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General Chemistry

General Chemistry. M. R. Naimi-Jamal Faculty of Chemistry Iran University of Science & Technology. فصل پنجم :. شیمی گرمایی. Contents. 5-1 Getting Started: Some Terminology 5-2 Heat 5-3 Heats of Reaction and Calorimetry 5-4 Work 5-5 The First Law of Thermodynamics

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General Chemistry

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  1. General Chemistry M. R. Naimi-Jamal Faculty of Chemistry Iran University of Science & Technology

  2. فصلپنجم: شیمی گرمایی

  3. Contents 5-1 Getting Started: Some Terminology 5-2 Heat 5-3 Heats of Reaction and Calorimetry 5-4 Work 5-5 The First Law of Thermodynamics 5-6 Heats of Reaction: U and H 5-7 The Indirect Determination of H: Hess’s Law 5-8 Standard Enthalpies of Formation 5-9 Fuels as Sources of Energy

  4. Thermochemistry: Basic Terms • Thermochemistry is the study of energy changes that occur during chemical reactions. • System: the part of the universe being studied. • Surroundings: the rest of the universe.

  5. Types of Systems • Open: energy and matter can be exchanged with the surroundings. • Closed: energy can be exchanged with the surroundings, matter cannot. • Isolated: neither energy nor matter can be exchanged with the surroundings. A closed system; energy (not matter) can be exchanged.

  6. Terminology • Energy, U • The capacity to do work. • Work • Force acting through a distance. • Kinetic Energy • The energy of motion.

  7. Work kg m m = J Fd w = [w ] = s2 Energy • Kinetic Energy 1 kg m2 = J mv2 ek= [ek ] = 2 s2

  8. Energy • Potential Energy • Energy due to condition, position, or composition. • Associated with forces of attraction or repulsion between objects. • Energy can change from potential to kinetic.

  9. Energy and Temperature • Thermal Energy • Kinetic energy associated with random molecular motion. • In general proportional to temperature. • An intensive property. • Heat and Work • q and w. • Energy changes.

  10. Thermal Equilibrium • Heat transfer ALWAYS occurs from a hotter object to a cooler object (directionality) • Transfer of heat continues until both objects are at the same temperature (thermal equilibrium) • The quantity of heat lost by a hotter object and the quantity of heat gained by a cooler object when they are in contact are numerically equal (required by the law of conservation of energy)

  11. Units of Heat • Calorie (cal) heat required to raise the temperature of 1.00 g of pure liquid water from 14.5 to 15.5oC. • Joule (J): SI unit for heat 1 cal = 4.184 J

  12. q = q = mcT CT Heat Capacity • The quantity of heat required to change the temperature of a system by one degree. • Specific heat capacity, c. • System is one gram of substance • Heat capacity, C. • C = Mass x specific heat (mc).

  13. Many metals have low specific heats. The specific heat of water is higher than that of almost any other substance.

  14. Example: How much heat, in joules and in kilojoules, does it take to raise the temperature of 225 g of water from 25.0 to 100.0 °C? Example: What will be the final temperature if a 5.00-g silver ring at 37.0 °C gives off 25.0 J of heat to its surroundings? The specific heat of silver is 0.235 Jg-1°C-1.

  15. Example: An Estimation Example Without doing detailed calculations, determine which of the following is a likely approximate final temperature when 100 g of iron at 100 °C is added to 100 g of 20 °C water in a calorimeter, the specific heat of iron is 0.449 Jg-1°C-1: (a) 20 °C (b) 30 °C (c) 60 °C (d) 70 °C.

