Announcements: Chapter 6: Thermochemistry Chem 6A, Section D Oct 11, 2011

Transcription

1 Chapter 6: Thermochemistry Chem 6A, Section D Oct 11, Announcements: Thurs Oct 13 quiz (#3) will cover chapter 4 Put a box around (or circle) your answers Bring student ID No notes, no calculators 2

2 Quiz 2 score histogram Grades so far (after quizzes 1-2) number of students F D C- C C+ B- B B+ A- A A+ 4

3 System vs Surroundings Internal energy = E Change in internal energy: ΔE = Efinal - Einitial System is what we are measuring Surroundings is everything outside the system Sign conventions: Always defined from the system s perspective Lose energy to the surroundings: ΔE = negative Gain energy from the surroundings: ΔE = positive system surroundings 2 Heat and Work Internal energy (E) takes the form of either heat (q) or work (w) Change in internal energy: ΔE = q + w Heat (q) = thermal energy; a change in temperature Work (w) = force applied over a distance; mechanical, electrical, etc.

4 Example: Heat and Work Your core body temperature increases by 2 C when sitting in a hot tub. If your body is the system and the hot tub is your surroundings, what are the signs of ΔE, q, and w? System Surroundings your body hot tub heat (q) q > 0 (+) work (w) w = 0 ΔE ΔE > 0 (+) Heat flows from the surroundings to the system and no work is performed Example: Heat and Work In one second, the engine in a steam locomotive delivers 2,240 kj to drive the train forward. In the process, it generates 3700 kj of heat. If the engine is the system and the locomotive is the surroundings, what are ΔE, q, and w? System steam engine Surroundings heat (q) work (w) ΔE locomotive q = -3,700 J w = -2,240 J ΔE -5,940 J Heat flows from the system to the surroundings and work is performed on the surroundings

5 Internal Energy vs Enthalpy Chemical reactions are usually performed in open containers, so it is difficult to measure w, the work performed on the atmosphere. ΔH (the change in We define the term enthalpy (H) to eliminate the need to consider enthalpy) PΔV work is the separately when heat calculating gained internal or lost energy changes in reactions: when a reaction is ΔH = ΔE + PΔV performed at constant pressure for reactions performed at constant pressure, ΔH = q Example: endothermic process Water evaporation is an endothermic process. Your body (the surroundings) gets colder when water on it (the system) evaporates. System Surroundings water your body heat (q) q > 0 (+) ΔH ΔH > 0 (+) Heat flows from the surroundings to the system: an endothermic process

6 Example: exothermic process Water condensation is an exothermic process. Your body (the surroundings) gets a burn when hot water vapor (the system) condenses on it. System Surroundings water your body heat (q) q < 0 (-) ΔH ΔH < 0 (-) Heat flows from the system to the surroundings: an exothermic process

7 Heat Capacity Heat Capacity is the relationship between heat absorbed and temperature change in a substance. Specific heat capacity: quantity of heat required to change the temperature of 1 gram of a substance by 1 K. How much heat does it take to warm up your jacuzzi? Heat capacity of water: J 1000L 1000g " " 4.184J 39 $ 20 " ( g -1 K -1 Volume of jacuzzi: 1000 L )K Initial temperature: L 20 C g# K Final temperature: 39 C = 79,500 kj Measuring the enthalpy of a reaction: calorimetry The chemical transformation involved in medical hot packs is the dissolution of calcium chloride in water. The enthalpy of solution, ΔH soln of CaCl 2(s) is kj/ mol. The hot pack contains g of water and 30.0 g of CaCl 2(s). If the pack is initially at room temperature (21.0 C), what temperature will the mixture become if the reaction goes to completion? Assume the specific heat capacity of the aqueous calcium chloride solution is 4.50 J K -1. g -1 and that the pack is held in a perfectly insulating container. Reaction Vessel Insulation Stirrer Vent

8 Calorimetry: Solution 81.3 kj mol CaCl 2 30 g CaCl 2 C g 1000 J mol g CaCl J g solution kj = 28.7 C So final temperature = = C 6

