Download Presentation
## Chapter 19: Thermodynamics

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -

**Chapter 19: Thermodynamics**Chemistry: The Molecular Nature of Matter, 6E Jespersen/Brady/Hyslop**Thermodynamics**• Study of energy changes and flow of energy • Answers several fundamental questions: • Is it possible for a given reaction to occur? • Will the reaction occur spontaneously (without outside interference) at a given T? • Will reaction release or absorb heat? • Tells us nothing about time frame of reaction • Kinetics • Two major considerations • Enthalpy changes, H (heats of reaction) • Heat exchange between system and surroundings • Nature's trend to randomness or disorder • Entropy**Review of First Law of Thermodynamics**• Internal energy, E • System's total energy • Sum of KE and PE of all particles in system • or for chemical reaction • E + energy into system • E– energy out of system**Two Methods of Energy Exchange Between System and**Surroundings • Heat qWork w • E = q + w • Conventions of heat and work**First Law of Thermodynamics**• Energy can neither be created nor destroyed • It can only be converted from one form to another • Kinetic Potential • Chemical Electrical • Electrical Mechanical • E is a state function • E is a change in a state function • Path independent • E = q + w**Work in Chemical Systems**• Electrical • Pressure-volume orPV • w = –PV • Where P = external pressure • If PV only work in chemical system, then**Heat at Constant Volume**• Reaction done at constantV • V = 0 • PV = 0, so • E = qV • Entire energy change due to heat absorbed or lost • Rarely done, not too useful**Heat at Constant Pressure**• More common • Reactions open to atmosphere • Constant P • Enthalpy • H = E + PV • Enthalpy change • H = E + PV • Substituting in first law for E gives • H = (q –PV)+ PV = qP • H = qP • Heat of reaction at constant pressure**Converting Between E and H For Chemical Reactions**• H E • Differ byH– E = PV • Only differ significantly when gases formed or consumed • Assume gases are ideal • Since P and T are constant**Converting Between E and H For Chemical Reactions**• When reaction occurs • V caused by n of gas • Not all reactants and products are gases • So redefine as ngas • Where ngas= (ngas)products– (ngas)reactants • Substituting into H = E + PV gives • or**Ex. 1 What is the difference between H and E for the**following reaction at 25 °C? 2 N2O5(g) 4 NO2(g) + O2(g) What is the % difference between H and E? Step 1: Calculate H using data (Table 7.2) Recall H° = (4 mol)(33.8 kJ/mol) + (1 mol)(0.0 kJ/mol) – (2 mol)(11 kJ/mol) H° = 113 kJ**Ex. 1. H and E (cont.)**Step 2: Calculate ngas ngas = (ngas)products–(ngas)reactants ngas = (4 + 1 – 2) mol = 3 mol Step 3: Calculate E using R = 8.31451 J/K mol T = 298 K E = 113 kJ – (3 mol)(8.314 J/K mol)(298 K)(1 kJ/1000 J) E = 113 kJ – 7.43 kJ = 106 kJ**Ex. 1. H and E (cont.)**Step 4: Calculate percent difference Bigger than most, but still small Note: Assumes that volumes of solids and liquids are negligible VsolidVliquid << Vgas**Is Assumption that VsolidVliquid << Vgas Justified?**• Consider CaCO3(s) + 2H+(aq) Ca2+(aq) + H2O+ CO2(g) 37.0 mL 2×18.0 mL 18 mL 18 mL 24.4 L • Volumes assuming each coefficient equal number of moles • So V = Vprod– Vreac = 24.363 L 24.4 L • Yes, assumption is justified Note: If no gases are present reduces to**Learning Check**• Consider the following reaction for picric acid: 8O2(g) + 2C6H2(NO2)3OH(l)→ 3 N2(g) + 12CO2(g) + 6H2O(l) • Calculate Η°, Ε° H° = 12 mol(–393.5 kJ/mol) + 6 mol(–241.83 kJ/mol) + 6 mol(0.00 kJ/mol) – 8 mol(0.00 kJ/mol) – 2 mol(3862.94 kJ/mol) ΔH° = –13,898.9 kJ Ε° = H° – ngasRT = H° – (15 – 8) mol × 298 K × 8.314 × 10–3 kJ/(mol K) Ε° = –13,898.9 kJ – 29.0 kJ =–13,927.9 kJ**Your Turn!