2012 General Chemistry I 1 Chapter 8. THERMODYNAMICS: THE SECOND AND THIRD LAW 2012 General Chemistry I ENTROPY 8.1 Spontaneous Change 8.2 Entropy and Disorder 8.3 Changes in Entropy 8.4 Entropy Changes Accompanying Changes in Physical State 8.5 A Molecular Interpretation of Entropy 8.6 The Equivalence of Statistical and Thermodynamic Entropies 8.7 Standard Molar Entropies 8.8 Standard Reaction Entropies

2012 General Chemistry I 2 ENTROPY (Sections ) - The 1 st law of thermodynamics says if a reaction takes place, then the total energy of the universe remains unchanged. It cannot be used to predict the directionality of a process. - The natural progression of a system and its surroundings (or “the universe”) is from order to disorder, from organized to random. - A new thermodynamic state function is needed to predict directionality and extent of disorder.

2012 General Chemistry I 3  Spontaneous change is a change that has a tendency to occur without needing to be driven by an external influence. - Spontaneous changes need not be fast: e.g. C(diamond)  C(graphite); H 2 (g) + 1/2O 2 (g)  H 2 O(l) 8.1 Spontaneous Change Heat flow Mixing of gases

2012 General Chemistry I Entropy and Disorder - Energy and matter tend to disperse in a disorderly fashion.  Entropy, S is defined as a measure of disorder.  The second law of thermodynamics: unit: J·K -1 - Entropy is a state function; the change in entropy of a system is independent of the path between its initial and final states. The entropy of an isolated system increases in any spontaneous change. At constant temperature

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6 8.3 Changes in Entropy - Thermal disorder: arising from the thermal motion of the molecules - Positional disorder: related to the locations of the molecules   S for a process with changing temperature: → (C V if V is constant, C P if P is constant)

2012 General Chemistry I 7   S for a reversible, isothermal expansion of an ideal gas

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2012 General Chemistry I 11 EXAMPLE 8.5 In an experiment, 1.00 mol Ar(g) was compressed suddenly (and irreversibly) from 5.00 L to 1.00 L by driving in a piston. and in the process its temperature was increased from 20.0 o C to 25.2 o C. What is the change in entropy of the gas? To solve this problem, we consider two reversible stages between initial and final states. Then  S(irrev) =  S(rev 1) +  S(rev 2).

2012 General Chemistry I Entropy Changes Accompanying Changes in Physical State - Phase transition: solid → liquid, T f (fusion or melting point) liquid → solid, T b (boiling point) - At the transition temperature (such as T b ), The temperature remains constant as heat is supplied. The transfer of heat is reversible. The heat supplied is equal to the enthalpy change due to the constant pressure (at 1 atm).  Entropy of vaporization,  S vap q rev =  H vap > 0 in all cases

2012 General Chemistry I 13 Some Standard entropies of vaporization at T b (Table 8.1) - Standard entropy of vaporization,  S vap o :  S vap at 1 bar

2012 General Chemistry I 14  Trouton’s rule:  S vap o = ~85 J·K -1 mol -1 There is approximately the same increase in positional disorder for most liquids when evaporating. - Exceptions; water, methanol, ethanol, ··· due to extensive hydrogen bonding in liquid phases - Standard entropy of fusion,  S fus o > 0 in all cases

2012 General Chemistry I 15  Temperature dependence of  S vap o -To determine the entropy of vaporization of water at 25 o C (not at T b ), we can use an entropy change cycle:

2012 General Chemistry I A Molecular Interpretation of Entropy  The third law of thermodynamics S → 0 as T → 0  Boltzmann formula - statistical entropy: S = k B ln W k B = × J·K -1 = R/N A - W : the number of microstates the number of ways that the atoms or molecules in the sample can be arranged and yet still give rise to the same total energy - When we measure the bulk properties of a system, we are measuring an average taken over the many microstates (ensemble) that the system has occupied during the measurement.

