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Chapter 14 Thermodynamics PART 2

Chapter 14 Thermodynamics PART 2. Standard Conditions. The standard state = state of a material at a defined set of conditions: pure gas at exactly ___________ pressure pure solid or liquid in its most stable form at exactly ___________ pressure and (usually) ______ ° C

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Chapter 14 Thermodynamics PART 2

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  1. Chapter 14 Thermodynamics PART 2

  2. Standard Conditions • The standard state = state of a material at a defined set of conditions: • pure gas at exactly ___________ pressure • pure solid or liquid in its most stable form at exactly ___________ pressure and (usually) ______ °C • substance in a solution with _____ M [ ]

  3. The standard enthalpy change, DH°, is enthalpy change when all reactants and products are in their standard states.

  4. Standard Conditions standard enthalpy of formation, DHf°, is enthalpy change for rxtn forming 1 mole of a pure compound from its constituent elements (elements must be in their standard states) DHf° for pure element in standard state = _____ kJ/molby definition

  5. Standard Enthalpies of Formation

  6. Formation Reactions Rxtns of elements in their standard state to form 1 mole of a pure compound if you are not sure what the standard state of an element is, find the form in Appendix Dthat has a DHf° = 0

  7. because the definition requires • 1 mole of compound be made, • the coefficients of reactants may be fractions H2 (g) + ½ O2 (g)  H2O (l) DH0f = -286 kJ/mol

  8. Writing Formation ReactionsWrite the Formation Rxtnfor CO(g) The formation reaction is the rxtnbtwn the elements in the compound, which are C and O: The elements must be in their standard state.

  9. Writing Formation ReactionsWrite the Formation Reaction for CO(g) C + O → CO(g) There are several forms of solid C, but the one with DHf° = 0 is ______________ oxygen’s standard state = _________________

  10. #n C(s, graphite) + #n O2(g) →1 mol CO(g) The equation must be balanced, but the coefficient of the product compound must be 1 mole Use whatever coefficient in front of reactants is necessary to make #atoms = on both sides withoutD’ing the product coefficient ___ C(s, graphite) + ___O2(g) →___CO(g)

  11. Write the formation reactions for the following: CO2(g) Al2(SO4)3(s) (use the rhombic form of S as S8)

  12. Calculating Standard DH for a Rxtn Any rxtn can be written as a sum of formation rxtns (or reverse of formation rxtns) for the reactants & products

  13. Calculating Standard DH for a Rxtn • The DH° for the reaction is then • the sum of the DHf° for the component rxtns • DH°reaction = SnDHf°(prod) − SnDHf°(react) • S means ________ • n is _____________ in the reaction

  14. CH4(g)+ 2O2(g)→ CO2(g) + 2H2O(g) C(s, graphite) + 2 H2(g) → CH4(g) DHf°= − 74.6 kJ/mol CH4 C(s, graphite) + O2(g) → CO2(g) DHf°= −393.5 kJ/mol CO2 2 H2(g) + O2(g) → 2 H2O(g) DH° = −483.6 kJ Look at each carefully. Which side gives you a reactant? Which side gives you a product?

  15. CH4(g)+ 2O2(g)→ CO2(g) + 2H2O(g)

  16. CH4(g)+ 2O2(g)→ CO2(g) + 2H2O(g)

  17. Example: Calculate the DH in the reaction: • 2 C2H2(g) + 5 O2(g) 4 CO2(g) + 2 H2O(l) • Write formation reactions for each compound and determine the DHf° for each • 2 C(s, gr) + H2(g)  C2H2(g) • DHf° = +227.4 kJ/mol • C(s, gr) + O2(g)  CO2(g) • DHf° = −393.5 kJ/mol • H2(g) + ½ O2(g)  H2O(l) • DHf° = −285.8 kJ/mol

  18. Example: Calculate the DHrxn • 2 C2H2(g) + 5 O2(g) 4 CO2(g) + 2 H2O(l) • Arrange eqtns so they add up to desired reaction

  19. Use Appendix D: Calculate the DH in the reaction 2 C2H2(g) + 5 O2(g)  4 CO2(g) + 2 H2O(l) DH°reaction = S n DHf°(products) − S n DHf°(reactants) DHrxn = [(4•DHCO2 + 2•DHH2O) – (2•DHC2H2 + 5•DHO2)] DHrxn = [(4•(−393.5) + 2•(−285.8)) – (2•(+227.4) + 5•(0))] DHrxn = −2600.4 kJ

