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Chapter 12 Thermodynamics

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Chapter 12 Thermodynamics Chapter 16 Creative Commons License Thermodynamics Images and tables in this file have been used from the following GCC CHM152 sources: OpenStax: Creative Commons Attribution License 4.0. ChemWiki (CC BY-NC-SA 3.0): Unless otherwise noted, the StatWiki is licensed under a Creative Commons Attribution- Noncommercial-Share Alike 3.0 United States License. Permissions beyond the scope of this license may be available at [email protected]. Principles of General Chemistry (CC BY-NC-SA 3.0): http://2012books.lardbucket.org/pdfs/principles-of-general- chemistry-v1.0.pdf Chemistry: OpenStax 2 Thermodynamics Thermochemistry Review • You are responsible for Thermo concepts from CHM 1st Law of Thermodynamics: Energy is conserved. 151. You may want to review Chapter 5, specifically Energy can be converted from one form to another sections 1-3. but it cannot be created or destroyed. • Thermodynamics: Study of energy changes in If a system gives off heat, the surroundings must chemical reactions and physical processes. absorb it (and vice versa). • Thermodynamics allows us to determine IF a reaction Heat flow (enthalpy, H) is defined with will occur. reference to the system • It also tells us the direction and extent of a reaction. • System absorbs heat, H > 0, endothermic • It does not tell us how fast a reaction occurs (kinetics • System gives off heat, H < 0, exothermic tell us this!) o o Predicting sign for H rxn Calculating H rxn Predict whether the following are endothermic The o symbol refers to the standard state, 1.00 atm (need heat) or exothermic (release heat): pressure, 25.0oC, 1.00 M for solutions Decomposition + Ho = Standard Molar Enthalpy of Formation Acid-Base Neutralization - f The enthalpy change when 1 mole of a compound is Combustion - formed from its elements in their standard states. Melting o o o + Appendix G lists values for H f, G f and S Ho = 0 for an element in its most stable form Freezing - f Know most stable forms for the elements! Boiling + o H f units: kJ/mol Remember breaking bonds requires energy = endo Ho = n Ho (products) - n Ho (reactants) but making new bonds releases energy = exo. rxn f f • n = moles (coefficient in balanced reaction) 1 1 o Heat of formation rxn & H rxn Spontaneous Processes Write heat of formation reaction for H2O(l). A spontaneous process proceeds on its own • Product = 1 mol H2O(l) without any external influence. The reverse • Reactants = elements in most stable form of a spontaneous process is always nonspontaneous –it does not occur naturally. H2(g) + ½ O2(g) H2O(l) Consider the expansion of a gas into a Calculate Hrxn for the following reaction. vacuum. This happens spontaneously. The reverse process does not! 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) o o H f [NO(g)] = 90.2 kJ/mol; H f [H2O(g)] = -241.8 kJ/mol o o H f [NH3(g)] = -46.1 kJ/mol; H f [O2(g)] = ? Answer: Hrxn = -905.6 kJ Spontaneous Chemical Reactions Enthalpy, Entropy and Spontaneity Spontaneous processes Nature tends toward state of lower energy. hot object cools Many spontaneous processes are exothermic (e.g. Ice melts at T > 0 C combustion reactions have H < 0), but some iron rusts spontaneous processes are endothermic (e.g. ice melting has H > 0). sodium reacts with H2O • Note that spontaneous does not mean fast! Nature tends to become more disordered Rusting of iron is spontaneous but occurs slowly! or random. Entropy of a system typically increases, S > 0 (+S = • What makes a process spontaneous? increase in disorder) Spontaneity is determined by both the enthalpy and Example: an office naturally becomes more messy entropy of a reaction. over time. Entropy (S) Entropy of Molecules Entropy is a measure of disorder or randomness. • Units of Entropy: J K mol • S: standard molar entropy of an element in its most stable form is not zero! The atoms, molecules, or ions that compose a chemical system can undergo several types of molecular motion, including translation, rotation, and vibration. The greater the molecular motion of a system, the greater the number of possible microstates and the higher the entropy. Image from Principles of General Chemistry http://2012books.lardbucket.org/books/principles-of-general-chemistry -v1.0/s19-chemical-equilibrium.html, CC-BY-NC-SA 3.0 license 12 2 2 Which has the greater entropy? Entropy and temperature Solid, liquid or gas? (ice, water, or steam) • Consider gases at different temperatures • High T: More motion & higher kinetic energy so there are more possible arrangements. S (s) < S (l) < S (g) Entropy and solubility Entropy Summary Dissolving a solute usually increases entropy since a solution • Temperature changes has more possible