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Chapter 5: thermochemistry

Chapter 5: thermochemistry. By Keyana Porter Period 2 AP Chemistry. What is thermochemistry?. Thermodynamics: the study of energy and its transformations Thermochemistry is one aspect of thermodynamics The relationship between chemical reactions and energy changes

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Chapter 5: thermochemistry

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  1. Chapter 5:thermochemistry By Keyana Porter Period 2 AP Chemistry

  2. What is thermochemistry? • Thermodynamics: the study of energy and its transformations • Thermochemistry is one aspect of thermodynamics • The relationship between chemical reactions and energy changes • Transformation of energy (heat) during chemical reactions

  3. 5.1Kinetic & Potential Energy • Energy is the capacity to do work or the transfer heat • Objects possess energy in 2 ways • Kinetic energy: due to motion of object • Potential/Stored energy: result of its composition or its position relative to another object • Kinetic energy (Ek) of an object depends on its mass (m) and speed (v): Ek= ½ mv2 • Kinetic energy increases as the speed of an object and its mass increases • Thermal energy: energy due to the substance’s temperature; associated with the kinetic energy of the molecules

  4. 5.1Kinetic & Potential Energy • Potential energy is a result of attraction & repulsion • Ex: an electron has potential energy when it is near a proton due to the attraction (electrostatic forces) • Chemical energy: due to the stored energy in the atoms of the substance

  5. 5.1Units of Energy • joule (J): SI unit for energy • 1 kJ= 1000 J • calorie (cal): non-SI unit for energy; amount of energy needed to raise the temperature of 1 g of water by 1oC • 1 cal = 4.184 J (exactly) • Calorie (nutrition unit) = 1000 cal = 1 kcal • A mass of 2 kg moving at a speed of 1 m/s = kinetic energy of 1 J • Ek= ½ mv2 = ½ (2 kg)(1 m/s)2 = 1 kg-m2/s2 = 1 J

  6. 5.1System & Surroundings • System (chemicals): portion that is singled out of the study • Surroundings (container and environment including you): everything else besides the system • Closed system: can exchange energy, in the form of heat & work, but not matter with the surroundings

  7. 5.1Transferring Energy: Work & Heat • Energy is transferred in 2 ways: • Cause the motion of an object against a force • Cause a temperature change • Force (F): any kind of push or pull exerted on an object • Ex: gravity • Work (w): energy used to cause an object to move against force • Work equals the product of the force and the distance (d) the object is moved: w = F x d • heat: the energy transferred from a hotter object to a colder one • Combustion reactions release chemical energy stored in the form of heat

  8. 5.2Internal Energy • The First Law of Thermodynamics: Energy is conserved; it is neither created nor destroyed • Internal energy (E): sum of ALL the kinetic and potential energy of all the components of the system • The change in internal energy = the difference between Efinal – Einitial ΔE = Efinal – Einitial • We can determine the value of ΔE even if we don’t know the specific values of Efinal and Einitial • All energy quantities have 3 parts: • A number, a unit, and a sign (exothermic versus endothermic)

  9. 5.2RelatingΔE to Heat & Work • A chemical or physical change on a system, the change in its internal energy is given by the heat (q) added to or given off from the system: ΔE = q + w • both the heat added to and the work done on the system increases its internal energy

  10. Sign Conventions Used and the Relationship Among q, w, and ΔE

  11. 5.2Endothermic & Exothermic Processes • Endothermic: system absorbs heat • Ex: melting of ice • Exothermic: system loses heat and the heat flows into the surroundings • Ex: freezing of ice • The internal energy is an example of a state function • Value of any state function depends only on the state or condition of the system (temperature, pressure, location), not how it came to be in that particular state • ΔE = q + w but, q and w are not state functions

