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Chemical Thermodynamics

Chemical Thermodynamics. Why Changes Take Place Temperature, Thermal Energy and Heat Law of Conservation of Energy Energy Units Heat Capacity and Specific Heat Measurement of Thermal Energy Changes Enthalpy Hess’s Law. Spontaneous. Nonspontaneous. Why changes take place.

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Chemical Thermodynamics

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  1. Chemical Thermodynamics • Why Changes Take Place • Temperature, Thermal Energy and Heat • Law of Conservation of Energy • Energy Units • Heat Capacity and Specific Heat • Measurement of Thermal Energy Changes • Enthalpy • Hess’s Law

  2. Spontaneous Nonspontaneous Why changes take place Spontaneous process Takes place ‘naturally’ with no apparent cause or stimulus. Nonspontaneous process Requires that something be done in order for it to occur.

  3. When will a reactionbe spontaneous? • Spontaneity of a reaction can be determined by a study of thermodynamics. • Thermodynamics can be used to calculate the amount of useful work that is produced by some chemical reactions. • The two factors that determine spontaneity are enthalpy and entropy.

  4. Energy • Energy - the ability to do work. • Work - when a force is applied to an object. • There are several types of energy: • Thermal - heat • Electrical • Radiant - including light • Chemical • Mechanical - like sound • Nuclear

  5. Energy • Energy can be classified as: • Potential energy • Stored energy - ability to do work. • Kinetic energy • Energy of motion - actually doing work. • Energy can be transferred from one object to another. It can also change form.

  6. Kinetic vs. potential energy Potential Energy

  7. Converting potentialto kinetic energy

  8. Kinetic vs. potential energy Kinetic Energy

  9. Energy and chemical bonds • During a chemical reaction • Old bonds break. • New bonds are formed. • Energy is either absorbed or released. • Exothermic Energy is released. • New bonds are more stable. • EndothermicEnergy is required. • New bonds are less stable.

  10. Exothermic Reactants Energy Products Since excess energy is released, the products are more stable.

  11. Endothermic Products Energy Reactants Additional energy is required because the products are less stable.

  12. Entropy • Entropy - a measure of the ‘disorder’ or randomness of a system. • Disorder is favored over order and may account for reaction occurring spontaneously even if it is endothermic. Increased entropy solid gas

  13. Rate of change • Not all spontaneous changes take place in a useful time period. • Some may require some initial energy to get them started. • 2H2 (g) + O2 (g) 2H2O (l) • Others can be made faster by adding a catalyst. • 2H2O2 (l) 2H2O (l) + O2 (g) • Kinetics - the study of the rate of a reaction. spark I-

  14. Temperature, energy and heat • Temperature. An intensive property of a material. • Thermal energy. Energy of motion of molecules, atoms or ions. All materials have this energy if at a temperature above 0 K. • Heat. Thermal energy transfer that results from a difference in temperature. Thermal energy flows from warm objects to cool ones.

  15. Law of conservation of energy • “Energy cannot be created or destroyed in a chemical reaction.” • During a reaction, energy can change from one form to another. • Example. Combustion of natural gas. • Chemical bonds can be viewed as potential energy. So during the reaction: • 2CH4 (g) + 3O2 (g) 2CO2 (g) + 2H2O (l) + thermal energy + light • some potential energy is converted to thermal energy and light.

  16. Energy units • Earlier, kinetic energy was defined as: • kinetic energy = mv2 • m = mass and v = velocity. • Joule (J) - the energy required to move a 2 kg mass at a speed of 1 m/s. It is a derived SI unit. • J = kinetic energy = (2 kg) (1 m/s)2 • = 1 kg m2 s-2 1 2 1 2

  17. Energy units • Calorie (cal) • Originally defined as the quantity of heat required to heat of one gram from 15 to 16 oC. • It is now defined as: 1 cal = 4.184 J • Dietary Calorie • This is what you see listed on food products. • It is actually a kilocalorie.

  18. Heat capacity • Every material will contain thermal energy. • Identical masses of substances may contain different amounts of thermal energy even if at the same temperature. • Heat capacity. The quantity of thermal energy required to raise the temperature of an object by one degree. • Specific heat. The amount of thermal energy required to raise the temperature of one gram of a substance by one degree.

  19. Substance SH Al(s) 0.90 Br2 (l) 0.47 C (diamond) 0.51 C (graphite) 0.71 CH2CH2OH (l) 2.42 CH3(CH2)6CH3 (l) 2.23 Substance SH Fe (s) 0.45 H2O (s) 2.09 H2O (l) 4.18 H2O (g) 1.86 N2 (g) 1.04 O2 (g) 0.92 Specific Heats at 25oC, 1 atm SH = specific heat, J g-1oC-1

  20. Heat capacity • Example. • How many joules must be added to a 50.0 g block of aluminum to heat it from 22oC to 85oC? • Heat required = mass x specific heat x DT • = 50.0 g x 0.90 J g-1oC-1 x (85-22)oC • = +2.8 kJ • This is an endothermic change - + sign.

  21. Measuring thermal energy changes • Thermal energy cannot be directly measured. • We can only measure differences in energy. • To be able to observe energy changes, we must be able to isolate our system from the rest of the universe. • Calorimeter - a device that is used to measure thermal energy changes and provide isolation of our system.

