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Chapter 5 Principles of Chemical Reactivity: Energy and Chemical Reactions

Chapter 5 Principles of Chemical Reactivity: Energy and Chemical Reactions. Energy & Chemistry. Questions that need to be addressed: How do we measure and calculate the energy changes that are associated with physical changes and chemical reactions?

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Chapter 5 Principles of Chemical Reactivity: Energy and Chemical Reactions

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  1. Chapter 5Principles of Chemical Reactivity: Energy and Chemical Reactions

  2. Energy & Chemistry Questions that need to be addressed: • How do we measure and calculate the energy changes that are associated with physical changes and chemical reactions? • What is the relationship between energy changes, heat, and work? • How can we determine whether a chemical reaction is product-favored or reactant-favored at equilibrium? • How can we determine whether a chemical reaction or physical change will occur spontaneously, that is, without outside intervention?

  3. Units of Energy James Joule 1818-1889 1 calorie = heat required to raise temp. of 1.00 g of H2O by 1.0 °C. 1000 cal = 1 kilocalorie = 1 kcal 1 kcal = 1 Calorie (a food “calorie”) SI units for energy: joule (J) 1 cal = exactly 4.184 J

  4. Energy & Chemistry Some Basic Principles • Energy as the capacity to do work or transfer heat (q). • Heat (q) is NOT temperature, temperature is a measure of kinetic energy. • Energy is divided into two basic categories: • Kinetic energy (the energy associated with motion) • Potential energy (energy that results from an object’s position). • The law of conservation of energy requires that energy can neither be created nor destroyed. • However, energy can be converted from one type into another.

  5. Energy & Chemistry Energy can be divided into two forms: Kinetic: Energy of motion: Thermal Mechanical Electrical Potential: Stored energy: Gravitational Electrostatic Chemical & Nuclear

  6. Potential & Kinetic Energy Kinetic energy— energy of motion • Translation

  7. Potential & Kinetic Energy Potential energy— energy a motionless body has by virtue of its composition and position.

  8. Burning peanuts supply sufficient energy to boil a cup of water. Burning sugar (sugar reacts with KClO3, a strong oxidizing agent) Energy & Chemistry

  9. Thermal Equilibrium • Energy transfer as heat will occur spontaneously from an object at a higher temperature to an object at a lower temperature. • Transfer of energy as heat continues until both objects are at the same temperature and thermal equilibrium is achieved. • At thermal equilibrium, the object with a temperature increase has gained thermal energy, the object with a temperature decrease has lost thermal energy.

  10. System & Surroundings • SYSTEM • The object under study • SURROUNDINGS • Everything outside the system • Energy flows between the two

  11. Directionality of Energy Transfer • When energy leaves the system and goes into the surroundings, the process is said to be EXOTHERMIC. • In the case of thermal energy, the temperature of the system decreases. (qsystem < 0) • Tsystem = (Tfinal – Tinitial) < 0

  12. Directionality of Energy Transfer • When energy enters the system and from the surroundings, the process is said to be ENDOTHERMIC. • In the case of thermal energy, the temperature of the system increases. (qsystem > 0) • Tsystem > 0

  13. Heat & Changes in Internal Energy • Heat flows between the system and surroundings. • U is defined as the internal energy of the system. • q = the heat absorbed or lost

  14. Heat & Changes in Internal Energy If heat enters the system: U > 0 therefore q is positive (+) If heat leaves the system: U < 0 therefore q is negative (–) The sign of q is a “convention”, it designates the direction of heat flow between the system and surroundings.

  15. Heat & Changes in Internal Energy Surroundings System Usystem Usystem heat in heat out qsystem < 0 (–) Usystem < 0 qsystem > 0 (+) Usystem > 0 decreases increases q = the heat absorbed or lost by the system.

  16. Heat Capacity When heat is absorbed or lost by a body, the temperature must change as long as the phase (s, g or l) remains constant. The amount of heat (q) transfer related to an object and temperature is given by: q = heat lost or gained (J) C = Heat Capacity of an object T = Tfinal  Tinitial is the temperature change (°C or K)

  17. Heat & Specific Heat Capacity  When heat is absorbed or lost by a body, the temperature must change as long as the phase (s, g or l) remains constant.  The amount of heat (q) transfer per unit mass of a substance is related to the mass and temperature by: m = mass of substance (g) q = heat lost or gained (J) C = the Specific Heat Capacity of a compound T = Tfinal  Tinitial is the temperature change (°C or K)

  18. Does it matter if we calculate a temperature change in Kelvin or degrees C? let Tin = 25.0 °C and Tf = 50.0 °C Recall that T = Tf –Tin 50.0 °C = 50.0 °C + 273.2 = 323.2 K Tf = –Tin =  25.0 °C =  (25.0 °C + 273.2) =  298.2 K = 25.0 °C T T = 25.0 K The are the same!

