Flashback!!

1. Gas Liquid Solid Flashback!! Heat Heat = AMOUNT of internal energy Temperature = a MEASURE of the average molecular kinetic energy

2. 100 g Pb T = 40 °C 10 g Pb T = 40 °C Both blocks are at the same temperature. Do they both contain the same amount of heat?

3. paraffin 100 g Pb 100g Specific heat capacity (Cp): amount of heat(q) required to raise 1 g of substance by 1 °C Which substance requires more heat to increase the temperature by 5 °C? Cp(Pb) = 0.126 J/g°C Cp(paraffin) = 2.1 J/g°C

4. If the same amount of heat was used to heat 100 g of water [Cp(liquid water) = 4.184 J/g°C], what would be the DT of the water? How much heat is required by the 100 g candle to increase the temperature by 5 °C? Cp(paraffin) = 2.1 J/g°C q = Cp(mass)(DT) q = (2.1 J/g°C)(100 g)(5 °C) q = 1050 J q = Cp(mass)(DT) 1050 J = (4.184 J/g°C)(100g)(DT) For the same amount of heat and mass, DT decreases as the specific heat of the substance increases DT = 2.5 °C

5. q = mCpT T = Tf - Ti -(1x105 g)(0.13 J/g°C)(Tf – 327°C) = (1x104 g)(4.18 J/g°C)(Tf – 20°C) 10 kg H2O If the temperature of the lead is 327°C before it hits the water, what is the final temperature of the lead after hitting the water? 100 kg Pb -qPb = qH2O 200 ft Tf = 93 °C Ti = 20 °C Cp(Pb) = 0.13 J/g°C Baltimore Shot Tower http://www.baltimore.to/ShotTower/ Cp (H2O) = 4.18 J/g°C

6. Urp! Soda Melting one 14-gram Al soda can requires 5.55 kJ of energy. What is its molar heat of fusion? 105,000 cans are recycled in the US every minute. How many kJ/s are being used in recycling Al cans? That’s equivalent to burning 2300 food Calories/s!

7. Experiment: Heat two beakers containing 18 g of water at the same rate, and monitor their temperatures. Question: Will their temperatures increase at the same rate? 100 °C 90 °C - 10 °C 0 °C 18 g H2O = 1 mole H2O

8. Experiment: Heat two beakers containing 18 g of ice and water at the same rate, and monitor their temperatures. Question: Will their temperatures increase at the same rate? Answer: It takes twice as long to increase the temperature of the liquid water by 10 °C than it does to increase the temperature of the ice by the same amount.

9. Heating curve of water Gas warming 100 Temperature (°C) liquid + gas present liquid warming solid + liquid present 0 solid warming Heat (kJ/s)

10. Heating curve of water 100 Temperature (°C) boiling/condensation point melting/freezing point 0 Heat (kJ/s) Temperature is constant during phase transitions!! All heat energy goes to changing the state of matter.

11. Heating curve of water 100 Temperature (°C) DHvap (heat of vaporization) 0 DHfus (heat of fusion) Heat (kJ/s) DHfus = the amount of heat needed to covert a solid into its liquid phase DHvap = the amount of heat needed to convert a liquid into its gaseous phase

12. H2O: H2PEw: DHfus = 6.01 kJ/mol DHfus = 20.2 kJ/mol DHvap = 40.7 kJ/mol DHvap = 10.3 kJ/mol Heating curve of water 100 Temperature (°C) And vice versa 0 A greater DHfus = more time to melt Heat (kJ/s)

13. Heating Curve Wrap Up: • The specific heat capacity (Cp)of a substance determines the temperature change observed when heat is added or withdrawn from the substance. • Temperature is INVARIANT during phase transitions. • The amount of heat required to convert one mole of the substance from one phase to another is its molar enthalpy of transition (DHfus, DHvap, DHsub). • The amount of heat given off for one mole of a substance during a phase transition while cooling is its molar enthalpy of transition (DHcond, DHsol, DHdep). • The shape of a heating curve depends upon the heating rate, specific heat capacities of the phases involved, and the enthalpies of transition. What is the sign for all three? +DH What is the sign for all three? -DH

14. Thermochemistry heat Stoichiometry with some

15. 1.0 gram of solid sodium metal is added to 100 g of water. The reaction produces sodium hydroxide and hydrogen gas. Calculate the molar heat of reaction if the water’s temperature increased by 2C. Step 1: Write out the chemical equation and balance it. +1 -1 2 Na (s) + H2O (l) 2 2 Na OH (aq) + H2(g) Step 2: Determine if there’s a limiting reagent.* Na is the limiting reagent. Only 1.26 mol of H2 will be formed. *Choose a product that has a coefficient of 1 for best results.

16. Step 3: Determine the amount of heat involved in the reaction. q = mCpT q = ? m = 100 g Cp = 4.184 g/JC T = 2 C Step 4: Calculate your molar heat of reaction. If a reaction that produced 1.26 moles of H2 also released 837 J of heat, then the molar enthalpy (heat) change for this reaction would be:

17. A simpler problem: How much heat is given off when 1.6 g of CH4 are burned in an excess of oxygen if Hcomb = -802 kJ/mol? Step 1: Write the reaction equation. CH4(g) + 2 O2(g) CO2(g) + 2 H2O (g) Step 2: Calculate molar amount involved Step 3: Calculate amount of heat given off DHrxn = (-802 kJ/mol)(0.100 mol CH4) = -80.2 kJ Q: Is this an exothermic or endothermic reaction?

