H 2 (g) + ½ O 2 (g) H 2 O(l) H o f [NO(g)] = 90.2 kj/mol; H o f [H 2 O(g)] = kj/mol H o f [NH 3 (g)] = kj/mol; H o f [O 2 (g)] =?

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1 Chapter 16 Thermodynamics GCC CHM152 Thermodynamics You are responsible for Thermo concepts from CHM 151. You may want to review Chapter 8, specifically 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!) Thermochemistry Review 1 st Law of Thermodynamics: Energy is conserved. Energy can be converted from one form to another but it cannot be created or destroyed. If a system gives off heat, the surroundings must absorb it (and vice versa). Heat flow (enthalpy, H) is defined with reference to the system System absorbs heat, H > 0, endothermic System gives off heat, H < 0, exothermic Predicting sign for H o rxn Predict whether the following processes are endothermic (need heat) or exothermic (release heat): Decomposition Acid-Base Neutralization Combustion Melting Freezing Boiling Calculating H o rxn The o symbol refers to the standard state, 1.00 atm pressure, 25.0 o C, 1.00 M for solutions H o f = Standard Molar Enthalpy of Formation The enthalpy change for forming 1 mole of a compound from its elements in their standard states. Appendix B lists values for H o f, G o f and S o H o f = 0 for an element in its most stable form Know most stable forms for the elements! H o f units: kj/mol H o rxn = n H o f (products) - n H o f (reactants) n = moles (coefficient in balanced reaction) Heat of formation rxn & H o rxn Write heat of formation reaction for H 2 O(l). Product = 1 mol H 2 O(l) Reactants = elements in most stable form H 2 (g) + ½ O 2 (g) H 2 O(l) Calculate H rxn for the following reaction. 4NH 3 (g) + 5O 2 (g) 4NO(g) + 6H 2 O(g) H o f [NO(g)] = 90.2 kj/mol; H o f [H 2 O(g)] = kj/mol H o f [NH 3 (g)] = kj/mol; H o f [O 2 (g)] =? Answer: H rxn = kj 1

2 16.1 Spontaneous Processes A spontaneous process proceeds on its own without any external influence. The reverse of a spontaneous process is always nonspontaneous it does not occur naturally. Consider the expansion of a gas into a vacuum. This happens spontaneously. The reverse process does not! Spontaneous Chemical Reactions Spontaneous processes hot object cools Ice melts at T > 0 C iron rusts Cs reacts with H 2 O Note that spontaneous does not mean fast! 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 Nature tends toward state of lower energy. Many spontaneous processes are exothermic (e.g. combustion reactions have H < 0), but some spontaneous processes are endothermic (e.g. ice melting has H > 0). Nature tends to become more disordered or random. Entropy of a system typically increases, S > 0 (+ S = increase in disorder) Example: an office naturally becomes more messy over time. Entropy (S) Entropy is a measure of disorder or randomness. Units of Entropy (S): J/K mol Entropy of element in most stable form is not zero! 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

3 Which has the greater entropy? Low temperature or high temperature? Consider gases at different temperatures High T: more motion, more possible arrangements Figure 16.7 Which has the greater entropy? Solute and solvent or solution? Solution has more arrangements, more motion Section 11.2 Which has the greater entropy? Reactants or products formed from them? N 2 (g) + O 2 (g) 2NO(g) Group Qz #17 Which has greater entropy? The same number of molecules are present & all are gases; therefore no difference in entropy. Worked example ; Problem 16.2, 16.3 Entropy Summary Temperature changes Increase: S > 0 (more energy, more positions, more possible arrangements) Phase changes Boiling, melting: S > 0 Creating more moles S > 0 Salts dissolving S > 0 (usually) Predict Entropy change Predict whether entropy increases ( S > 0) or decreases ( S < 0) for the following processes. Explain why! 2H 2 (g) + O 2 (g) 2H 2 O(l) KBr (s) K + (aq) + Br - (aq) 4 Al(s) + 3 O 2 (g) 2Al 2 O 3 (s) I 2 (s) I 2 (g) PCl 5 (s) PCl 3 (l) + Cl 2 (g) 3

4 Entropy and Temperature Entropy and Temperature 3 rd 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 25 o C (J/mol K) Calculate like H o rxn (products - reactants) Standard Molar Entropies Notice which phases have lowest values? Highest? Entropy Changes Calculate S o rxn for: 2Na(s) + Cl 2 (g) 2NaCl(s) Na(s) = J/mol K, Cl 2 (g) = J/mol K, NaCl(s) = J/mol K Worked example 16.3, Problem 16.5 Entropy, 2 nd Law of Thermo. Entropy 2 nd Law of Thermodynamics: in any spontaneous process, the total entropy of a system and its surroundings always increases. The system is the chemical reaction itself. S tot is positive for a spontaneous process; S tot is 0 at equilibrium S tot = S system + S surroundings S total = S system + S surroundings S total > 0, the reaction is spontaneous S total < 0, the reaction is nonspontaneous S total = 0, the reaction is at equilibrium All reactions proceed spontaneously in a direction that increases the entropy of the system plus the surroundings. 4