  12. 5.3Enthalpy • Enthalpy (H): state function; the heat absorbed or released under constant pressure • The change in enthalpy equals the heat (qP) gained or lost by the system when the process occurs under constant pressure: ΔH = Hfinal – Hinitial = qP • only under the condition of constant pressure is the heat that is transferred equal to the change in the enthalpy • The sign on ΔH indicated the direction of heat transfer • + value of ΔH means it is endothermic • - value of ΔH means it is exothermic

  13. 5.4Enthalpies of Reaction • Enthalpy of reaction (ΔHrxn): the enthalpy change that accompanies a reaction • The enthalpy change for a chemical reaction is given by the enthalpy of the products minus the reactants: ΔH = H(products) – H(reactants) • Thermochemical equations: balanced chemical equations that show the associated enthalpy change • The magnitude of ΔH is directly proportional to the amount or reactant consumed in the process

  14. 5.4Enthalpies of Reaction • The enthalpy change for the reaction is equal in magnitude but opposite in sign to ΔH for the reverse reaction CO2(g) + 2H2O(l)  CH4(g) + 2O2(g) ΔH= 890 kJ CH4(g) + 2O2(g) Enthalpy  Reversing a reaction changes the sign but not the magnitude of the enthalpy change: ΔH2 = - ΔH1 ΔH1= - 890 kJ ΔH2= 890 kJ CO2(g) + 2H2O(l)

  15. 5.4Enthalpies of Reaction • The enthalpy change for a reaction depends on the state of the reactants and products • Ex: CH4 (g) + 2O2 (g)  CO2 (g) + 2H2O (l) ΔH = -890 kJ • If the product was H2O (g) instead of H2O (l), the ΔH would be - 820 kJ instead of - 890 kJ

  16. 5.5Calorimetry/Heat Capacity & Specific Heat • Calorimetry: the measure of heat flow • Calorimeter: measures heat flow • Heat capacity: the amount of heat required to raise its temperature by 1 K • The greater the heat capacity of a body, the greater the heat required to produce a given rise in temperature • Molar heat capacity: the heat capacity of 1 mol of a substance • Specific heat: the heat capacity of 1 g of a substance; measured by temperature change (ΔT) that a known mass (m) of the substance undergoes when it gains or loses a specific quantity of heat (q):

  17. 5.5Calorimetry/Heat Capacity & Specific Heat specific heat = quantity of heat transferred (grams of substance) x (temperature change) = q m x ΔT Practice Exercise (B&L page 160) Calculate the quantity of heat absorbed by 50 kg of rocks if their temperature increases by 12.0 OC. (Assume the specific heat of the rocks is .82 J/g-K.)

  18. 5.5Calorimetry/Heat Capacity & Specific Heat • Solving the problem q = (specific heat) x (grams of substance) x ΔT = (0.82 J/g-K)(50,000 g)(285 K) = 4.9 x 105 J

  19. 5.5Constant-Pressure Calorimetry • The heat gained by the solution (qsoln) is equal in magnitude and opposite in sign from the heat of the reaction • qsoln = (specific heat of solution) x (grams of solution) x ΔT = - qrxn • For dilute aqueous solutions, the specific heat of the solution is approx. the same as water (4.18 J/g-K) Practice Exercise (B&L page 161) When 50 mL of .100M AgNO3 and 50.0 mL of .100 M HCl are mixed in a constant-pressure calorimeter, the temperature of the mixture increases from 22.30oC to 23.11oC. The temperature increase is caused by this reaction: AgNO3 + HCl  AgCl + HNO3 Calculate ΔH for this reaction, assuming that the combined solution has a mass of 100 g and a specific heat of 4.18 J/g-oC

  20. 5.5Constant-Pressure Calorimetry • Solving the problem qrxn = -(specific heat of solution) x (grams of solution) x ΔT = - (4.18 J/g-oC)(100 g)(0.8 K) = - 68,000 J/mol

