substance in J/g∙∘C? • Convert units and find intermediates, if needed ∆T = Tfinal – Tinitial ∆T = 100. ∘C – 25 ∘C = 75 ∘C • Plug in and solve cp = cp = 0.20 J/g∙∘C • Make sure the answer is reasonable

Ex) A 21.0 g sample of a substance with a specific heat of 0.449 J/g∙∘C is heated from 288 K to 434 K. How much heat, in joules, was absorbed by the substance? • Specific Heat • Convert units and find intermediates, if needed • Tinital = 288 K – 273 = 15 ∘C • Tfinal = 434 K – 273 = 161 ∘C • ∆T = Tfinal – Tinitial • ∆T = 161 ∘C – 15 ∘C = 146 ∘C • Plug in and solve Q = (21.0 g)(0.449 J/g∙∘C )(146 ∘C) Q = 1380 J • Make sure the answer is reasonable • Write unknown and givens cp = 0.449 J/g∙∘C Q = ? J m = 21.0 g Tinitial = 288 K Tfinal= 434 K • Identify the formula and rearrange, if needed • Q = m cp∆T

Lesson Objective (11E) • Calculate the heat of a chemical process using calorimetry • Calorimetry

Calorimetry – measurement of the heat that is transferred during a reaction or physical process • Calorimeter – equipment used to measure the heat that is transferred during a reaction or physical process • Calorimetry Basic Calorimeter Setup

Simple Calorimeter Bomb Calorimeter • Calorimetry

Law of conservation of energy – energy is always conserved • Exothermic process • Heat lost by the system is gained by the water • Endothermic process • Heat gained by the system is lost by the water • Heat gained = heatlost • At constant pressure Q = ∆H • Exothermic Qwater = –Qprocess • Endothermic –Qwater= Qprocess • Calorimetry

Calorimetry Calculation • Write unknown and givens mw = 1000. g cw = 4.18 J/g∙∘C Tinitial = 25.0 ∘C Tfinal= 56.5 ∘C Qw = ? Qp= ? mp = 2.80 g • Identify the formula and rearrange, if needed Qw= –Qp Qw = mwcw ∆Tw Ex) A calorimeter contains 1000. g of water. When 2.80 g of gasoline are combusted, it changes the temperature of the water in the calorimeter from 25.0 ∘C to 56.5 ∘C. How much heat is released by this chemical process? What is the heat of combustion of gasoline in kJ/g? • Convert units and find intermediates, if needed ∆Tw = Tfinal – Tinitial ∆Tw = 56.5 ∘C – 25.0 ∘C = 31.5 ∘C • Plug in and solve Qw= (1000. g)(4.18 J/g∙∘C)(31.5 ∘C) Qw = 132,000 J –Qp = 132,000 J Qp = – 132,000 J • Make sure the answer is reasonable × = –132 kJ Qp = ∆H = = –47.1 kJ/g

Lesson Objectives Classify reactions as endothermic or exothermic ( 11C )