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Introduction to Thermochemistry Thats Me! Okaynot really Or is it?

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Why study thermochemistry? We depend on energy for nearly every aspect of our lives Think about how you use energy every day Electricity Cooking Automobiles Heating/Cooling Batteries Cars, cell phones, iPods, watches Human Body Temperature regulation Conversion of food into usable cellular energy

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Seriously, why are we studying this? Energy crisis in the world! Over-use/dependence on fossil fuels (petroleum, gasoline, diesel, methane, propane) Research/Development of alternative energy resources Nuclear Fuel Cells (H 2, Methanol, Solid oxide) Hydro Wind Solar NEWS FLASH: This is going to be OUR problem!

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Major Sources of Energy over the last 150 years

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An interesting quote So called global warming is just a secret ploy by wacko tree-huggers to make America energy independent, clean our air and water, improve the fuel efficiency of our vehicles, kick-start 21 st century industries, and make our cities safer and more livable. Dont let them get away with it! - Chip Giller Founder of Grist.com, where environmentally minded people gather online.

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Organizing the Universe Any process can be defined by the following Universe = everything, the big enchilada, the whole kit and caboodle, all that and a bag of chips, the whole shabang, all that jazz, tout, todos, sarra, alles, everything-ay, Chuck Norris, etc. System = the part of the universe we are studying i.e. a particular chemical reaction, etc. System Surroundings Universe Surroundings = everything else in the universe that is NOT being studied i.e. the environment, the lab, the students, your mom, etc.

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Internal Energy Total Energy = Kinetic Energy + Potential Energy Endothermic process: energy enters the system (from the surroundings) Exothermic process: energy leaves the system (to the surroundings) System Surroundings Energy System Surroundings Energy due to motionstored energy

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Kinetic Energy in Atoms & Molecules Temperature average KE of particles in a sample Components of Kinetic Energy: KE tot = all components added together is proportional to TranslationRotation Vibration

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Think of potential energy in terms of Coulombs Law Q = charge r = distance Components of Potential Energy: Nucleus nucleus interactions: + (repulsive, less stable) Electron electron interactions: + (repulsive, less stable) Nucleus electron interactions: (attractive, more stable) Intermolecular forces:between molecules (attractive, more stable) PE total = all interactions added together Potential Energy in Atoms & Molecules - + + + + - + +

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Potential Energy (PE) Chemical Change: intramolecular potential energy Breaking and forming chemical bonds (PE nucleus-electron ) Example: 2H 2 + O 2 2H 2 O Physical Change: intermolecular potential energy Disrupting or forming intermolecular attractions (PE IMF ) Example: H 2 O ( ) H 2 O (g)

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The Law of Conservation of Energy is also known as The First Law of Thermodynamics Energy cannot be created or destroyed E universe = 0 (reminder: = change in) E universe = E system + E surroundings = 0 E system = - E surroundings Energy can be transferred as:heat (q) work (w) E sys = q + w

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The First Law of Thermodynamics Endothermic process: Exothermic process: System Surroundings Energy System Surroundings Energy In either case, the energy of the universe remains constant Universe

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Conservation of Energy in a Chemical Reaction Surroundings System Surroundings System Energy Before reaction After reaction In this example, the energy of the reactants and products increases, while the energy of the surroundings decreases. In every case, however, the total energy does not change. Myers, Oldham, Tocci, Chemistry, 2004, page 41 Endothermic Reaction Reactant + Energy Product

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Conservation of Energy in a Chemical Reaction Surroundings System Surroundings System Energy Before reaction After reaction In this example, the energy of the reactants and products decreases, while the energy of the surroundings increases. In every case, however, the total energy does not change. Myers, Oldham, Tocci, Chemistry, 2004, page 41 Exothermic Reaction Reactant Product + Energy

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Heat (q) Heat: the transfer of energy between objects due to a temperature difference Flows from higher-temperature object to lower-temperature object System (T 1 ) Surroundings (T 2 ) Heat If T 1 > T 2 q system = - q surroundings = + exothermic System (T 1 ) Surroundings (T 2 ) Heat If T 1 < T 2 q system = + q surroundings = - endothermic

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work = force distance force = pressure area = P m 2 distance = m work = P m 2 m = P V Assuming external pressure (P ext ) is constant, the volume can change (V) w system = -P ext V (in L atm) Where V = V f V i (in L) Pressure/Volume Work (PV work)

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Sign Conventions In chemistry, always take the systems perspective (physics takes the surrs view) Heat (q) q > 0heat is added to the sys by surr (q is +) q < 0heat is leaving sys to surr (q is -) Work (w) w > 0work done on sys by surr (work added to sys by surr, so w is +) w < 0work done by sys on surr (work is leaving surr, so w is -)

