1. Bell Work  Big Idea 5 Reading / Viewing Assignment Due Today!

2. Learning Target Essential Knowledge: o When two systems are in contact with each other, they will exchange energy. o The energy that comes out of one system is equal to the energy that goes into the other system o The quantity of thermal energy transferred from one system to another is called heat. Calorimetry Experimental Process To Measure Heat Exchange Calorimetry Experimental Process To Measure Heat Exchange

3. Learning Target ΔH = q = mcΔT ΔH rxn = ΣΔH products – ΣΔH reactants

4. Heat Of Reactions All reactions involve a change of energy so heat is either a reactant or product Endothermic Exothermic Heat Is A ReactantHeat Is A Product

5. Laws of Thermodynamics 1 st Law Of Thermodynamics = law of conservation of energy o During a reaction the total amount of energy remains the same but it can o change forms, PE chemical bonds  Heat o be transferred, rxn system  surroundings PE Heat PE Heat Heat Exchange Is Enthalpy of Rxn ∆H Heat Exchange Is Enthalpy of Rxn ∆H Amount Of Heat Depends On Amounts That React! Amount Of Heat Depends On Amounts That React! Yup, You Guessed It! Limiting Reactant Stoichiometry! Yup, You Guessed It! Limiting Reactant Stoichiometry!

6. Calculating Heat Of Reactions  Limiting Stoich = Enthalpy change depends on amount of chemicals that react  Treat heat as if it were a chemical and create mole ratios with it just as you do with other chemicals Solid aluminum reacts with solid iron III oxide in a single replacement rxn, commonly known as the thermite rxn. The reaction is a highly exothermic and produces 847.6 kJ of heat along with iron and aluminum oxide solids. If 90.0g of aluminum reacts with 180.0g of iron III oxide how many kJ of heat will be evolved? Write The Balanced Rxn Equation! Determine The Limiting Reactant! Write The Balanced Rxn Equation! Determine The Limiting Reactant!

7. Calculating Heat Of Reactions 2Al (s) + Fe 2 O 3(s)  2Fe (s) + Al 2 O 3 (s) + 847.6kJ 90g 180 g 90.0 g Al ( 1 mol Al / 26.98 g Al) = 3.34 mole Al HAVE 180.0 g Fe 2 O 3 ( 1 mol / 159.68 Fe 2 O 3 g ) = 1.127 mol Fe 2 O 3 HAVE 3.34 mol Al (1 mol Fe 2 O 3 / 2 mol Al) = 1.67 mol Fe 2 O 3 NEED iron III oxide limits 1.127 mol Fe 2 O 3 (847.6 kJ/ 1 mol Fe 2 O 3 ) = 955.2 kJ heat produced On the basis of the thermodynamic data given above, compare the sum of the bond strengths of the reactants to the sum of the bond strengths of the products. Justify your answer. On the basis of the thermodynamic data given above, compare the sum of the bond strengths of the reactants to the sum of the bond strengths of the products. Justify your answer. THINK about meaning of endo & exo… What if the rxn was reversed… What if the rxn was reversed… ←

8. Enthalpy change for a reaction ΔH˚ rxn = Σ n p ΔH˚ f (products) - Σ n r ΔH˚ f (reactants ) Δ Enthalpy rxn = Difference between sum total, for all moles, enthalpy of formation for products – sum total, for all moles, enthalpy of formation for reactants o f = standard enthalpy of formation Formed from its elements understand standard conditions

9. Standard Enthalpies of Formation ΔH˚ f Use known enthalpies of formation (appendix/provided) PAY ATTENTION TO STATE OF MATTER!  Standard conditions = room temp, 1 atm gases, 1 M solutions  ΔH˚ f for elements is “0”  ΔH˚ f for compounds is usually negative Why? Forming Vs. Breaking Bonds Forming Vs. Breaking Bonds

10. Sample ΔH˚ rxn Decomposition of sodium chlorate into oxygen gas and solid sodium chloride salt… 2 NaClO 3 (s)  2 NaCl (s) + 3 O 2 (g) ∆H˚f Values: NaClO 3 (s) = -365.4 kJ/mol NaCl (s) = -411.1 kJ/mol O 2 (g) = 0 kJ/mol ΔH rxn = ∑ (n p ΔH˚ f [NaCl] + n p ΔH˚ f [O 2 ]) - ∑ n r ΔH˚ f [NaClO 3 ] = [2 (-411.1 kJ/mol) + 3mol ( 0 kJ/mol)] – 2 (-365.4 kJ/mol) = - 91.4 kJ Give Two Reasons Why This Rxn Is Favored … Likely to Naturally Happen Give Two Reasons Why This Rxn Is Favored … Likely to Naturally Happen THINK…THINGS THAT ARE FAVORED SPREAD ENERGY SPREAD MATTEER THINK…THINGS THAT ARE FAVORED SPREAD ENERGY SPREAD MATTEER

