2. HeatHeat SolidLiquidGas Heat = AMOUNT of internal energy Temperature = a MEASURE of the average molecular kinetic energy

3. 10 g Pb T = 40 °C 100 g Pb T = 40 °C Both blocks are at the same temperature. Do they both contain the same amount of heat?

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

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

6. Baltimore Shot Tower http://www.baltimore.to/ShotTower/ 200 ft 100 kg Pb 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? C p (Pb) = 0.13 J/g°C C p (H 2 O) = 4.18 J/g°C q = mC p  T  T = T f - T i T f = 93 °C -q Pb = q H2O -(1x10 5 g)(0.13 J/g°C)(T f – 327°C) = (1x10 4 g)(4.18 J/g°C)(T f – 20°C) 10 kg H 2 O T i = 20 °C

7. 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? 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!

8. 18 g H 2 O = 1 mole H 2 O - 10 °C 90 °C 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? 0 °C 100 °C

9. 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.

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

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

12. Temperature (°C) 0 100 Heating curve of water  H fus = the amount of heat needed to covert a solid into its liquid phase  H fus  H vap (heat of fusion) (heat of vaporization)  H vap = the amount of heat needed to convert a liquid into its gaseous phase

13. Temperature (°C) 0 100 Heating curve of water Heat (kJ/s) H 2 O:  H fus = 6.01 kJ/mol  H vap = 40.7 kJ/mol  H fus = 20.2 kJ/mol  H vap = 10.3 kJ/mol H 2 PEw: A greater  H fus = more time to melt And vice versa

14. Heating Curve Wrap Up: The specific heat capacity (C p )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 (  H fus,  H vap,  H sub ). The amount of heat given off for one mole of a substance during a phase transition while cooling is its molar enthalpy of transition (  H cond,  H sol,  H dep ). 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?+H+H -H-H

16. 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. Na (s)+ H 2 O (l)NaOH +1 + H 2 (g)(aq)222 Step 2: Determine if there’s a limiting reagent.* Na is the limiting reagent. Only 1.26 mol of H 2 will be formed. *Choose a product that has a coefficient of 1 for best results.

17. Step 3: Determine the amount of heat involved in the reaction. q = mC p  T q = ?m = 100 g C p = 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 H 2 also released 837 J of heat, then the molar enthalpy (heat) change for this reaction would be:

18. A simpler problem: How much heat is given off when 1.6 g of CH 4 are burned in an excess of oxygen if  H comb = -802 kJ/mol? Step 1: Write the reaction equation. CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O (g) Step 2: Calculate molar amount involved Step 3: Calculate amount of heat given off  H rxn = (-802 kJ/mol)(0.100 mol CH 4 ) = -80.2 kJ Q: Is this an exothermic or endothermic reaction?

19. EX 3: What is the molar heat of combustion of propene (C 3 H 6 ) if burning 3.2 g releases 156 kJ of heat? Step 1: Write the reaction equation. 2 C 3 H 6 (g) + 9 O 2 (g)  6 CO 2 (g) + 6 H 2 O (g) This reaction equation involves the combustion of 2 moles of C 3 H 6 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. C 3 H 6 (g) + 4.5 O 2 (g)  3 CO 2 (g) + 3 H 2 O (g) Step 2: Convert grams of propene to moles. Step 3: Divide the heat released by moles of propene.

20. Active Ingredients KCl NaCl NH 2 CONH 2 (urea) C 7 O 6 H 14 (methyl-D-glucopyranoside; a surfactant) Ace Ice Melter RAPIDLY MELTS ICE AND SNOW AND PREVENTS RE ‑ FREEZING Ace Ice Melter features a special custom blend of superior ice melting ingredients. Together they melt even the most stubborn ice and snow and work to prevent re-freezing. Melts ice down to 0°F (-18 °C).

21. 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?

22. 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.