5 2 nd Law of Thermodynamics We know how to calculate S system (previous calculations); now calculate S surr Figure 16.9 Calculating S surr Exothermic: heat leaves system, enters surroundings, surroundings have more energy which gives them more disorder S surr > 0 when exothermic: H < 0 Endothermic: heat enters system, leaves surroundings, surroundings have less energy which gives them more order S surr < 0 when endothermic: H > 0 S surr - H rxn and S surr 1/T Therefore: S surr = - H rxn /T S Tot = S sys - H rxn /T Free Energy Change in Gibbs Free Energy Section 8.14: Gibbs Free Energy Change ( G) Definition: maximum amount of energy available to do work on the surroundings Takes into account enthalpy and entropy to predict spontaneity of a reaction. G = H - T S ( G in units of kj/mol) H is enthalpy (kj/mol), T is temperature (K), S is change in entropy (J/mol K) In any spontaneous process (constant T and P), the free energy of the system always decreases! Gibbs free energy change or free energy change = G E or H: some energy is used to do work within the system (rearrange particles, new bonds,...) and the rest is available to do work on the surroundings ( G) G < 0 Spontaneous process G > 0 Nonspontaneous process G = 0 Process is at equilibrium Free Energy Free Energy Notice that the T S term is temperature dependent. Temperature plays a part in predicting spontaneity. Endothermic processes are spontaneous at higher temps (T S > H) Exothermic processes are spontaneous at lower temps (T S < H) 5

6 Free Energy Using G = H T S, we can predict the sign of G from the sign of H and S. If H is negative and S is positive, G will always be negative. If both H and S are negative, G will be at low T. If H is positive and S is negative, G will always be positive. If both H and S are positive, G will be at high T. Entropy of Phase Changes Phase changes occur at equilibrium ( G = 0) H 2 O(s) H 2 O(l) H = 6.01 kj/mol; S = 22.0 J/K mol Solid and liquid are in equilibrium G = H - T S = 0, solving for temperature tells us the melting point! T = H/ S = 273 K ; this is the melting point of water Melting Ice: 263K (-10 o C): G = kj/mol (nonspontaneous T< 0 C) 273K: G = 0.00 kj/mol (equilibrium; T = 0 C) 283K: G = kj/mol (spontaneous, T > 0 C) Calculating Entropy of Vaporization Phase changes occur at eq: S = H /T The boiling point of water is 100 o C and the enthalpy change for the conversion of water to steam is H vap = kj/mol. What is the entropy change for vaporization, S vap, in J/(K mol)? S vap = H vap / T kj 1000J mol S vap = 1kJ = 109 J 373K K mol Group Quiz 18 Calculate the boiling point of ethanol if entropy of vaporization is J/K and H vap = 39.3 kj Watch units! Find T for spontaneous rxn Calculate G given H= -227 kj, S = -309 J/K, T = 1450 K. Is this process spontaneous at this temperature? If not, calculate the temperature (in o C) at which this reaction becomes spontaneous. Standard Free Energy Changes The free energy of a substance depends on temperature, pressure, and physical states (like enthalpy and entropy). We must look at standard-state conditions: Solids, liquids, gases: pure form at 1 atm Solutions at 1 M concentration, gases at 1 atm Room temperature: 25 o C (298 K) Standard free energy change ( G o ) is free energy change with reactants and products in their standard states. 6

7 Standard Free Energy Changes Can get G o from G o = H o - T S o Use G o to predict spontaneity in the standard state Can also get values of G o from free energies of formation: G o f (formation from the elements) G o f = 0 for an element in its stable form G o rxn = n G o fproducts - n G o freactants What are the characteristics of a formation reaction? 1 mole of product formed Reactants are elements in standard state. G o f Which one of these reactions corresponds to G o f of H 2 O(g)? 2H 2 (g) + O 2 (g) 2H 2 O(g) H 2 (g) + ½O 2 (g) H 2 O(g) H 2 (g) + ½ O 2 (s) H 2 O(g) G o = n G o f products - n G o f reactants Values of G o f, H o f, S o are listed for in App B G o G o calc f in kj/mol CaO(s) Ca(OH) 2 (s) H 2 O(g) H 2 O(l) Calculate G o for these reactions and predict whether they will be spontaneous. 2H 2 (g) + O 2 (g) 2H 2 O(g) CaO(s) + H 2 O(l) Ca(OH) 2 (s) Relating G o to G and K G is not standard. It changes as a chemical reaction proceeds, as concentrations and/or temperatures change. G o IS standard and does NOT change during a reaction. G = G o + RT lnq Q (from Ch not necessarily at equil.) RT lnq is correction for non-standard conditions; R = J/mol K Example 16.7, Problem G o and Equilibrium Figure G < 0 (- G ), product-favored Free Energy Curve for + G G > 0, Reactant favored Eq mix 7

8 G o and K G o and Equilibrium G = G o + RT ln Q At equilibrium, G = 0 and Q = K Thus, G o = - RT ln K G o > 0 when K << 1 (lies toward reactants) G o < 0 when K >>1 (lies toward products) G o = 0 when K = 1 (roughly equal amounts of reactants and products) See Table 16.4 Calculating K G o is kj/mol for the formation of methanol. C(s) + ½ O 2 (g) + 2H 2 (g) CH 3 OH(g) Calculate the equilibrium constant, K, at 25 C for this reaction. Group Quiz #19 Calculate K (the equilibrium constant) for the following unbalanced reaction at 25.0 o C. Is this reaction reactant- or product-favored? H 2 (g) + Br 2 (l) HBr(g) G o f (HBr) = kj/mol Calculations Practice At 25 C, K a for acetic acid is a) Predict the sign of G o for CH 3 COOH(aq) + H 2 O(l) H 3 O + (aq) + CH 3 COO - (aq). b) Calculate G o at 25 C. c) Calculate G at 25 C for the acetic acid equilibrium reaction, when [H 3 O + ] = M, [CH 3 COO - ] = M and [CH 3 COOH] = 0.10 M. (Use G o from part b.) Calculations Practice G for the reaction H 2 (g) + I 2 (g) 2 HI(g) is 2.60 kj/mol at 25 C. In one experiment, the initial pressures are P H2 = 4.3 atm, P I2 = 0.34 atm, and P HI = 0.23 atm. Calculate G and predict the direction that this reaction will proceed. 8