  19. Determine the final temperature of a 25.0 g block of metal that absorbs 255 cal of energy. The initial temperature of the block was 17.0 C The specific heat capacity of the metal is: Strategy Map: Find Tfinal of the metal after it absorbs the energy Data Information: Mass, initial temp, heat capacity of metal, heat (q) absorbed. Solve q = mCT for Tfinal, plug in data. Final temperature of the metal is determined.

  20. Determine the final temperature of a 25.0 g block of metal that absorbs 255 cal of energy. The initial temperature of the block was 17.0 C The specific heat capacity of the metal is: 32.7 °C rearranging: +255 cal Tf = + 17.0 °C = 25.0 g Tf > Tin as expected

  21. Example Problem: • 55.0 g of iron at 99.8 °C is placed into into 225 g water at initially at 21.0 °C. • At thermal equilibrium, the water and iron are both at 23.1 °C • What is the specific heat capacity of the metal?

  22. Example Problem: • 55.0 g of iron at 99.8 °C is placed into into 225 g water at initially at 21.0 °C. • At thermal equilibrium, the water and iron are both at 23.1 °C • What is the specific heat capacity of the metal? Solution: Heat is transferred from the hot metal to the colder water. Energy is conserved so:

  23. Example Problem: • 55.0 g of iron at 99.8 °C is placed into into 225 g water at initially at 21.0 °C. • At thermal equilibrium, the water and iron are both at 23.1 °C • What is the specific heat capacity of the metal? Solution:

  24. Example Problem: • 55.0 g of iron at 99.8 °C is placed into into 225 g water at initially at 21.0 °C. • At thermal equilibrium, the water and iron are both at 23.1 °C • What is the specific heat capacity of the metal? Solution:

  25. Example Problem: • 55.0 g of iron at 99.8 °C is placed into into 225 g water at initially at 21.0 °C. • At thermal equilibrium, the water and iron are both at 23.1 °C • What is the specific heat capacity of the metal? Solution:

  26. Energy & Changes of State • When matter absorbs heat, its temperature will rise until it undergoes a Phase Change. • The matter will continue to absorb energy, however during the phase change its temperature remains constant: Phase changes are “Isothermal” processes.

  27. Energy Transfer & Changes of State • Some changes of state (phase changes) are endothermic: • When you perspire, water on your skin evaporates. • This requires energy. • Heat from your body is absorbed by the water as it goes from the liquid state to the vapor state, as a result you cool down. + energy

  28. Energy Transfer & Changes of State • Some changes of state (phase changes) are exothermic: • When it is muggy outside, water condenses on your skin. • This releases energy. • Heat from condensation is absorbed by your skin as water in the vapor state coverts to the liquid state. • As a result you feel hot. + energy

  29. Heating/Cooling Curve for Water Note the isotherms at each phase transition.

  30. Heat & Changes of State • The energy associated with a change of state is given by the Enthalpy or Heat of the phase change. • Since there is no temperature change associated with the process, the units are most often in J/g or J/mol. Sublimation: subH > 0 (endothermic) Vaporization: vapH > 0 (endothermic) Melting or Fusion: fusH > 0 (endothermic) Deposition: depH < 0 (exothermic) Condensation: conH < 0 (exothermic) Freezing: freH < 0 (exothermic) Where H refers to the “Heat” of a phase change

  31. Energy Change Calculations Heating & Cooling: q(heating or cooling)= Heat absorbed or lost in a Phase change: q(phase change) = (n = moles or grams)

  32. +333 J/g +2260 J/g Problem: What quantity of heat is required to melt 500. g of ice at 0 °C then heat the resulting water to steam at 100 °C? melt the ice  form liquid water at 0 °C  heat the water to 100 °C  boil water  fusH  Cwater  VapH

  33. Problem: What quantity of heat is required to melt 500. g of ice at 0 °C then heat the resulting water to steam at 100 °C? Equations: qphase change = mH qheat = mCT