  21. 5.5Bomb Calorimetry (Constant-Volume Calorimetry) • Bomb calorimeter: used to study combustion reactions • Heat is released when combustion occurs, absorbed by the calorimeter contents, raising the temperature of the water (measured before and after the reaction) • To calculate the heat of combustion from the measured temperature increase in the bomb calorimeter, you must know the heat capacity of the calorimeter (Ccal) qrxn = - Ccal x ΔT

  22. 5.6Hess’s Law • Hess’s Law: if a reaction is carried out in a series of steps, ΔH for the reaction will be equal to the sum of the enthalpy changes for the individual steps CH4 (g) + 2O2 (g)  CO2 (g) + 2H2O (g) ΔH = -802 kJ (ADD)2H2O (g)  2H2O (l) ΔH = -88 kJ CH4 (g) + 2O2 (g) + 2H2O (g) CO2 (g) + 2H2O (g) + 2H2O (l) ΔH = -890 kJ

  23. 5.6Hess’s Law • Practice Exercise 5.8 (B&L page 165) Calculate ΔH for the reaction: 2C (s) + H2 C2H2 (g) Given the following reactions and their respective enthalpy changes: C2H2 (g) + 5/2 O2  2CO2 (g) + H2O (l) ΔH = -1299.6 kJ C (s) + O2 (g)  CO2 (g) ΔH = -393.5 kJ H2 (g) + ½ O2 (g)  H2O (l) ΔH = -285.8 kJ

  24. 5.6Hess’s Law • Solving the problem 2CO2 (g) + H2O (l)  C2H2 (g) + 5/2 O2ΔH = 1299.6 kJ 2C (s) + 2O2 (g)  2CO2 (g) ΔH = -393.5 kJ (2) -787.0 kJ H2 (g) + ½ O2 (g)  H2O (l) ΔH = -285.8 kJ 2C (s) + H2 C2H2 (g) ΔH = 226.8 kJ • if the reaction is reversed, the sign of ΔH changes • if reaction is multiplied, so is ΔH

  25. Enthalpy Diagram The quantity of heat generated by combustion of 1 mol CH4 is independent of whether the reaction takes place in one or more steps: ΔH1 = ΔH2 + ΔH3

  26. 5.7Enthalpies of Formation • Enthalpies of vaporization: ΔH for converting liquids to gases • Enthalpies of fusion: ΔH for melting solids • Enthalpies of combustion: ΔH for combusting a substance in oxygen • Enthalpy of formation (ΔHf): enthalpy change where the substance has been formed from its elements • Standard enthalpy (ΔH o): enthalpy change when all reactants and products are at 1 atm pressure and specific temperature (298 K) • Standard enthalpy of formation (ΔHof): the enthalpy change for the reaction that forms 1 mol of the compound from its elements, with all substances in their standard states • ΔHof = 0 for any element in its purest form at 295 K and 1 atm pressure

  27. 5.7Enthalpies of Formation • The standard enthalpy change for any reaction can be calculated from the summations of the reactants and products in the reaction ΔH orxn = nΔH of (products) - mΔH of (reactants) Practice Problem 5.9 (B&L page 169) Calculate the standard enthalpy change for the combustion of 1 mol of benzene, C6H6 (l), to CO2 (g) and H2O (l).

  28. 5.7Enthalpies of Formation • Solving the problem C6H6 (l) + 15/2 O2 (g)  6CO2 (g) + 3H2O (l) ΔH orxn= [6Δ H of (CO2) + 3Δ H of (H2O)] – [ΔH of (C6H6) + 15/2 ΔH of (O2)] = [6(-393.5 kJ) + 3(-285.8 kJ)] – [(49.0 kJ) + 15/2 (0 kJ)] = (-2361 – 857.4 – 49.0) kJ = -3267 kJ

  29. Extra Equations • Force = mass x 9.8 m/s2 • Internal energy… ΔE = Efinal – Einitial • Entropy… Δ S = Sfinal – Sinitial • Enthalpy… Δ H = Hfinal – Hinitial • Gibbs Free Energy… Δ G = Gfinal – Ginitial • Δ S = Δ H/ T

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