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Chapter 15

Chapter 15. Chemical Thermodynamics 化學熱力學. Outline. Heat Changes and Thermochemistry The First Law of Thermodynamics 熱力學第一定律 Some Thermodynamic Terms Enthalpy Changes 焓的變化 Calorimetry 熱量的測量 Thermochemical Equations 熱化學方程式 Standard States and Standard Enthalpy Changes

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Chapter 15

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  1. Chapter 15 Chemical Thermodynamics 化學熱力學

  2. Outline Heat Changes and Thermochemistry • The First Law of Thermodynamics 熱力學第一定律 • Some Thermodynamic Terms • Enthalpy Changes 焓的變化 • Calorimetry 熱量的測量 • Thermochemical Equations 熱化學方程式 • Standard States and Standard EnthalpyChanges • Standard Molar Enthalpies of Formation, Hfo • Hess’s Law • Bond Energies • Changes in Internal Energy,  E • Relationship of H and E

  3. Outline Spontaneity of Physical and Chemical Changes • The Two Aspects of Spontaneity • Dispersal of Energy and Matter • Entropy熵, S, and Entropy Change, DS • The Second Law of Thermodynamics • Free Energy Change, DG, and Spontaneity • The Temperature Dependence of Spontaneity

  4. The First Law of Thermodynamics熱力學第一定律 • Thermodynamics is the study of the changes in energy and transfers of energy that accompany chemical and physical processes.所謂的熱力學是指在化學及物理反應中能量的改變及轉移 • In this chapter we will address 3 fundamental questions. • Will two (or more) substances react when they are mixed under specified conditions?當兩個物質在特定狀態混合時，是否會發生反應? • If they do react, what energy changes and transfers are associated with their reaction?如果發生反應，那反應中有那些能量的改變及轉移? • If a reaction occurs, to what extent does it occur?

  5. The First Law of Thermodynamics Energy is the capacity to do work or to transfer heat. • Exothermic reactions release energy in the form of heat.(放熱反應以熱的形式釋出能量) • For example, the combustion of propane is exothermic.(丙烷的燃燒為放熱反應) • C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(l) + 2.22x103 kJ • The combustion of n-butane is also exothermic (正丁烷的燃燒亦為放熱反應) • 2C4H10(g) + 13 O2(g) 8 CO2(g) + 10 H2O(l)+ 5.78x103 kJ • A process that absorbs energy from its surroundings is called endothermic. • H2O(s)+ 6.02kJ H2O(l) 烷類和過量的氧作用會形成二氧化碳和水

  6. The First Law of Thermodynamics • Exothermic reactions generate specific amounts of heat.(放熱反應產生熱) • This is because the potential energies of the products are lower than the potential energies of the reactants.(此乃因產物的位能比反應物未能來得低) 活化能 (activation energy) Kinetic energy 動能 v.s. Potential energy 位能 (包含化學能)

  7. The First Law of Thermodynamics • There are two basic ideas of importance for thermodynamic systems.(熱力學系統有兩個最基本的概念) • Chemical systems tend toward a state of minimum potential energy.(系統傾向最小位能) • Some examples of this include: • H2O flows downhill. • Objects fall when dropped. • The energy change for these two examples is: • Epotential = mgh • Epotential = mg(h)

  8. The First Law of Thermodynamics • Chemical systems tend toward a state of maximum disorder.(系統傾向最大亂度) • Common examples of this are: • A mirror shatters when dropped and does not reform. • It is easy to scramble an egg and difficult to unscramble it. • Food dye when dropped into water disperses.

  9. The First Law of Thermodynamics • This law can be stated as, “The combined amount of matter and energy in the universe is constant.”(熱力學第一定律可定義為物質和能量的總和不會改變) • The first law is also known as the Law of Conservation of Energy(即為能量守恆定律). • Energy is neither created nor destroyed in chemical reactions and physical changes. 在化學及物理反應時，能量既不會憑空消失，也不會憑空產生，只能從一種形式轉化成另一種形式，或者從一個物體轉移到另一個物體，而總量保持不變

  10. Some Thermodynamic Terms • The substances involved in the chemical and physical changes under investigation are called the system. • In chemistry lab, the system is the chemicals inside the beaker. • The environment around the system is called the surroundings. • The surroundings are outside the beaker. • The system plus the surroundings is called the universe. • The set of conditions that specify all of the properties of the system is called the thermodynamic state of a system. • For example the thermodynamic state could include: • The number of moles and identity of each substance. • The physical states of each substance. • The temperature of the system. • The pressure of the system.

