2.  Chemical rxns involve changes in energy – Breaking bonds requires energy – Forming bonds releases energy  The study of the changes in energy in chem rxns is called thermochemistry.  The energy involved in chemistry is real and generally measurable, and can be thought of as heat – Energy units are numerous, but our focus will be Joule (SI base unit) and the calorie – 1 calorie = 4.184 Joules  Chemical rxns involve changes in energy – Breaking bonds requires energy – Forming bonds releases energy  The study of the changes in energy in chem rxns is called thermochemistry.  The energy involved in chemistry is real and generally measurable, and can be thought of as heat – Energy units are numerous, but our focus will be Joule (SI base unit) and the calorie – 1 calorie = 4.184 Joules INTRO TO THERMOCHEMISTRY

3. WHAT IS HEAT?  Hot & cold, are automatically associated with the words heat and temperature –Heat & temperature are NOT synonyms –The temperature of a substance is directly related to the vibrational energy of its particles, specifically its:  Hot & cold, are automatically associated with the words heat and temperature –Heat & temperature are NOT synonyms –The temperature of a substance is directly related to the vibrational energy of its particles, specifically its:  The Kinetic Energy defines the temperature –Particles vibrating fast = hot –Particles vibrating slow = cold

4.  Kinetic energy is transferred from one particle to the next (a.k.a. conduction) – Sometimes this energy can be transferred from one object to another and influence physical properties – The more energy an object has the more energy is transferred  Kinetic energy is transferred from one particle to the next (a.k.a. conduction) – Sometimes this energy can be transferred from one object to another and influence physical properties – The more energy an object has the more energy is transferred An Ice Cold Spoon A Hot Spoon

5. 2 Hot Spoons  Thermal energy (q) is the total energy of all the particles that make up a substance – Kinetic energy from vibration of particles – Potential energy from molecular attraction (within or between the particles)  Thermal energy is dependent upon the amount or mass of material present (KE =½mv 2 )  Thermal energy is also related to the type of material

6.  Different types of materials may have the same temp, same mass, but different connectivity – Affected by the potential energy stored in chemical bonds or the IMFs holding molecules together  So it is possible to be at same temp (same KE) but have very different thermal energies  The ability to hold onto or release thermal (heat) energy is referred to as the substance’s heat capacity

7.  Thermal energy can be transferred from object to object through direct contact – Molecules collide, transferring energy from molecule to molecule – The flow of thermal energy is called heat

8. DEFINITION THE FLOW OF THERMAL ENERGY FROM SOMETHING WITH A HIGHER TEMP TO SOMETHING WITH A LOWER TEMP UNITS MEASURED IN JOULES OR CALORIES TYPES THROUGH WATER OR AIR = CONVECTION THROUGH SOLIDS = CONDUCTION TRANSFERRED ENERGY BY COLLISION WITH PHOTON = RADIANT ENERGY

9. HEAT CAPACITY  The measure of how well a material absorbs or releases heat energy is its heat capacity – It can be thought of as a reservoir to hold heat, how much it holds before it overflows is its capacity  Heat capacity is a physical property unique to a particular material – Water takes 1 calorie of energy to raise temp 1 °C – Steel takes only 0.1 calorie of energy to raise temp 1 °C  The measure of how well a material absorbs or releases heat energy is its heat capacity – It can be thought of as a reservoir to hold heat, how much it holds before it overflows is its capacity  Heat capacity is a physical property unique to a particular material – Water takes 1 calorie of energy to raise temp 1 °C – Steel takes only 0.1 calorie of energy to raise temp 1 °C

10. SPECIFIC HEAT CAPACITY  The amount of energy it takes to raise the temp of 1 gram of an object 1°C is that object’s specific heat capacity (C or s)  Specific heats can be listed on data tables – Smaller the specific heat  the less energy it takes the substance to feel hot They heat up quickly and cool down quickly – Larger the specific heat  the more energy it takes to heat a substance up (bigger the heat reservoir) They heat up slowly and cool down slowly  The amount of energy it takes to raise the temp of 1 gram of an object 1°C is that object’s specific heat capacity (C or s)  Specific heats can be listed on data tables – Smaller the specific heat  the less energy it takes the substance to feel hot They heat up quickly and cool down quickly – Larger the specific heat  the more energy it takes to heat a substance up (bigger the heat reservoir) They heat up slowly and cool down slowly