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Sign Conventions: Heat and Work System + q (endo) Heat - q (exo) Work - w + w Surroundings

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Practice with Work and Heat P ext = 2.0 atm V 1 = 5.0 L V 2 = 12.0 L An ideal gas expands from 5.0 L to 12.0 L against a constant external pressure of 2.0 atm. How much work is done? w = -P ext V w = -(P ext )(V final V initial ) w = -(2.0 atm)(12.0 L 5.0 L) w = -14 L atm 1 L atm = 101.325 J w = -14 L atm (101.325 J/1 L atm) w = -1400 J If E = 0 for this process, what is q? E = q + w = q + (-1.42 10 3 J) = 0 q = 1400 J

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Practice with Work and Heat Calculate the change in energy for a system that loses 15 kJ of heat and expands from a volume of 10. L to 200. L under a constant external pressure of 2.0 atm. E = q + w w = -P ext V w = -(2.0 atm)(200L -10L) = w = -380 L atm (101.325 J/1 L atm) (1 kJ/1000 J) = -38.5035 kJ E = 15 kJ + -38.5035 kJ = -53.5035 kJ -380 L atm - Remember: 101.325 J = 1 L atm 1000 J = 1 kJ -54 kJ

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Calorimetry (Calorie Measuring) Calorimetry: the science of measuring heat C = heat capacity the heat required to raise temp by 1C (or 1 K) Units Joules/ o C (J/ o C or J/K) Because the temperature increase depends on the amount of stuff Specific heat capacity (C p ) heat capacity per gram = J/g o C or J/gK Molar heat capacity (C m ) heat capacity per mole = J/mol o C or J/mol K

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Various Heat Capacities Substance Specific heat capacity (J/K g) Molar heat capacity (J/K mol) Molar mass (g/mol) Gold Silver Copper Iron Aluminum H 2 O (l) H 2 O (s) H 2 O (g) 0.129 0.235 0.385 0.449 0.897 4.184 2.03 1.998 197.0 107.9 63.55 55.85 26.98 18.02 25.4 24.5 25.1 24.2 75.3 36.6 36.0

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q = m C T q (J) = mass (g) C (J/g o C) T ( o C) q = joules (J) q (J) = moles (mol) C (J/K mol) T (K) q = joules (J) Mnemonic device: q = m CAT Using heat capacities

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Triple Point Diagrams Show the phases of a substance at different temperatures and pressures. Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

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Heating Curve of water Temperature ( o C) 40 20 0 -20 -40 -60 -80 -100 120 100 80 60 140 Energy Added Melting Point Boiling Point Solid Liquid Gas s l l g

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Heating Curve of Water Temperature ( o C) 40 20 0 -20 -40 -60 -80 -100 120 100 80 60 140 Energy Added

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Heating Curves Temperature Change within phase change in KE (molecular motion) depends on heat capacity of phase C H 2 O (l) = 4.184 J/g o C C H 2 O (s) = 2.03 J/g o C C H 2 O (g) = 1.998 J/g o C Phase Changes (s l g) change in PE (molecular arrangement) temperature remains constant overcoming intermolecular forces (requires the most heat) (requires the least heat)

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Heating Curves Temperature ( o C) 40 20 0 -20 -40 -60 -80 -100 120 100 80 60 140 Energy Added Melting - PE Solid - KE Liquid - KE Boiling - PE Gas - KE

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Calculating Energy Changes Temperature ( o C) 40 20 0 -20 -40 -60 -80 -100 120 100 80 60 140 Energy Added q = ? q = mass x C liquid x T q = mass x C gas x T q = mass x C solid x T Same Formulas Different C Values

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Phase Transitions During phase transitions, added heat is used to overcome intermolecular forces rather than to increase temperature H fusion = energy needed to convert solid to liquid For water, H fusion = 6.02 kJ/mol For liquid to solid, H = - 6.02 kJ/mol H vaporization = energy needed to convert liquid to gas For water, H vap = 40.7 kJ/mol For gas to liquid, H = - 40.7 kJ/mol

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Depending on the process occurring, all qs are calculated individually and then added together Note: steps 2 and 4 kJ steps 1, 3, 5 J Temperature ( o C) 40 20 0 -20 -40 -60 -80 -100 120 100 80 60 140 Energy Added 2. q = mol x H fus 4. q = mol x H vap 3. q = mass x C liquid x T 5. q = mass x C gas x T 1. q = mass x C solid x T Unit Conversions!