**Given the following: 3H2(g) + N2(g) → 2NH3(g) Η°= –46.19 kJ mol–1 Determine E for the reaction. A. –51.14 kJ mol–1 B. –41.23 kJ mol–1 C. –46.19 kJ mol–1 D. –46.60 kJ mol–1 • Η = E + nRTE = Η – nRT • E = –46.19 kJ mol – (–2 mol)(8.314 J K–1mol–1)(298 K)(1 kJ/1000 J) • E = –51.14 kJ mol–1**Enthalpy Changes and Spontaneity**• What are relationships among factors that influence spontaneity? • Spontaneous Change • Occurs by itself • Without outside assistance until finished • e.g. • Water flowing over waterfall • Melting of ice cubes in glass on warm day**Nonspontaneous Change**• Occurs only with outside assistance • Never occurs by itself: • Room gets straightened up • Pile of bricks turns into a brick wall • Decomposition of H2O by electrolysis • Continues only as long as outside assistance occurs: • Person does work to clean up room • Bricklayer layers mortar and bricks • Electric current passed through H2O**Nonspontaneous Change**• Occur only when accompanied by some spontaneous change • You consume food, spontaneous biochemical reactions occur to supply muscle power • to tidy up room or • to build wall • Spontaneous mechanical or chemical change to generate electricity**Direction of Spontaneous Change**• Many reactions which occur spontaneously are exothermic: • Iron rusting • Fuel burning • H and E are negative • Heat given off • Energy leaving system • Thus, H is one factor that influences spontaneity**Direction of Spontaneous Change**• Some endothermicreactions occur spontaneously: • Ice melting • Evaporation of water • Expansion of CO2 gas into vacuum • H and E are positive • Heat absorbed • Energy entering system • Clearly other factors influence spontaneity**Your Turn!**We can expect the combustion of propane to be: A. spontaneous B. non-spontaneous C. neither**Entropy (Symbol S)**• Thermodynamic quantity • Describes number of equivalent ways that energy can be distributed • Quantity that describes randomness of system • Greater statistical probability of particular state means greater the entropy! • Larger S, means more possible ways to distribute energy and that it is a more probable result**Fig. 19.6 - Entropy Distribution**• Low Entropy – (a) • A absorbs E in units of 10 • Few ways to distribute E • ●represent E’s of molecules of A • High Entropy – (b) • More ways to distribute E • B absorbs E in units of 5 • ●represent E’s of molecules of B**Entropy**• If Energy = money • Entropy (S) describes number of different ways of counting it**Examples of Spontaneity**• Spontaneous reactions • Things get rusty spontaneously • Don't get shiny again • Sugar dissolves in coffee • Stir more—it doesn't undissolve • Ice liquid water at RT • Opposite does NOT occur • Fire burns wood, smoke goes up chimney • Can't regenerate wood • Common factor in all of these: • Increase in randomness and disorder of system • Something that brings about randomness more likely to occur than something that brings order**Entropy, S**• State function • Independent of path • S = Change in entropy • For chemical reactions or physical processes**Effect of Volume on Entropy**• For gases, entropy increases as volume increases • Gas separated from vacuum by partition • Partition removed, more ways to distribute energy • Gas expands to achieve more probable particle distribution • More random, higher probability, more positive S**Effect of Temperature on Entropy**• As Tincreases, entropy increases (a) T = 0 K, particles in equilibrium lattice positions and S relatively low (b) T> 0 K, molecules vibrate, Sincreases (c) Tincreases further, more violent vibrations occur and Shigher than in (b)**Effect of Physical State on Entropy**• Crystalline solid very low entropy • Liquid higher entropy,molecules can move freely • More ways to distribute KE among them • Gas highest entropy,particles randomly distributed throughout container • Many, many ways to distribute KE**Entropy Affected by Number of Particles**• Adding particles to system • Increase number of ways energy can be distributed in system • So all other things being equal • Reaction that produces more particles will have positive S**Your Turn!