  20. How many kg of octane must be combusted to supply 1.0 x 1011 kJ of energy? DHf°’s DHrxn° kJ mol C8H18 g C8H18 kg C8H18 Given: Find: 1.0 x 1011 kJ mass octane, kg DH°rxn = −5074.1 kJ Write the balanced equation per mole of octane Conceptual Plan: Relationships: MMoctane = 114.2 g/mol, 1 kg = 1000 g Solution: C8H18(l) + 25/2 O2(g) → 8 CO2(g) + 9 H2O(g) the units and sign are correct, the large value is expected Check:

  21. How many kg of octane must be combusted to supply 1.0 x 1011 kJ of energy? DHf°’s DHrxn° kJ mol C8H18 g C8H18 kg C8H18 Given: Find: 1.0 x 1011 kJ mass octane, kg Write the balanced equation per mole of octane Conceptual Plan: Relationships: MMoctane = 114.2 g/mol, 1 kg = 1000 g from above Solution: C8H18(l) + 25/2 O2(g) → 8 CO2(g) + 9 H2O(g) Look up the DHf° for each material in Appendix D Check:

  22. How many kg of octane must be combusted to supply 1.0 x 1011 kJ of energy? DHf°’s DHrxn° kJ mol C8H18 g C8H18 kg C8H18 Given: Find: 1.0 x 1011 kJ mass octane, kg Write the balanced equation per mole of octane Conceptual Plan: Relationships: MMoctane = 114.2 g/mol, 1 kg = 1000 g from above Solution: C8H18(l) + 25/2 O2(g) → 8 CO2(g) + 9 H2O(g)

  23. How many kg of octane must be combusted to supply 1.0 x 1011 kJ of energy? DHf°’s DHrxn° kJ mol C8H18 g C8H18 kg C8H18 Given: Find: 1.0 x 1011 kJ mass octane, kg DH°rxn = −5074.1 kJ Write the balanced equation per mole of octane Conceptual Plan: Relationships: MMoctane = 114.2 g/mol, 1 kg = 1000 g from above Solution: the units and sign are correct, the large value is expected Check:

  24. Calculate the DH° for decomposing 10.0 g of limestone, CaCO3, under standard conditions.CaCO3(s) →CaO(s) + O2(g)

  25. DHf°’s DH°rxn per mol CaCO3 g CaCO3 mol CaCO3 kJ Practice – Calculate the DH° for decomposing 10.0 g of limestone, CaCO3(s) → CaO(s) + O2(g) Given: Find: 10.0 g CaCO3 DH°, kJ DH°rxn = +572.7 kJ Conceptual Plan: Relationships: MMlimestone = 100.09 g/mol, Solution:

  26. DHf°’s DH°rxn per mol CaCO3 g CaCO3 mol CaCO3 kJ Practice – Calculate the DH° for decomposing 10.0 g of limestone, CaCO3(s) → CaO(s) + O2(g) Given: Find: 10.0 g CaCO3 DH°, kJ DH°rxn = +572.7 kJ Conceptual Plan: Relationships: MMlimestone = 100.09 g/mol, Solution: Check: the units and sign are correct

  27. Heat Exchange Heatis exchange of thermal E btwn the system & surroundings Occurs when system & surroundings have a diff in temp Temperature is measure of amount of thermal E within a sample of matter

  28. Heat Exchange • Heat flows from matter with high temp • to matter with low temp • until both objects reach same temp: • = thermal equilibrium

  29. Quantity of Heat Energy Absorbed:Heat Capacity When a system absorbs heat, its temp increases. Increase in temp a amount of heat absorbed

  30. Heat Capacity Increase in temp a amount of heat absorbed Proportionality constant = heat capacity, C units of C are J/°C or J/K q = C x DT

  31. Heat Capacity q = C x DT The larger the C of the obj being studied, the smaller the temp rise will be for a given amount of heat

  32. Factors Affecting Heat Capacity • Heat capacity (extensive property) depends on its amount of matter • often measured by its mass. • 200 g of water requires 2x heat to raise its temp by 1°C as does 100 g of water

