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Energy – Capacity to do work or transfer heat Types of Energy 1. Kinetic – Energy of motion

Types of Energy. Energy – Capacity to do work or transfer heat Types of Energy 1. Kinetic – Energy of motion a. KE = ½ mv 2 b. Unit – Joules (kg –m 2 /s 2 ) 1 calorie = 4.184 J 1000 cal = 1 Cal (nutr). F = ma W = Fd K = ½ mv 2 U = mgh. Types of Energy.

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Energy – Capacity to do work or transfer heat Types of Energy 1. Kinetic – Energy of motion

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  1. Types of Energy • Energy – Capacity to do work or transfer heat • Types of Energy 1. Kinetic – Energy of motion a. KE = ½ mv2 b. Unit – Joules (kg –m2/s2) 1 calorie = 4.184 J 1000 cal = 1 Cal (nutr)

  2. F = ma W = Fd K = ½ mv2 U = mgh

  3. Types of Energy c. Macroscale KE – energy of movement (car, baseball) d. Microscale KE – Temp. measure of average KE of molecules

  4. Types of Energy 2. Potential Energy – stored energy a. Macroscale PE – energy of position (rollercoaster ex.) b. Microscale PE – Energy in chemical bonds. (Food example)

  5. Types of Energy System and Surroundings 1. System – What we are studying 2. Surroundings – rest of the universe a. Open system – Car, earth b. Closed system – Vacuum

  6. The First Law The First Law of Thermodynamics Energy is neither created nor destroyed. It only changes form. Energy is conserved. A. DE = q + w DE = change in energy q = heat w = work

  7. The First Law B. Examples 1. Car 2. Vacuum Cleaner

  8. The First Law C. Transfer of Energy 1. Car – Chemical to heat to mechanical D. Consequences of the First Law 1. No machine is 100% efficient 2. Always lose some energy to heat

  9. Exo and Endothermic

  10. Enthalpy – Heat Energy • Heat – Energy transferred because of a difference in temperature (Cooking a turkey example) • Extensive Property – depends on amount of material Glass of water vs. iceberg “Which has more heat”

  11. Calorimetry • Calorimetry – Measuring heat by measuring a temperature change • Specific Heat – Amount of heat to raise temperature of one gram of a substance one degree Celsius or Kelvin 1. Unit – J/goC 2. Higher the sp. heat, more energy needed to raise temp

  12. Calorimetry • 3. Example – Wooden Spoon (~2 J/goC) vs. metal spoon (~0.50 J/goC) • Water • a. Very high specific heat - 4.18 J/goC • b. Oceans • c. Our bodies

  13. Calorimetry C. Heat for one substance 1. Formula q = mCpDT q = heat m = mass (grams) Cp = specific heat (J/goC) DT = Tfinal – Tinitial

  14. Calorimetry • How much heat is needed to warm 250.0 grams of water from 22.0oC to 98.0oC to make tea? (Ans: 79.5 kJ) • Calculate the specific heat of copper if 12.0 grams of copper cools from 98.0oC to 96.2 oC and tranfers 8.32 J of heat. (0.385 J/goC)

  15. Calorimetry • Calculate how much heat is absorbed by 50.0 kg of rocks if their temperature increases by 12.0 oC? Assume the specific heat is 0.82 J/g oC. (492 kJ) • Calculate how many calories of heat they absorbed. (117 kcal) • Calculate the final temperature of the rocks if they absorbed 450 kJ of heat and began at a temperature of 26.9 oC. (37.9 oC) • 50 kg of Copper (0.385 J/goC)absorbs the same amount of heat as the rocks in part c. Would the temperature change be greater or less for copper?

  16. A student heats 60.0 grams of iron to 98.0oC in a hot water bath. They then submerge the metal in 150.0 g of water (Cp = 4.18 J/goC) at 20.0oC. After a few minutes, the cup and metal reach a final temperature of 23.2oC. • Calculate the heat gained by the water. • Calculate the specific heat (Cp) of the metal.

  17. Calorimetry 3. Heat Capacity – heat to raise temperature of a particular object by 1K; unit of J/K a. Used only for one object b. Car part c. heat capacity= mCp

  18. Calorimetry • How many Joules of heat are required to raise 1 gram of water by 1 degree Celsius? • Calculate the molar heat capacity of water? (Ans: 75.2 J/ oC) • Calculate the heat capacity of 250.0 grams of water? (Ans: 1050 J/K) • Calculate the heat capacity of 6.20 mole of iron? (Ans: 156 J/oC)

  19. Heat of Reaction • Calorimeter – Coffee Cup • Used to measure heats of a chemical reaction • DH = -q DH = -mCpDT m = mass of BOTH reactants Cp = often 4.18 J/goC for aqueous solns

  20. Heat of Reaction Example 1. When a student mixes 50.0 mL of 1.00 M HCl and 50.0 mL of 1.00 M NaOH in a calorimeter, the temperature rises from 21.0oC to 27.5oC. Calculate the change in enthalpy for the reaction. (Ans: -2.7 kJ and –54 kJ/mol NaOH)

