TOPIC 5 ENERGETICS/THERMOCHEMISTRY - PowerPoint PPT Presentation

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TOPIC 5 ENERGETICS/THERMOCHEMISTRY

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  1. TOPIC 5ENERGETICS/THERMOCHEMISTRY 5.3 BOND ENTHALPIES

  2. ESSENTIAL IDEA Energy is absorbed when bonds are broken and is released when bonds are formed. NATURE OF SCIENCE (2.2) Models and theories – measured energy changes can be explained based on the model of bonds broken and bonds formed. Since these explanations are based on a model, agreement with empirical data depends on the sophistication of the model and data obtained can be used to modify theories where appropriate.

  3. UNDERSTANDING/KEY IDEA 5.3.A Bond-forming releases energy and bond-breaking requires energy.

  4. UNDERSTANDING/KEY IDEA 5.3.B Average bond enthalpy is the energy needed to break one mole of a bond in a gaseous molecule averaged over similar compounds.

  5. GUIDANCE Average bond enthalpies are only valid for gases so calculations involving bond enthalpies may be inaccurate because they do not take into account intermolecular forces.

  6. A common error is to fail to indicate that all species have to be in their gaseous state.

  7. APPLICATION/SKILLS Be able to sketch and evaluate potential energy profiles in determining whether reactants or products are more stable and if the reaction is exothermic or endothermic.

  8. Chemical reactions involve the breaking of bonds in reactants and the making of bonds in products. • To understand energy changes in chemical reactions, we need to look at the energy needed to break bonds in reactants and the energy released when bonds are formed in products.

  9. Endothermic processes involve the separation of particles which are held together by a force of attraction. • Exothermic processes involve the bringing together of particles which have an attractive force between them.

  10. Exothermic reactionEnthalpy diagram reactants enthalpy H ΔH = negative products extent of reaction

  11. Endothermic reactionEnthalpy diagram products enthalpy H ΔH = positive reactants extent of reaction

  12. Sample problem • Which of the following processes are endothermic? • 2Cl Cl2 • Na Na+ + e- • Na+ + Cl- NaCl • Na(g) Na(s) • The answer is the 2nd equation which is the only one separating into particles.

  13. A covalent bond is due to the electrostatic attraction between the shared pair of electrons and the positive nucleus. • Energy is needed to separate the atoms in a bond.

  14. GUIDANCE Average bond enthalpies are only valid for gases and calculations involving bond enthalpies may be inaccurate because they do not take into account intermolecular forces.

  15. Average Bond Enthalpies • Bond enthalpies are calculated by separating the atoms. • Cl2(g) 2Cl(g) ∆H⁰ = +242 kJ/mol • It is more complicated in molecules that contain more than 2 atoms. • H2O(g) H(g) + OH(g) ∆H⁰ = +502 kJ/mol • OH(g) H(g) + O(g) ∆H⁰ = +427 kJ/mol • In order to compare bond enthalpies, average bond enthalpies are tabulated.

  16. All tabulated bond enthalpies refer to reactions in their gaseous state so that the enthalpy changes caused by the formation and breaking of intermolecular forces can be ignored. • Multiple bonds have higher bond enthalpies and are shorter than single bonds.

  17. LIMITATIONS OF USING AVERAGE BOND ENTHALPIES • As said before, everything has to be in the gaseous state. • If water were a liquid as a product, then more heat would be evolved because it would also have to include the heat of vaporization of water. • Average bond enthalpies are used which are obtained considering a number of similar compounds. • In reality, the energy of a particular bond varies slightly in different compounds and it is affected by neighboring atoms. • So ∆H values obtained from using average bond enthalpies will not necessarily be very accurate.

  18. Explain, in terms of bond enthalpies, why some reactions are exothermic and some are endothermic.

  19. The reaction is exothermic overall if the bonds which are formed are stronger than the bonds which are broken. • The reaction is endothermic when the bonds broken are stronger than the bonds that are formed.

  20. APPLICATION/SKILLS Be able to calculate the enthalpy changes from known bond enthalpy values and compare them to experimentally measured values.

  21. GUIDANCE Bond enthalpy values are given in the data booklet in section 11.

  22. Sample problem • Use bond enthalpies to calculate the heat of combustion of methane. First write and balance the equation. CH4 + 2O2 CO2 + 2H2O Next figure out the structural bonds. 4 C-H + 2O=O 2C=O + 4O-H

  23. Very important: When using bond enthalpies, you use reactants minus products, not products minus reactants as in the standard heat of formation formula. • ∆H = ∑(bonds broken) – ∑(bonds formed) reactants products

  24. Now look up the values in a table or they will be given to you. Be very careful not to confuse double or triple bonds with single bonds. • ∆H = [4C-H + 2O=O] – [2C=O + 4O-H] = [4(412) + 2(496)]-[2(743)+4(463)] = (2640)-(3338) = -698 kJ/mol

  25. The literature value is -890 kJ/mol which is the value for standard conditions. • The standard state for water is liquid and the bond enthalpy calculation assumed the gaseous state so this is not the most accurate method.

  26. APPLICATION/SKILLS Be able to discuss the bond strength of ozone relative to oxygen in its importance to the atmosphere.

