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Chapter 6 Thermochemistry

CHEMISTRY. Chapter 6 Thermochemistry. Thermodynamics. - is the study of energy and the transformations it undergoes in chemical reactions. Units: Joules (J) calorie = 4.184 J calorie – the amount of E needed to raise the temperature of 1 gram of water 1 o C. Energy.

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Chapter 6 Thermochemistry

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  1. CHEMISTRY Chapter 6Thermochemistry Chapter 5

  2. Thermodynamics - is the study of energy and the transformations it undergoes in chemical reactions. Units: Joules (J) calorie = 4.184 J calorie – the amount of E needed to raise the temperature of 1 gram of water 1 oC Chapter 5

  3. Energy • Examples of energy: • Potential E • Kinetic E • Work – E needed to move an object against a force • Heat – E transferred from hot to cold objects - capacity to do work or transfer heat Chapter 5

  4. Today’s Topics • 4 Thermodynamic Functions • Definition of State Function • Internal Energy • Point of Views – System, Surroundings Universe • Sign Conventions Chapter 5

  5. 4 Thermodynamic Functions • E - Internal Energy • H - Enthalpy • S - Entropy • G - Gibb’s Free Energy Chapter 5

  6. What is a state function? • A state function is a property that depends on the present condition and not on how the change occurs. • Derived from calculating the change: D = [Final value – initial value] Chapter 5

  7. 4 State Functions • Therefore: • DE = Ef -Ei DH = Hf - Hi DS = Sf - Si DG = Gf - Gi Chapter 5

  8. Internal Energy (E) • Internal Energy - is the sum of the kinetic and potential energy of a system • Is the sum of heat (q) and work (w) DE = Efinal – Einitial DE = q + w Chapter 5

  9. Point of Views • System - the reaction we are studying • Surroundings – anything else besides the reaction For example: the container, you, etc…. • Universe – system + surroundings Chapter 5

  10. q = heat and w = work • If q is (+), system gaining heat from surroundings (endo) • If q is (-), system giving up heat to the surroundings (exo) • If w is (+), system is the recipient of work from the surroundings. (in short, surroundings is doing work on the system. Chapter 5

  11. Thermodynamic Functions • Have value, unit and magnitude. • The sign of DE depends on the magnitude of q and w. • Knowing the value of DE does not tell us which variable is larger, q or w. Chapter 5

  12. Problem 1 • Calculate the change in internal energy of the system for a process in which the system absorbs 140 J of heat from the surroundings and does 85 J of work on the surroundings. Chapter 5

  13. Problem 2 • Consider the reaction of hydrogen and oxygen gases to produce water. As the reaction occurs, the system loses 1150 J of heat to the surroundings. The expanding gas does 480 J of work on the surroundings as it pushes against the atmosphere. Calculate the change in the internal energy of the system? Chapter 5

  14. Problem 3 • Calculate DE and determine whether the process is endothermic or exothermic. • 1.) q = 1.62 kJ and w = -874 J • 2.) The system releases 113 kJ of heat to the surroundings and does 39 kJ of work. • 3.) The system absorbs 77.5 kJ of heat while doing 63.5 kJ of work on the surroundings. Chapter 5

  15. Thermodynamic Functions • Have value, unit and magnitude. • The sign of DE depends on the magnitude of q and w. • Knowing the value of DE does not tell us which variable is larger, q or w. Chapter 5

  16. Enthalpy • Enthalpy – accounts for heat flow in chemical reactions that occur at constant P when nothing other than P-V work are performed DH = Hfinal - Hinitial • If: H = E + PDV • Then: DH = DE + PDV Chapter 5

  17. Ways of Measuring DH • Calorimetry • Hess’s Law • Heats of Formation (DHof) Chapter 5

  18. Calorimetry • - Is the measurement of heat flow • Specific Heat capacity (C) - the amount of heat needed to raise the temperature of 1 g of substance 1 oC. • The greater the heat capacity, the greater the heat required to produce a rise in temp. Chapter 5