  33. Factors Affecting Heat Capacity • Heat capacity depends on type of material 1000 J heat E raises temp of 100 g sand 12.0 °C, but only raises temp of 100 g water by 2.4 °C

  34. Specific Heat Capacity Measures substance’s intrinsic ability to absorb heat

  35. Specific heat capacity (Cs) = amount of heat E required to raise temp of 1 g (mass) of a subst 1°C • Csunits are J/(g∙°C) • q = (m) x (Cs) x (DT)

  36. Molar heat capacity (Cp) = amount of heat E required to raise temp of one moleof a substance 1°C q = (m) x (Cs) x (DT) q = (n) x (Cp) x (DT)

  37. Specific Heat of Water • the Standard • H2O has high specific heat • Which means: Absorbs a lot of heat E • without large temp increase • Without H2O, Earth’s temp would be ~ = moon’s temp on side facing sun (avg 107 °C or 225 °F)

  38. Specific Heat of Water Water - used as a coolant Why? it can absorb a lot of heat and remove heat from important mechanical parts to keep them from overheating (or melting) Can transfer heat to something else because it is a fluid

  39. Quantifying Heat Energy heat capacity of an object is a to mass & specific heat of the material.

  40. Quantifying Heat Energy if we know the mass, specific heat, and temp D of the object, we can calculate the quantity of heat absorbed by an object: Heat= (mass) x (specific heat) x (temp D)

  41. Cs m, DT q How much heat is absorbed by a Cu penny with mass 3.10 g & temp rises from −8.0 °C to 37.0 °C? • Sort Information Given: Find: T1 = −8.0 °C, T2= 37.0 °C, m = 3.10 g q,J q = m ∙ Cs ∙ DT Cs = 0.385 J/g•ºC (Table 6.4) • Strategize Conceptual Plan: Relationships: • Follow the conceptual plan to solve the problem Solution: • Check Check: the unit is correct, the sign is reasonable as the penny must absorb heat to make its temp rise

  42. Calculate the amount of heat released when 7.40 g of water cools from 49° to 29 °C(water’s specific heat is 4.18 J/gºC) What equation will you choose?

  43. Cs m, DT q Calculate the amount of heat released when 7.40 g of water cools from 49° to 29 °C • Sort Information Given: Find: T1= 49 °C, T2 = 29 °C, m = 7.40 g q inJoules q = m ∙ Cs ∙ DT Cs = 4.18 J/gC • Strategize Conceptual Plan: Relationships: • Follow the concept plan to solve the problem Solution: • Check Check: the unit is correct, the sign is reasonable as the water must release heat to make its temperature fall

  44. Heat Transfer & Final Temp two objects at diff temps are placed in contact, heat flows from material at higher temp to the material at lower temp. Heat flows until both materials reach the same final temperature. Amount of heat E lost by hot material = amount of heat gained by cold material

  45. Heat lost by the metal > heat gained by water Heat gained by water > heat lost by the metal Heat lost by the metal > heat lost by the water Heat lost by the metal = heat gained by water More information is required Clicker Qstn: A piece of metal at 85 °C is added to water at 25 °C, the final temperature of both metal & water is 30 °C. Which of the following is true? 47

  46. Cs, AlmAl, Cs, H2O mH2O DTAl = kDTH2O A 32.5-g cube of Al initially at 45.8 °C is submerged into 105.3 g of water at 15.4 °C. What is the final temperature of both substances at thermal equilibrium? Given: Find: mAl = 32.5 g, Tal = 45.8 °C, mH20 = 105.3 g, TH2O= 15.4 °C Tfinal,°C q = m ∙ Cs ∙ DT Cs, Al = 0.903 J/g•ºC, Cs, H2O = 4.18 J/g•ºC(Table 6.4) Conceptual Plan: Relationships: qAl = −qH2O Tfinal Solution: the unit is correct, the number is reasonable as the final temperature should be between the two initial temperatures Check:

  47. Specific heat capacities of some common substances Clicker Qstn: A hot piece of metal weighing 350.0 g is heated to 100.0 °C. It is then placed into a coffee cup calorimeter containing 160.0 g of water at 22.4 °C. The water warms and the copper cools until final T is 35.2 °C. Calculate the specific heat of the metal & identify the metal.

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