1. THERMOCHEMISTRY

2. Law of conservation of energy Energy can be neither created or destroyed but can be converted from one form to another. Energy in = Energy out. Sometimes the energy is lost as heat, but it’s still considered party of the energy of the system.

3.  Chemical potential energy- is the energy stored in chemicals bonds because of their compositions—different substances have different atoms/bonds/arrangements, and so have different amounts of energy  Nuclear Energy  Thermal Energy  Kinetic energy

4.  Baking soda and HCl  NaHCO 3 + HCl→  Copper(II) chloride and Al  Al + CuCl 2 →

5.  Endothermic reactions -absorb energy from the surroundings. The container gets cold.  Heat flows into the system, so heat is given a positive ∆H value.  Exothermic reactions release energy from the system. The container gets hot.  Heat flows out of the system, so it is given a negative ∆H value.

6. In studying energy changes, you can define a system as the part of the universe on which you focus your attention. Everything else in the universe makes up the surroundings.

7.  The products have more energy than the reactants. The products have absorbed energy.  2NaHCO 3 (s)  Na 2 CO 3 (s) + H 2 O(l) + CO 2 (g) ∆H = 85 kJ  2NaHCO 3 (s) + 85kJ  Na 2 CO 3 (s) + H 2 O(l) + CO 2 (g)

8.  The products have less energy than the reactants. The products have given off energy.  2H 2 (g) + O 2 (g) → 2H 2 O (g)  ΔH = −483.6 kJ/mol of O 2  2H 2 (g) + O 2 (g) → 2H 2 O (g) +483.6 kJ

9.  Enthalpy is the amount of energy absorbed or lost by a system during a process at constant pressure.  Types of enthalpy › Heat of Reaction › Heat of Formation › Heat of Combustion › Heat of Vaporization › Heat of Fusion Enthalpy=ΔH

10. Thermochemical Equation A thermochemical equation is a balanced chemical equation that includes the physical states of all reactants and products and the energy change, usually expressed as the change in enthalpy, ∆ H. 2H 2(g) + O 2(g)  2H 2 O (g) + 967.2 kJ

11. Enthalpy and Change in Enthalpy Although you cannot measure the actual energy or enthalpy of a substance, you can measure the change in enthalpy, which is the heat absorbed or released during a chemical reaction. The change in enthalpy for a reaction is called the Heat of Reaction (∆ H ).

12. Heat of Reaction ∆ H rxn is the difference between the enthalpy of the substances that exist at the end of the reaction and the enthalpy of the substances present at the start. The formula for Heat of Reaction is ΔH = Δ H° f (products) - Δ H° f (reactants)

13.  The Heat of Combustion is the amount of energy released as heat by the complete combustion of one mole of a substance.  The formula for Heat of Combustion is ΔH c = H° f products - H° f reactants  As you can see, it is the same as Heat of reaction except for the beginning: ΔH c

14.  Heat of formation (H° f ) is the energy needed to form one mole of a compound from its elements.  Elements have 0 H° f

15.  When heat is produced in a reaction, you can actually indicate it in the balanced equation:  CaO (s) + H 2 O (l) → Ca(OH) 2 (s) + 65.2 kJ  This is an exothermic reaction, because the heat is being formed (ie, released)  we can also write:  CaO (s) + H 2 O (l) → Ca(OH) 2 (s) ∆H = -65.2 kJ

16.  ΔH° f CO 2(gas) = -393.0 kJ/mol  ΔH° f H 2 O (liquid) = -285.8 kJ/mol  ΔH° f C 5 H 12(gas) = -145.7 kJ/mol  Calculate Heat of Combustion for this reaction C 5 H 12 (g) + 8O 2(g)  5CO 2(g) + 6 H 2 O (l) ΔH c = H° f products - H° f reactants ΔH c = [(5)(-393.0 kJ/mol) + (6)(-285.8 kJ/mol)] – [-145.7 kJ/mol + 0 kJ/mol] = -3534.1 kJ or -3530 kJ You do not include 8O 2 because it is not a compound

17. Changes in Enthalpy

19. Phase Changes Heating Curves and

21. Heat can also be absorbed or released during a change of state. The heat required to vaporize one mole of a liquid is called its heat of vaporization (∆ H vap ). The molar heat of condensation (∆ H cond) is the amount of heat released when one mole of a vapor condenses at its normal boiling point. The quantity of heat absorbed by a vaporizing liquid is exactly the same as the quantity of heat released when the vapor condenses; ∆ H vap = –∆ H cond. ∆H vap and ∆H cond

22.  The molar heat of solidification (∆H solid ) is the heat lost when one mole of a liquid substance solidifies at a constant temperature. › The quantity of heat absorbed by a melting solid is exactly the same as the quantity of heat released when the liquid solidifies; that is, ∆H fus = –∆H solid. The heat required to melt one mole of a solid is its heat of fusion (∆Hfus).

