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Chemistry: Atoms First Julia Burdge & Jason Overby

Chapter 10. Chemistry: Atoms First Julia Burdge & Jason Overby. Energy Changes in Chemical Reactions. Kent L. McCorkle Cosumnes River College Sacramento, CA. 10. Thermochemistry. 10.1 Energy and Energy Changes 10.2 Introduction to Thermodynamics States and State Functions

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Chemistry: Atoms First Julia Burdge & Jason Overby

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  1. Chapter 10 Chemistry: Atoms First Julia Burdge & Jason Overby Energy Changes in Chemical Reactions Kent L. McCorkle Cosumnes River College Sacramento, CA

  2. 10 Thermochemistry 10.1 Energy and Energy Changes 10.2 Introduction to Thermodynamics States and State Functions The First Law of Thermodynamics Work and Heat 10.3 Enthalpy Reactions Carried Out at Constant Volume or at Constant Pressure Enthalpy and Enthalpy Changes Thermochemical Equations 10.4 Calorimetry Specific Heat and Heat Capacity Constant-Pressure Calorimetry Constant-Volume Calorimetry 10.5 Hess’s Law 10.6 Standard Enthalpies of Formation 10.7 Bond Enthalpy and the Stability of Covalent Molecules 10.8 Lattice Energy and the Stability of Ionic Compounds The Born-Haber Cycle Comparison of Ionic and Covalent Compounds

  3. Energy and Energy Changes 10.1 System The system is a part of the universe that is of specific interest. The surroundings constitute the rest of the universe outside the system. The system is usually defined as the substances involved in chemical and physical changes. Surroundings Universe = System + Surroundings

  4. Energy and Energy Changes Thermochemistry is the study of heat (the transfer of thermal energy) in chemical reactions. Heat is the transfer of thermal energy. Heat is either absorbed or released during a process. heat Surroundings

  5. System 2H2(g) + O2(g) 2H2O(l) + energy Energy and Energy Changes An exothermic processoccurs when heat is transferred from the system to the surroundings. “Feels hot!” heat Surroundings Universe = System + Surroundings

  6. System energy + 2HgO(s) 2Hg(l) + O2(g) Energy and Energy Changes An endothermic processoccurs when heat is transferred from the surroundings to the system. “Feels cold” heat Surroundings Universe = System + Surroundings

  7. Introduction to Thermodynamics 10.2 Thermodynamics is the study of the interconversion of heat and other kinds of energy. In thermodynamics, there are three types of systems: An open systemcan exchange mass and energy with the surroundings. A closed system allows the transfer of energy but not mass. An isolated systemdoes not exchange either mass or energy with its surroundings.

  8. Introduction to Thermodynamics State functionsare properties that are determined by the state of the system, regardless of how that condition was achieved. The magnitude of change depends only on the initial and final states of the system. • Energy • Pressure • Volume • Temperature

  9. ΔUsys + ΔUsurr = 0 ΔUsys = –ΔUsurr The First Law of Thermodynamics The first law of thermodynamicsstates that energy can be converted from one form to another, but cannot be created or destroyed. ΔU is the change in the internal energy. “sys” and “surr” denote system and surroundings, respectively. ΔU = Uf – Ui; the difference in the energies of the initial and final states.

  10. ΔU = q + w Work and Heat The overall change in the system’s internal energy is given by: q is heat • q is positive for an endothermic process (heat absorbed by the system) • q is negative for an exothermic process (heat released by the system) w is work • w is positive for work done on the system • w is negative for work done by the system

  11. ΔU = q + w Work and Heat

  12. Worked Example 10.1 Calculate the overall change in internal energy, ΔU, (in joules) for a system that absorbs 188 J of heat and does 141 J of work on its surroundings. StrategyCombine the two contributions to internal energy using ΔU = q + w and the sign conventions for q and w. SolutionThe system absorbs heat, so q is positive. The system does work on the surroundings, so w is negative. ΔU = q + w = 188 J + (-141 J) = 47 J Think About ItConsult Table 10.1 to make sure you have used the proper sign conventions for q and w.