  16. Conservation of Energy • In interactions between a system and its surroundings the total energy remains constant. energy is neither created nor destroyed. qsystem + qsurroundings = 0 qsystem = -qsurroundings

  17. First Law: Sign Convention • Energy entering a system carries a positive sign: • heat absorbed by the system, or • work done on the system • Energy leaving a system carries a negative sign • heat given off by the system • work done by the system

  18. Determination of Specific Heat

  19. Example: Determining Specific Heat from Experimental Data. Use the data presented on the last slide to calculate the specific heat of lead. qlead = -qwater qwater = mcT = (50.0 g)(4.184 J/g °C)(28.8 - 22.0)°C qwater = 1.4x103 J qlead = -1.4x103 J = mcT = (150.0 g)(c)(28.8 - 100.0)°C clead = 0.13 Jg-1°C-1

  20. Heats of Reaction and Calorimetry • Chemical energy. • Contributes to the internal energy of a system. • Heat of reaction, qrxn. • The quantity of heat exchanged between a system and its surroundings when a chemical reaction occurs within the system, at constant temperature.

  21. Heats of Reaction (qrxn) • An exothermic reaction gives off heat • In an isolated system, the temperature increases. • The system goes from higher to lower energy; • qrxnis negative. • An endothermicreaction absorbs heat • In an isolated system, the temperature decreases. • The system goes from lower to higher energy; • qrxnispositive.

  22. Define the heat capacity of the calorimeter: qcal = ∑miciT = CT heat i Bomb Calorimeter qrxn = -qcal qcal = qbomb + qwater + qwires +…

  23. Example: • Using Bomb Calorimetry Data to Determine a Heat of Reaction. • The combustion of 1.010 g sucrose, in a bomb calorimeter, causes the temperature to rise from 24.92 to 28.33°C. The heat capacity of the calorimeter assembly is 4.90 kJ/°C. • What is the heat of combustion of sucrose, expressed in kJ/mol C12H22O11 • Verify the claim of sugar producers that one teaspoon of sugar (about 4.8 g) contains only 19 calories.

  24. Calculate qrxn: qrxn = -qcal = -16.7 kJ Example: Calculate qcalorimeter: qcal = CT = (4.90 kJ/°C)(28.33-24.92)°C = (4.90)(3.41) kJ = 16.7 kJ per 1.010 g

  25. 343.3 g = -16.5 kJ/g qrxn 1.00 mol (a) = -5.65 x 103 kJ/mol 1.00 cal 4.8 g )= -19 kcal/tsp qrxn = (-16.5 kJ/g)( )( (b) 4.184 J 1 tsp Example: Calculate qrxn in the required units: -16.7 kJ qrxn = -qcal = = -16.5 kJ/g 1.010 g Calculate qrxn for one teaspoon: Note: in food industry we say calorie instead of kcal!

  26. Internal Energy • Internal Energy, U. • Total energy (potential and kinetic) in a system. • Translational kinetic energy. • Molecular rotation. • Bond vibration. • Intermolecular attractions. • Chemical bonds. • Electrons.

  27. First Law of Thermodynamics • A system contains only internal energy. • A system does not contain heat or work. • These only occur during a change in the system. • Law of Conservation of Energy • The energy of an isolated system is constant q = heat gained by system w = work gained (+) by or done on system (-) U = q + w

  28. Example: A gas does 135 J of work while expanding, and at the same time it absorbs 156 J of heat. What is the change in internal energy? U = q + w = 156 – 135 = 21 J

  29. State Functions • Thestate of a system: its exact condition at a fixed instant. • State is determined by the kinds and amounts of matter present, the structure of this matter at the molecular level, and the prevailing pressure and temperature. • A state function is a property that has a unique value that depends only on the present state of a system, and does not depend on how the state was reached (does not depend on the history of the system).

  30. State Functions • Water at 293.15 K and 1.00 atm is in a specified state. • d = 0.99820 g/mL • This density is a unique function of the state. • It does not matter how the state was established.

  31. Functions of State • U is a function of state. • Not easily measured. • U has a unique value between two states. • Is easily measured.

