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Thermodynamics Chapter 16 • You Are Responsible for Thermo Concepts from CHM Thermodynamics 151

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Thermodynamics Chapter 16 • You Are Responsible for Thermo Concepts from CHM Thermodynamics 151 Thermodynamics Chapter 16 • You are responsible for Thermo concepts from CHM Thermodynamics 151. You may want to review Chapter 8, specifically GCC CHM152 sections 2, 5, 6, 7, 9, and 10 (except “work”). • Thermodynamics: Study of energy changes in chemical reactions and physical processes. • Thermodynamics allows us to determine IF a reaction will occur. • It also tells us the direction and extent of a reaction. • It does not tell us how fast a reaction occurs (kinetics tell us this!) o Thermochemistry Review Predicting sign for H rxn • 1st Law of Thermodynamics: Energy is Predict whether the following processes are endothermic (need heat) or exothermic (release conserved. Energy can be converted from one form heat): to another but it cannot be created or destroyed. Decomposition If a system gives off heat, the surroundings must Acid-Base Neutralization absorb it (and vice versa). Combustion Heat flow (enthalpy, H) is defined with reference to the system Melting • System absorbs heat, H > 0, endothermic Freezing • System gives off heat, H < 0, exothermic Boiling o o Calculating H rxn Heat of formation rxn & H rxn • The o symbol refers to the standard state, 1.00 atm • Write heat of formation reaction for H2O(l). pressure, 25.0oC, 1.00 M for solutions • Product = 1 mol H2O(l) o H f = Standard Molar Enthalpy of Formation • Reactants = elements in most stable form • The enthalpy change for forming 1 mole of a H2(g) + ½ O2(g) H2O(l) compound from its elements in their standard states. o o o Calculate H for the following reaction. Appendix B lists values for H f, G f and S • rxn Ho = 0 for an element in its most stable form f 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) Know most stable forms for the elements! o o o H f [NO(g)] = 90.2 kJ/mol; H f [H2O(g)] = -241.8 kJ/mol H f units: kJ/mol o o H f [NH3(g)] = -46.1 kJ/mol; H f [O2(g)] = ? o o o • H rxn = n H f (products) - n H f (reactants) • n = moles (coefficient in balanced reaction) • Answer: Hrxn = -905.6 kJ 1 1 16.1 Spontaneous Processes Spontaneous Chemical Reactions • A spontaneous process proceeds on its own • Spontaneous processes without any external influence. The reverse hot object cools of a spontaneous process is always Ice melts at T > 0 C nonspontaneous –it does not occur naturally. iron rusts Cs reacts with H2O Consider the expansion of a gas into a vacuum. This happens spontaneously. The reverse • Note that spontaneous does not mean fast! process does not! Rusting of iron is spontaneous but occurs slowly! • What makes a process spontaneous? Spontaneity is determined by both the enthalpy and entropy of a reaction. Enthalpy, Entropy and Spontaneity Entropy (S) Nature tends toward state of lower energy. • Entropy is a measure of disorder or Many spontaneous processes are exothermic (e.g. combustion reactions have H < 0), but some randomness. spontaneous processes are endothermic (e.g. ice melting has H > 0). • Units of Entropy (S): J/Kmol Nature tends to become more disordered or random. Entropy of a system typically increases, S > 0 (+S = • Entropy of element in most stable form is increase in disorder) not zero! Example: an office naturally becomes more messy over time. Which has the greater entropy? Enthalpy & Entropy: H & S • Solid, liquid or gas? (ice, water, or steam) S (s) < S (l) < S (g), Section 10.4 Melting: H ?, S ? Boiling: H ?, S ? 2 2 Which has the greater entropy? Which has the greater entropy? • Low temperature or high temperature? • Solute and solvent or solution? • Consider gases at different temperatures • Solution has more arrangements, more motion • High T: more motion, more possible • Section 11.2 arrangements • Figure 16.7 Which has the greater entropy? Group Qz #17 • Reactants or products formed from them? Which has greater entropy? N2(g) + O2(g) 2NO(g) • The same number of molecules are present & all are gases; therefore no difference in entropy. • Worked example 16.1-3; Problem 16.2, 16.3 Entropy Summary Predict Entropy change • Temperature changes Increase: S > 0 (more energy, more • Predict whether entropy increases (S > positions, more possible arrangements) 0) or decreases (S < 0) for the following • Phase changes processes. Explain why! 2H (g) + O (g) 2H O(l) Boiling, melting: S > 0 • 2 2 2 + - • Creating more moles • KBr (s) K (aq) + Br (aq) • 4 Al(s) + 3 O (g) 2Al O (s) S > 0 2 2 3 • I (s) I (g) • Salts dissolving 2 2 • PCl (s) PCl (l) + Cl (g) S > 0 (usually) 5 3 2 3 3 Entropy and Temperature Entropy and Temperature • 3rd Law of Thermodynamics: The entropy of a perfectly ordered crystalline substance at 0 K is zero. • This allows us to calculate entropy (S) and changes in entropy (S); unlike enthalpy where we can only measure change! • Use Standard Molar Entropies (Table 