2012 General Chemistry I 17 EXAMPLE 8.7 Calculate the entropy of a tiny solid made up of four diatomic molecules of a compound such as carbon monoxide, CO, at T = 0 when (a) the four molecules have formed a perfectly ordered crystal in which all molecules are aligned with their C atoms on the left and (b) the four molecules lie in random orientations, but parallel. (a) 4 CO molecules perfectly ordered: (b) 4 CO in random, but parallel: (c) 1 mol CO in random, but parallel:

2012 General Chemistry I 18 - Residual entropy at T = 0, arising from positional disorder 4.6 J·K -1 for the entropy of 1 mol CO < 5.76 J·K -1 Nearly random arrangement due to a small electric dipole moment - Solid HCl; S ~ 0 at T = 0 due to the bigger dipole moment leading strict head-to-tail arrangement

2012 General Chemistry I 19 EXAMPLE 8.8 The entropy of 1.00 mol FClO 3 (s) at T = 0 is 10.1 J·K -1. Interpret it. 4 orientations possible nearly random arrangement

2012 General Chemistry I The Equivalence of Statistical and Thermodynamic Entropies Thermodynamic entropy  S = q rev /T behavior of bulk matter Statistical entropy S = k ln W behavior of molecules = - Consider a one-dimensional box, for the statistical entropy, At T = 0, only the lowest energy level occupied → W = 1 and S = 0 At T > 0, W > 1 and S > 0 When the box length is increased at constant T, the molecules are distributed across more levels. → W and S increase.

2012 General Chemistry I 21 - W = constant × V - For N molecules, - The change when a sample expands isothermally from V 1 to V 2 is, = nR ln V2V2 V1V1 - By raising the temperature, The molecules have access to larger number of energy levels → W and S increase.

2012 General Chemistry I 22 - The equations used to calculate changes in the statistical entropy and the thermal entropy lead to the same result. 1. Both are state functions. 3. Both increase in a spontaneous change. 2 × no. of molecules = entropy changes from k ln W to 2k ln W Number of microstates depends only on its current state. 2. Both are extensive (dependent on “extent”) properties. 4. Both increase with temperature. In any irreversible change, the overall disorder increases → no. of microstates increases. When T increases, more microstates become accessible.

2012 General Chemistry I Standard Molar Entropies For heating at constant P, C P and C P /T → 0 as T → 0 Molar entropy, S(T), can be determined from measurement of C p at different temperatures.

2012 General Chemistry I 24  Standard molar entropy, S m o is the molar entropy of the pure substance at 1 bar.

2012 General Chemistry I 25 - Diamond (2.4 J·K -1 ) vs. lead (64.8 J·K -1 ): rigid bonds vs. vibrational energy levels - H 2 (130.7 J·K -1 ) vs. N 2 (191.6 J·K -1 ): the greater the mass, the closer energy levels lightheavy - CaCO 3 (92.9 J·K -1 ) vs. CaO (39.8 J·K -1 ): large, complex vs. smaller, simpler - In general, S m o : gases >> liquids > solids Related to freedom of movement and disordered state  Standard molar entropy, S m o :

2012 General Chemistry I Standard Reaction Entropies  Standard reaction enthalpy,  S o, is the difference between the standard molar entropies of the products and those of the reactants, taking into account their stoichiometric coefficients.

2012 General Chemistry I 27 EXAMPLE 8.9 Calculate  S o for N 2 (g) + 3H 2 (g) → 2NH 3 (g) at 25 o C.

2012 General Chemistry I 28 Chapter 8. THERMODYNAMICS: THE SECOND AND THIRD LAW 2012 General Chemistry I GLOBAL CHANGES IN ENTROPY 8.9The Surroundings 8.10The Overall Change in Entropy 8.11Equilibrium GIBBS-FREE ENERGY 8.12Focusing on the System 8.13Gibbs Free Energy of Reaction 8.14The Gibbs Free Energy and Nonexpansion Work 8.15The Effect of Temperature 8.16Impact on Biology: Gibbs Free Energy Changes in Biological Systems

2012 General Chemistry I 29 GLOBAL CHANGES IN ENTROPY (Sections ) 8.9 The Surroundings - The second law refers to an isolated system (system + surroundings = universe). - Only if the total entropy change is positive will the process be spontaneous.