  21. Heat of Reaction Example 2. When 50.0 mL of 0.100 M AgNO3 and 50.0 mL of 0.100 M HCl are mixed, the temperature rises from 22.30oC to 23.11oC. Calculate the change in enthalpy for the reaction per mole of AgNO3. (Ans: –68 kJ/mol AgNO3)

  22. Standard State • Standard State – State of an element @25oC and 1 atm. 1. Standard Enthalpy of formation of an element in its st. state is zero 2. Takes no energy to make an element assume st. state. “Exists in this form naturally”

  23. Standard State • ExamplesDHfo • O2(g) 0 • H2(g) 0 • N2(g) 0 • F2(g), Cl2(g), Br2(l), I2(s) 0 C(graphite) 0 • Fe(s) 0

  24. Standard State • Carbon Allotropes – different physical forms of the same element • a. Graphite • b. Diamond • c. Buckyball (Buckminster Fuller) • C(gr) 0 • C(dia) 1.88 kJ/mol • C60(s) 2269 kJ/mol

  25. Standard State • Writing Standard State Equations • Examples: •  CO(g) •  H2O(g) •  KClO3(s)

  26. Standard State  NaNO3(s)  NH4CN(s)  Al2(CO3)3(s)

  27. Standard State  Fe(NO3)3(s)  CH3COOH(l)  NaOH (s)

  28. DH of Reaction 6. Using Standard Enthalpies of Formation DHor = SnDHfoprod – S mDHforeactants n,m = coefficients DHfo = values in text

  29. DH of Reaction Example 1. Calculate DH for the combustion of liquid C6H6 to CO2 and water (Ans:-3267 kJ/mol C6H6)

  30. DH of Reaction Example 2. Calculate DH for the combustion of liquid C2H5OH to CO2 and water (Ans:-1367 kJ/mol C2H5OH)

  31. DH of Reaction Example 3. Calculate DHfo for CaCO3(s) if you are given: CaCO3(s)  CaO(s) + CO2(g) DHrxno = +178.1 kJ (Ans:-1207.1 kJ/mol CaCO3)

  32. DH of Reaction Example 4. Calculate DHfo for CuO(s) if you are given: CuO(s) + H2(g)  Cu(s) + H2O(l) DHrxno = -129.7 kJ (Ans:-156.1 kJ/mol CuO)

  33. 20.0 mL of 0.100 M NaOH(aq) is neutralized with 0.0950 M HNO3(aq). • Write the balanced chemical reaction for this process. • Calculate the volume of HNO3 required.(21.1 mL) • When these two volumes are mixed, the temperature rises from 22.3oC to 23.0 oC. Calculate the DHrxn/mole of NaOH. (-60.1 kJ/mol) • Calculate heat of reaction the using the heats of formation values found in the appendix (not the pancreas). (-55.83 kJ/mol) • How do your two answers compare?

  34. Enthalpy – Heat Energy • Example 2H2(g) + O2(g)  2H2O(g) DH = -483.6 kJ 1. Hindenberg 2. Heat is released to the surroundings. 3. -483.6 kJ released per 2 mole of H2, 1 mole O2, or 2 mole of H2O

  35. Enthalpy – Heat Energy • Example 1. • How much heat is released from the combustion of 10.0 g of H2? • 2H2(g) + O2(g)  2H2O(g) DH = -483.6 kJ • (Ans: -1210 kJ)

  36. Enthalpy – Heat Energy Example 2. How much heat is released when 5.00 g of H2O2 decomposes? 2H2O2 2H2O + O2DH= -196 kJ Answer: -14.4 kJ

  37. Enthalpy – Heat Energy Example 3 How much heat is required to produce 25.0 g of H2O2? Reverse above reaction Answer: +72.1 kJ

  38. Enthalpy – Heat Energy Example 4 How much H2O2 is produced if 300.0 kJ of heat are absorbed? 2H2O + O2  2H2O2DH= +196 kJ Answer: 104 grams

  39. Enthalpy – Heat Energy Example 5 How many grams of water are produced if 8437 kJ of heat are released? C6H6 + 15/2O2  6CO2 +3H2O DH= -3267 kJ Answer: 139.5grams

  40. Hess’s Law • Hess’s Law - If a reaction is carried out in a series of steps, you add the DH’s for the individual steps • Can calculate DH without having to do the experiment. • Rules 1. If you flip the reaction, flip the sign of DH 2. If you multiply the reaction, multiply the DH.

  41. Hess’s Law • Example 1. • C(gr)  C(diamond) • C(gr) + O2(g)CO2(g) DH = -393.5 kJ • C(dia) + O2(g)CO2(g) DH = -395.4 kJ • (Ans: +1.9 kJ)

  42. Hess’s Law • Example 2. • NO (g) + O(g)  NO2(g) • NO(g) + O3(g) NO2(g) + O2 (g) DH=-198.9 kJ • O3(g) 3/2 O2 (g) DH=-142.3kJ • O2(g) 2 O(g) DH=+495.0kJ • (Ans: -304.1 kJ)

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