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# Energy ch 16

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### Energy ch 16

1. 1. Energy Energy is an ability possessed by matter to do work or exchange heat. That is, the capacity to move matter or supply heat. Energy exists in different forms which can be converted The S.I. units are the Joule (J)
2. 2. Heat Heat is the transfer of thermal energy from one object to another because they are at different temperatures. We usually use the letter q to represent this quantity Heat is a method of energy transfer from a hot object to a cold one Thermal energy is the energy associated with the random motion of the atoms and or molecules Temperature change is a change in the average kinetic energy of an object
3. 3. Calorie The amount of heat required to raise the temperature of one gram of pure water by one degree Celsius one calorie = 4.184 Joules or one Joule = 0.239 calories 0ne Joule of energy is about the energy needed for a single heartbeat
4. 4. Potential energy (PE) energy an object has as a result of its position or composition. The energy stored in chemical bonds is called chemical potential energy Natural gas, coal and other major sources of heat have potential (chemical) energy due to their composition
5. 5. Kinetic energy (KE) Kinetic energy - energy as object has as a result of its motion. K.E. = 1/2 mv2
6. 6. Law of Conservation of Energy Energy can be converted from one form to another, but the total quantity of energy is assumed to remain constant in ordinary processes
7. 7. Measuring heat transfer Specific heat Defined as the quantity of heat required to raise the temperature of 1.00 gram of the substance by 1.00 oC – the book uses the symbol C for water, C = 4.184 J/goC
8. 8. Specific Heat Formula Changing the temperature of a sample from Ti to Tf requires heat (q) = m C ∆T ∆T = Tf - Ti C = specific heat for a particular substance m = mass of the sample of matter always have Tf first as the sign of ∆ T is important q is + for an endothermic process - Tf > Ti q is - for an exothermic process - Tf < Ti
9. 9. Specific Heat Formula Loss of heat for one object = gain in heat of the second - q lost = q gained - m C ∆T = m C ∆T**Tip Hot side vs. Cold side** Ti – Tf Tf - Ti
10. 10. Sample problem How much heat is required to raise the temperature of 100.0 g of Cu from 15.0oC to 35.0oC? C for Cu is 0.384 J/goC q= m C ∆T q = 100.0 g *0.384 J/goC* (35.0 - 15.0)oC q = 768 J Notice that the value is ( + ) since heat was added to the system
11. 11. Sample problem A 4.50 g nugget of pure gold absorbed 276 J of heat. What was the final temperature of the gold if the initial temperature was 25.0 oC? C for Au is 0.129 J/goC q = m C ∆T The final temperature is 500. oC
12. 12. Thermochemistry The study of the quantity of heat absorbed or given off by chemical reactions. Or the study of heat change in chemical reactions. All chemical reactions either give off or absorb energy
13. 13. First Law of Thermodynamics The total energy of the universe is a constant. Energy can be converted from one form to another or transferred from a system to the surroundings or vice versa.
14. 14. First Law of Thermodynamics Universe = system + surroundings System = the specific part of the universe that contains the reaction or process you wish to study Surroundings = everything in the universe other than the system
15. 15. The system is the specific part of the universe that isof interest in the study. SURROUNDINGS SYSTEM open closed isolated Exchange: mass & energy energy nothing 6.2
16. 16. Exothermic process is any process thatgives off heat or transfers thermal energyfrom the system to the surroundings. 2H2 (g) + O2 (g) 2H2O (l) + energy H2O (g) H2O (l) + energy
17. 17. Endothermic process is any process inwhich heat has to be supplied to thesystem from the surroundings.energy + 2HgO (s) 2Hg (l) + O2 (g) energy + H2O (s) H2O (l)
18. 18. Enthalpy (H) is used to quantify the heat flow into or out of asystem in a process that occurs at constant pressure. H = H (products) – H (reactants) H = heat given off or absorbed during a reaction at constant pressure
19. 19. The heat supplied is equal to the change inanother thermodynamic property called enthalpy(H)i.e. H = q• this relation is only valid at constant pressure As most reactions in chemistry take place at constant pressure we can say that: A change in enthalpy = heat supplied
20. 20. Thereforethe processis Exothermic Hproducts < Hreactants H<0 Therefore H is negative
21. 21. Thereforethe processis Endothermic Hproducts > Hreactants H>0 Therefore H is positive
22. 22. Thermochemical Equations Is H negative or positive? System gives off heat Exothermic H<0 890.4 kJ are released for every 1 mole of methane that is combusted at 25.00C and 1 atm.CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) H = -890.4 kJ
23. 23. Thermochemical Equations Is H negative or positive? System absorbs heat Endothermic H>06.01 kJ are absorbed for every 1 mole of ice that meltsat 00C and 1 atm. H2O (s) H2O (l) H = 6.01 kJ
24. 24. Thermochemical Equations• The stoichiometric coefficients always refer to the number of moles of a substance H2O (s) H2O (l) H = 6.01 kJ• If you reverse a reaction, the sign of H changes H2O (l) H2O (s) H = -6.01 kJ• If you multiply both sides of the equation by a factor n, then H must change by the same factor n. 2H2O (s) 2H2O (l) H = 2 x 6.01 = 12.0 kJ
25. 25. Thermochemical Equations• The physical states of all reactants and products must be specified in thermochemical equations. H2O (s) H2O (l) H = 6.01 kJ H2O (l) H2O (g) H = 44.0 kJ
26. 26. Thermochemical EquationsHow much heat is evolved when 266 g of whitephosphorus (P4) burn in air? P4 (s) + 5O2 (g) P4O10 (s) H = -3013 kJ 1 mol P4 266 g P4 x x 3013 kJ = 6470 kJ 123.9 g P4 1 mol P4
