6/14/20161 Chapter 10- Causes of Change Heat and Chemical change

6/14/ The flow of energy -heat: Energy and heat: Thermochemistry- concerned w/ the _____ _____ that occur during _______ ___. Energy- capacity for _____ ___ or ________ _____ ex. coke in a fridge ex. coke in a fridge Heat ( )- energy that transfers from one object to another due to temperature difference (ex. coke in a fridge)ex. coke in a fridge only changes in heat can be detected! flows from warmer  cooler object until objects have the same temperature this energy that transfers is thermal energy Joule (J) the SI unit of heat and energy heat changechemical rxn doing worksupplying heat q

6/14/ Exothermic and Endothermic Processes Essentially all chemical reactions, and changes in physical state, involve either: release of heat (exothermic), or absorption of heat (endothermic)

6/14/ C. Endothermic Reaction reaction that absorbs energy reactants have lower PE than products 2Al 2 O 3 + energy  4Al + 3O 2 energy absorbed

6/14/ B. Exothermic Reaction reaction that releases energy products have lower PE than reactants 2H 2 (l) + O 2 (l)  2H 2 O(g) + energy energy released

6/14/ Exothermic - heat flowing out of a system into its surroundings: defined as negative q has a negative value system loses heat as the surroundings heat up Ex. water freezing, raining, & CH 4 + 2O 2  CO 2 + 2H 2 O + Heat Endothermic - heat flowing into a system from its surroundings: defined as positive q has a positive value system gains heat as the surroundings cool down Ex. ice melting, sweating, & CaCO 3 + heat  CaO + CO 2

Question: 6/14/2016 7

8 Heat Capacity Heat Capacity: the amount of heat needed to increase the temperature of an object exactly 1 o C Heat Capacity of a substance depends on two factors: 1) A 20 g iron bar and 20 g of water (both at room temp) are set outside on a sunny day for 20 minutes. After the 20 minutes elapse which will have a higher temperature? 2) Two cups of room temp water are set outside on a sunny day for 1 hour. Cup A: 20 g of water Cup B: 200 g of water Which will be warmer after the one hour? Ans:Fe Factor 1: Chemical Structure Ans: Cup A Factor 2: Mass

6/14/ Heat Capacity and Specific Heat A calorie is defined as the quantity of heat needed to raise the temperature of 1 g of pure water 1 o C a Calorie, written with a capital C, always refers to the energy in food 1 Calorie = 1 kilocalorie = 1000 cal. 1 cal = J

6/14/ the amount of heat it takes to raise the temperature of 1 gram of the substance by 1 o C specific heat capacity ( C p )- the amount of heat it takes to raise the temperature of 1 gram of the substance by 1 o C at constant pressure also called “Specific Heat” cal/g°CJ/g°C (or J/gK) water aluminum copper silver gold

6/14/ Heat Capacity and Specific Heat For water, C = 4.18 J/(g o C), and also C = 1.00 cal/(g o C) Thus, for water: it takes a long time to heat up, and it takes a long time to cool off!

6/14/ Visual Concepts Specific Heat Chapter 10

6/14/ To solve for heat capacity (q) … To calculate, use the formula: “q” is heat, unit- J or cal “m” is mass, unit- g “  T” = change in temperature, (  T = T f -T i ) unit- o C or K “C p ”or “C”= Specific Heat Capacity units: J/(g o C) or cal/(g o C) q = mC p T q = mC p T

6/14/ Same eq’n, different look! (q is always on top and by itself) q = m Cp  T q = m · Cp ·  T Cp =Cp =Cp =Cp =  T =  T = (  T = T f -T i ) m = q m ·  T q m Cp m · Cp q Cp Cp·TCp Cp·T

6/14/ Example Problems It takes 24.3 calories to heat 15.4 g of a metal from 22 ºC to 33ºC. What is the specific heat of the metal? given: q = 24.3 cal  T = T f -T i =33 – 22 = 11 ºC m = 15.4 g Cp = ?? q = m Cp  T Cp = = = q m  T (24.3 cal) (15.4 g)(11 ºC) 0.14 cal/(g o C)

6/14/ Example Problems 2.Iron has a specific heat of 0.39 J/gºC. How much will the temperature change when 520 J of energy is absorbed by 48.3 g of Cu? given: q= 520 J  T =?? m = 48.3 gCp = 0.39 J/gºC q = mCp  T  T = q m Cp m Cp  T = (520 J) (48.3 g)(0.39 J/gºC)  T = 28 ºC