  27. OZONE REVIEW • Ozone, O3, has a bent shape with a bond angle of 117◦ • It has 2 resonance structures. • The double bond consists of one pi and one sigma bond. • The electrons in the pi bond are held less tightly so they become delocalized giving rise to the resonance structure. • The bond order is 1.5 which means the length is intermediate and the strength is between a double and single bond. • www.chemwiki.ucdavis.edu

  28. The ozone molecule is polar which is explained by formal charge and the uneven distribution of electrons. • The lower part of the stratosphere, known as the ozone layer, contains 90% of the atmospheric ozone. • Ozone levels are maintained through a cycle of reactions involving the formation and breakdown of oxygen and ozone. • Oxygen and ozone form a protective screen which ensures that the radiation that reaches the earth is different from that emitted by the sun.

  29. Ref: schooltutoring.com

  30. OZONE CYCLE There are 2 key steps in the ozone cycle. The O. represents a free radical that has an unpaired electron so it is highly reactive. Oxygen dissociation O2(g) O.(g) + O.(g) light wavelength <242 nm Ozone dissociation • O3(g)fast O.(g) +2O2(g) light wavelength <330 nm 2. O3(g) + O.(g) slow 2O2(g) exothermic reaction H=neg

  31. The O2 bonds are stronger and harder to break than the O3 bonds of ozone. • The stronger O2 bonds require UV light energy of shorter wavelengths because they need the higher energy radiation to break the bonds. • The bond energy of O2 with a bond order of 2 is 498 kJ/mol. • The bond energy of O3 with a bond order of 1.5 is 364 kJ/mol.

  32. The fact that ozone absorbs radiation of wavelengths of 200 nm to 315 nm is very important. • This corresponds to the higher range of ultraviolet light, known as UV-B and UV-C which causes damage to living tissue. • The ozone layer protects us from this radiation. • The absorption of UV radiation by ozone is also a major source of heat in the stratosphere and is the reason why the temperature in the stratosphere rises with height.

  33. CHAPMAN CYCLE STEP 1: Oxygen is broken into two free radicals with uv light with wavelengths less than 242 nm. (Endo) STEP 2: The free radical reacts with oxygen to form ozone. (Exothermic and the heat given out heats up the stratosphere.) STEP 3: Ozone is broken into oxygen and a free radical with uv light of lower energy less than 330nm. (Endo) STEP 4: The oxygen free radical then reacts with another ozone molecule to form two oxygen molecules. (Exothermic so provides more heat to the stratosphere.) The level of ozone stays at a constant level as the rate of formation of ozone is balanced by the rate of its removal. This cycle of reactions is significant because dangerous uv light has been absorbed and the stratosphere has become warmer. Both processes are essential for the survival of life on earth.

  34. CHAPMAN CYCLE STEP 1: O2(g) O.(g) + O.(g)λ<242 nm STEP 2: O.(g) +2O2(g) O3(g) exorxn STEP 3: O3(g)fast O.(g) +2O2(g) λ <330 nm STEP 4: O3(g) + O.(g) slow 2O2(g) exorxn

  35. APPLICATION/SKILLS Be able to describe the mechanism of the catalysis of ozone depletion when catalyzed by CFCs and NOx.

  36. Ozone’s ability to absorb UV radiation also means that it is unstable. • It reacts easily with compounds found in the stratosphere that have been created by human activity. • There are two types of compounds that produce highly reactive free radicals that catalyze the decomposition of ozone to oxygen. • Nitrogen oxides, NOx • Chlorofluorocarbons, CFC’s

  37. Nitrogen oxides • Nitrogen monoxide is produced in vehicle engines. • It is a free radical as it has an odd number of electrons. • Nitrogen dioxide forms from the oxidation of NO and is also a free radical. • The reactions of nitrogen oxides with ozone are: NO.(g) + O3(g) NO2.(g) + O2(g) NO2.(g) + O.(g) NO.(g) + O2(g) NO. has acted as a catalyst because it regenerated during the reaction and the net change is the breakdown of O3. O3(g) + O.(g) 2O2(g)

  38. Chlorofluorocarbons • Chlorofluorocarbons are widely used in aerosols, refrigerants, solvents and plastics due to their low reactivity and low toxicity in the troposphere. • However, when they reach the stratosphere, the higher energy UV radiation breaks them down releasing free chlorine atoms which are reactive free radicals. • The reaction of the CFC freon is: CCl2F2(g) CClF2.(g) + Cl.(g) Cl.(g) + O3(g) O2(g) + ClO.(g) ClO.(g) + O.(g) O2(g) + Cl.(g)

  39. Here the chlorine radical acts as the catalyst and the net reaction is again: O3(g) + O.(g) 2O2(g) • These reactions upset the balance of the ozone cycle and lead to the thinning of the ozone layer. • The UV radiation reaching the Earth is most pronounced in the polar regions. • This has been a global concern since the 1970’s.

  40. Citations Brown, Catrin, and Mike Ford. Higher Level Chemistry. 2nd ed. N.p.: Pearson Baccalaureate, 2014. Print. Most of the information found in this power point comes directly from this textbook. The power point has been made to directly complement the Higher Level Chemistry textbook by Catrin and Brown and is used for direct instructional purposes only.