23. 2 H 2 + O 2 2 H 2 O Energy 2 H 2 + O 2 2 H 2 O If the products contain less energy, energy must have been given off during the reaction. Energy barrier

24. 2 H 2 + O 2 2 H 2 O Energy 2 H 2 + O 2 2 H 2 O If the products contain more energy, energy must have been absorbed during the reaction. Energy barrier

25. 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.

26. 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.

27. 2 AlBr 3 + 3 Cl 2 2 AlCl 3 + 3 Br 2 Energy 2 AlBr 3 + 3 Cl 2 2 AlCl 3 + 3 Br 2  H rxn = Heat content of products – heat content reactants  H rxn < 0 Reaction is exothermic But how do we determine the heat content in the first place?

28. Heat of formation,  H f The  H f of all elements in their standard state equals zero. The  H f of all compounds is the molar heat of reaction for synthesis of the compound from its elements  H f (AlBr 3 ): 2 Al + 3 Br 2 2 AlBr 3  H rxn = 2  H f (AlBr 3 )  H rxn 2  H f (AlBr 3 ) = Since the  H rxn can be used to find  H f, this means that  H f can be used to find  H rxn WITHOUT having to do all of the calorimetric measurements ourselves!! The Law of Conservation of Energy strikes again!!

29. Hess’s Law:  H rxn =  H f (products) –  H f (reactants) 6 CO 2 (g) + 6 H 2 O (l)C 6 H 12 O 6 (s) + 6 O 2 (g)  H rxn = [  H f (C 6 H 12 O 6 ) + 6  H f (O 2 )] – [6  H f (CO 2 ) + 6  H f (H 2 O)] From  H f tables:  H f (C 6 H 12 O 6 ) = -1250 kJ/mol  H f (CO 2 ) = -393.5 kJ/mol  H f (H 2 O) = -285.8 kJ/mol  H rxn = [-1250 kJ/mol] – [6(-393.5 kJ/mol) + 6(-285.8 kJ/mol)]  H rxn = +2825.8 kJ/mol

30. Using Hess’ Law with  H rxn What is the  H comb for ethane? 2 C 6 H 6 (g) + 7 O 2 (g) → 4 CO 2 (g) + 6 H 2 O (g) C 2 H 4 (g) + 3 O 2 (g) → 2 CO 2 (g) + 2 H 2 O (g)  H rxn = -1323 kJ/mol C 2 H 4 (g) + H 2 (g) → C 2 H 6 (g)  Hrxn = -137 kJ/mol

31. H 2 O (l)H 2 O (g) Energy H 2 O (l) H 2 O (g)  H vap = +40.7 kJ/mol Water will spontaneously evaporate at room temperature even though this process is endothermic. What is providing the uphill driving force?

32. 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 S products > S reactants,  S is > 0 Predict the sign of  S: ClF (g) + F 2 (g)ClF 3 (g)  S < 0 CH 3 OH (l)CH 3 OH (aq)  S > 0

33. Are all +  S reactions spontaneous? 2 H 2 O (l) 2 H 2 (g) + O 2 (g)  S is large and positive… …but  H is large and positive as well. Gibb’s Free Energy,  G, allows us to predict the spontaneity of a reaction using  H AND  S. If –  G  spontaneous reaction

34. 2 H 2 O (l) 2 H 2 (g) + O 2 (g) What is  G for this reaction at 25  C?  H rxn =  H f (products) –  H f (reactants)  H rxn = [2(0) + 0] - 2(-285.83 kJ/mol ) = 571.66 kJ/mol  S rxn = [2(130.58 J/molK ) + 205.0 J/molK ] - 2(69.91 J/molK )  S rxn =  S f (products) –  S f (reactants)  S rxn = 326.34 J/molK = 0.32634 kJ/molK  G rxn =  H rxn – T  S rxn = 571.66 kJ/mol - 298K(0.32634 kJ/molK )  G rxn = +474.41 kJ/mol

35. 2 H 2 O (l) 2 H 2 (g) + O 2 (g)  G rxn =  H rxn – T  S rxn = 571.66 kJ/mol - T(0.32634 kJ/molK ) What is the minimum temperature needed to make this reaction spontaneous? Set  G rxn = 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