11.  Specific Heat Equation Q = MC  T  Q = Quantity of heat (joules)  M = Mass of substance (grams)  C = Specific heat capacity   T = Change in temperature T final – T initial = Change in temp. http://www.youtube.com/watch?v=XLWP03 pwTYY http://www.youtube.com/watch?v=XLWP03 pwTYY  Specific Heat Equation Q = MC  T  Q = Quantity of heat (joules)  M = Mass of substance (grams)  C = Specific heat capacity   T = Change in temperature T final – T initial = Change in temp. http://www.youtube.com/watch?v=XLWP03 pwTYY http://www.youtube.com/watch?v=XLWP03 pwTYY SPECIFIC HEAT CAPACITY

12. SUBSTANCE SPECIFIC HEAT CAPACITY, C P WATER 4.18 J/g°C OR 1 cal/g°C ICE 2.10 J/g°C OR.502 cal/g°C STEAM 1.87 J/g°C OR.447 cal/g°C MERCURY, Hg.139 J/g°C OR.033 cal/g°C ALCOHOL (Ethyl) 2.40 J/g°C OR.580 cal/g°C CALCIUM, Ca.647 J/g°C OR.155 cal/g°C ALUMINUM, Al.992 J/g°C OR.237 cal/g°C TABLE SALT, NaCl.865 J/g°C OR.207 cal/g°C AMMONIA, NH 3 2.09 J/g°C OR.500 cal/g°C SILVER, Ag.235 J/g°C OR.056 cal/g°C LEAD, Pb.129 J/g°C OR.031 cal/g°C

13.  There are three methods used to transfer heat/thermal energy – Conduction – transfer of heat through direct contact – Convection – transfer of heat through a medium like air or water – Radiant – transfer of heat by electromagnetic radiation  There are three methods used to transfer heat/thermal energy – Conduction – transfer of heat through direct contact – Convection – transfer of heat through a medium like air or water – Radiant – transfer of heat by electromagnetic radiation

15. CHANGE IN HEAT ENERGY (ENTHALPY)  The energy used or produced in a chem rxn is called the enthalpy of the rxn (  H rxn ) – Burning a 15 gram piece of paper produces a particular amount of thermal energy or heat energy (enthalpy)  Enthalpy is a value that also contains a component of direction (energy in or energy out) – Heat gained by the surroundings is the out-of/exo direction – Heat lost by the surroundings is the in-to/endo direction

16. HEATHEATHEATHEATHEATHEATHEATHEAT

17.  Chemical rxns can be classified as either: –Exothermic  a reaction in which heat energy is generated (a product) –Endothermic  reaction in which heat energy is absorbed (a reactant)  Exothermic rxns typically feel warm as the rxn proceeds (from the perspective of the surroundings) –Give off heat energy (light, fire, heat)  Endothermic rxns typically feel cooler the longer the rxn proceeds (from the perspective of the surroundings) –Absorb heat energy, sometimes enough to get very cold http://www.youtube.com/watch?v=XgiCn1Ipvz M&list=PL65159266CFC74682 http://www.youtube.com/watch?v=XgiCn1Ipvz M&list=PL65159266CFC74682

18. C3H8C3H8 C3H8C3H8 + + 5O 2 2043kJ   3CO 2 4H 2 O + + + +  Exothermic rxn –To a cold camper, the important product here is the heat energy C 3 H 8 + O 2

19. In an exothermic process the amount of energy given off is more than the initial energy invested. So the products are always lower in energy than the reactants.

20. NH 4 NO 3 +H 2 O+ 752kJ  NH 4 OH + HNO 3  Endothermic rxn –Similar system as what is found in cold packs NH 4 NO 3 + H 2 O NH 4 OH + HNO 3

21. In an endothermic process more energy is required to cause the rxn to proceed than obtained in return. So the products are always higher in energy than the reactants.

22.  Most common measurement of the energy or enthalpy in a reaction is actually a change in enthalpy (  H) –  H rxn = ∑H products - ∑H reactants  The enthalpy absorbed or gained (changed) in a rxn is dependent on the number of moles of material reacting – We can stoichiometrically calculate how much energy a rxn uses or produces –  H values can be provided with a rxn equn and have magnitude & direction of transfer (+ or -) CHANGE IN ENTHALPY

23. (For Example) How much heat will be absorbed for 1.0g of H 2 O 2 to decompose in a bombardier beetle to produce a defensive spray of steam 2H 2 O 2 +190kJ  2H 2 O + O 2 USING  H IN CALCULATIONS  Chemical reaction equations are very powerful tools. – Given a rxn equation with an energy value, We can calculate the amount of energy produced or used for any given amount of reactants.