  32. Path Dependent Functions For example: Fuel-Consume to go from one point to the another one is NOT a function of state, but a path dependent function

  33. Internal Energy Change at Constant Volume • For a system where the reaction is carried out at constant volume, DV = 0 and DU = qV. • All the thermal energy produced by conversion from chemical energy is released as heat; no P-V work is done.

  34. Work • In addition to heat effects chemical reactions may also do work. • Gas formed pushes against the atmosphere. • Volume changes. • Pressure-volume work.

  35. Heats of Reaction: U and H Reactants → Products Ui Uf U = Uf - Ui U = qrxn + w In a system at constant volume: PDV=0 U = qrxn + 0 = qrxn = qv But we live in a constant pressure world! How does qp relate to qv?

  36. lnternal Energy Change at Constant Pressure • For a system where the reaction is carried out at constant pressure, DU = qP – PDVor DU + PDV = qP • Most of the thermal energy is released as heat. • Some work is done to expand the system against the surroundings (push back the atmosphere).

  37. Heats of Reaction

  38. Heats of Reaction U = qP + w We know that w = - PV (minus because done by system) therefore: U = qP - PV qP = U + PV These are all state functions, so define a new function. Let H = U + PV Then H = Hf – Hi = U + PV= U + PV + VP If we work at constant pressure and temperature: VP = 0 H = U + PV = qP

  39. Standard States and Standard Enthalpy Changes • Define a particular state as a standard state. • Standard enthalpy of reaction, H° • The enthalpy change of a reaction in which all reactants and products are in their standard states. • Standard State • The pure element or compound at a pressure of 1 atm and at the temperature of interest (usually 25 °C).

  40. Enthalpy Diagrams

  41. Indirect Determination of H:Hess’s Law • H is an extensive property. • Enthalpy change is directly proportional to the amount of substance in a system. N2(g) + O2(g) → 2 NO(g) H = +180.50 kJ ½N2(g) + ½O2(g) → NO(g) H = +90.25 kJ • H changes sign when a process is reversed NO(g)→ ½N2(g) + ½O2(g)H = -90.25 kJ

  42. ½N2(g) + ½O2(g) → NO(g)H = +90.25 kJ NO(g) + ½O2(g) → NO2(g)H = -57.07 kJ Hess’s Law • Hess’s law of constant heat summation • If a process occurs in stages or steps (even hypothetically), the enthalpy change for the overall process is the sum of the enthalpy changes for the individual steps. ½N2(g) + O2(g) → NO2(g)H = +33.18 kJ

  43. Example: Calculate the enthalpy change for reaction (a) given the data in equations (b), (c), and (d). (a) 2 C(graphite) + 2 H2(g)  C2H4(g) DH = ? (b) C(graphite) + O2(g)  CO2(g) DHb= –393.5 kJ (c) C2H4(g) + 3 O2  2 CO2(g) + 2 H2O(l) DHc= –1410.9 kJ (d) H2(g) + ½ O2  H2O(l) DHd = –285.8 kJ • = 2 (b) + 2 (d) – (c) & DHa = 2DHb + 2DHd - DHc

  44. Standard Enthalpies of Formation Hf° • The enthalpy change that occurs in the formation of one mole of a substance in the standard state from the reference forms of the elements in their standard states. • The standard enthalpy of formation of a pure element in its reference state is 0.

  45. Standard Enthalpy of Formation When we say: “The standard enthalpy of formation of CH3OH(l) is 238.7 kJ”, we are saying that the reaction: C(graphite) + 2 H2(g) + ½ O2(g)  CH3OH(l) has a value of ΔH of –238.7 kJ. We can treat ΔHf° values as though they were absolute enthalpies, to determine enthalpy changes for reactions. Question: What is ΔHf° for an element in its standard state [such as O2(g)]? Hint: since the reactants are the same as the products …

  46. Standard Enthalpies of Formation

  47. Enthalpy of Reaction Hrxn = ∑Hf°products - ∑Hf°reactants

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