  13. Enthalpy 10.3 2NaN3(s) 2Na(s) + 3N2(g) Sodium azide detonates to give a large quantity of nitrogen gas. Under constant volume conditions, pressure increases:

  14. 2NaN3(s) 2Na(s) + 3N2(g) Enthalpy Sodium azide detonates to give a large quantity of nitrogen gas. Under constant volume conditions, pressure increases:

  15. w = −PΔV Enthalpy Pressure-volume, or PV work, is done when there is a volume change under constant pressure. P is the external opposing pressure. ΔV is the change in the volume of the container.

  16. 101.3 J 1 L∙atm 1 L 1000 mL 101.3 J 1 L∙atm 1 L 1000 mL Worked Example 10.2 Determine the work done (in joules) when a sample of gas extends from 552 mL to 891 mL at constant temperature (a) against a constant pressure of 1.25 atm, (b) against a constant pressure of 1.00 atm, and (c) against a vacuum (1 L∙atm = 101.3 J). StrategyDetermine change in volume (ΔV), identify the external pressure (P), and use w = −PΔV to calculate w. The result will be in L∙atm; use the equality 1 L∙atm = 101.3 J to convert to joules. SolutionΔV = (891 – 552)mL = 339 mL. (a) P = 1.25 atm, (b) P = 1.00 atm, (c) P = 0 atm. (a) w = -(1.25 atm)(339 mL) (b) w = -(1.00 atm)(339 mL) = -42.9 J = -34.3 J

  17. 101.3 J 1 L∙atm 1 L 1000 mL Worked Example 10.2 (cont.) Solution (c) w = -(0 atm)(339 mL) = 0 J Think About ItRemember that the negative sign in the answers to part (a) and (b) indicate that the system does work on the surroundings. When an expansion happens against a vacuum, no work is done. This example illustrates that work is not a state function. For an equivalent change in volume, the work varies depending on external pressure against which the expansion must occur.

  18. qV = ΔU Enthalpy Pressure-volume, or PV work, is done when there is a volume change under constant pressure. w = −PΔV substitute ΔU = q + w ΔU = q − PΔV When a change occurs at constant volume, ΔV = 0 and no work is done.

  19. qP = ΔU + PΔV Enthalpy Under conditions of constant pressure: ΔU = q + w ΔU = q − PΔV

  20. H = U + PV Enthalpy The thermodynamic function of a system called enthalpy (H) is defined by the equation: A note about SI units: Pressure: pascal; 1Pa = 1 kg/(m . s2) Volume: cubic meters; m3 PV: 1kg/(m . s2) x m3 = 1(kg . m2)/s2 = 1 J Enthalpy: joules U, P, V, and H are all state functions.

  21. If pressure is constant: ΔH = ΔU + PΔV (2) Rearrange to solve for ΔU: ΔU = ΔH + PΔV (3) Remember, qp: qp = ΔU + ΔV (4) Substitute equation (3) into equation (4) and solve: qp = (ΔH − PΔV) + PΔV (5) for a constant-pressure process qp = ΔH Enthalpy and Enthalpy Changes For any process, the change in enthalpy is: ΔH = ΔU + Δ(PV) (1)

  22. ΔH = H(products) – H(reactants) Enthalpy and Enthalpy Changes The enthalpy of reaction (ΔH)is the difference between the enthalpies of the products and the enthalpies of the reactants: Assumes reactions in the lab occur at constant pressure ΔH > 0 (positive) endothermic process ΔH < 0 (negative) exothermic process

  23. H2O(s) H2O(l) ΔH = +6.01 kJ/mol Thermochemical Equations Concepts to consider: • Is this a constant pressure process? • What is the system? • What are the surroundings? • ΔH > 0 endothermic

  24. CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ΔH = −890.4 kJ/mol Thermochemical Equations Concepts to consider: • Is this a constant pressure process? • What is the system? • What are the surroundings? • ΔH < 0 exothermic

  25. Units refer to mole of reaction as written Thermochemical Equations Enthalpy is an extensive property. Extensive properties are dependent on the amount of matter involved. H2O(l) → H2O(g) ΔH = +44 kJ/mol 2H2O(l) → 2H2O(g) ΔH = +88 kJ/mol Double the amount of matter Double the enthalpy

  26. CH4(g) + 2O2(g) CO2(g) + 2H2O(g) ΔH = −802.4 kJ/mol different states different enthalpies CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ΔH = +890.4 kJ/mol Thermochemical Equations The following guidelines are useful when considering thermochemical equations: 1) Always specify the physical states of reactants and products becausethey help determine the actual enthapy changes.