  19. Calorimetry Equation • q = mCDT • - qsystem = qsurroundings • - qsystem = qwater + qcalorimeter • - qsubstance = (mCDTsolution + mCDTcalorimeter) • Simplifies to: • - qsubstance = (mCDTsolution + CDTcalorimeter) Chapter 5

  20. Calorimetry Problem • A 30.0 gram sample of water at 280 K is mixed with 50.0 grams of water at 330 K. Calculate the final temperature of the mixture assuming no heat loss to the surroundings. The specific heat capacity of the solution is 4.18 J/g-oC. Chapter 5

  21. Calorimetry Problem • A 46.2 gram sample of copper is heated to 95.4 oC and then placed in a calorimeter containing 75.0 gram of water at 19.6 oC. The final temperature of the metal and water is 21.8 oC. Calculate the specific heat capacity of copper, assuming that all the heat lost by the copper is gained by the water. Chapter 5

  22. Calorimetry Problem • A 15.0 gram sample of nickel metal is heated to 100.0 oC and dropped into 55.0 grams of water, initially at 23 oC. Assuming that no heat is lost to the calorimeter, calculate the final temperature of the nickel and water. The specific heat of nickel is 0.444 J/g-oC. The specific heat of water is 4.18 J/g-oC. Chapter 5

  23. Problem 4 • The specific heat of water is 4.18 J/g-K. • How much heat is needed to warm 250 g of water from 22 oC to 98 oC? • What is the molar heat capacity of water? • Molar heat Capacity = C x molar mass Chapter 5

  24. Problem 5 • Large beds of rocks are used in solar heated homes to store heat. Assume that the specific heat of rocks is 0.82 J/g-K. • A. Calculate the amount of heat absorbed by 50 kg. of rocks if their temperature rose by 12.0 oC. • B. What temperature change would these rocks undergo if they emitted 450 kJ of heat? Chapter 5

  25. Work – E needed to move an object against a force • When P is constant, P-V work is given by w = - PDV Chapter 5

  26. Enthalpy • Enthalpy of a reaction or heat of Reaction: DH = Hproducts - Hreactants • 1. sign of DH depends on the amount of reactant consumed • 2. DH sign is opposite for backwards reaction • 3. DHrxn depends on the physical state of the reactants and products. Chapter 5

  27. Problem 1 • Given the reaction: • 2H2 (g) + O2 (g)  2 H2O (g) DH = -483 kJ • Calculate the DH value for: • 2 H2O (g)  2H2 (g) + O2 (g) Chapter 5

  28. Problem 2 • Given the reaction: • 2H2 (g) + O2 (g)  2 H2O (g) DH = -483 kJ • How much heat is released when 10.5 grams of H2 is burned in a constant-pressure system? Chapter 5

  29. Ways of Measuring DH • Calorimetry • Hess’s Law • Heats of Formation (DHof) Chapter 5

  30. Hess’ Law • If a reaction is carried out in steps, DH for the reaction will equal the sum of the enthalpy changes for the individual steps. Chapter 5

  31. Problem 1 • Given: • C (s) + O2 (g)  CO2 (g) DH = -393.5 kJ • CO(g) + ½ O2 (g)  CO2 (g) DH = -283.5 kJ • Calculate the enthalpy of combustion for: C(s) + ½ O2 (g)  CO (g) Chapter 5

  32. Problem 2 • Given: • C(graphite) + O2 (g)  CO2 (g) DH = -393.5 kJ • C(diamond) + O2 (g)  CO2 (g) DH = -395.4 kJ • Calculate the enthalpy of combustion for: C(graphite)  C (diamond) Chapter 5

  33. Problem 3 • Given: • C2H2 (g) + 5/2 O2 (g)  2CO2 (g) + H2O (l) DH = -1299.6 kJ • C (s) + O2 (g)  CO2 (g) DH = -395.4 kJ • H2 (g) + 1/2 O2 (g)  H2O (l) DH = -285.8 kJ • Calculate the enthalpy of combustion for: 2C (s) + H2 (g)  C2H2 (g) Chapter 5