6/14/ Practice Problem ex. When 435 J of heat is added to 3.4 g of olive oil at 21 ºC, the temperature increases to 85 ºC. What is the specific heat of olive oil? 2.0 J/g·°C

6/14/ Turn to page Pg 342 q = n C  T q = n · C ·  T n = moles (not grams) C = molar specific heat (in moles not grams) n = 10.0 mol  T = 7.5 K C = 27.8 J/ K mol C = 27.8 J/ K · mol q = (10.0 mol)(27.8 J/ K mol)(7.5 K) q = (10.0 mol)(27.8 J/ K · mol)(7.5 K) = 2100 J = 2100 J

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6/14/ Visual Concepts Enthalpy Change Chapter 10

6/14/ Measuring and Expressing Heat changes: Calorimetry- Calorimetry- - Calorimetry- accurate and precise measurement of heat change for chemical and physical processes Calorimetry- enthalpy ( )- - enthalpy change ( ): ∆H = H products – H reactants - q = exothermic exothermic  H is negative for exothermic rxnsexothermic endothermic endothermic  H is positive for endothermic rxnsendothermic H heat content at constant P ∆ H change in enthalpy m C  T ∆ H = m · C ·  T

6/14/ Calorimeter Foam cups are excellent heat insulators, and are commonly used as simple calorimeters The heat absorbed by the water is equal to the heat released by the substance

6/14/ Visual Concepts Calorimeter Chapter 10

6/14/ Heating Curve of water

6/14/ Heat in Changes of State (Heating Curve) Is there a phase change ….? 1. Heat of Fusion and Solidification ( _____ _____ ) - latent heat of fusion ( )- - latent heat of solidification ( )- ∆H fus = - ∆H solid - (Let’s think! ) if molar heat of fusion is heat for one mole, how would I determine heat involved for 2 moles? ∆H = solid liquid ∆H f heat absorbed by 1 g in constant temp. ∆H solid heat released by 1 g in constant temp. m(∆H )fus or solid or any other state change

6/14/ Heats of Vaporization and Condensation( _____ _____ ) - latent heat of vaporization ( )- - latent heat of condensation ( )- ∆H∆H ∆H vap = - ∆H cond ∆Hcond heat lost when 1 g of vapor condenses gas liquid ∆Hvap heat necessary to vaporize 1 g of a given liquid

6/14/ Heat of Solution - molar heat of solution ( )- -ex: NaOH(s)  Na + (aq) + OH - (aq) ∆H soln = kJ/mol -ex. How much heat is released when 2.500mol NaOH is dissolved in water? ∆Hsoln heat change when 1 mol of solid dissolves in water mol NaOH 1 mol NaOH kJ = kJ

6/14/ Some values for specific latent heats of fusion and vaporization: Substance Specific latent heat of fusion kJ.kg -1 ºC Specific latent heat of vaporization kJ.kg -1 ºC Water Ethanol Ethanoic acid Chloroform Mercury Sulphur Hydrogen Oxygen Nitrogen

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6/14/ CaCO 3  CaO + CO 2 Energy ReactantsProducts  CaCO 3 CaO + CO kJ CaCO kJ  CaO + CO 2

6/14/ Chemistry Happens in MOLES An equation that includes energy is called a thermochemical equation CH 4 + 2O 2  CO 2 + 2H 2 O kJ 1 mole of CH 4 releases kJ of energy. When you make kJ you also make 2 moles of water

6/14/ Thermochemical Equations A heat of reaction is the heat change for the equation, exactly as written The physical state of reactants and products must also be given. Standard conditions for the reaction is kPa (1 atm.) and 25 o C

6/14/ CH O 2  CO H 2 O kJ If grams of CH 4 are burned completely, how much heat will be produced? (just like stoichiometry) g CH g CH 4 1 mol CH kJ = 515 kJ

6/14/ CH O 2  CO H 2 O kJ How many liters of O 2 at STP would be required to produce 23 kJ of heat?(22.4L/mol) 23 kJ 2 mol oxygen22.4 L oxygen kJ1 mol oxygen =1.3 L oxygen How many grams of water would be produced with 506 kJ of heat? 506 kJ 2 mol water g water kJ1 mol water =22.7 g water

6/14/ You’ve Finally Met Your Match

6/14/ Ex: You put your can of coke into the refrigerator so that you can get a cold can of coke. Why is it inaccurate to say that the fridge is cooling the coke? - Refrigerator Special see thru door