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Thermochemistry: Chemical Energy

Thermochemistry: Chemical Energy. Chemistry 4th Edition McMurry/Fay. Thermodynamics 01. Energy: is the capacity to do work, or supply heat. Energy = Work + Heat Kinetic Energy: is the energy of motion. E K = 1 / 2 mv 2 (1 Joule = 1 kg m 2 /s 2 ) (1 calorie = 4.184 J)

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Thermochemistry: Chemical Energy

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  1. Thermochemistry: Chemical Energy Chemistry 4th Edition McMurry/Fay

  2. Thermodynamics 01 • Energy:is the capacity to do work, or supply heat. • Energy = Work + Heat • Kinetic Energy:is the energy of motion. • EK = 1/2mv2(1 Joule = 1 kgm2/s2) • (1 calorie = 4.184 J) • Potential Energy:is stored energy. Chapter 08

  3. Thermodynamics 03 • In an experiment:Reactants and products are the system; everything else is the surroundings. • Energy flowfrom the systemto the surroundings has a negative sign (loss of energy). • Energy flow from thesurroundingsto the system has a positive sign (gain of energy). Chapter 08

  4. Thermodynamics 04 • Closed System: Only energy can be lost or gained. • Isolated System: No matter or energy is exchanged. Chapter 08

  5. Thermodynamics 05 • The law of the conservation of energy: Energy cannot be created or destroyed. • The energy of an isolated system must be constant. • The energy change in a system equals the work done on the system + the heat added. • DE = Efinal – Einitial = E2 – E1 = q + w • q = heat, w = work Chapter 08

  6. State Functions 01 • State Function: A function or property whose value depends only on the present state (condition) of the system. • The change in a state function is zero when the system returns to its original condition. • For nonstate functions, the change is not zero if the path returns to the original condition. Chapter 08

  7. State Functions 02 • State and Nonstate Properties: The two paths below give the same final state: • N2H4(g) + H2(g)  2 NH3(g) + heat (188 kJ) • N2(g) + 3 H2(g)  2 NH3(g) + heat (92 kJ) • State properties include temperature, total energy, pressure, density, and [NH3]. • Nonstate properties include the heat. Chapter 08

  8. Enthalpy Changes 01 • Enthalpies of Physical Change: Chapter 08

  9. Enthalpy Changes 02 • Enthalpies of Chemical Change:Often called heats of reaction (DHreaction). • Endothermic:Heat flows into the system from the surroundings and DH has a positive sign. • Exothermic:Heat flows out of the system into the surroundings and DH has a negative sign. Chapter 08

  10. Enthalpy Changes 03 • Reversing a reaction changes the sign of DHfor a reaction. • C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l) DH = –2219 kJ • 3 CO2(g) + 4 H2O(l)  C3H8(g) + 5 O2(g) DH = +2219 kJ • Multiplying a reaction increases DHby the same factor. • 3 C3H8(g) + 15 O2(g)  9 CO2(g) + 12 H2O(l) DH = –6657 kJ Chapter 08

  11. Enthalpy Changes 04 • How much heat (in kilojoules) is evolved or absorbed in each of the following reactions? • Burning of 15.5 g of propane: C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l) DH = –2219 kJ • Reaction of 4.88 g of barium hydroxide octahydrate with ammonium chloride: Ba(OH)2·8 H2O(s) + 2 NH4Cl(s)  BaCl2(aq) + 2 NH3(aq) + 10 H2O(l)DH = +80.3 kJ Chapter 08

  12. Enthalpy Changes 05 • Thermodynamic Standard State:Most stable form of a substance at 1 atm pressure and 25°C; 1 M concentration for all substances in solution. • These are indicated by a superscript ° to the symbol of the quantity reported. • Standard enthalpy change is indicated by the symbol DH°. Chapter 08

  13. Hess’s Law 01 • Hess’s Law:The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for the individual steps in the reaction. • 3 H2(g) + N2(g)  2 NH3(g) DH° = –92.2 kJ Chapter 08