16.1) - entropy of 1 mole of a pure substance at 1 atm pressure and 25oC (J/mol•K) o Calculate like H rxn (products - reactants) Standard Molar Entropies Entropy Changes o • Notice which phases have lowest values? Highest? • Calculate S rxn for: • 2Na(s) + Cl2(g) 2NaCl(s) • Na(s) = 51.05 J/mol•K, Cl2(g) = 223.0 J/mol•K, NaCl(s) = 72.38 J/mol•K • Worked example 16.3, Problem 16.5 Entropy, 2nd Law of Thermo. Entropy nd • 2 Law of Thermodynamics: in any • Stotal = Ssystem + Ssurroundings spontaneous process, the total entropy of a system and its surroundings always • Stotal > 0, the reaction is spontaneous increases. • Stotal < 0, the reaction is nonspontaneous The system is the chemical reaction itself. • Stotal = 0, the reaction is at equilibrium • Stot is positive for a spontaneous process; S is 0 at equilibrium • All reactions proceed spontaneously in a tot direction that increases the entropy of the Stot = Ssystem + Ssurroundings system plus the surroundings. 4 4 nd 2 Law of Thermodynamics Calculating Ssurr • Exothermic: heat leaves system, enters • We know how to calculate Ssystem (previous surroundings, surroundings have more energy which gives them more disorder calculations); now calculate Ssurr • Figure 16.9 Ssurr > 0 when exothermic: H < 0 • Endothermic: heat enters system, leaves surroundings, surroundings have less energy which gives them more order Ssurr < 0 when endothermic: H > 0 Ssurr -Hrxn and Ssurr 1/T • Therefore: Ssurr = -Hrxn/T • STot = Ssys -Hrxn/T Free Energy Change in Gibbs Free Energy • Section 8.14: Gibbs Free Energy Change (G) • Gibbs free energy change or free energy change = G Definition: maximum amount of energy available to do work on the surroundings • E or H: some energy is used to do work Takes into account enthalpy and entropy to predict within the system (rearrange particles, new spontaneity of a reaction. bonds, ...) and the rest is available to do work on the surroundings (G) • G = H - TS (G in units of kJ/mol) • G < 0 Spontaneous process H is enthalpy (kJ/mol), T is temperature (K), S is change in entropy (J/mol·K) • G > 0 Nonspontaneous process • In any spontaneous process (constant T and P), • G = 0 Process is at equilibrium the free energy of the system always decreases! Free Energy Free Energy • Notice that the TS term is temperature dependent. Temperature plays a part in predicting spontaneity. Endothermic processes are spontaneous at higher temps (TS > H) Exothermic processes are spontaneous at lower temps (TS < H) 5 5 Free Energy Entropy of Phase Changes • Phase changes occur at equilibrium (G = 0) • Using ∆G = ∆H – T∆S, we can predict the sign of H2O(s) H2O(l) ∆G from the sign of ∆H and ∆S. H = 6.01 kJ/mol; S = 22.0 J/K•mol Solid and liquid are in equilibrium • If ∆H is negative and ∆S is positive, ∆G will always be G = H - TS = 0, solving for temperature tells us the negative. melting point! • If both ∆H and ∆S are negative, ∆G will be – at low T. • T = H/S = 273 K ; this is the melting point of water • If ∆H is positive and ∆S is negative, ∆G will always be positive. • Melting Ice: • If both ∆H and ∆S are positive, ∆G will be – at high T. • 263K (-10oC): G = +0.22 kJ/mol (nonspontaneous T< 0C) • 273K: G = 0.00 kJ/mol (equilibrium; T = 0 C) • 283K: G = -0.22 kJ/mol (spontaneous, T > 0 C) Calculating Entropy of Vaporization Group Quiz 18 • Phase changes occur at eq: S = H /T • The boiling point of water is 100 oC and the • Calculate the boiling point of ethanol if enthalpy change for the conversion of entropy of vaporization is 111.9 J/K and water to steam is Hvap = 40.67 kJ/mol. Hvap = 39.3 kJ What is the entropy change for • Watch units! vaporization, S , in J/(K·mol)? vap • Svap = Hvap / T kJ 1000J 40.67 • Svap = mol 1 kJ = 109 J 373K K mol Find T for spontaneous rxn Standard Free Energy Changes • Calculate G given H= -227 kJ, • The free energy of a substance depends on S = -309 J/K, T = 1450 K. temperature, pressure, and physical states (like enthalpy and entropy). We must look at standard-state conditions: • Is this process spontaneous at this Solids, liquids, gases: pure form at 1 atm temperature? If not, calculate the Solutions at 1 M concentration, gases at 1 atm o temperature (in oC) at which this reaction Room temperature: 25 C (298 K) becomes spontaneous. • Standard free energy change (Go) is free energy change with reactants and products in their standard states. 6 6 o Standard Free Energy Changes G f • Can get Go from Go = Ho - TSo Use Go to predict spontaneity in the standard state • Which one of these reactions corresponds • Can also get values of Go from free energies o o to G f of H2O(g)? of formation: G f (formation from the elements) o 2H2(g) + O2(g) 2H2O(g) • G f = 0 for an element in its stable form o o o H (g) + ½O (g) H O(g) G rxn = n G fproducts - n G freactants 2 2 2 • What are the characteristics of a formation H2(g) + ½ O2(s) H2O(g) reaction? o o o • G = n G f products - n G f reactants 1 mole of product formed Reactants are elements in standard state.
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