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Chapter 8

Chapter 8. Thermochemistry: Chemical Energy. Energy. Energy – capacity to supply heat or do work Energy = Heat + Work E = q + w 2 types of Energy Potential Energy Kinetic Energy. Energy. Two fundamental kinds of energy. Potential energy is stored energy.

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Chapter 8

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  1. Chapter 8 Thermochemistry: Chemical Energy

  2. Energy • Energy – capacity to supply heat or do work Energy = Heat + Work E = q + w • 2 types of Energy • Potential Energy • Kinetic Energy

  3. Energy • Two fundamental kinds of energy. • Potential energy is stored energy. • Kinetic energy is the energy of motion. • Law of Conservation of Energy • Energy can be converted from one kind to another but never destroyed

  4. Energy • Units • SI Unit – Joule (J) • Additional units • Calorie (Cal) – food calorie • calorie (cal) – scientific calorie • Conversions • 1 cal = 4.184 J • 1000 cal = 1 Cal

  5. Energy and Chemical Bonds • Chapter 6 • Kept a careful accounting of atoms as they rearranged themselves • Reactions also involve a transfer of energy

  6. Energy and Chemical Bonds • A chemical • Potential - attractive forces in an ionic compound or sharing of electrons covalent compound • Kinetic – (often in form of heat) occurs when bonds are broken and particles allowed to move • To determine the energy of a reaction it is necessary to keep track of the energy changes that occur during the reaction

  7. Internal Energy and State Functions • 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).

  8. Internal Energy and State Functions • Tracking energy changes • Energy changes are measured from the point of view of the system (Internal Energy - IE) • Change in Energy of the system – ΔE • ΔE = Efinal -Einitial

  9. Internal Energy and State Functions • IE depends on • Chemical identity, sample size, temperature, etc. • Does not depend on the system’s history • Internal Energy is a state function • A function or property whose value depends only on the present state (condition) of the system, not on the path used to arrive at that condition

  10. Expansion Work • E = q + w • In physics w = force (F) x distance (d) • Force – energy that produces movement of an object • In chemistry w = expansion work • Force - the pressure that the reaction exerts on its container against atmospheric pressure hence it is negative • Distance – change in volume of the reaction • w = -PΔV

  11. Energy and Enthalpy • ΔE = q – PΔV • The amount of heat exchanged between the system and the surroundings is given the symbolq. q = DE + PDV • At constant volume (DV = 0):qv = DE • At constant pressure: Energy due to heat and work but work minimal compared to heat energy • qp = DE + PDV = DH • Enthalpy change (heat of reaction):DH = Hproducts – Hreactants

  12. The Thermodynamic Standard State • ΔH = amount of energy absorbed or released in the form of heat • DH = Hproducts – Hreactants • Important factors • States of matter • Thermodynamic standard state – most stable form of a substance at 1 atm and at a specified temperature, usually 25oC; and 1 M concentration for all substances in solution • DH – valid for the reaction as written including exact # of moles of substances • N2H4(g) + H2(g)  2 NH3(g) + heat (188 kJ)

  13. Enthalpies of Physical and Chemical Change

  14. Enthalpies of Physical and Chemical Changes • 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. Unfavorable Process • Exothermic:Heat flows out of the system into the surroundings and DH has a negative sign. Favorable process

  15. Enthalpies of Physical and Chemical Changes • Reversinga 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

  16. Problems • 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

  17. Determination of Heats of Reaction • Experimentally – calorimetry • Hess’s Law • Standard Heat’s of Formation • Bond Dissociation Energies

  18. Calorimetry and Heat Capacity • 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.

  19. Calorimetry and Heat Capacity

  20. q C = T D Calorimetry and Heat Capacity • 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.

  21. Problems • 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 the heat absorbed or released for this reaction.

  22. Hess’s Law • Allows the enthalpy to be determined for: • Reactions that occur too quickly or take too long to use calorimetry • Reactions that are too dangerous • Works like the Haber process in chapter 6 • Take reactions for which the heat is known and manipulate them to give the desired reaction

  23. Standard Heats of Formation • 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

  24. Standard Heats of Formation 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

  25. Standard Heats of Formation • Calculating DH° for a reaction: DH° = Σ[DH°f (products) x moles] – Σ[DH°f (Reactants) x moles] • 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)]

  26. Problems • 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.

  27. Energy Calculations • Other methods for calculating enthalpies • Bond dissociation energies – measures the energy given off by the formation of bonds in the products and substracts the energy required to break bonds in the reactants

  28. Why do chemical reactions occur? • A chemical reaction will move from less stability to greater stability. • Achieved by giving off more energy than is absorbed by the reactants • This indicates that exothermic reactions occur by why do endothermic reactions occur? • Gibb’s Free Energy • DG = DH – TDS • DH – enthalpy, T – temperature, DS - entropy

  29. An Introduction to Entropy • Second Law of Thermodynamics: Reactions proceed in the direction that increases the entropy of the system plus surroundings. (increases the degree of disorder) • 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.

  30. An Introduction to Entropy

  31. An Introduction to Entropy

  32. An Introduction to Entropy • 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 (favorable). • Negative value of DS indicates decreased disorder (unfavorable).

  33. Problems • 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)

  34. An Introduction to Free Energy • 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).

  35. An Introduction to Free Energy • 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 (favorable) • DG = 0 Process is at equilibrium • DG > 0 Process is nonspontaneous (unfavorable)

  36. Problems • 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

  37. An Introduction to Free Energy • 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

  38. Problem • 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?

  39. Optional Homework • Text - 8.28, 8.32, 8.50, 8.52, 8.56, 8.58, 8.66, 8.70, 8.74, 8.82, 8.88, 8.90 • Chapter 8 Homework from website

  40. Required Homework • Chapter 8 Assignment

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