  14. Hess’s Law 02 • Reactants and products in individual steps can be added and subtracted to determine the overall equation. • (a) 2 H2(g) + N2(g) N2H4(g) DH°1 = ? • (b) N2H4(g) + H2(g) 2 NH3(g) DH°2 = –187.6 kJ • (c) 3 H2(g) + N2(g) 2 NH3(g) DH°3 = –92.2 kJ • DH°1 = DH°3 – DH°2 = (–92.2 kJ) – (–187.6 kJ) = +95.4 kJ Chapter 08

  15. Hess’s Law 03 • The industrial degreasing solvent methylene chloride (CH2Cl2, dichloromethane) is prepared from methane by reaction with chlorine: • CH4(g) + 2 Cl2(g) CH2Cl2(g) + 2 HCl(g) • Use the following data to calculate DH° (in kilojoules) for the above reaction: • CH4(g) + Cl2(g) CH3Cl(g) + HCl(g) • DH° = –98.3 kJ • CH3Cl(g) + Cl2(g) CH2Cl2(g) + HCl(g) • DH° = –104 kJ Chapter 08

  16. Standard Heats of Formation 01 • Standard Heats of Formation (DH°f): The enthalpy change for the formation of 1 mole of substance in its standard state from its constituent elements in their standard states. • The standard heat of formation for any element in its standard state is defined as being ZERO. • DH°f = 0 for an element in its standard state Chapter 08

  17. Standard Heats of Formation 02 H2(g) + 1/2 O2(g)  H2O(l) DH°f = –286 kJ/mol 3/2 H2(g) + 1/2 N2(g)  NH3(g) DH°f = –46 kJ/mol 2 C(s) + H2(g)  C2H2(g) DH°f = +227 kJ/mol 2 C(s) + 3 H2(g) + 1/2 O2(g)  C2H5OH(g) DH°f = –235 kJ/mol Chapter 08

  18. Standard Heats of Formation 03 • Calculating DH° for a reaction: • DH° = DH°f (Products) – DH°f (Reactants) • For a balanced equation, each heat of formation must be multiplied by the stoichiometric coefficient. • aA + bB cC + dD • DH° = [cDH°f(C) + dDH°f(D)] – [aDH°f(A) + bDH°f(B)] Chapter 08

  19. CO(g) -111 C2H2(g) 227 Ag+(aq) 106 CO2(g) -394 C2H4(g) 52 Na+(aq) -240 H2O(l) -286 C2H6(g) -85 NO3-(aq) -207 NH3(g) -46 CH3OH(g) -201 Cl-(aq) -167 N2H4(g) 95.4 C2H5OH(g) -235 AgCl(s) -127 HCl(g) -92 C6H6(l) 49 Na2CO3(s) -1131 Standard Heats of Formation 04 Some Heats of Formation, Hf° (kJ/mol) Chapter 08

  20. Standard Heats of Formation 05 • Calculate DH° (in kilojoules) for the reaction of ammonia with O2 to yield nitric oxide (NO) and H2O(g), a step in the Ostwald process for the commercial production of nitric acid. • Calculate DH° (in kilojoules) for the photosynthesis of glucose from CO2 and liquid water, a reaction carried out by all green plants. Chapter 08

  21. Bond Dissociation Energy 01 • Bond Dissociation Energy:Can be used to determine an approximate value for DH°f. • DH = D (Bonds Broken) – D (Bonds Formed) • For the reaction between H2 and Cl2 to form HCl: • DH = D(H–Cl) – ∑ {D(H–H) + D(O=O)} Chapter 08

  22. Bond Dissociation Energy 02 Chapter 08

  23. Bond Dissociation Energy 03 • Calculate an approximate DH° (in kilojoules) for the synthesis of ethyl alcohol from ethylene: • C2H4(g) + H2O(g)  C2H5OH(g) • Calculate an approximate DH° (in kilojoules) for the synthesis of hydrazine from ammonia: • 2 NH3(g) + Cl2(g)  N2H4(g) + 2 HCl(g) Chapter 08

  24. Calorimetry and Heat Capacity 01 • Calorimetry is the science of measuring heat changes (q) for chemical reactions. There are two types of calorimeters: • Bomb Calorimetry: A bomb calorimeter measures the heat change at constant volume such that q = DE. • Constant Pressure Calorimetry: A constant pressure calorimeter measures the heat change at constant pressure such that q = DH. Chapter 08