  11. Some Thermodynamic Terms • The properties of a system that depend only on the state of the system are called state functions.(狀態函數指一個系統的特質只與狀態有關而與途徑無關, 通常以大寫表示) • The value of a state function is independent of pathway. • An analog to a state function is the energy required to climb a mountain taking two different paths.(走兩條不同的路而到達山頂的位能差相同，但路徑不同) • E1 = energy at the bottom of the mountainE1 = mgh1 • E2 = energy at the top of the mountainE2 = mgh2 • E = E2-E1 = mgh2 – mgh1 = mg(h) • Notice that the energy change in moving from the top to the bottom is independent of pathway but the work required may not be! • Some examples of state functions are: • T, P, V, E, H, and S • Examples of non-state functions are: • n, q, w

  12. Some Thermodynamic Terms • In thermodynamics we are often interested in changes in functions. • We will define the change of any function X as: • X = Xfinal – Xinitial • If X increases X > 0 • If X decreases X < 0.

  13. Enthalpy Change焓變化 (反應熱) • Chemistry is commonly done in open beakers on a desk top at atmospheric pressure. • Because atmospheric pressure only changes by small increments, this is almost at constant pressure. • The enthalpy change, H, is the change in heat content at constant pressure.焓變化是指定壓下熱含量的變化 • H=qP

  14. Enthalpy Change • Hrxn is the heat of reaction. 反應熱 • This quantity will tell us if the reaction produces or consumes heat.(Hrxn的量表示反應產生熱或消耗熱) • If Hrxn < 0 the reaction is exothermic. • If Hrxn > 0 the reaction is endothermic. • Hrxn = Hproducts - Hreactants • Hrxn = Hsubstancesproduced - Hsubstancesconsumed • Notice that this is Hrxn = Hfinal – Hinitial

  15. Calorimetry 熱量的測量 • A coffee-cup calorimeter (熱卡計)is used to measure the amount of heat produced (or absorbed) in a reaction at constant P • This is one method to measure qP for reactions in solution.

  16. If an exothermic reaction is performed in a calorimeter, the heat evolved by the reaction is determined from the temperature rise of the solution. This requires a two part calculation. Amount of heat gained by calorimeter is called the heat capacity of the calorimeter or calorimeter constant. The value of the calorimeter constant is determined by adding a specific amount of heat to calorimeter and measuring the temperature rise. Calorimetry + = Amount of heat absorbed by calorimeter Amount of heat Released by reaction Amount of heat absorbed by solution Heat capacity (C) = Increase in temperature (oC) Heat absorbed (J) 所謂熱容量(heat capacity)是指使一個物體上升1oC所需的熱量

  17. Calorimetry Example 15-1: When 3.425 kJ of heat is added to a calorimeter containing 50.00 g of water the temperature rises from 24.00oC to 36.54oC. Calculate the heat capacity of the calorimeter in J/oC. The specific heat of water is 4.184 J/g oC. • 水的溫度變化 T=(36.54 – 24.00)oC = 12.54oC • 水所吸收的熱 qp = mCT • = (50.00g) (4.184 J/goC) (12.54oC) • = 2623.37 J • 熱卡計所吸收的熱量 = 3425J – 2623J • = 802 J • 熱卡計的熱容量 =802J / 12.54oC • = 64.0 J/oC