23.  The conversion of 1 mol of ice at 0ºC to 1 mol of liquid water at 0ºC requires the absorption of 6.01 kJ of heat. › This quantity of heat is the molar heat of fusion of water.  The conversion of 1 mol of liquid water at 0ºC to 1 mol of ice at 0ºC releases 6.01 kJ of heat. › This quantity of heat is the molar heat of solidification of water.  H 2 O(s)  H 2 O(l)∆H fus = 6.01 kJ/mol  H 2 O(l)  H 2 O(s)∆H solid = –6.01 kJ/mol

26.  Day 3

27. Heat, which is represented by the symbol q, is energy that is in the process of flowing from a warmer object to a cooler object. Heat When the warmer object loses heat, its temperature decreases. When the cooler object absorbs heat, its temperature rises.

28. Measuring heat In the metric system, the amount of heat required to raise the temperature of one gram of pure water by one degree Celsius (1°C) is defined as a calorie (cal). The SI unit of heat and energy is the Joule (J). One Joule is the equivalent of 0.2390 calories, or one calorie equals 4.184 Joules.

29.  Temperature is a measure of the average kinetic energy of the particles in a sample of matter.  Temperature is measured in °C or K Temperature

30. Heat and temperature are not the same thing

31. Specific heat (c) is the amount of energy that is required to raise the temperature of one gram of substance by one degree Celsius (1°C). The specific heat of water is 4.18 J/g∙°C Specific Heat

32. Specific Heats of Some Common Substances Substance Specific heat J/(g · ºC)cal/(g · ºC) Liquid water4.181.00 Ethanol2.40.58 Ice2.10.50 Steam1.90.45 Chloroform0.960.23 Aluminum0.900.21 Iron0.460.11 Silver0.240.057 Mercury0.140.033

33. The heat absorbed or released (∆H) by a substance during a change in temperature depends not only upon the specific heat of the substance, but also upon the mass of the substance and the amount by which the temperature changes. Q= mc ∆T ∆H= mc ∆T Calculating Specific Heat

34.  The formula for calculating specific heat is:  ∆H = C x m x ∆T  C= q___ mx ΔT

35. q or ∆ H = the heat absorbed or released or enthalpy change c = the specific heat of the substance m = the mass of the sample in grams ∆ T is the change in temperature in °C ∆ T is the difference between the final temperature and the initial temperature or, T final – T initial. Calculating Specific Heat

36.  How much heat is given off when an 869 g iron bar cools from 940.0°C to 50.0°C? c Fe = 0.444 J/g °C m = 869 g ΔT = T final – T initial = 50.0°C – 940.0°C = -890°C q = c m ΔT q = 0.444 J/g °C x 869 g x –890.0°C q = -343,000 J or -3.43 x 10 5 J

37.  The temperature of a 95.4-g piece of copper increases from 25.0ºC to 48.0ºC when the copper absorbs 849 J of heat. Calculate the specific heat of copper. q=ΔH= 849 Jm = 94.5g ΔT = T final – T initial = 48.0°C – 25.0°C = 23°C c =__q___ m ΔT Ccu = ___849 J/g____ 95.4g x 23°C C cu = 0.387 J/(g · ºC)

38.  For a phase change to occur, energy must be added to or taken away from the substance.  you must know the physical states of reactants and products, because it makes a difference in the heat of reaction:  H 2 O (l) → H 2 (g) + ½ O 2 (g)∆H = 285.8 kJ  H 2 O (g) → H 2 (g) + ½ O 2 (g)∆H = 241.8 kJ  it takes less heat to decompose water that’s already in a vapor state!!