24. Analyze: we know that if we had 2 mols of H 2 O 2 decomposing we would use 190kJ of heat, but how much would it be if only 1.0 g of H 2 O 2 Therefore: we have to convert our given 1.0 g of H 2 O 2 to moles of H 2 O 2 1.0g H 2 O 2 1mol H 2 O 2 34g H 2 O 2 2H 2 O 2 +190kJ  2H 2 O + O 2 =.02941 mol Molar mass

25. Therefore: with 2 moles of H 2 O 2 it requires the use of 190 kJ of energy, but we don’t have 2 moles we only have.02941 mols of H 2 O 2, so how much energy would the bug require? 190kJ 2molH 2 O 2 = 2.8kJ.02941 mol Rxn equation 2H 2 O 2 +190kJ  2H 2 O + O 2

26. How much heat will be released when 4.77 g of ethanol (C 2 H 5 OH) react with excess O 2 according to the following equation: C 2 H 5 OH+3O 2  2CO 2 +3H 2 O  H  = -1366.7kJ How much heat will be released when 4.77 g of ethanol (C 2 H 5 OH) react with excess O 2 according to the following equation: C 2 H 5 OH+3O 2  2CO 2 +3H 2 O  H  = -1366.7kJ Example #2 4.77g C 2 H 5 OH 1mol C 2 H 5 OH 46g C 2 H 5 OH -1366.7kJ 1mol C 2 H 5 OH = -142 kJ

27. 1.Ethanol, C 2 H 5 OH, is quite flammable and when 1 mole of it burns it has a reported  H of -1366.8 kJ. How much energy is given off in the combustion of enough ethanol to produce 12.0 L of Carbon dioxide @ 755 mmHg and 25.0°C? Classroom Practice 1 1 C 2 H 5 OH+ 3 O 2  2 CO 2 + 3 H 2 O  H= -1366.8 kJ 1 C 2 H 5 OH+ 3 O 2  2 CO 2 + 3 H 2 O  H= -1366.8 kJ

28.  H = FINAL TEMP – INITIAL TEMP FINAL TEMP – INITIAL TEMP SPECIFIC HEAT SPECIFIC HEAT MASS  We can also track energy changes due to temp changes, using  H=mC  T:  If the temp difference is positive – The rxn is exothermic because the final temp is greater than the initial temp – So the enthalpy ends up positive  If the temp difference is positive – The rxn is exothermic because the final temp is greater than the initial temp – So the enthalpy ends up positive  if the temp change is negative – the enthalpy ends up negative – the rxn absorbed heat into the system, so it’s endothermic  if the temp change is negative – the enthalpy ends up negative – the rxn absorbed heat into the system, so it’s endothermic

29. Example: If you drink 4 glasses of ice water at 0°C, how much heat energy is transferred as this water is brought to body temp? Each glass contains 250 g of water & body temp is 37°C.  mass of 4 glasses of water: –m = 4 x 250g = 1000g H 2 O  change in water temp: –T f – T i = 37°C - 0°C  specific heat of water: –C H2O = 4.18 J/g C°(from previous slide)  mass of 4 glasses of water: –m = 4 x 250g = 1000g H 2 O  change in water temp: –T f – T i = 37°C - 0°C  specific heat of water: –C H2O = 4.18 J/g C°(from previous slide)  H = mC H2O  T  H = (1000g)(4.18J/g°C)(37°C)  H = 160,000J

30. Example 2: 500 g of a liquid is heated from 25°C to 100°C. The liquid absorbs 156,900 J of energy. What is the specific heat of the liquid and identify it. Example 2: 500 g of a liquid is heated from 25°C to 100°C. The liquid absorbs 156,900 J of energy. What is the specific heat of the liquid and identify it.  H = mC  T C  =  H/m  T C = 156,900J/(500g)(75°C) C = 4.184 J/g°C H2OH2O H2OH2O

31. 2.An orange contains 445 kJ of energy. What volume of water could this same amount of energy raise from a temp of 25.0°C to the boiling point? 3.Water at 0.00°C was poured into 30.0g of water in a cup at 45.0°C. The final temp of the water mixture was 19.5°C. What was the mass of the 0.00°C water? 2.An orange contains 445 kJ of energy. What volume of water could this same amount of energy raise from a temp of 25.0°C to the boiling point? 3.Water at 0.00°C was poured into 30.0g of water in a cup at 45.0°C. The final temp of the water mixture was 19.5°C. What was the mass of the 0.00°C water? Classroom Practice 2

32. 2.An orange contains 445 kJ of energy. What volume of water could this same amount of energy raise from a temp of 25.0°C to the boiling point?  H= 445 KJ x 1000 = 445000J  T = 100.0 – 25.0 = 75.0 C = 4.18 J/g°C M =  H / C  T 445000J / (4.18 x 75.0) = 1420 g H 2 O = 1420 ml H 2 O