  18. Example 15-2: A coffee-cup calorimeter is used to determine the heat of reaction for the acid-base neutralization CH3COOH(aq) + NaOH(aq) NaCH3COO(aq) + H2O() When we add 25.00 mL of 0.500 M NaOH at 23.000oC to 25.00 mL of 0.600 M CH3COOH already in the calorimeter at the same temperature, the resulting temperature is observed to be 25.947oC. (The heat capacity of the calorimeter : 27.8 J/0C. The specific heat of the mixture is 4.18 J/g0C and the density: 1.02 g/mL.) 溫度變化 T= 25.947 – 23.000 = 2.947 oC 熱卡計所吸收的熱量 q= (2.947oC)(27.8 J/oC) = 81.9 J 混合溶液的重量 =(25.00ml + 25.00ml) x1.02 g/ml = 51.0g 溶液所吸收的熱量 q= mCT =(51.0g)(4.18 J/goC)(2.947oC) = 628 J 反應所產生的總熱量為 = 81.9J + 628J = 709.9J CH3COOH(aq) + NaOH(aq) NaCH3COO(aq) + H2O() 1 1 1 25.00mlx0.5 M =12.5mmol 25.00mlx0.6 M =15.0mmol 12.5mmol 12.5mmol = 0.0125 mol 指每莫耳所放出的熱量 Hrxn = 709.9J / 0.0125 mol = 56792 J/mol = 5.68kJ/mol

  19. Thermochemical equationsare a balanced chemical reaction plus the H value for the reaction. (熱化學方程式為平衡的化學式再加上反應所需的熱量H.) For example, this is a thermochemical equation. C5H12(g) + 8 O2(g) 5 CO2(g) + 6 H2O(l) + 3523kJ 1 mol of C5H12 reacts with 8 mol of O2 to produce 5 mol of CO2, 6 mol of H2O, and releasing 3523 kJ is referred to as one mole of reactions. This is an equivalent method of writing thermochemical equations. ThermochemicalEquations 熱化學方程式 1mole 8moles 5moles 6moles C5H12(g) + 8 O2(g) 5 CO2(g) + 6 H2O(l) Horxn= - 3523kJ • H < 0 designates an exothermic reaction. • H > 0 designates an endothermic reaction

  20. Thermochemical Equations Example 15-3: When 2.61g of dimethyl ether (CH3OCH3) is burned at constant pressure, 82.5 kJ of heat is given off. Find H for the reaction. CH3OCH3(l) + 8 O2(g) 2 CO2(g) + 3 H2O(l) CH3OCH3的分子量為 46 CH3OCH3的 mole數 = 2.61g/ 46 = 5.67x10-2 mole 此反應為放熱反應 Hrxn = - (82.5 kJ/ 5.67x10-2 mol) = -1455 kJ/ mol rxn

  21. Thermochemical Equations Example 15-4: When aluminum metal is exposed to atmospheric oxygen, it is oxidized to form aluminum oxide. How much heat is released by the complete oxidation of 24.2 g of aluminum at 25oC and 1atm? 4 Al(s) + 3 O2(g) 2 Al2O3(s) H = -3352 kJ/mol rxn Al 的原子量為 27 Al 的 mole數 = 24.2g/ 27 = 0.896 mole Hrxn = - 3352 kJ/mol rxn 表示1mole的反應所釋出的熱量 而1mole的反應需4mole的Al 0.896 mol x kJ x= -751kJ = 4 mol Al -3352 kJ

  22. Standard States and Standard Enthalpy Changes 標準狀態及標準焓變 • Thermochemical standard state conditions • The thermochemical standard T = 298.15 K.(25oC) • The thermochemical standard P = 1.00 atm. • 例如標準狀態下(1大氣壓, 25oC)，H2(g), Hg(l), Na(s), C2H5OH(l), CaCO3(s), CO2(g) • Thermochemicalstandard states of matter • For pure substances in their liquid or solid phase the standard state is the pure liquid or solid. • For gases the standard state is the gas at 1.00 atm of pressure. • For gaseous mixtures the partial pressure must be 1.00 atm. • For aqueous solutions the standard state is 1.00 M concentration. • Standard enthalpy change, HorxnAll at standard states (只看反應前、反應後，不管反應過程狀態是否改變)