27. 27. Phase Changes Melting phase change from solid to liquid Freezing phase change from liquid to solid freezing point/melting point temperature at which a liquid converts into the solid state & vice versa
28. 28. Melting Energy is supplied to a solid to enable it to vibrate more enthusiastically until molecules can move past each other and flow as a liquid endothermic process ( H positive)
29. 29. Freezing Liquid releases energy and allows molecules to settle into a lower energy state and form a solid exothermic process ( H negative) (we remove heat from water when making ice in freezer)
30. 30. Enthalpy of Fusion Enthalpy of fusion (Hfus) energy necessary to break enough IMF’s to turn a solid into a liquid heat absorbed by the substance in changing solid into a liquid without raising its temperature Energy released when turning a liquid into a solid = - Hfus
31. 31. Enthalpy of Fusion Calculations Enthalpy of fusion is used in calculating energy needed to melt or released when freezingq = m × H fus q = energy m = mass of sample or mol depends upon information given on the chart Hfus = enthalpy of fusion (melting) (use (–Hfus ) for freezing)
32. 32. Sample problem Find the enthalpy of fusion of water if it takes 4175 J to melt 12.5 g of water. m = 12.5 g H2O q = 4175 J q = m × H fus 4175 J = (12.5 g) × H fus Hfus = 4175 J / 12.5 g Hfus = 334 J/g
33. 33. Phase Changes Vaporization phase change from liquid to a gas Condensation phase change from gas to liquid boiling point/condensation point temperature at which a liquid converts into the gas state & vice versa
34. 34. Vaporization Energy has to be supplied to a liquid to enable it to overcome forces that hold molecules together endothermic process ( H positive)
35. 35. Condensation Energy has to be released from a gas to enable it to hold molecules close together exothermic process ( H negative)
36. 36. Enthalpy of Vaporization Enthalpy of vaporization (Hvap) The energy necessary to break the rest of the IMF’s and turn a liquid into a gas The amount of heat required to convert a liquid at its boiling point into vapor without an increase in temperature Energy released when turning a gas into a liquid = - Hvap
37. 37. Enthalpy of Vaporization Calculations Enthlapy of vaporization is used in calculating energy needed to boil or released when condensingq = m × H vap q = energy m = mass of sample or mol depends upon information given on the chart Hvap = enthalpy of vaporization (boiling) (use –Hvap for condensing)
38. 38. Sample Problem How much energy is required to vaporize 10.0 g of water at its boiling point? m = 10.0 g H2O Hvap = 2260 J/g q = m x Hvap q = 10.0 g x 2260 J / g q = 22600 J
39. 39. Heating Curve for H2O Each step along the line requires a certain amount of energy Step 1  q1 = m C ∆T Step 2  q2 = m H fus Step 3  q3 = m C ∆T Step 4  q4 = m H vap Step 5  q5 = m C ∆TTotal energy  qtot= q1+q2+q3+q4+q5
40. 40. Sample Problem See packet
41. 41. Phase Changes Sublimation conversion of a solid directly to a gas Deposition conversion of a gas directly to a solid ΔHsublimation = -ΔHdeposition
42. 42. Standard Enthalpies of Formation The standard enthalpy of formation of a substance, denoted Hfo, is the enthalpy change for the formation of one mole of a substance in its standard state from its component elements in their standard state. – Note that the standard enthalpy of formation for a pure element in its standard state is zero. Standard conditions = 25 0 C and 1.0 atm
43. 43. Examples
44. 44. Standard Enthalpies of Formation The law of summation of heats of formation states that the enthalpy of a reaction is equal to the total formation energy of the products minus that of the reactants. o o o H n H f (products ) m H f (reactants ) is the mathematical symbol meaning “the sum of”, and m and n are the coefficients of the substances in the chemical equation.
45. 45. A Problem to Consider You record the values of Hfo under the formulas in the equation, multiplying them by the coefficients in the equation.4NH 3 (g ) 5O 2 (g ) 4NO(g ) 6H 2O(g )4( 45.9) 5(0) 4(90.3) 6( 241.8) - You can calculate Ho by subtracting the values for the reactants from the values for the products.
46. 46. Calorimetry Process of measuring the heat of reaction of a physical or chemical change
47. 47. Coffee cup calorimetry
48. 48. •A chemical reaction is spontaneous if it is accompanied by an increase in thetotal entropy of the system and the surroundings• Spontaneous exothermic reactions are common (e.g. hot metal blockspontaneously cooling) because they release heat that increases the entropyof the surroundings.•Endothermic reactions are spontaneous only when the entropy of the systemincreases enough to overcome the decrease in entropy of the surroundings
49. 49. A spontaneous change is a change that has a tendency to occur without been driven by an external influence e.g. the cooling of a hot metal block to the temperature of its surroundingsA non-spontaneous change is a change that occursonly when drivene.g. forcing electric current through a metal blockto heat it
50. 50. Total entropy entropy change of entropy change ofchange = system + surroundings Dissolving disorder of solution disorder of surroundings • must be an overall increase in disorder for dissolving to occur
51. 51. Second Law of Thermodynamics: the disorder (or entropy) of a system tends toincrease ENTROPY (S) •Entropy is a measure of disorder • Low entropy (S) = low disorder •High entropy (S) = greater disorder• hot metal block tends to cool• gas spreads out as much as possible
52. 52. •A change in internal energy can be identified with the heat supplied at constantvolume ENTHALPY (H)(comes from Greek for “heat inside”) • the change in internal energy is not equal to the heat supplied when the system is free to change its volume • some of the energy can return to the surroundings as expansion work U < q