11. Calorimetry The study of the heat released or absorbed during a change or rxn The amount of heat energy lost or gained is based on o The amount of substances that react (mass, m) o The type of substances (specific heat capacity, c) o Magnitude of the temperature change (∆T )

12. Useful Equations Change in heat (ENTHALPY) q(H) = m x c x ∆T Specific heat capacity Change in temperature

13. Which will have a greater temp change when 5000.0 J of energy is added to both metals at room temp (23˚C)? Why? ΔT = q / (mc) ΔT = 5000.0 J/[(1000.0 g)(0.385 J/g  ˚ C)] ΔT = 13.0 ˚ C ΔT = q / (mc) ΔT = 5000.0 J/[(1000.0 g)(0.902 J/g  ˚C)] ΔT = 5.54 ˚C Calculate The Temp Changes Calculate The Temp Changes

14. Calorimeter: a device used to experimentally determine the heat associated with a chemical rxn or substance Assumptions Law of conservation Heat change of calorimeter = Heat of rxn No heat is lost or to the calorimeter No path to surroundings Heat capacity is insignificant Solutions can be treated as water Specific heat = 4.184 J/g˚C 1 mL = 1 g

15. Heat Lost By Rxn/Object = Heat Gained By Calorimeter 1 st Metal is heated in a boiling water bath 2 nd Hot metal is transferred to calorimeter 3 rd Metal transfers its heat energy to calorimeter 4 th Calorimeter gains heat energy, temp increases

16. A 9.888g piece of manganese metal is heated in a boiling water bath of 100.0°C and then transferred into a calorimeter that contained 50.0mL of water at 19.4°C. The temp of the calorimeter rose to 21.3°C. Calculate the specific heat for manganese. ∆T calorimeter = T f – T i = 21.3 – 19.4 = 1.9 °C q cal = c cal m cal ∆T cal = 4.184 x 50.0g x 1.9°C = 400J gained by the water Heat gained by calorimeter = Heat lost by the metal = - 400J from the metal ∆T metal = T f – T i = 21.3 - 100 = -78.7°C c metal = q metal /m metal ∆T metal = - 400J / 9.888 g x -78.7 °C = 0.51 J/g°C Sample Specific Heat Problem

17. Calculating Heat Of Reactions 1) Balance equation noting heat as reactant or product 2) Calc heat for desired chemical using stoich or calorimetry 3) Determine moles of desired chemical using stoich 4) Determine ratio of heat / mole = kJ/mol

18. 50.0mL of 0.100M CuSO 4 and 50.0mL of 0.100M KOH are mixed to produce a ppt of Cu(OH) 2 and aqueous soln of K 2 SO 4. The solutions were both initially at 23.5°C and the final temp after they were mixed was 29.7°C. Calculate the heat that accompanies this reaction in kJ/mol of ppt made. c = 4.184 J/g˚C solutions are aqueous m = 50mL + 50 mL = 100 mL = 100g reacted aq solutions ∆T = 29.7 – 23.5 = 6.2 1)CuSO 4 (aq) + 2KOH (aq)  Cu(OH) 2 (s) + K 2 SO 4 (aq) + Heat 2)∆H = q = cm∆T = 4.184 J/g °C x 100.0 g x 6.2 °C = 2600J 3)KOH is Limiting: MV same for both, n=MV so equal moles, rxn needs more KOH 0.00500mol KOH (1 mol Cu(OH) 2 /2 mol KOH) = 0.002500mol Cu(OH) 2 4 ) H rxn = 2600 J / 0.002500 mol = 1,040,000 J/mol = 1040 kJ/ mol = 1000 kJ/mol

19. Your Turn - A neutralization reaction is performed in a calorimeter with two aqueous solutions, 75.0mL of 0.50M H 2 SO 4 and 75.0mL of 0.80M of NaOH. Both solutions were at room temp, 23.0˚C when placed mixed in the calorimeter and mixed. Upon completion of the reaction, the temperature of the calorimeter was 28.3 ˚C. Calculate the enthalpy change per mole of water formed by the rxn. 1 st - Balanced Equation H 2 SO 4 + 2 NaOH  2 H 2 O + Na 2 SO 4 + Heat 2 nd - Determine heat using calorimetry ∆H = q = cm∆T = 4.184 J/g°C x 150.0 g x 5.3°C = 3300 J = 3.3 kJ 3 rd – Determine moles of desired chemical n=MV = 0.50M x 0.0750L = 0.038 mol acid 0.80M x 0.0750L = 0.060 mol base 0.038 mol acid ( 2 mol NaOH / 1 mol acid) = 0.076 mol base needed NaOH Limits 0.060mol NaOH (2mol H 2 O / 2 mol NaOH) = 0.060 mol H 2 O 4 th – Determine ratio of heat to moles for rxn 3.3 kJ/0.060 mol H 2 O = 55 kJ heat /mol H 2 O 110 kJ

20. Things To Do  Free Response Question  Get FRQ Answer Approved  Finish Exam Corrections  Do Chromatography Pre-Lab

21. Homework Thurs = 7:30 am Pre-Lab Due