  25. Calorimetry and Heat Capacity 02 Constant Pressure Bomb Chapter 08

  26. q C = T D Calorimetry and Heat Capacity 03 • Heat capacity (C)is the amount of heat required to raise the temperature of an object or substance a given amount. • Specific Heat:The amount of heat required to raise the temperature of 1.00 g of substance by 1.00°C. • Molar Heat:The amount of heat required to raise the temperature of 1.00 mole of substance by 1.00°C. Chapter 08

  27. Calorimetry and Heat Capacity 04 • What is the specific heat of lead if it takes 96 J to raise the temperature of a 75 g block by 10.0°C? • When 25.0 mL of 1.0 M H2SO4 is added to 50.0 mL of 1.0 M NaOH at 25.0°C in a calorimeter, the temperature of the solution increases to 33.9°C. Assume specific heat of solution is 4.184 J/(g–1·°C–1), and the density is 1.00 g/mL–1, calculate DH for the reaction. Chapter 08

  28. Calorimetry and Heat Capacity 05 Chapter 08

  29. Introduction to Entropy 01 • Second Law of Thermodynamics: Reactions proceed in the direction that increases the entropy of the system plus surroundings. • A spontaneous process is one that proceeds on its own without any continuous external influence. • A nonspontaneous processtakes place only in the presence of a continuous external influence. Chapter 08

  30. Introduction to Entropy 02 • The measure of molecular disorder in a system is called the system’s entropy; this is denoted S. • Entropy has units of J/K (Joules per Kelvin). • DS = Sfinal – Sinitial • Positive value of DS indicates increased disorder. • Negative value of DS indicates decreased disorder. Chapter 08

  31. Introduction to Entropy 03 Chapter 08

  32. Introduction to Entropy 04 • To decide whether a process is spontaneous, both enthalpy and entropy changes must be considered: • Spontaneous process: Decrease in enthalpy (–DH). Increase in entropy (+DS). • Nonspontaneous process: Increase in enthalpy (+DH). Decrease in entropy (–DS). Chapter 08

  33. Introduction to Entropy 05 • Predict whether DS° is likely to be positive or negative for each of the following reactions. Using tabulated values, calculate DS° for each: • a. 2 CO(g) + O2(g)  2 CO2(g)b. 2 NaHCO3(s)  Na2CO3(s) + H2O(l) + CO2(g)c. C2H4(g) + Br2(g)  CH2BrCH2Br(l)d. 2 C2H6(g) + 7 O2(g)  4 CO2(g) + 6 H2O(g) Chapter 08

  34. Introduction to Free Energy 01 • Gibbs Free Energy Change (DG): Weighs the relative contributions of enthalpy and entropy to the overall spontaneity of a process. • DG = DH – TDS • DG < 0 Process is spontaneous • DG = 0 Process is at equilibrium • DG > 0 Process is nonspontaneous Chapter 08

  35. Introduction to Free Energy 02 • Situations leading to DG < 0: • DH is negative and TDS is positive • DH is very negative and TDS is slightly negative • DH is slightly positive and TDS is very positive • Situations leading to DG = 0: • DH and TDS are equally negative • DH and TDS are equally positive • Situations leading to DG > 0: • DH is positive and TDS is negative • DH is slightly negative and TDS is very negative • DH is very positive and TDS is slightly positive Chapter 08

  36. Introduction to Free Energy 03 • Which of the following reactions are spontaneous under standard conditions at 25°C? • a. AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq) DG° = –55.7 kJ • b. 2 C(s) + 2 H2(g)  C2H4(g)DG° = 68.1 kJ • c. N2(g) + 3 H2(g)  2 NH3(g) DH° = –92 kJ; DS° = –199 J/K Chapter 08

  37. Introduction to Free Energy 04 • Equilibrium (DG° = 0): Estimate the temperature at which the following reaction will be at equilibrium. Is the reaction spontaneous at room temperature? • N2(g) + 3 H2(g)  2 NH3(g) DH° = –92.0 kJ DS° = –199 J/K • Equilibrium is the point where DG° = DH° – TDS° = 0 Chapter 08

  38. Introduction to Free Energy 05 • Benzene, C6H6, has an enthalpy of vaporization, DHvap, equal to 30.8 kJ/mol and boils at 80.1°C. What is the entropy of vaporization, DSvap, for benzene? Chapter 08

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