arrangements and is more dispersed but some ionic compounds have a negative entropy change upon Increase: S > 0 (more energy, more dissolution. positions, more possible arrangements) Ionic compounds: entropy can decrease or increase • Phase changes NaCl (s): Ssoln = +43.4 J/K AlCl3 (s): Ssoln = -253.2 J/K Boiling, melting: S > 0 • Creating more moles S > 0 • Salts dissolving: S +/- but usually • Also entropy usually as complexity of molecule increases. Predict Entropy change Answers • 2H (g) + O (g) 2H O(l): Predict whether entropy increases (S > 0) 2 2 2 S < 0; fewer particles, phase change or decreases (S < 0) for the following • KBr(s) K+(aq) + Br-(aq): processes. Explain why! S > 0; more particles/phase change • 2H2(g) + O2(g) 2H2O(l) • 4 Al(s) + 3 O2(g) 2Al2O3(s): • KBr (s) K+(aq) + Br-(aq) S < 0; fewer particles/phase change • I2(s) I2(g): • 4 Al(s) + 3 O2(g) 2Al2O3(s) S > 0; phase change • I (s) I (g) 2 2 • PCl (s) PCl (l) + Cl (g): PCl (s) PCl (l) + Cl (g) 5 3 2 • 5 3 2 S > 0; more particles/phase changes 3 3 o Standard Molar Entropy Standard Entropy Values (S ) Which phase has the lowest entropy? Highest? rd 3 Law of Thermodynamics: The entropy of a Substance So (J mol-1 K-1) Substance So (J mol-1 K-1) perfectly ordered crystalline substance at 0 K is C (s, diamond) 2.38 H (g) 114.6 zero. C (s, graphite) 5.740 H2 (g) 130.57 Al (s) 28.3 CH4 (g) 186.3 • This allows us to calculate entropy (S) and Al2O3 (s) 50.92 HCl (g) 186.8 changes in entropy (S); unlike enthalpy where BaSO4 (s) 132.2 H2O (g) 188.8 we can only measure changes! CO (g) 197.7 Standard Molar Entropies (Table 16.2) - entropy of H2O (l) 70.0 CO2 (g) 213.8 1 mole of a pure substance at 1 atm pressure and Hg (l) 75.9 C H (g) 219.3 25oC (J/mol•K) 2 4 H2O2 (l) 109.6 C2H6 (g) 229.5 o S rxn = nS(products) - nS(reactants) Br2 (l) 152.23 CH3OH (g) 239.9 CCl (l) 214.4 CCl (g) 309.7 4 4 Note: The standard entropy for a pure element in its most stable o form is NOT zero (like for H f). 20 Entropy Change for a Reaction 2nd Law of Thermodynamics • 2nd Law of Thermodynamics: in any Calculate So for 2Na(s) + Cl (g) 2NaCl(s) rxn 2 spontaneous process, the total entropy of a S values: Na(s) = 51.05 J/mol•K, Cl2(g) = system and its surroundings always 223.0 J/mol•K, NaCl(s) = 72.38 J/mol•K increases. (system = reaction) o • Stotal = Ssystem + Ssurroundings • S rxn = -180.3 J/K • Stotal > 0, the reaction is spontaneous • S < 0, the reaction is nonspontaneous • Entropy change is < 0. There is more order due total • Stotal = 0, the reaction is at equilibrium to the phase change. • All reactions proceed spontaneously in a direction that increases the entropy of the system plus the surroundings. 2nd Law of Thermodynamics Calculating Ssurr • Ssys is the entropy change for the reaction that we just calculated. • Exothermic: heat flows from system to surroundings, surroundings have more energy = • Ssurr is related to the enthalpy of the more disorder. (Hotter particles have more motion) system; if Heat flows into the Ssurr > 0 when exothermic: H < 0 surroundings then their entropy increases • Endothermic: heat flows into system from due to increased motion. surroundings, surroundings have less energy which gives them more order. Ssurr < 0 when endothermic: H > 0 • Therefore: S = -H /T Entropy surr sys Entropy Heat Heat • STot = Ssys -Hsys/T 4 4 Gibbs free energy (G) or free energy Predicting sign for Free Energy Change • G < 0 Spontaneous process G is the maximum amount of energy available to do work • G > 0 Nonspontaneous process on the surroundings. • G = 0 Process is at equilibrium Takes into account enthalpy and entropy to predict spontaneity of a reaction. G = H - TS • G = H - TS (G units: kJ/mol) T must be in Kelvin! Watch units! The actual amount of work (wmax) obtained is always less than the maximum available because of energy lost in carrying out a process (given off as heat, light, sound, etc. energy). • In any spontaneous process (constant T and P), the free energy of the system always decreases! Free Energy sign is T dependent when Calculating Entropy of Vaporization enthalpy and entropy have same sign! • Phase changes occur at eq: S = H /T • Notice that the TS term is temperature • The boiling point of water is 100 oC and the dependent. Temperature plays a part in enthalpy change for the conversion of water to predicting spontaneity. steam is Hvap = 40.67 kJ/mol. What is the entropy change for vaporization, Svap, in Endothermic processes are spontaneous at J/(K·mol)? higher temps (TS > H) • S = H / T Exothermic processes are spontaneous at vap vap kJ 1000J lower temps (TS < H) 40.67 J • Svap = mol 1 kJ = 109 K mol 373K Find T for spontaneous rxn Find T spontaneous rxn • Calculate G given H= -227 kJ, S = -309 • G = +221 kJ; not spontaneous J/K, T = 1450 K.
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