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CHAPTER 17

CHAPTER 17. THERMOCHEMISTRY. ENERGY. Thermochemistry is the study of the heat changes that occur during chemical reactions and physical changes of state . Heat (q) is a form of energy that always flows from a warmer object to a cooler object .

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CHAPTER 17

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  1. CHAPTER 17 THERMOCHEMISTRY

  2. ENERGY • Thermochemistry is the study of the heat changes that occur during chemical reactions and physical changes of state. • Heat(q) is a form of energy that always flows from a warmer object to a cooler object. • Energyis the capacity to do work or to supply heat. • Kinetic energy and Chemical Potential Energy • energy stored in chemicals because of their compositions

  3. ENERGY Remember the law of conservation of energy? • In any chemical or physical process, energy is neither created nor destroyed. • All energy can be accounted for as work, stored energy, or heat. System + Surroundings = Universe Constant Includes everything outside the system. Any specific part of the universe that attention is focused on

  4. HEAT CHANGES Endothermic Process • A process that absorbs heat from the surroundings • Surroundings cool down • System heats up • Heat change > 0 (positive) Endothermic Process • A process that loses heat to the surroundings • Surroundings heat up • System cools down • Heat change < 0 (negative)

  5. HEAT CAPACITY • The SI unit of heat (q) and energy (E) is the joule (J). • The amount of heat it takes to change an object’s temperature by 1ºC is the heat capacity(J/ºC) of that object. • The amount of heat that it takes to change the temperature of 1g of a substance by 1ºC is the specific heat of that object. • One calorie is the quantity of heat that raises the temperature of 1 gram of water 1ºC. 1 Calorie = 1000 calories and 1 Joule = 0.239 cal

  6. HEAT CAPACITY VS. SPECIFIC HEAT Specific Heat (J/g.oC) • The amount of heat it would take to raise the temperature of a 1g of a substance by 1oC • Independent of mass • The specific heat of steel is about 0.5 J/g.oC Heat Capacity (J/ºC) • The amount of heat it would take to raise the temperature of a sample of a substance by 1oC • Dependent on Mass • It would take much more heat to raise the temperature of a steel girder by 1oC than it would to raise the temperature of a steel nail by the same amount .002 kg = 1J 200 kg = 1 x 105J

  7. HEAT CAPACITY VS. SPECIFIC HEAT PROBLEMS Heat Capacity (J/oC) How much heat is required to raise 150g of silver by 1oC? (Specific Heat of Ag is 0.24J/g.oC) What is the specific heat of silver if takes 36J to increase a 150g sample of silver from 25oC to 26oC?

  8. CALORIMETRY • Calorimetryis the accurate and precise measurement of the heat change for chemical and physical processes. • The enthalpychange (△H), is the heat change(q) for a process at constant pressure. • q = △H = mC△T Constant Volume Calorimetry

  9. CALORIMETRY CALCULATIONS To measure heat changes using calorimetry... 1. Measure initial temperature of water(SURROUNDINGS) 2. Perform reaction inside insulated container. 3. Record final water temperature. 4. Use specific heat of water to measure enthalpy change with the formula qsys=△H= -qsurr = -mC△T For endothermic reactions qsys is positive qsurris negative For exothermic reactions qsys is negative qsurr is positive

  10. EXAMPLE PROBLEM The following acid-base reaction is performed in a coffee cup calorimeter: H+(aq) + OH-(aq) --> H2O(l) 55mL of water containing 0.1mol of HCl and 55mL of water containing 0.1mol of NaOH are combined in a calorimeter. The temperature of the water rises from 25.0°C to 26.2°C when 0.10 mol of each substance react. • Calculate qwater • Calculate ΔH for the reaction • Calculate ΔH if 1.00 mol OH- reacts with 1.00 mol H+

  11. SOLUTION

  12. THERMOCHEMICAL EQUATIONS • A chemical equation that includes the amount of heat produced or absorbed during the reaction is the thermochemical equation. Exothermic Reaction • The heat of reactionexplains the heat released or absorbed during a chemical reaction (the change in enthalpy, △H). + 2220 kJ Endothermic Reaction NH4NO3(s) + H2O(l) + 26.2KJ --> NH4OH(aq) + HNO3(aq)

  13. THERMOCHEMICAL EQUATIONS • The heat of reaction for the complete burning of one mole of a substanceheat of combustion • The heat absorbed by one mole of a substance while melting molar heat of fusion (△Hfus) • The heat lost when one mole of a substance freezes molar heat of solidification(△Hsolid) • The heat absorbed when one mole of a substances vaporizes molar heat of vaporization (△Hvap) • The heat absorbed when one mole of a substances condenses molar heat of condensation (△Hcond) • The heat change caused by dissolution of one mole of a substance molar heat of solution(△Hsoln)

  14. MOLAR HEAT CALCULATIONS The heat of combustion of methane is -890KJ/mol. How much heat is released when 45g of methane are combusted? How much heat would be required to melt 523g of osmium is its molar heat of fusion is 31.79 KJ/mol?

  15. Potentially Useful Information • For water… • ΔHfus = 6.01 KJ/mol • ΔHvap= 40.7 KJ/mol • What about… • ΔHsolid • ΔHcond

  16. PHASE CHANGES AND ENTHALPY To calculate the amount of heat absorbed or released as a substance changes state, q=mcDT for pure states and molar heats for phase changes. FOR WATER • ΔHvap= 40.7 KJ/mol • ΔHfus = 6.01 KJ/mol solid

  17. Sample Problem • Calculate the total enthalpy change when 42.0 grams of ice at -10oC is warmed to water at 20oC.

  18. HESS’S LAW • Hess’s law of heat summationstates that if you add two or more thermochemical equations to give a final equation, then you can also add the heat changes to give the final heat change. EXAMPLE Calculate the value of Ho/kJ for the following reaction using the listed thermochemical equations: 2 F2(g) + 2 H2O(l) 4 HF(g) + O2(g) H2(g) + F2(g) 2 HF(g) Ho/kJ = - 542 kJ 2 H2(g) + O2(g) 2 H2O(l) Ho/kJ = - 572 kJ H2(g) + F2(g) 2 HF(g) Ho/kJ = - 542 kJ 2 H2O(l) 2 H2(g) + O2(g) Ho/kJ = + 572 kJ 2 F2(g) + 2 H2O(l) 4 HF(g) + O2(g) Ho/kJ = +30 kJ

  19. STANDARD HEATS OF FORMATION • The standard heat of formation(△Hfo) of a compound is the change in enthalpy that accompanies the formation of one mole of the compound from its elements with all substances in their standard states at 25 oC. • For free elements △Hf°= 0 in their standard states. • Standard heat of reaction... △H°= S△Hfo(products) - S△Hfo(reactants) *Page 530 – standard heats of formation table

  20. EXAMPLE Calculate the heat of reaction (kJ) using the listed heats of formation: CH4(g) + 4 F2(g) CF4(g) + 4 HF(g) Hof CH4 = -74.86 kJ/mole Hof CF4 = -925.0 kJ/mole Hof HF = -271.1 kJ/mole DHfo = ((-925kJ/mol) + (4moles x -271.1kJ/mol)) - ((1 mole x -74.86kJ/mol) + (1 x 0)) = 1935kJ

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