39.  Day 4

40.  Accurate and precise measurement of the heat change for chemical/physical processes  In a calorimeter a reaction takes place (often in water) and the temperature change of the water is measured  You can calculate the heat absorbed or released by the surroundings using the formula:  q sys = ∆H = – q surr = m  C  ∆T CALORIMETRY

41.  In a bomb calorimeter, a sample of a compound is burned in a constant- volume chamber in the presence of oxygen at high pressure. › By measuring the temperature increase of the water, it is possible to calculate the quantity of heat released during the combustion reaction.

42. › When 25.0 mL of water containing 0.025 mol HCl at 25.0ºC is added to 25.0 mL of water containing 0.025 mol NaOH at 25.0ºC in a foam-cup calorimeter, a reaction occurs. Calculate the heat absorbed or released (in kJ) during this reaction if the highest temperature observed is 32.0ºC. › C water = 4.18 J/(g · ºC) › V final = V HCl + V NaOH = 25.0 mL + 25.0 mL = 50.0 mL= 50.0g › T i = 25.0ºC T f = 32.0ºC ∆ T=  Unknown › ∆H = ? kJ  ∆Hrxn= 1.5kJ ∆Hwater= -1.5kJ

43.  19. A ‘coffee-cup’ calorimeter contains 150g of water at 24.6 °C. A 110 g block of molybdenum metal is heated to 100 °C and then placed in the water in the calorimeter. The contents of the calorimeter come to an average temperature of 28.0 °C. The specific heat of the water is 4.184 J/°Cg? Calculate the amount of heat released by the molybdenum. ( Hint: It equals the amount of heat absorbed by the water.)  ΔHwater = ΔHMo=  20. Using your ΔH from the question above what is the specific heat capacity of the molybdenum metal?

45.  Use the following thermochemical equation to calculate the energy change (in kJ) that occurs when 2.24 mol of NaHCO 3 ( s ) decomposes. 2NaHCO 3 ( s ) + 85 kJ  Na 2 CO 3 ( s ) + H 2 O( l ) + CO 2 ( g )  Knowns › amount of NaHCO 3 ( s ) that decomposes = 2.24 mol › ∆ H = 85 kJ for 2 mol NaHCO 3 ( s )  Unknown › ∆ H = ?kJ for 2.24 mol NaHCO 3 ( s )

46.  2. Calculate Energy Changes The production of iron and carbon dioxide from iron(III) oxide and carbon monoxide is an exothermic reaction. Use the following thermochemical equation to calculate the energy change (in kJ) when 3.40 mol Fe2O3 reacts with an excess of CO.  Fe 2 O 3 (s) + 3CO(g)  2Fe(s) + 3CO 2 (g) + 26.3 kJ

47.  3. The production of iron and carbon dioxide from iron(III) oxide and carbon monoxide is an exothermic reaction. Use the following thermochemical equation to calculate the amount of grams produced when CO 2 releases 72.0 kJ of heat.  Fe 2 O 3 (s) + 3CO(g)  2Fe(s) + 3CO 2 (g) + 26.3 kJ

49.  Entropy is a measure of the degree of randomness of the particles in a system.  Gases have high entropy and solids have low entropy. Does your bedroom have high Entropy?

50.  Find the number of moles of ice that can be melted by the addition of 2.25 kJ of heat. Convert moles of ice to grams of ice.  Knowns › Initial and final temperatures are 0ºC. › ∆H fus = 6.01 kJ/mol › ∆H = 2.25 kJ  Unknown › m ice = ? g

51.  Calculate the heat (in kJ) that is absorbed when 24.8 g H 2 O(l) at 100ºC and 101.3 kPa are converted to H 2 O(g) at 100ºC.  Knowns › Initial and final conditions are 100ºC and 101.3 kPa. › mass of liquid water converted to steam = 24.8 g › ∆H vap = 40.7 kJ/mol  Unknown › ∆H = ? kJ

52. Reaction Rate Reaction rate is the change in concentration of reactants per unit time as a reaction proceeds. Factors that influence reaction rate: 1.Nature of reactants – some substances tend to react faster than other. 2. Surface area – The more surface area, the faster a reaction will occur. 3. Temperature – Higher temperatures make a reaction rate increase. 4. Concentration – Higher concentrations make a reaction rate increase. 5. Presence of a catalyst – Can speed up or slow down a reaction depending on the catalyst.