  23. Standard Molar Enthalpies of Formation,標準莫耳生成焓,Hfo • The standard molar enthalpy of formation is defined as the enthalpy for the reaction in which one mole of a substance is formed from its constituent elements.(在標準狀態下由各元素形成一莫耳物質所需的熱量及稱之) • The symbol for standard molar enthalpy of formation (簡稱heat of formation) is Hfo. • The standard molar enthalpy of formation for MgCl2 is: Mg(s) + Cl2(g) MgCl2(s) + 64.18 kJ HofMgCl2 = -641.8 kJ/mol MgCl2(s) • The standard molar enthalpy of formation for HBr is: H2(g) + Br2(l) 2HBr(g)Horxn = -72.8 kJ/mol rxn  ½ H2(g) + ½ Br2(l) HBr(g)Horxn = -36.4 kJ/mol HofHBr = -36.4 kJ/mol HBr(g)

  24. Standard Molar Enthalpies of Formation 標準莫耳生成焓, Hfo • Standard molar enthalpies of formation have been determined for many substances and are tabulated in Table 15-1 (Table6.2) and Appendix 4 in the text. • Standard molar enthalpies of elements in their most stable forms at 298.15 K and 1.000 atm are zero. • Example 15-5: The standard molar enthalpy of formation for phosphoric acid is -1281 kJ/mol. Write the equation for the reaction for which Horxn= -1281 kJ. P in standard state is P4 Phosphoric acid in standard state is H3PO4(s) 3/2 H2(g) + 2 O2(g) + 1/4 P4(s) H3PO4(s) + 1281 kJ HofH3PO4 = -1281 kJ/mol

  25. Standard Molar Enthalpies of Formation, Hfo Example 15-6: Calculate the enthalpy change for the reaction of one mole of H2(g) with one mole of F2(g) to form two moles of HF(g) at 25oC and one atmosphere. H2(g) + F2(g) 2HF(g) • Std. state Std. state Std. state for this rxn HoHF = 2 Hof 查表得知:Hof =-271 kJ/mol 所以 HoHF=(2mol)(-271kJ/mol) = -542 kJ

  26. Standard Molar Enthalpies of Formation, Hfo • Example 15-6: Calculate the enthalpy change for the reaction in which 15.0 g of aluminum reacts with oxygen to form Al2O3 at 25oC and one atmosphere. HofAl2O3 = -1676 kJ/mol Al2O3 2 Al(s) + 3/2 O2(g) Al2O3(s) Al mole=15.0g/ 27 = 0.556 mole 0.556 mol x kJ x= -466kJ = 2 mol Al -1676 kJ

  27. Hess’s Law • Hess’s Law of Heat Summation states that the enthalpy change for a reaction is the same whether it occurs by one step or by any (hypothetical) series of steps. • Hess’s Law is true because H is a state function. 赫士定律(Hess’s law,又名反應熱加成性定律):若一反應為二個反應式的代數和時, 其反應熱為此二反應熱的代數和。 Horxn = Hoa + Hob + Hoc + …….

  28. Hess’s Law • If we know the following Ho’s [1] 4 FeO(s) + O2(g) 2 Fe2O3(s)Ho=-560 kJ [2] 2 Fe(s) + O2(g) 2 FeO(g)Ho=-544 kJ [3] 4 Fe(s) + 3 O2(g) 2 Fe2O3(s)Ho=-1648 kJ • For example, we can calculate the Ho for reaction [1] by properly adding (or subtracting) the Ho’s for reactions [2] and [3]. • Notice that reaction [1] has FeO and O2 as reactants and Fe2O3 as a product. (2x[-2]) + [3] = [1] • Arrange reactions [2] and [3] so that they also have FeO and O2 as reactants and Fe2O3 as a product. • Each reaction can be doubled, tripled, or multiplied by half, etc. • The Ho values are also doubled, tripled, etc. • If a reaction is reversed the sign of the Ho is changed. • 2x [-2] 2 (2 FeO(g)2 Fe(s) + O2(g))Ho=2x (-(-544 kJ) ) • [3] 4 Fe(s) + 3 O2(g) 2 Fe2O3(s)Ho=-1648 kJ • [1] 4 FeO(s) + O2(g) 2 Fe2O3(s)Ho=-560 kJ