33.  Enthalpy is dependent on the conditions of the rxn – It’s important to have a standard set of conditions, which allows us to compare Cthe affect of temps, pressures, etc. On different substances  Chemist’s have defined a standard set of conditions – Stand. Temp = 298K or 25°C – Stand. Press = 1atm or 760mmHg  Enthalpy produced in a rxn under standard conditions is the standard enthalpy (  H°)

34.  Standard enthalpies can be found on tables measured as standard enthalpies of formations, enthalpies of combustion, enthalpies of solution, enthalpies of fusion, and enthalpies of vaporization – Enthalpy of formation (  H  f ) is the amount of energy involved in the formation of a compound from its component elements. – Enthalpy of combustion (  H  comb ) is the amount of energy produced in a combustion rxn. – Enthalpy of solution (  H  diss ) is the amount of energy involved in the dissolving of a compound

35. – Enthalpy of fusion (  H  fus ) is the amount of energy necessary to melt a substance. – Enthalpy of vaporization (  H  vap ) is the amount of energy necessary to convert a substance from a liquid to a gas.  All of these energies are measured very carefully in a laboratory setting under specific conditions – At 25 °C and 1atm of pressure  These measured energies are reported in tables to be used in calculations all over the world.

36.  Calorimetry is the process of measuring heat energy – Measured using a device called a calorimeter – Uses the heat absorbed by H 2 O to meas- ure the heat given off by a rxn or an object  The amount of heat soaked up by the water is equal to the amount of heat released by the rxn  H SYS =-  H SUR  H sys is the system or what is taking place in the main chamber (rxn etc.) And  H sur is the surroundings which is generally water.

37. A COFFEE CUP CALORIMETER A BOMB CALORIMETER used when trying to find the amount of heat produced by burning something. Used for a reaction In water, or just a transfer of heat.

38.  With calorimetry we use the sign of what happens to the water – When the water loses heat into the system it obtains a negative change (-  H surr ) – Endothermic (+  H sys ) – When the water gains heat from the system it obtains a positive change (+  H surr ) – Exothermic (-  H sys ) CALORIMETRY

39. HEATHEATHEATHEAT HEATHEATHEATHEAT -  H sys = =  H surr - SIGN MEANS HEAT WAS RELEASED BY THE RXN + SIGN MEANS HEAT WAS ABSORBED BY WATER  H sys = = -  H surr + SIGN MEANS HEAT WAS ABSORBED BY THE RXN - SIGN MEANS HEAT WAS RELEASED BY WATER

40.  You calculate the amount of heat absor- bed by the water (using  H= mC  T)  Which leads to the amount of heat given off by the rxn – you know the mass of the water (by weighing it) – you know the specific heat for water (found on a table) – and you can measure the change in the temp of water (using a thermometer) CALORIMETRY

41. A chunk of Al that weighs 72.0g is heated to 100°C is dropped in a calorimeter containing 120ml of water at 16.6°C. the H 2 O’s temp rises to 27°C. -mass of Al = 72g -T initial of Al = 100°C -T final of Al = 27°C -C Al =.992J/g°C (from table)  H Al  H Al = 72g.992J/g°C27°C-100°C  H Al = =-5214J

42.  We can do the same calc with the water info HH2OHH2O HH2OHH2O HH2OHH2OHH2OHH2O=5216J Equal but opposite, means that the Al decreased in temp, it released its stored heat into the H 2 O, causing the temp of the H 2 O to increase. HH2OHH2OHH2OHH2O = 120g 4.18J/g°C 4.18J/g°C27°C-16.6°C – Mass of H 2 O= 120g – T initial of H 2 O= 16.6°C – T final of H 2 O = 27°C – C H2O = 4.18J/g°C (from table)

43. When a 4.25 g sample of solid NH 4 NO 3 dissolves in 60.0 g of water in a calori- meter, the temperature drops from 21.0°C to 16.9°C. Calculate the energy involved in the dissolving of the NH 4 NO 3.  H water = (m water )(C water )(  T water )  H water = (60g)(4.18J/g°C)(16.9°C-21.0°C)  H water = -1.03 x 10 3 J  H water =  H NH4NO3  H NH4NO3 = 1.03 x 10 3 J

45. 4.A coffee-cup calorimeter with a mass of 4.8 g is filled with water to mass of 250 g. The water temperature was 24.2  C before 3.2 g of NaOH pellets was added to the water. After the NaOH pellets had dissolved the temp of the water registered 85.8  C. How much heat did the H 2 O absorb, and how much heat did the NaOH produce? 5.41.0g of glass at 95°C is placed in 175 g of Water at 19.5°C in a calorimeter. The temps are allowed to equalize. What is the final temp of the glass/water mixture? (Water = 4.18J/g°C; Glass = 8.78J/g°C) Classroom Practice 3