9 Problem: Energy and reaction stoichiometry The thermochemical equation for the combustion of octane (gasoline) is shown below. If it takes 444 kj of energy to get a pound car up to a speed of 70.0 mph, what volume of gasoline must be used to obtain this? Assume the engine is 100% efficient and the density of octane is 0.70 g/ml. 2 C 8 H 18(l) + 25 O 2(g) 16 CO 2(g) + 18 H 2 O (l) (ΔH = -10,942 kj) 7 Solution: 444 kj 2 mol C 8 H g C 8 H 18 ml 10,942 kj mol 0.7 g = 13.2 ml 8

10 Problem: Hess Law Given the standard enthalpy of formation of SO 2(g) is kj/mol and the standard enthalpy of formation of SO 3(g) is kj/ mol, what is the reaction enthalpy for the following reaction: 2SO 2(g) + O 2(g) 2SO 3(g) 9 Solution: ΔH (kj/mol) S (s) + O 2(g) SO 2(g) S (s) + 3/2O 2(g) SO 3(g) Need: 2SO 2(g) + O 2(g) 2SO 3(g)? 10

11 Solution: ΔH (kj/mol) 2SO 2(g) 2S (s) + 2O 2(g) 2( ) 2S (s) + 3O 2(g) 2SO 3(g) 2( ) 2SO 2(g) + O 2(g) 2SO 3(g) ΔH reaction = kj/mol 11 Problem: The solid fuel in the booster stage of the space shuttle is a mixture of ammonium perchlorate and aluminum powder. Upon ignition, one of the reactions that occurs is: NH 4 ClO 4(s) + Al (s) Al 2 + N 2(g) + HCl (g) + H 2 O (g) Balance this skeletal equation 6NH 4 ClO 4(s) + 10Al (s) 5Al 2 + 3N 2(g) +6 HCl (g) + 9H 2 O (g) 2. If Buzz Aldrin has 20 moles of Al powder and Neal Armstrong has 13 moles of ammonium perchlorate, which reactant will be limiting? Al How many moles of the excess reactant will be left over if the reaction goes to completion? 1 mole of NH 4 ClO 4 will be left. Identify the oxidizing agent and the reducing agent in this reaction Al is the reducing agent, NH 4 ClO 4 is the oxidizing agent 14 and the reducing Chem 6A agent Michael J. (NH + Sailor, UC 4 San is Diego oxidized, ClO - 4 is reduced)

12 Problem: Hess Law applied to the Thermite reaction Fe 2 + 2Al (s) 2Fe (s) + Al 2 Calculate the standard change in enthalpy for the Thermite reaction, given the thermochemical data on the next page. 24 Data: ΔH f (Fe 2 ) = 826kJ/mol ΔH f (Al 2 ) = 1676kJ/mol ΔH f (Fe (s) ) = 0kJ/mol ΔH f (Al (s) ) = 0kJ/mol 2Al (s) + Fe 2 2Fe (s) + Al 2 ΔH reaction = n p (products) - n r (reactants) ΔH reaction = 2(0)+( 1676) [2(0)+( 826)] = 850kJ 25

13 Solution: (Fe 2 ) = 826kJ/mol (Al 2 ) = 1676kJ/mol (Fe (s) ) = 0kJ/mol (Al (s) ) = 0kJ/mol 2Al (s) + Fe 2 2Fe (s) + Al 2 ΔH reaction ΔH reaction = n p (products) - n r (reactants) = 2(0)+( 1676) [2(0)+( 826)] = 850kJ 26 Solution: ΔH f (Fe 2 ) = 826kJ/mol ΔH f (Al 2 ) = 1676kJ/mol ΔH f (Fe (s) ) = 0kJ/mol ΔH f (Al (s) ) = 0kJ/mol 2Al (s) + Fe 2 2Fe (s) + Al 2 ΔH reaction = n p (products) - n r (reactants) ΔH reaction = 2(0)+( 1676) [2(0)+( 826)] = 850kJ 27