  22. Conduct a reaction and look at the temperature change. Coffee cup calorimeter

  23. Calorimetry example • You are given the two solutions listed below. Each has an initial temperature of 20.0 oC. • 50 ml of 0.50 M NaOH • 50 ml of 0.50 M HCl • Both are rapidly added to a coffee cup calorimeter and stirred. The reaction takes place rapidly. The highest temperature is 23.3 oC. Solution density is 1.0 g/ml. • Determine the heat of reaction if the specific heat of the solution is 4.18 J g-1oC-1

  24. Calorimetry example • First, determine the energy given off. • = 100.0 g (4.18 J g-1oC-1) (23.3 - 20.0) oC • = - 1.4 x103 J (use ‘-’ because heat is given off) • Next, determine the moles of HCl or NaOH involved in the reaction -- both are the same. • molHCl = (0.5 ) ( 0.05 L) • = 0.025 mol HCl mol L

  25. Calorimetry example • The heat of neutralization for the reaction: • HCl (aq) + NaOH (aq) NaCl(aq) + H2O (l) • is • = -1.4 x103 J / 0.025 mol • = 5.6 x 104 J/mol • = 56 kJ/mol

  26. Enthalpy • The energy gained or lost when a change takes place under constant pressure. • DH = Hfinal - Hinitial • Subscripts are used to show the type of change. • DHvap heat of vaporization • DHneut heat of neutralization • DHfusion heat of fusion • DHsol heat of solution • DHrxn heat of reaction

  27. Stoichiometry • Many reactions are conducted simply for the thermal energy that is released. • Combustion of gasoline, coal, natural gas. • The thermal energy released can be shown as a product in a reaction. • CH4 (g) + 2O2 (g) CO2 (g) + H2O (l) + 890.32 kJ • or • CH4 (g) + 2O2 (g) CO2 (g) + H2O (l)DHrxn = -890.32 kJ • When given for a reaction, DH is interpreted in terms of moles.

  28. Stoichiometry • Determine the thermal energy released when 50.0 grams of methane is burned in an excess of oxygen. • First, determine the number of moles of methane (MM = 16.043 u). • mol CH4 = (50.0 g) / (16.043 g/mol) • = 3.12 mol CH4

  29. Stoichiometry • Now look at the balanced thermochemical equation. • CH4 (g) + 2O2 (g) CO2 (g) + H2O (l)DHrxn = -890.32 kJ • DHrxn = -890.32 kJ / mol CH4 so: • Thermal energy released • = (3.12 mol CH4(g)) (-890.32 kJ / mol CH4 ) • = - 2.78 x 103 kJ

  30. Hess’s law • The thermal energy given off or absorbed in a given change is the same whether it takes place in a single step or several steps. • This is just another way of stating the law of conservation of energy. • If the net change in energy were to differ based on the steps taken, then it would be possible to create energy -- this cannot happen!

  31. A + B A + B F energy E + D + B C C Hess’s law

  32. State functions • Depend only on the initial and final states of a system. They are independent of how the system gets from one state to another. • State functions include: • Pressure • Volume • Temperature • Enthalpy

  33. Calculating enthalpies • Thermochemical equations can be combined to calculate DHrxn. • Example. • 2C(graphite) + O2 (g) 2CO (g) • This cannot be directly determined because CO2 is always formed. • However, we can measure the following: • C(graphite) + O2 (g) CO2(g) DHrxn= -393.51 kJ • 2CO (g) + O2 (g) 2CO2 (g)DHrxn= -565.98 kJ

  34. Calculating enthalpies • By combining the two equations, we can determine the DHrxn we want. • 2 [ C(graphite) + O2 (g) CO2(g) ] DHrxn= -787.02 kJ • 2CO2 (g) 2CO (g) + O2 (g)DHrxn= +565.98 kJ • Note. • Because we need 2 moles of CO2 to be produced in the top reaction, the equation and its DHrxn were doubled.

  35. Calculating enthalpies • Now all we need to do is to add the two equations together. • 2 C(graphite) + 2O2 (g) 2CO2(g)DHrxn= -787.02 kJ • 2CO2 (g) 2 CO (g) + O2 (g)DHrxn= +565.98 kJ • 2 C(graphite) + O2 (g) 2 CO (g)DHrxn= -221.04 kJ • Note. • The 2CO2 cancel out, as does one of the O2 on the right-hand side.

  36. Calculating enthalpies • The real problem with using Hess’s law is figuring out what equations to combine. • The most often used equations are those for formation reactions. • Formation reactions • Reactions in which compounds are formed from elements. • 2 H2 (g) + O2 (g) 2 H2O (l)DHrxn = -571.66 kJ

  37. Standard enthalpy of formation • DHfo • Enthalpy change that results from one mole of a substance being formed from its elements. • All elements are at their standard states. • The DHfo of an element in its standard state has a value of zero.

  38. Standard enthalpies of formation • Substance DHfo, kJ/mol • CaCO3 (s) -1206.92 • CaO (s) -635.09 • CH4 (g) -74.85 • C2H6 (g) -84.67 • CH3OH (l) -238.64 • CH3OH (g) -201.2 • CO (g) -110.52 • CO2 (g) -393.51 • HCl (g) -92.31 • H2O (l) -285.83 • H2O (g) -238.92 • NaCl (s) -411.12 • SO2 (g) -296.83 Standard enthalpy of formation values are available for a wide range of substances. In addition, separate values for a substance in different states will also be given where appropriate.

  39. Phase change • We can use DHof values to determine the energy required to change from one phase to another. • Example. Conversion of methanol from a liquid to a solid. • kJ • C (s) + 2 H2 + O2 (g) CH3OH (g)DHorxn = -201.2 • C(s) + 2 H2 + O2 (g) CH3OH(l)DHorxn = -238.6 1 2 1 2

  40. Phase change • kJ • C (s) + 2H2 + O2 (g) CH3OH (g)DHorxn = -201.2 • CH3OH (l)C (s) + 2H2 (g) + O2 (g)DHorxn = +238.6 • CH3OH (l) CH3OH (g)DHorxn = +37.4 • This is not DHovap because the values are at 25 oC. • DHovap would be the thermal energy required at the boiling point of methanol. 1 2 1 2

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