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THERMOCHEMISTRY

THERMOCHEMISTRY. CHAPTER 5. WHY ENERGY?. Life requires energy From plants – photosynthesis To equipment – electricity Vehicles – fossil fuels To animals - food. ENERGY FROM WHERE?. Most of our energy as humans From food i.e. from chemical reactions In the world From chemical reactions.

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THERMOCHEMISTRY

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  1. THERMOCHEMISTRY CHAPTER 5

  2. WHY ENERGY? • Life requires energy • From plants – photosynthesis • To equipment – electricity • Vehicles – fossil fuels • To animals - food

  3. ENERGY FROM WHERE? • Most of our energy as humans • From food • i.e. from chemical reactions • In the world • From chemical reactions

  4. WHAT IS THERMOCHEMISTRY? Quantitative study of chemical reactionsinvolving heat changes Consumeenergy Produce energy

  5. Examples: Food we eat - provides us with energy Burning coal- to produce electrical energy Energy from the sun - responsible for chemical reactions in plants which make them grow

  6. RECALL • FORCE(F) – push/pull exerted on an object • WORK(w) – energy used to cause an object to move against a force (w = Fd) • HEAT(q) – energy transfer as a result of a temperature difference. Heat always flows from a hotter to a colder object until they have the same temperature • ENERGY(E) – capacity to do work or to transfer heat

  7. HOT STUFF? • Can something transfer cold?

  8. Does ice make a drink cold?

  9. When you blow hot soup do you make it cold?

  10. SO IN FACT: • In theory stirring soup long enough and fast enough could make it boil

  11. Heat always flows from a hotter to a colder object until they have the same temperature

  12. BUT I KNOW IT’S COLD!! Heat from our body is transferred, making something feel cold because we lose heat from the area touching the cold object

  13. NATURE OF ENERGY 2 types of energy Kinetic energy Potential energy Energy due toposition of object relative to others(Stored energy) Energy due to motion of object

  14. KINETIC ENERGY Note: Ek increases as velocity increases and Ek increases as mass increases. Atoms and molecules have mass and are in motion, thus they have kinetic energy!

  15. energy an object possesses because of its temperature. Thermal energy Associated with kinetic energy of atoms and molecules. The higher the temperature,  the faster theatoms and moleculesmove the more kinetic energy molecules and atoms have  the greater the thermal energy of the object. The total thermal energy of the object is the sum of the individual energies of the atoms or molecules.

  16. MISSING: KINETIC ENERGY, PLEASE HELP • What is it to have no kinetic energy? • Absolute zero! • Can we detect that? • No! • Why?! • Because in order to detect it we would have to interact with it which would heat it

  17. POTENTIAL ENERGY We know Ep = mgh Forces other than gravity that lead to potential energy = ELECTROSTATIC forces Attractions and repulsions due to oppositely charged objects (e.g. positive and negative ions)

  18. Recall: Eel = electrostatic energy k = Coulomb constant Q = charge d = distance between charges Describes the electrostatic energy between 2 charges. UNITS OF ENERGY SI unit = J (Joule) 1 J = 1 kg.m2.s-2 (Others 1 cal (calorie) = 4.184 J) Ek=0.5mv2 =(kg)(m.s-1)2 (The amount of heat energy to raise temp of 1g H2O by 1oC)

  19. A BRAINTEASER • If everything has kinetic energy • And chemical potential energy • Why is everything not moving around constantly? • Why are solids not felt to vibrate?

  20. SYSTEM AND SURROUNDINGS Everything else Portion singled out for study System Surroundings System+surroundings=universe

  21. Open system – can exchange both energy and matter with the surroundings Closed system – exchanges energy but NOT matter with the surroundings Isolated system – neither energy nor matter can be exchanged with the surroundings

  22. A QUOTE NOT TO LIVE BY “Thermodynamics is a funny subject. The first time you go through it, you don't understand it at all. The second time you go through it, you think you understand it, except for one or two small points. The third time you go through it, you know you don't understand it, but by that time you are so used to it, it doesn't bother you anymore.” http://www.eoht.info/page/Thermodynamics+humor

  23. FIRST LAW OF THERMODYNAMICS Energy is neither created nor destroyed, it is merely converted from one form into another. Energy is transferred between the system and the surroundings in the form of work and heat. The total energy of the universe remains constant.

  24. INTERNAL ENERGY (U) Also given as E Internal energy (of the system) = sum of all Ek and Ep of all components of the system Symbol for change =  Change in internal energy of the system U U = UFinal - Uinitial Number & Unit  Magnitude Sign  Direction(in which energy is transferred)

  25. U = UFinal - Uinitial Usyst > 0 Uf > Ui GAINED energy from surroundings Usyst < 0 Uf < Ui LOST energy to surroundings

  26. RELATING U TO WORK AND HEAT General ways to change ENERGY of a closed system: WORK Done by/on system HEAT lost/gained by system When a system undergoes a chemical/physical change Relationship: U = q + w We can’t measure U, but we can determine U.

  27. SIGN CONVENTION q<0 Heat transferredFROMsystem q>0 Heat transferredTOsystem SYSTEM w<0 Work doneBYsystem w>0 Work doneONsystem The sign of U will depend on the sign and magnitude of q and w since U = q + w.

  28. Example: System Calculate the change in internal energy for a process in which the system absorbs 120 J of heat from the surroundings and does 64 J of work on the surroundings. 120 J heat 64 J work U = q + w = (120 J) + (- 64 J) = 56 J

  29. SOME DEFINITIONS Extensive property – dependent on the amount of matter in the system. E.g. mass, volume etc Intensive Property – NOT dependent on the amount of matter in the system. E.g. density, temperature etc

  30. A B C State function – a function that depends on the state or conditions of the system and NOT on the details of how it came to be in that state.