  29. Hess’s Law [1] 2N2(g) + O2(g) 2 N2O(g)Ho= 164.1 kJ [2] N2(g) + O2(g) 2 NO(g)Ho= 180.5 kJ [3] N2(g) + 2 O2(g) 2NO2(g)Ho= 66.4 kJ Example 15-7: Given the following equations and Hovalues calculate Ho for the reaction below. N2O (g) + NO2(g) 3 NO(g)Ho= ? ½ [-1] + 3/2 [2] + ½ [-3] • ½ [-1] N2O(g)  N2(g)+ ½ O2(g)Ho= ½ x (-164.1) kJ 3/2 [2] 3/2 N2(g)+ 3/2 O2(g) 3 NO(g)Ho= 3/2 x (180.5) kJ • ½ [-3] NO2(g)  ½ N2(g)+ O2(g)Ho= ½ (-66.4) kJ N2O (g)+ NO2(g) 3 NO(g) Ho= (-82.05) + 270.75 + (-33.2) = 155.5 kJ (endothermic reaction)

  30. Hess’s Law Horxn=  nHofproducts- Hofreactants

  31. Hess’s Law Example 15-8: Calculate the H o298 forthe following reaction. (Appendix 4, A20) C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O (l) Horxn= [3 HofCO2(g)+ 4 HofH2O(l)]-[HofC3H8(g)+ 5 HofO2(g)] Horxn= [3 (-393.5) + 4 (-286)]-[(-104) + 5 (0.0)] Horxn= -2220.5 kJ Exothermic reaction

  32. Hess’s Law Example 15-9: Given the following information, calculate Hfo for H2S(g). 2 H2S(g) + 3 O2(g) 2 SO2(g) + 2 H2O (l)Ho298 = -1124 kJ Hfo? 0 -296.8 -285.8 Horxn= [2 HofSO2(g)+ 2 HofH2O(l)]-[2 HofH2S(g)+ 3 HofO2(g)] -1124 kJ = [2x(-296.8) + 2 (-285.8)]-[2HofH2S(g) + 5 (0.0)] 2 HofH2S(g)= -41.2 kJ HofH2S(g)= -20.6 kJ

  33. Bond Energies 鍵能 • Bond energy is the amount of energy required to break the bond and separate the atoms in the gas phase.(一莫耳氣態原子之化學鍵斷裂時吸收的能量即稱為鍵能) • To break a bond always requires an absorption of energy! A-B(g) + bond energy  A(g) + B(g) H-Cl(g) + 432 kJ/mol  H(g) + Cl(g)

  34. Bond Energies • Bond energies can be calculated from otherHo298values Example 15-10: Calculate the bond energy for hydrogen fluoride, HF. H-F(g) + Bond Energy (BEHF)  H(g) + F (g) (atoms NOT ions) or H-F(g) H(g) + F (g) Ho298= BEHF Ho298= [HofH(g)+ HofF(g)]-[ HofHF(g)] Ho298= [218.0 kJ+ 78.99 kJ]-[ -271 kJ] Ho298= 568.0 kJ Bond energy for HF

  35. Bond Energies Example 15-11: Calculate the average N-H bond energy in ammonia, NH3. NH3(g) N(g) + 3H (g) Ho298= 3 BEN-H Ho298= [HofN(g)+ 3 HofH(g)]-[ HofNH3(g)] Ho298= [472.7 kJ+ 3x(218) kJ]-[ -46.11 kJ] Ho298= 1173 kJ 3 N-H Bond energy for NH3 Average BEN-H = 1173/3 = 391 kJ/mol N-H bonds

  36. Bond Energies Ho298=  BEreactants-  BEproducts • In gas phase reactions Ho values may be related to bond energies of all species in the reaction.

  37. Bond Energies Example 15-12: Use the bond energies listed in Tables 15-2 and 15-3 to estimate the heat of reaction at 25oC for the reaction below. CH4(g) + 2 O2(g) CO2(g) + 2 H2O (g) Ho298= [4 BEC-H+ 2 BEO=O]-[2 BEC=O+4 BEO-H] Ho298= [4x(414 kJ)+ 2x(498 kJ)]-[2x(741 kJ) + 4x(464 kJ)] Ho298= -686 kJ

  38. Changes in Internal Energy, E (內能的改變) • The internal energy, E, is all of the energy contained within a substance. • This function includes all forms of energy such as kinetic, potential, gravitational, electromagnetic, etc.(包括所有形式的能量如動能、位能、重力、電磁能等) • The First Law of Thermodynamics states that the change in internal energy, E, is determined by the heat flow, q, and the work, w.