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Heat q 1 : Heat the ice to 0C q 2 : Melt the ice into a liquid at 0C q 3 : Heat the water from 0C to 100C q 4 : Boil the liquid into a gas at 100C q 5 : Heat the gas above 100C Heating Curve of Water From Ice to Steam in Five Easy Steps q = m C ice T ice q = mol H fus q = m C water T water q = mol H vap q = m C steam T steam 1 2 3 4 5

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5 4 3 2 1 Heating Curve Practice How much energy (J) is required to heat 12.5 g of ice at 10 o C to water at 0.0 o C? 4,430 J q 1 : Heat the ice from -10 to 0C q 2 : Melt the ice at 0C to liquid at 0 o C q = 12.5 g (2.03 J/g o C)(0.0 - -10.0 o C) = q = 12.5 g ice 6.02 kJ 18.016 g 1 mol = q tot = q 1 + q 2 253.75 J 4.177 kJ = 253.75 J + 4,177 J =

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5 4 3 2 1 Heating Curve Practice How much energy (J) is required to heat 25.0 g of ice at 25.0 o C to water at 95.0 o C? 19,560 J q 1 : Heat the ice from -25 to 0C q 2 : Melt the ice at 0C to liquid at 0 o C q 3 : Heat the water from 0C to 95 C q = 25.0 g (2.03 J/g o C)(0.0 - -25.0 o C) = q = 25.0 g ice 6.02 kJ 18.016 g 1 mol = q tot = q 1 + q 2 + q 3 1268.75 J 8.352 kJ = 1268.75 J + 8,352 J + 9937 J = q = 25.0 g (4.184 J/g o C)(95.0 0.0 o C) = 9937 J Notice that your q values are positive because heat is added

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5 4 3 2 1 Heating Curve Practice How much energy (J) is removed to cool 50.0 g of steam at 115.0 o C to ice at -5.0 o C? -153,000 J q 5 : Cool the steam from 115.0 to 100C q 4 : Condense the steam into liquid at 100C q 3 : Cool the water from 100C to 0 C q 2 : Freeze the water into ice at 0 C q = 50.0 g H 2 O q 1 : Cool the ice from 0C to 5.0 C q = 50.0 g (1.998 J/g o C)(100.0 - 115.0 o C) = q = 50.0 g H 2 O - 40.7 kJ 18.016 g 1 mol = q tot = q 1 + q 2 + q 3 + q 4 + q 5 -1498.5 J -112.96 kJ = -1498.5 J + -112,960 J + -20920 J + -16,710 J + -507.5 J = q = 50.0 g (4.184 J/g o C)(0.0 100.0 o C) = -20920 J - 6.02 kJ 18.016 g 1 mol = -16.71 kJ q = 50.0 g (2.03 J/g o C)(- 5.0 0.0 o C) = -507.5 J Notice that your q values are negative because heat is removed

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Heating Curve Challenge Problems 1. A sample of ice at -25 o C is placed into 75 g of water initally at 85 o C. If the final temperature of the mixture is 15 o C, what was the mass of the ice? Temperature ( o C) 40 20 0 -20 -40 -60 -80 -100 120 100 80 60 140 Time H = mol x H fus H = mol x H vap Heat = mass x t x C p, liquid Heat = mass x t x C p, gas Heat = mass x t x C p, solid 2.A 38 g sample of ice at -5 o C is placed into 250 g of water at 65 o C. Find the final temperature of the mixture assuming that the ice sample completely melts. 3.A 35 g sample of steam at 116 o C are bubbled into 300 g water at 10 o C. Find the final temperature of the system, assuming that the steam condenses into liquid water. 52.8 g ice 45.6 o C 76.6 o C

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C H 2 O (l) = 4.184 J/g o C C H 2 O (s) = 2.03 J/g o C C H 2 O (g) = 1.998 J/g o C Temperature ( o C) 40 20 0 -20 -40 -60 -80 -100 120 100 80 60 140 Energy Added H fusion = 6.02 kJ/mol H vap = 40.7 kJ/mol Heating Curve of H 2 O Constants and Graph

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What will happen over time? Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 291

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Lets take a closer look Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 291

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Eventually, the temperatures will equalize Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 291

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Calorimetry Allows us to measure the flow of heat Think back to the H fus expt.. q system = - q surroundings In other words, whatever energy is lost by one is gained by the other (direct E transfer) Major Assumption: No E is lost to other parts of the surroundings

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Thermometer Styrofoam cover Styrofoam cups Stirrer Typical Apparatus A Coffee Cup Calorimeter Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 302 Used for calorimetry under constant pressure i.e. normal lab conditions

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Example: 5.8 g of NaCl dissolves in 50.0 mL H 2 O NaCl (s) Na + (aq) + Cl - (aq) T i = 22.2 C T f = 20.5 C What is q rxn ? q rxn = q system q surr = m C T q surr = (55.8 g)(4.184 J/C g)(20.5 C 22.2C) q surr = -397 J q system = -q surr = 397 J Constant-Pressure Calorimetry Coffee cup calorimeter q rxn ? We cant directly measure q sys = -q surroundings In this case, of the surroundings