  27. CH4(g) + 2O2(g) CO2(g) + 2H2O(g) ΔH = −802.4 kJ/mol 2CH4(g) + 4O2(g) 2CO2(g) + 4H2O(g) ΔH = −1604.8 kJ/mol CH4(g) + 2O2(g) CO2(g) + 2H2O(g) ΔH = −802.4 kJ/mol CO2(g) + 2H2O(g) CH4(g) + 2O2(g) ΔH = +802.4 kJ/mol Thermochemical Equations The following guidelines are useful when considering thermochemical equations: 2)When multiplying an equation by a factor (n), multiply the ΔH value by same factor. 3)Reversing an equation changes the sign but not the magnitude of ΔH.

  28. 1 mol C6H12O6 180.2 g C6H12O6 Worked Example 10.3 Given the thermochemical equation for photosynthesis, 6H2O(l) + 6CO2(g) → C6H12O6(s) + 6O2(g) ΔH = +2803 kJ/mol calculate the solar energy required to produce 75.0 g of C6H12O6. StrategyThe thermochemical equation shows that for every mole of C6H12O6 produced, 2803 kJ is absorbed. We need to find out how much energy is absorbed for the production of 75.0 g of C6H12O6. We must first find out how many moles there are in 75.0 g of C6H12O6. The molar mass of C6H12O6 is 180.2 g/mol, so 75.0 g of C6H12O6 is 75.0 g C6H12O6 × We will multiply the thermochemical equation, including the enthalpy change, by 0.416, in order to write the equation in terms of the appropriate amount of C6H12O6. = 0.416 mol C6H12O6

  29. Worked Example 10.3 (cont.) Solution (0.416 mol)[6H2O(l) + 6CO2(g) → C6H12O6(s) + 6O2(g)] and (0.416 mol)(ΔH) = (0.416 mol)(2803 kJ/mol) gives 2.50H2O(l) + 2.50CO2(g) → 0.416C6H12O6(s) + 2.50O2(g) ΔH = +1.17×103 kJ Therefore, 1.17×103 kJ of energy in the form of sunlight is consumed in the production of 75.0 g of C6H12O6. Note that the “per mole” units in ΔH are canceled when we multiply the thermochemical equation by the number of moles of C6H12O6. Think About ItThe specified amount of C6H12O6 is less than half a mole. Therefore, we should expect the associated enthalpy change to be less than half that specified in the thermochemical equation for the production of 1 mole of C6H12O6.

  30. Calorimetry 10.4 Calorimetry is the measurement of heat changes. Heat changes are measured in a device called a calorimeter. The specific heat(s) of a substance is the amount of heat required to raise the temperature of 1 g of the substance by 1°C.

  31. heat capacity of 1 kg water Specific heat capacity of water Specific Heat and Heat Capacity The heat capacity (C) is the amount of heat required to raise the temperature of an object by 1°C. The “object” may be a given quantity of a particular substance. Specific heat capacity has units of J/(g• °C) Heat capacity has units of J/°C

  32. q = msΔT q = CΔT Specific Heat and Heat Capacity The heat associated with a temperature change may be calculated: m is the mass. s is the specific heat. ΔT is the change in temperature (ΔT = Tfinal – Tinitial). C is the heat capacity.

  33. 4.184 J g∙°C Worked Example 10.4 Calculate the amount of heat (in kJ) required to heat 255 g of water from 25.2°C to 90.5°C. StrategyUse q = msΔT to calculate q. s = 4.184 J/g∙°C, m = 255 g, ΔT = 90.5°C – 25.2°C = 65.3°C. Solution q = × 255 g × 65.3°C = 6.97×104 J or 69.7 kJ Think About ItLook carefully at the cancellation of units and make sure that the number of kilojoules is smaller than the number of joules. It is a common error to multiply by 1000 instead of dividing in conversions of this kind.

  34. Calorimetry Calculate the amount of heat required to heat 1.01 kg of water from 0.05°C to 35.81°C. Solution: Step 1: Use the equation q = msΔT to calculate q.