2012 General Chemistry I 30 Sometimes  S surr can be difficult to compute, but in general it can be obtained from the enthalpy change for the process.

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2012 General Chemistry I The Overall Change in Entropy - To use the entropy to judge the direction of spontaneous change, we must consider the change in the entropy of the system plus the entropy change in the surroundings:

2012 General Chemistry I 33 EXAMPLE 8.11 Is the reaction spontaneous? 2 Mg(s) + O 2 (g) → 2 MgO(s)  S o = -217 J·K -1  H o = kJ The reaction is spontaneous

2012 General Chemistry I 34 - Spontaneous endothermic (  H>0) reactions: There can still be an overall increase in entropy if the disorder of the system increases enough. Summary

2012 General Chemistry I 35 - A process produces maximum work if it takes place reversibly.  Clausius inequality  S = > SS - For an isolated system (universe), q = 0 The entropy of an isolated system cannot decrease. - For two given states of the system,ΔS is a state function (path-independent) but ΔS tot is not. (See EXAMPLE 8.12)

2012 General Chemistry I 36 EXAMPLE 8.12 Calculate  S,  S surr, and  S tot for (a) the isothermal, reversible expansion and (b) the isothermal, free expansion of 1.00 mol of ideal gas molecules from 8.00 L to L at 292 K. Explain any differences between the two paths. (a) Isothermal reversible expansion at 292 K

2012 General Chemistry I 37 (b) Isothermal free expansion 292 K

2012 General Chemistry I Equilibrium  Dynamic equilibrium is one where there is no net tendency to change but microscopic forward and reverse processes occur at matching rates. Thermal equilibrium: no net flow of energy as heat Mechanical equilibrium: no tendency to expand or contract Chemical equilibrium: no net change in composition at thermodynamic equilibrium

2012 General Chemistry I 39 GIBBS FREE ENERGY (Sections ) 8.12 Focusing on the System - at constant T and P  Gibbs free energy, G G = H - TS - Change in Gibbs free energy at constant T and P - The direction of spontaneous change is the direction of decreasing Gibbs free energy.

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2012 General Chemistry I 41  G sys < 0 Spontaneous, irreversible  G sys = 0 Reversible  G sys > 0 Nonspontaneous at constant T and P - the condition for equilibrium,  S tot = 0, and  G = 0 at constant T and P

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2012 General Chemistry I 43 EXAMPLE 8.13 Calculate the change in molar Gibbs free energy,  G m, for the process H 2 O(s) → H 2 O(l) at 1 atm and (a) 10 o C; (b) 0 o C. Decide for each temperature whether melting is spontaneous or not. Treat  H fus and  S fus as independent of temperature. (a) At 10 o C, = kJ·mol -1 < 0; spontaneous melting

2012 General Chemistry I 44 (b) At 0 o C, equilibrium (reversible)

2012 General Chemistry I 45 - G decreases as its T is raised at constant P. G↓ = H – T↑S ; H and S vary little with T, S > 0 - Decreasing rate of G m : vapor >> liquid > solid S m (vapor) >> S m (liquid) > S m (solid) Variation of G with temperature: phase transitions In most cases (opposite), heating leads to melting, then boiling.

2012 General Chemistry I 46 In some cases, at certain pressures, G for the liquid may never be lower than those of the other two phases. The liquid phase is unstable and the phase transition is solid vapor (sublimation).