  34. Cwater fusH vapH Problem: What quantity of heat is required to melt 500. g of ice at 0 °C then heat the resulting water to steam at 100 °C? Ice H2O(s) (0 °C) water H2O(l) (0 °C) water H2O(l) (100 °C) steam H2O(g) (100 °C)

  35. melt ice boil water heat water Cwater fusH vapH Problem: What quantity of heat is required to melt 500. g of ice at 0 °C then heat the resulting water to steam at 100 °C? Ice H2O(s) (0 °C) water H2O(l) (0 °C) water H2O(l) (100 °C) steam H2O(g) (100 °C) qtotal =

  36. Cwater fusH vapH Problem: What quantity of heat is required to melt 500. g of ice at 0 °C then heat the resulting water to steam at 100 °C? Ice H2O(s) (0 °C) water H2O(l) (0 °C) water H2O(l) (100 °C) steam H2O(g) (100 °C)

  37. First Law of Thermodynamics The total internal energy of an isolated system is constant. The “system” is that which we are interested in. The “surroundings” are everything in contact with the system. Together: System + Surroundings = Universe

  38. First Law of Thermodynamics Energy Energy Any energy lost by the system is transferred to the surroundings and vice versa. Any change in energy is related to the final and initial states of the system. ∆Usystem = Ufinal – Uinitial The same holds for the surrounding!

  39. Relating U to Heat and Work Energy cannot be created or destroyed.  Energy of the universes (system + surroundings) is constant.  Any energy transferred from a system must be transferred to the surroundings (and vice versa). From the first law of thermodynamics: When a system undergoes a physical or chemical change, the change in internal energy is given by the heat added to or absorbed by the system plus the work done on or by the system: change in system energy heat lost or gained by the system work done by or on the system = +

  40. energy transfer in (endothermic), +q energy transfer out (exothermic), -q w transfer in (+w) w transfer out (-w) SYSTEM ∆U = q + w

  41. work out work in Work q out q in Heat work and heat can balance! Energy in Energy out Uf > Ui Uf < Ui U Final U initial energy U Final U initial Usystem < 0 (–) Usystem > 0 (+) Usystem = 0

  42. Energy is the capacity to do work. Work equals a force applied through a distance. When you do work, you expend energy. When you push down on a bike pump, you do work. change in volume V = x3 P × V = work Therefore work is equal to a change of volume at constant pressure.

  43. Enthalpy the subscript “p” indicates constant pressure 0 Enthalpy, “H” is the heat transferred between the system and surroundings under conditions of constant pressure. H = qp if no “PV” work is done by the system, V = 0 the change in enthalpy is the change in internal energy at constant pressure H = Up

  44. Enthalpy is a “StateFunction” B A No matter which path is taken (AB) the results are the same: Final – Initial Since individual Enthalpies cannot be directly measured, we only deal with enthalpy changes (H = Hf – Hi) For a “system” the overall change in Enthalpy is path independent.

  45. Enthalpy Conditions • Since Enthalpies are state functions, one must specify the conditions at which they are measured. • H(T,P): Enthalpy is a function of temperature and pressure. • “H°” indicates that the Enthalpy is taken at Standard Stateconditions. • Standard State Conditions are defined as: • 1 atm = 760 mm Hg or 760 torr & 298.15 K or 25 °C

  46. Enthalpies & Chemical Reactions: rH Reactants Products H = Hfinal  Hinitial Enthalpy of reaction = rH = Hproducts  Hreactants H (Products) H (Reactants) rH < 0 rH > 0 Energy H (Reactants) H (Products) Exothermic Endothermic

  47. Thermochemical Equations Just like a regular chemical equation, with an energy term. CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) rHo = 802 kJ CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) + 802 kJ energy out… Exothermic Energy is a product just like CO2 or H2O! What does this imply? CONVERSION FACTORS!!! From the equation:

  48. Problem: How many kJ of energy are released when 128.5 g of methane, CH4(g) is combusted? rHo = 802 kJ CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) g  mols  J  molar mass  Reaction enthalpy 6.43103 kJ

  49. Thermochemical Equations CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) rHo = 802 kJ When the reaction is reversed , the sign of DH reverses: Exothermic CO2(g) + 2H2O(g)  CH4(g) + 2O2(g) rHo = +802 kJ Endothermic

  50. Thermochemical Equations CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) rHo =  802 kJ DH scales with the reaction: [ ] [ ] rHo =  401 kJ Yes you can write the reaction with fractions, so long as you are writing it on a mole basis…

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