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Bomb Calorimeter Constant V calorimetry (as opposed to P) Bomb has own heat capacity (kJ/ 0 C) Multiply heat capacity by the T to find the total heat transferred Factor heat into # of g or mol burned (kJ/g or kJ/mol)

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Bomb Calorimeter Example A 1.800 g sample of sugar, C 12 H 22 O 11, was burned in a bomb calorimeter whose total heat capacity is 15.34 kJ/ o C. Once burned, the temperature of the calorimeter plus contents increased from 21.36 o C to 28.78 o C. What is the heat of combustion per gram of sugar? per mole of sugar? q = C bomb x T q = 15.34 kJ/ o C (28.78 o C 21.36 o C) q = 113.8 kJ 113.8 kJ released per 1.800 g sugar More sugar = more energy! 113.8 kJ / 1.800 g = 63.23 kJ/g 63.23 kJ g 342.3 g mol = 21640 kJ/mol

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Food and Energy Caloric Values Food joules/grams calories/gram Calories/gram Protein 17,000 4,000 4 Fat 38,000 9,000 9 Carbohydrates 17,000 4,000 4 Smoot, Smith, Price, Chemistry A Modern Course, 1990, page 51 1000 calories = 1 Calorie "science" "food" 1 calorie = 4.184 joules or 1 Kcal = 1 Calorie

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Does water have negative calories? How many Calories (nutritional) will you burn by drinking 1.0 L of water, initially at 36.5 o F (standard refrigeration temperature)? Assume that the body must expend energy to heat the water to body temperature at 98.6 o F. 37 o C 2.5 o C 1 L = 1000 mL 1 mL = 1 g q = 1.0 x 10 3 g (4.184 J/g o C)(37 o C - 2.5 o C) = 144,348 J 1000 calories = 1 Calorie 1 calorie = 4.184 joules 144348 J1 cal 4.184 J = 35 Cal 1 Cal 1000 cal

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150. mL 20.0 C Heat Transfer Experiments - q Cu = q water q = m x C x T for both cases, although specific values differ Plug in known information for each side Density of water = 1 g/mL(150 mL H 2 O) = 150 g H 2 O -m Cu C Cu T = m water C water T -20 g (0.385 J/g o C)(T f 250 o C) = 150 g (4.184 J/g o C)(T f 20 o C) Solve for T f... T f = 22.8 C Cu 20.0 g 250.0 C C = 0.385 J/C g What is the final temperature, T f, of the mixture?

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240. g of water (initially at 20.0 o C) are mixed with an unknown mass of iron initially at 500.0 o C (C Fe = 0.4495 J/g o C). When thermal equilibrium is reached, the mixture has a temperature of 42.0 o C. Find the mass of the iron. T = 500 o C mass = ? grams T = 20 o C mass = 240 g LOSE heat = GAIN heat - - [ (mass) (C Fe ) ( T)] = (mass) (C H2O ) ( T) - [ (X g) (0.4495 J/g o C) (42 o C - 500 o C)] = (240 g) (4.184 J/g o C) (42 o C - 20 o C)] - [ (X) (0.4495) (-458)] = (240 g) (4.184) (22) 205.9 X = 22091 X = 107 g Fe -q 1 = q 2 Fe

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A 97.0 g sample of gold at 785 o C is dropped into 323 g of water, which has an initial temperature of 15.0 o C. If gold has a specific heat of 0.129 J/g o C, what is the final temperature of the mixture? Assume that the gold experiences no change in state of matter. T = 785 o C mass = 97.0 g T = 15.0 o C mass = 323 g LOSE heat = GAIN heat - - [(C Au) (mass) ( T)] = (C H 2 O) (mass) ( T) - [(0.129 J/g o C) (97 g) (T f - 785 o C)] = (4.184 J/g o C) (323 g) (T f - 15 o C)] - [(12.5) (T f - 785 o C)] = (1.35) (T f - 15 o C)] -12.5 T f + 9.82 x 10 3 = 1.35 x 10 3 T f - 2.02 x 10 4 3 x 10 4 = 1.36 x 10 3 T f T f = 22.1 o C Au

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If 59.0 g of water at 13.0 o C are mixed with 87.0 g of water at 72.0 o C, find the final temperature of the system. T = 13.0 o C mass = 59.0 g LOSE heat = GAIN heat - - [ (mass) (C H 2 O) ( T)] = (mass) (C H 2 O) ( T) - [ (59 g) (4.184 J/g o C) (T f - 13 o C)] = (87 g) (4.184 J/g o C) (T f - 72 o C)] - [(246.8) (T f - 13 o C)] = (364.0) (T f - 72 o C)] -246.8 T f + 3208 = 364 T f - 26208 29416 = 610.8 T f T f = 48.2 o C T = 72.0 o C mass = 87.0 g