  35. Calorimetry A coffee-cup calorimeter may be used to measure the heat exchange for a variety of reactions at constant pressure: Heat of neutralization: HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq) Heat of ionization: H2O(l) → H+(aq) + OH‒(aq) Heat of fusion: H2O(s) → H2O(l) Heat of vaporization: H2O(l) → H2O(g)

  36. qsys = −msΔT qsurr = msΔT qsys = −qsurr Calorimetry Concepts to consider for coffee-cup calorimetry: qP = ΔH System: reactants and products (the reaction) Surroundings: water in the calorimeter For an exothermic reaction: • the system loses heat • the surroundings gain (absorb) heat The minus sign is used to keep sign conventions consistent.

  37. Worked Example 10.5 A metal pellet with a mass of 100.0 g, originally at 88.4°C, is dropped into 125 g of water originally at 25.1°C. The final temperature of both pellet and the water is 31.3°C. Calculate the heat capacity C (in J/°C) of the pellet. StrategyWater constitutes the surroundings; the pellet is the system. Use qsurr = msΔT to determine the heat absorbed by the water; then use q = CΔT to determine the heat capacity of the metal pellet. mwater = 125 g, swater = 4.184 J/g∙°C, and ΔTwater = 31.3°C – 25.1°C = 6.2°C. The heat absorbed by the water must be released by the pellet: qwater = -qpellet, mpellet = 100.0 g, and ΔTpellet = 31.3°C – 88.4°C = -57.1°C.

  38. 4.184 J g∙°C Worked Example 10.5 (cont.) Solution qwater = Thus, qpellet = -3242.6 J From q = CΔT we have -3242.6 J = Cpellet × (-57.1°C) Thus, Cpellet = 57 J/°C × 125 g × 6.2°C = 3242.6 J Think About ItThe units cancel properly to give appropriate units for heat capacity. Moreover, ΔTpellet is a negative number because the temperature of the pellet decreases.

  39. qcal = −qrxn Constant-Volume Calorimetry Constant volume calorimetry is carried out in a device known as a constant-volume bomb. A constant-volume calorimeter is an isolated system. Bomb calorimeters are typically used to determine heats of combustion.

  40. qrxn = −CcalΔT Constant-Volume Calorimetry To calculate qcal, the heat capacity of the calorimeter must be known. qcal = CcalΔT qrxn = −qcal

  41. Think About ItAccording to the label on the cookie package, a service size is four cookies, or 29 g, and each serving contains 150 Cal. Convert the energy per gram to Calories per serving to verify the result. 21.5 kJ g 1 Cal 4.184 kJ 29 g serving × = 1.5×102 Cal/serving × 1.559×102 kJ 7.25 g Worked Example 10.6 A Famous Amos bite-sized chocolate chip cookie weighing 7.25 g is burned in a bomb calorimeter to determine its energy content. The heat capacity of the calorimeter is 39.97 kJ/°C. During the combustion, the temperature of the water in the calorimeter increases by 3.90°C. Calculate the energy content (in kJ/g) of the cookie. StrategyUse qrxn = -CcalΔT to calculate the heat released by the combustion of the cookie. Divide the heat released by the mass of the cookie to determine its energy content per gram Ccal = 39.97 kJ/°C and ΔT = 3.90°C. Solution qrxn = -CcalΔT = -(39.97 kJ/°C)(3.90°C) = -1.559×102 kJ Because energy content is a positive quantity, we write energy content per gram = = 21.5 kJ/g

  42. Hess’s Law 10.5 CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ΔH = −890.4 kJ/mol 2H2O(l) 2H2O(g) ΔH = +88.0 kJ/mol CH4(g) + 2O2(g) CO2(g) + 2H2O(g) ΔH = −802.4 kJ/mol Hess’s lawstates that the change in enthalpy for a stepwise process is the sum of the enthalpy changes for each of the steps. CH4(g) + 2O2(g) ΔH = −802.4 kJ ΔH = −890.4 kJ CO2(g) + 2H2O(g) ΔH = +88.0 kJ CO2(g) + 2H2O(l)

  43. Hess’s Law When applying Hess’s Law: 1) Manipulate thermochemical equations in a manner that gives the overall desired equation. 2)Remember the rules for manipulating thermochemical equations: 3)Add theΔH for each step after proper manipulation. 4) Process is useful for calculating enthalpies that cannot be found directly. Always specify the physical states of reactants and products because they help determine the actual enthalpy changes. When multiplying an equation by a factor (n), multiply the ΔH value by same factor. Reversing an equation changes the sign but not the magnitude of ΔH.