2012 General Chemistry I Gibbs Free Energy of Reaction  Gibbs free energy of reaction  Standard Gibbs free energy of reaction (standard state: pure form at 1 bar) -  G o is fixed for a given reaction and temperature. -  G depends on the composition of the reaction mixture and so it varies – and might even change sign as the reaction proceeds.

2012 General Chemistry I 48  Standard Gibbs free energy of formation,  G f o, is the standard Gibbs free energy of reaction per mole for the formation of a compound from its elements in their most stable form. - For most stable form of elements,  G f o = 0 E.g.  G f o (I 2, s) = 0;  G f o (I 2, g) > 0 Examples of most stable forms of elements (Table 8.6)

2012 General Chemistry I 49 Some Standard Gibbs free Energies of Formation at 25 o C (kJ mol -1 ) (Table 8.7)

2012 General Chemistry I 50 EXAMPLE 8.14 Calculate the standard Gibbs free energy of formation of HI(g) at 25 o C from its standard molar entropy and standard enthalpy of formation.

2012 General Chemistry I 51 - Thermodynamically stable compound; - Thermodynamically unstable compound; Stable and unstable: thermodynamic tendency to decompose into its elements Labile, nonlabile, and inert: the rate at which a thermodynamic tendency to react is realized but nonlabile or even inert - Another solution for standard Gibbs free energies of reaction: Thermodynamic stability and reactivity

2012 General Chemistry I 52 EXAMPLE 8.15 Calculate the standard Gibbs free energy of the reaction 4 NH 3 (g) + 5 O 2 (g) → 4 NO(g) + 6 H 2 O(g) and decide whether the reaction is spontaneous under standard conditions at 25 o C. The reaction is spontaneous

2012 General Chemistry I The Gibbs Free Energy and Nonexpansion Work - The Gibbs free energy is a measure of the energy free to do nonexpansion work.

2012 General Chemistry I 54  E.g. Bioenergetics of glucose oxidation - The maximum nonexpansion work obtainable from 1 mol of glucose is kJ at 1 bar g of glucose can be used to build 170 (= 2879/17) mole of peptide links. In practice, only about 10 moles of peptide links can be built. - If we know the change in Gibbs free energy of a process taking place at constant T and P, then we immediately know how much nonexpansion work it can do.

2012 General Chemistry I The Effect of Temperature - The values of  H o and  S o do not change much with temperature. - However,  G o does depend much on temperature.  G o =  H o - T  S o 1) For an exothermic reaction (  H o <0) with  S o <0,  G o <0 at low T but it may become >0 at high T.

2012 General Chemistry I 56 3) For an endothermic reaction (  H o >0) with  S o <0,  G o >0 at all T and the reaction is never spontaneous. 2) For an endothermic reaction (  H o >0) with  S o >0,  G o >0 at low T but it may become <0 at high T.

2012 General Chemistry I 57 4) For an exothermic reaction (  H o <0) with  S o >0,  G o <0 at all T and the reaction is always spontaneous. The Gibbs free energy increases with T for reactions with a negative  S o and decreases with T for reactions with a positive  S o.

2012 General Chemistry I 58 EXAMPLE 8.16 Estimate T at which it is thermodynamically possible for carbon to reduce iron(III) oxide to iron under standard conditions by the endothermic reaction., above 565 o C

2012 General Chemistry I Impact on Biology: Gibbs Free Energy Changes in Biological Systems - A reaction that produces a lot of entropy can drive another nonspontaneous reaction forward. - A process may be driven uphill in Gibbs free energy by another reaction that rolls downhill.

2012 General Chemistry I 60  Hydrolysis of adenosine triphosphate (ATP) to adenosine diphosphate (ADP): the reaction used for driving nonspontaneous biochemical reactions - The nonspontaneous reaction restoring of ATP from ADP is driven by the food we eat.

2012 General Chemistry I 61 The End!

2012 General Chemistry I 62 Thank you for listening to this lecture, and please continue to General Chemistry II to know a variety of Green World!