  31. DEFINITIONS APPLIED TO INTERNAL ENERGY Recall: Total internal energy of a system= sum of all Ek and Ep of all components of the system Thus total internal energy of a system  total quantity of matter in system U is an extensive property

  32. e.g. if a gas sample undergoes: ApressureXheatB, or AheatYpressureB U is the same in both cases. Internal energy depends on the state or conditions of the system (e.g. pressure, temperature, location) Does not depend on how it came to be in that state. state function U only depends on Ui and Uf and not how the change occurred. NOTE: Heat and work are not state functions!

  33. ENTHALPY (H) We cannot measure enthalpy (H), we can only measure change in enthalpy (H). Change in enthalpy (H) is the heat gained or lost by the system when a process occurs under constant pressure. E.g. atm pressure H = Hfinal - Hinitial = qp H is a state function and an extensive property

  34. NOTE: At constant pressure, most of the energy lost / gained is in the form of heat. Very little workis done for the expansion / contraction against the atmospheric force, especially if the reaction does not involve gases.

  35. qp>0 qp<0 SYSTEM H>0 H<0 Exothermic process – systemevolves heat Endothermicprocess - system absorbs heat

  36. Example of an exothermic reaction:

  37. Example of an endothermic reaction:

  38. FINDING THE RELATIONSHIP BETWEEN U AND H –CONSIDER PV WORK We know that: U = q + w Assume we do PV work only (e.g. expanding gases in cylinder of a car does PV work on the piston)

  39. Gas expands and moves piston distance d W = Fd But F = PA W = PAd W = PV Sign: system is doing work on the piston W = -PV

  40. U = q + w and w = -PV U = q - PV At constant volume, V = 0 U = qv At constant pressure: U = qp - PV But H = qp U = H - PV The volume change in many reactions is very small, thus PV is very small and hence the difference between U and H is small.

  41. ENTHALPIES OF REACTION enthalpy change that accompanies a reaction (heat of reaction) Hrxn = H(products) – H(reactants) final initial Thermochemical equation: 2H2(g) + O2(g)  2H2O(g) H = -483.6 kJ exothermic Balanced equation H associated with the reaction Coefficients in balanced equation = no. of moles of reactant/product producing associated H (extensive)

  42. 2H2(g) + O2(g)  2H2O(g) H = -483.6 kJ Note: Since enthalpy is an extensive property, H depends on the amount of reactant consumed. 4H2(g) + 2O2(g) 4H2O(g) H = -967.2 kJ Also, H for a reaction is equal in magnitude, but opposite in sign to H for the reverse reaction. 2H2O(g)  2H2(g) + O2(g) H = +483.6 kJ

  43. Notes are available on http://hobbes.gh.wits.ac.za/chemnotes/fmain-chem1033-35.htm Go to the section for Mrs C Billing Lectures Chapter 7, Chapter 8 and Chapter 9 are this work We are busy with Chapter 7 at the moment

  44. m 25 g n = = 1.4 mol = M 18.016 g/mol Example: Calculate the heat needed to convert 25 ml water to steam at atmospheric pressure if the enthalpy change is 44 kJ/mol. (Assume the density of water is 1.0 kg/l) =1.0 g/ml H2O(l)  H2O(g) H = 44 kJ/mol 25 ml m = v = (1.0 g/ml)(25 ml) = 25 g qp = H = 44 kJ/mol x 1.4 mol = 62 kJ Check - endothermic

  45. Example: m 2.50 g nH2O2 = = = 0.0735 mol 34.02 g/mol M Hydrogen peroxide can decompose to water and oxygen by the reaction: 2H2O2 (l)  2H2O (l) + O2 (g) H = -196 kJ Calculate the heat produced when 2.50 g H2O2 decomposes at constant pressure. qp = H = -196 kJ for 2 mol H2O2 = -7.2 kJ for 0.0735 mol

  46. Example – burning money: • Ethanol burns in air to give water vapour and carbon dioxide • Calculate the enthalpy of reaction given enthalpy of formations of: ethanol (l) = -277.7 kJ/mol • water (l) = -285.8 kJ/mol carbon dioxide (g)= -393.5 kJ/mol • 2. Calculate the minimum amount of water needed to prevent the paper from burning after be soaked in 1g of ethanol First: find the balanced chemical equation C2H5OH (l) + 3O2 3H2O (l) + 2CO2 (g) Enthalpy of reaction: H = H°f(products) - H°f(reagents) = 2(-393.5) + 3(-285.8) – (-277.7 + 3x0) kJ/mol = -1365.8 kJ/mol

  47. Example – burning money: m 1.00 g n = = 0.0217 mol = M 46.07 g/mol H2O(l)  H2O(g) H = 44 kJ/mol Recall: -29.6 kJ n = = 0.67 mol = H 44 kJ/mol Ethanol burns in air to give water vapour and carbon dioxide 2. Calculate the minimum amount of water needed to prevent the paper from burning after be soaked in 1.00g of ethanol Find the heat released by burning 1.00 g of ethanol qp = H = -1365.8 kJ/mol x 0.0217 mol = -29.6 kJ qp m = nM = (0.67 mol)(18.0148 g/mol) = 12 g

  48. CALORIMETRY The experimental determination of heat flow associated with a chemical reaction by measuring the temperature changes it produces.

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