  34. Enthalpies of Formation • - known as heat of formation • - gives the energy needed for a compound to form • Standard enthalpy - is the enthalpy change (DH) when the reactants and products are in their standard state, usually 1 atm and 25 oC - denoted by DHo ( ex. DHof) Chapter 5

  35. Standard Enthalpy of Formation, (DHof) • By definition: • The standard enthalpy of formation ( ex. DHof) of the most stable form of any element is ZERO because there is no formation reaction needed when the element is in its standard state. • - important for diatomic molecules • - need knowledge of standard states of compounds Chapter 5

  36. Enthalpies of Formation • If there is more than one state for a substance under standard conditions, the more stable one is used. • Standard enthalpy of formation of the most stable form of an element is zero. • Using Enthalpies of Formation of Calculate Enthalpies of Reaction • We use Hess’ Law to calculate enthalpies of a reaction from enthalpies of formation. Chapter 5

  37. Enthalpies of Formation Chapter 5

  38. Enthalpies of Formation • Using Enthalpies of Formation of Calculate Enthalpies of Reaction • For a reaction Chapter 5

  39. Next Topic • DS = Entropy = disorder Chapter 5

  40. Thermodynamic Question • Can a process occur? • Spontaneous or Non-spontaneous? • Forward vs. Reverse reactions • Example: Gas Expansion • Reversible or irreversible? Chapter 5

  41. Entropy • Entropy, S, is a measure of the disorder of a system. • Spontaneous reactions proceed to lower energy or higher entropy. • In ice, the molecules are very well ordered because of the H-bonds. • Therefore, ice has a low entropy. Chapter 5

  42. As ice melts, the intermolecular forces are broken (requires energy), but the order is interrupted (so entropy increases). • Water is more random than ice, so ice spontaneously melts at room temperature. • Conclusion: The higher the entropy the more spontaneous the reaction. Chapter 5

  43. Generally, when an increase in entropy in one process is associated with a decrease in entropy in another, the increase in entropy dominates. • Entropy is a state function. • For a system, S = Sfinal - Sinitial. • If S > 0 the randomness increases, if S < 0 the order increases. Chapter 5

  44. Entropy and the Second Law of Thermodynamics • Entropy • Suppose a system changes reversibly between state 1 and state 2. Then, the change in entropy is given by • at constant T where qrev is the amount of heat added reversibly to the system. (Example: a phase change occurs at constant T with the reversible addition of heat.) Chapter 5

  45. Entropy and the Second Law of Thermodynamics • The Second Law of Thermodynamics • Spontaneous processes have a direction. • In any spontaneous process, the entropy of the universe increases. • Suniv = Ssys + Ssurr: the change in entropy of the universe is the sum of the change in entropy of the system and the change in entropy of the surroundings. • Entropy is not conserved: Suniv is increasing. Chapter 5

  46. Entropy and the Second Law of Thermodynamics • The Second Law of Thermodynamics • Reversible process: Suniv = 0. • Spontaneous process (i.e. irreversible): Suniv > 0. • Ssys for a spontaneous process can be less than 0as long asSsurr > 0.This would makeSuniv still (+). • For an isolated system, Ssys = 0 for a reversible process and Ssys > 0 for a spontaneous process. Chapter 5

  47. Third Law of Thermodynamics • The entropy of a perfect crystal at 0 K is zero. • Entropy changes dramatically at a phase change. • As we heat a substance from absolute zero, the entropy must increase. Chapter 5

  48. Chapter 5

  49. Entropy Changes in Chemical Reactions • Absolute entropy can be determined from complicated measurements. • Standard molar entropy, S: entropy of a substance in its standard state. Similar in concept to H. • Units: J/mol-K. Note units of H: kJ/mol. • Standard molar entropies of elements (S ) are not 0. • For a chemical reaction which produces n moles of products from m moles of reactants: Chapter 5

  50. Chapter 5

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