  39. Changes in Internal Energy, E E =Eproducts– Ereactants E = q+w q > 0 if the heat is absorbed by the system q < 0 if the heat is absorbed by the surroundings w > 0 if the surrounding do work on the system w < 0 if the system does work on the surroundings

  40. Changes in Internal Energy, E • E is negative when energy is released by a system undergoing a chemical or physical change. • Energy can be written as a product of the process. • C5H12(l) + 8 O2(g) 5 CO2(g) + 6 H2O(l) + 3.516x103 kJ • E = - 3.516x103 kJ • E ispositivewhen energy is absorbed by a system undergoing a chemical or physical change. • Energy can be written as a reactant of the process. • 5 CO2(g) + 6 H2O(l) + 3.516x103 kJ C5H12(l) + 8 O2(g) • E = + 3.516x103 kJ

  41. Changes in Internal Energy, E Example 15-13: If 1200 joules of heat are added to a system in energy state E1, and the system does 800 joules of work on the surroundings, what is the : 1. energy change for the system, Esys E =Eproducts– Ereactants = q+w E =1200 J + (-800J) E sys=+400 J • 2. energy change of the surroundings, Esurr E surr=-400 J • 3. energy of the system in the new state, E2 E surr=E2 – E1 E2 = E1 + Esys = E1 +400J

  42. Changes in Internal Energy, E • In most chemical and physical changes, the only kind of work is pressure-volume work. • Pressure is force per unit area. force F P = = • Volume is distance cubed. area d2 d2 V = d3 • PV is a work term, i.e., the same units are used for energy and work. F PV= x d3= Fxd (所做的功)

  43. Changes in Internal Energy, E A system that absorbs heat and does work.

  44. Using the ideal gas law PV = nRT, we can look at volume changes of ideal gases at constant T and P due to changes in the number of moles of gas present, ngas. ngas = (number of moles of gaseous products) - (number of moles of gaseous reactants) Changes in Internal Energy, E PV = nRT P(V)=(ngas)RT • Work is defined as a force acting through a specified distance. w= F x d = -PV = -(ngas)RT  w = -(ngas)RT at constant T and R

  45. Changes in Internal Energy, E • Consequently, there are three possibilities for volume changes: WhenThenExamples • V2 = V1 PV = 0 ngas = 0 • CO(g) + H2O(g) H2(g) + CO2(g) 2 mol gas 2 mol gas • Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g) 2. V2 > V1 PV > 0 ngas > 0 1 mol gas 3 mol gas • N2(g) + 3 H2(g) 2 NH3(g) 3. V2 < V1 PV < 0 ngas < 0 2 mol gas 4 mol gas

  46. Changes in Internal Energy, E • Consider the following gas phase reaction at constant pressure at 200oC. • 2 NO(g) + O2(g) 2 NO2(g) • V2 < V1 thus V<0 and PV<0 • w= -PV >0 • Works is done on system by surroundings • Consider the following gas phase reaction at constant pressure at 1000oC. PCl5(g)  PCl3(g) + Cl2(g) 2 mol gas 1 mol gas 3 mol gas 2 mol gas • V2 > V1 thus V>0 and PV>0 • w= -PV <0 • Works is done by the system on the surroundings

  47. Relationship of H and E • The total amount of heat energy that a system can provide to its surroundings at constant temperature and pressure is given by H= E + PV • which is the relationship betweenH and E. H = change in enthalpy of system E = change in internal energy of system PV = work done by system

  48. Relationship of H and E • At the start of Chapter 15 we defined H = qP. • Here we define H = E + PV. • Are these two definitions compatible? • Remember E = q + w. • We have also defined w = -PV . • Thus E = q + w = q -PV • Consequently, H = q- PV + PV = q • At constant pressureH = qP.

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