  44. 3 2 Worked Example 10.7 Given the following thermochemical equations, determine the enthalpy change for the reaction NO(g) + O(g) → NO2(g) NO(g) + O3(g) → NO2(g) + O2(g) ΔH = –198.9 kJ/mol ΔH = –142.3 kJ/mol O3(g) → O2(g) ΔH = +495 kJ/mol O2(g) → 2O(g) StrategyArrange the given thermochemical equations so that they sum to the desired equation. Make the corresponding changes to the enthalpy changes, and add them to get the desired enthalpy change.

  45. 3 2 1 2 3 2 1 2 Worked Example 10.7 (cont.) Solution The first equation has NO as a reactant with the correct coefficients, so we will use it as is. The second equation must be reversed so that the O3 introduced by the first equation will cancel (O3 is not part of the overall chemical equation). We also must change the sign on the corresponding ΔH value. These two steps sum to give: NO(g) + O3(g) → NO2(g) + O2(g) ΔH = –198.9 kJ/mol ΔH = +142.3 kJ/mol O2(g) → O3(g) NO(g) + O3(g) → NO2(g) + O2(g) ΔH = –198.9 kJ/mol ΔH = +142.3 kJ/mol + O2(g) O2(g) → O3(g) NO(g) + O2(g) → NO2(g) ΔH = –56.6 kJ/mol

  46. 1 2 1 2 3 2 1 2 Worked Example 10.7 (cont.) Solution We then replace O2 on the left with O by incorporating the last equation. To do so, we divide the third equation by 2 and reverse its direction. As a result, we must also divide ΔH value by 2 and change its sign. Finally, we sum all the steps and add their enthalpy changes. O(g) → O2(g) ΔH = –247.5 kJ/mol NO(g) + O3(g) → NO2(g) + O2(g) ΔH = –198.9 kJ/mol ΔH = +142.3 kJ/mol O2(g) → O3(g) O(g) → O2(g) ΔH = –247.5 kJ/mol + ΔH = –304 kJ/mol NO(g) + O(g) → NO2(g) Think About ItDouble-check the cancellation of identical items–especially where fractions are involved.

  47. Standard Enthalpies of Formation 10.6 C(graphite) + O2(g) CO2(g) ΔHf° = −393.5 kJ/mol The standard enthalpy of formation (ΔHf°) is defined as the heat change that results when 1 mole of a compound is formed from its constituent elements in their standard states. Elements in standard states 1 mole of product

  48. Standard Enthalpies of Formation The standard enthalpy of formation (ΔHf°) is defined as the heat change that results when 1 mole of a compound is formed from its constituent elements in their standard states. The superscripted degree sign denotes standard conditions. • 1 atm pressure for gases • 1 M concentration for solutions “f” stands for formation. ΔHf° for an element in its most stable form is zero. ΔHf° for many substances are tabulated in Appendix 2 of the textbook.

  49. ΔH°rxn = [cΔHf°(C) + dΔHf°(D) ] – [aΔHf°(A) + bΔHf°(B)] ΔH°rxn = ΣnΔHf°(products) – ΣmΔHf°(reactants) Standard Enthalpies of Formation The standard enthalpy of reaction (ΔH°rxn) is defined as the enthalpy of a reaction carried out under standard conditions. aA + bB → cC + dD n and m are the stoichiometric coefficients for the reactants and products.

  50. Worked Example 10.8 Using data from Appendix 2, calculate ΔH°rxn for Ag+(aq) + Cl-(aq) → AgCl(s). StrategyUse ΔH°rxn = ΣnΔHf°(products) – ΣmΔHf°(reactants) and ΔHf° values from Appendix 2 to calculate ΔH°rxn. The ΔHf° values for Ag+(aq), Cl-(aq), and AgCl(s) are +105.9, –167.2, and –127.0 kJ/mol, respectively. Solution ΔH°rxn = ΔHf°(AgCl) – [ΔHf°(Ag+) + ΔHf°(Cl-)] = –127.0 kJ/mol – [(+105.9 kJ/mol) + (–167.2 kJ/mol)] = –127.0 kJ/mol – (–61.3 kJ/mol) = –65.7 kJ/mol Think About ItWatch out for misplaced or missing minus signs. This is an easy place to lose track of them.

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