**Which represents an increase in entropy? A. Water vapor condensing to liquid B. Carbon dioxide subliming C. Liquefying helium gas D. Proteins forming from amino acids**Entropy Changes in Chemical Reactions**Reactions without gases • Calculate number of mole molecules n = nproducts – nreactants • If n is positive, entropy increases • More molecules, means more disorder • Usually the side with more molecules, has less complex molecules Reactions involving gases • Calculate change in number of moles of gas, ngas • If ngas is positive , S is positive • ngas is more important than nmolecules**Entropy Changes in Chemical Reactions**Ex. N2(g) + 3H2(g) 2NH3(g) nreactant = 4 nproduct = 2 n = 2 – 4 = –2 Predict Srxn < 0 Lower positional probability Higher positional probability**Ex. 2 Predict Sign of S for Following Reactions**CaCO3(s) + 2H+(aq) Ca2+(aq) + H2O + CO2(g) • ngas = 1 mol – 0 mol = 1 mol • sincengas is positive, S is positive 2 N2O5(g) 4 NO2(g) + O2(g) • ngas = 4 mol + 1 mol – 2 mol = 3 mol • sincengas is positive, S is positive OH–(aq) + H+(aq) H2O • ngas = 0 mol • n = 1 mol – 2 mol = –1 mol • sincengas is negative, S is negative**Predict Sign of S in the Following:**Dry ice → carbon dioxide gas Moisture condenses on a cool window AB → A + B A drop of food coloring added to a glass of water disperses 2Al(s) + 3Br2(l) → 2AlBr3(s) CO2(s) → CO2(g) positive H2O(g) → H2O(l) negative positive positive negative**Your Turn!**Which of the following has the most entropy at standard conditions? • H2O(l) • NaCl(aq) • AlCl3(s) • Can’t tell from the information**Your Turn!**Which reaction would have a negative entropy? A. Ag+(aq) + Cl–(aq) → AgCl(s) B. N2O4(g) → 2NO2(g) C. C8H18(l) + 25/2 O2(g) → 8CO2(g) + 9H2O(g) D. CaCO3(s) → CaO(s) + CO2(g)**Both Entropy and Enthalpy Affect Reaction Spontaneity**• Sometimes they work together • Building collapses • PE decreases His negative • Stones disordered S is positive • Sometimes work against each other • Ice melting (ice/water mix) • Endothermic • H is positive nonspontaneous • Increase in disorder of molecules • S is positive spontaneous**Which Prevails?**• Hard to tell—depends on temperature! • At 25 °C, ice melts • At –25 °C, water freezes • So three factors affect spontaneity: • H • S • T • Next few slides will develop the relationship between H, S,and T that defines a spontaneous process**Second Law of Thermodynamics**• When a spontaneous event occurs, total entropy of universe increases • (Stotal > 0) • In a spontaneous process, Ssystem can decrease as long as total entropy of universe increases • Stotal = Ssystem + Ssurroundings • It can be shown that**Spontaneous Reactions (cont.)**Law of Conservation of Energy • Says q lost by system must be gained by surroundings • qsurroundings= –qsystem • If system at constant P, then • qsystem = H • So • qsurroundings = –Hsystem • and**Thus Entropy for Entire Universe is**Multiplying both sides by T we get TStotal = TSsystem – Hsystem or TStotal = – (Hsystem– TSsystem) • For reaction to be spontaneous • TStotal > 0 (entropy must increase) So, (Hsystem– TSsystem) < 0 must be negative for reaction to be spontaneous**Gibbs Free Energy**• Would like one quantity that includes all three factors that affect spontaneity of a reaction • Define new state function, G • Gibbs Free Energy • Maximum energy in reaction that is "free" or available to do useful work GH – TS • At constant P and T, changes in free energy G = H – TS**G = H – TS**G state function Made up of T, H and S = state functions Has units of energy Extensive property G = Gfinal– Ginitial Gibbs Free Energy**Criteria for Spontaneity?**• At constant P and T, process spontaneous only if it is accompanied by decrease in free energy of system**Summary**• When H and S have same sign, T determines whether spontaneous or nonspontaneous • Temperature-controlledreactions are spontaneous at one temperature and not at another**Your Turn!**At what temperature (K) will a reaction become nonspontaneous when H = –50.2 kJ mol–1 and S = +20.5 J K–1 mol–1? A. 298 K B. 1200 K C. 2448 K D. The reaction cannot become non-spontaneous at any temperature**Third Law of Thermodynamics**• At absolute zero (0 K) • Entropy of perfectly ordered, pure crystalline substance is zero • S = 0 at T = 0 K • Since S = 0 at T = 0 K • Define absolute entropy of substance at higher temperatures • Standard entropy, S° • Entropy of 1 mole of substance at 298 K (25 °C) and 1 atm pressure • S° = S for warming substance from 0 K to 298 K (25 °C)**Consequences of Third Law**• All substances have positive entropies as they are more disordered than at 0 K • Heating increases randomness • S° is biggest for gases—most disordered • For elements in their standard states • S° 0 (but Hf° = 0) • Units of S° J/(mol K) Standard Entropy Change • To calculate S° for reaction, do Hess's Law type calculation • Use S° rather than entropies of formation