18. EX 3: What is the molar heat of combustion of propene (C3H6) if burning 3.2 g releases 156 kJ of heat? Step 1: Write the reaction equation. 2 C3H6(g) + 9 O2(g) 6 CO2(g) + 6 H2O (g) C3H6(g) + 4.5 O2(g) 3 CO2(g) + 3 H2O (g) • This reaction equation involves the combustion of 2 moles of C3H6 and we want to find out what it is for one mole. • Save yourself a headache and simplify future calculations by dividing the reaction equation through by 2. Step 2: Convert grams of propene to moles. Step 3: Divide the heat released by moles of propene.

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20. What have we learned? Sometimes heat is given off during a chemical reaction. This makes it feel hotter. Sometimes heat is absorbed during a chemical reaction. This makes it feel colder. What causes it to be different?

21. Chemical bonds contain energy. • Add the energy of all of the bonds in the reactants together to find their total energy. • Add the energy of all of the bonds in the products together to find their total energy. • If the two numbers aren’t the same (and they almost never are), then there will be heat energy given off or taken in.

22. 2 H2 + O2 2 H2 + O2 2 H2O Energy barrier 2 H2O Energy • If the products contain less energy, energy must have been given off during the reaction.

23. 2 H2 + O2 2 H2O 2 H2O 2 H2 + O2 Energy barrier Energy • If the products contain more energy, energy must have been absorbed during the reaction.

24. If heat energy is given off during a reaction, it is called an EXOTHERMIC REACTION. Heat exits = exothermic Exothermic reactions can be recognized by a temperature INCREASE.

25. If heat energy is absorbed during a reaction, it is called an ENDOTHERMIC REACTION. Heat enters = endothermic Endothermic reactions can be recognized by a temperature DECREASE.

26. 2 AlBr3 + 3 Cl2 2 AlCl3 + 3 Br2 2 AlBr3 + 3 Cl2 2 AlCl3 + 3 Br2 Energy But how do we determine the heat content in the first place? DHrxn = Heat content of products – heat content reactants DHrxn < 0 Reaction is exothermic

27. 2 AlBr3 2 Al + 3 Br2 DHrxn DHf(AlBr3) = 2 Heat of formation, DHf • The DHf of all elements in their standard state equals zero. • The DHf of all compounds is the molar heat of reaction for synthesis of the compound from its elements DHf (AlBr3): DHrxn = 2DHf(AlBr3) • Since the DHrxn can be used to find DHf, this means that DHf can be used to find DHrxnWITHOUT having to do all of the calorimetric measurements ourselves!! The Law of Conservation of Energy strikes again!!

28. Hess’s Law: DHrxn = S DHf(products) – S DHf(reactants) 6 CO2 (g) + 6 H2O (l) C6H12O6(s) + 6 O2(g) From DHf tables: DHf(C6H12O6) = -1250 kJ/mol DHf(CO2) = -393.5 kJ/mol DHf(H2O) = -285.8 kJ/mol DHrxn = [DHf(C6H12O6) + 6 DHf(O2)] – [6 DHf(CO2) + 6 DHf(H2O)] DHrxn = [-1250 kJ/mol] – [6(-393.5 kJ/mol) + 6(-285.8 kJ/mol)] DHrxn = +2825.8 kJ/mol

29. Using Hess’ Law with DHrxn What is the DHcomb for ethane? 2 C6H6 (g) + 7 O2 (g) → 4 CO2 (g) + 6 H2O (g) C2H4 (g) + 3 O2 (g) → 2 CO2 (g) + 2 H2O (g) DHrxn = -1323 kJ/mol C2H4 (g) + H2 (g) → C2H6 (g) DHrxn = -137 kJ/mol

30. H2O (l) H2O (g) H2O (g) DHvap = +40.7 kJ/mol H2O (l) Energy Entr o p y • Water will spontaneously evaporate at room temperature even though this process is endothermic. • What is providing the uphill driving force?

31. ClF (g) + F2(g) ClF3(g) CH3OH (l) CH3OH (aq) Entr o p y a measure of the disorder or randomness of the particles that make up a system • Water will spontaneously evaporate at room temperature because it allows the disorder of the water molecules to increase. • The entropy, S, of gases is >> than liquids or solids. • If Sproducts > Sreactants, DS is > 0 Predict the sign of DS: DS < 0 DS > 0

32. 2 H2O (l) 2 H2 (g) + O2 (g) This is a very nonspon- taneous process!! G = H T S Are all +DS reactions spontaneous? DS is large and positive… …but DH is large and positive as well. • Gibb’s Free Energy, DG, allows us to predict the spontaneity of a reaction using DH AND DS. If –DG  spontaneous reaction

33. 2 H2O (l) 2 H2 (g) + O2 (g) DHrxn = [2(0) + 0] - 2(-285.83 kJ/mol) = 571.66 kJ/mol DSrxn = 326.34 J/molK = 0.32634 kJ/molK What is DG for this reaction at 25C? DHrxn = SHf(products) –SHf(reactants) DSrxn = SSf(products) –SSf(reactants) DSrxn = [2(130.58 J/molK) + 205.0 J/molK] - 2(69.91 J/molK) DGrxn = DHrxn – TDSrxn = 571.66 kJ/mol - 298K(0.32634 kJ/molK) DGrxn = +474.41 kJ/mol

34. 2 H2O (l) 2 H2 (g) + O2 (g) What is the minimum temperature needed to make this reaction spontaneous? DGrxn = DHrxn – TDSrxn = 571.66 kJ/mol - T(0.32634 kJ/molK) Set DGrxn = 0 to find minimum temperature 0 = 571.66 kJ/mol - T(0.32634 kJ/molK) T = (571.66 kJ/mol)/(0.32634 kJ/molK) = 1751 K T > 1479 C