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### Ch%2010%20Entropy%20Free%20Energy

Course: CHEM 1b, Winter 2006
School: UCSB
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Word Count: 1290

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10: Chapter Spontaneity, Entropy and Free Energy VanKoppen First Law of Thermodynamics Energy is conserved. The energy of the universe is constant. E = q + w Energy of the system increases when heat is added to the system (q &gt; 0) and when work is done on the system (w &gt; 0). The First Law accounts for energy changes: How much energy is involved in a process? What is the direction of flow: into or...

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10: Chapter Spontaneity, Entropy and Free Energy VanKoppen First Law of Thermodynamics Energy is conserved. The energy of the universe is constant. E = q + w Energy of the system increases when heat is added to the system (q > 0) and when work is done on the system (w > 0). The First Law accounts for energy changes: How much energy is involved in a process? What is the direction of flow: into or out of the system? What is the final form of energy? Chapter 10: Focuses on why a process occurs in a given direction (the first law does not tell us why). A spontaneous process is a process which occurs without outside intervention. For example, a gas expands to fill an entire container spontaneously. The gas molecules in a container will not all move to one side of the container spontaneously. Why not? A spontaneous process may be fast or slow. Thermodynamics tells us the direction in which a process will occur but says nothing about the speed (rate) of a process. Chemical kinetics (Chapter 15) is concerned with the rate of reaction. Entropy is a measure of randomness or disorder. It is the driving force for all spontaneous processes. Natural Progression: Order ----> Disorder Low Entropy ----> High Entropy Entropy (S) corresponds to the number of arrangements (positions and/or energy levels) that are available to a system in a given state. It is closely related to probability. The more ways a particular state can be achieved, the greater is the likelihood (probability) that state will occur. (Nature spontaneously proceeds towards the states that have the highest probabilities of existing). Generally: Ssolid < Sliquid << Sgas Because one mole of gas will occupy a greater volume then one mole of liquid or solid, the number of positions available to the gas molecules is greater and the entropy of a gas is therefore greater. For a reversible, isothermal, expansion or compression of n moles of an ideal gas at a given temperature T: reversible => taking an infinite number of steps, isothermal => constant temperature, T = 0 wrev = nRT ln(V2/V1) qrev = wrev ( since T = 0 and E = n cvT = 0, E = q + w = 0 => qrev = wrev ) qrev = nRT ln(V2/V1) => E = 0 because T = 0 E = n cvT H = n cpT => H = 0 because T = 0 In a reversible, cyclic process both the system and the surroundings are returned exactly to their original conditions. In an irreversible process, even if the process is cyclic (state 1 --> state 2 --> state 1), the surroundings are changed in a permanent way. Work is converted to heat in the surroundings. All real processes are irreversible. A reversible process gives us the maximum work obtainable from the gas. wmax = wrev qmax = qrev This is true for all work: e.g. pressure-volume work or electrical work done by an electrochemical cell (Chapter 11). S = kB ln where kB = Boltzmann's constant (R = kBNA) = number of microstates corresponding to a given state (including both position and energy). Definition of Entropy: Discovered by Ludwig Boltzman (1868). It provides a link between the microscopic world of atoms and molecules and the macroscopic world of bulk matter and thermodynamics. This is shown as follows: Consider an isothermal expansion of n moles of gas from V1 to V2: S = n R ln(2/1) at constant T When the gas expands from volume V1 to 2V1 each particle has double the number of positions available to it. This means that the number of microstates increases from 1 to 21. Thus, 2/1 = V2/V1 and S = nR ln(V2/V1) Considering that at constant T qrev /T = nR ln(V2/V1) at constant T which is the thermodynamic definition of S. it implies that S = qrev/T An example of a constant temperature reversible process is a phase transition such as the melting of a solid at constant pressure. This occurs reversibly the at normal freezing temperature (Tf ) because an infinitesimal change in external conditions (e.g. a lowering of the temperature) serves to reverse the process. The reversible heat when 1 mole of substance melts is qrev = Hfus. S fus = qrev Tf = H fus Tf Similarly, S vap = qrev Tb = H vap Tb where Svap is the molar entropy of vaporization of a liquid at constant pressure at the boiling temperature, Tb. Amazingly, most liquids have about the same molar entropy of vaporization. Trouton's rule states that the magnitude of this entropy change: Svap = 88 5 J mol-1 -1 (for most liquids) S = n cv ln (T2/T1) at constant V S = n cp ln (T2/T1) at constant P The Second Law of Thermodynamics: In any spontaneous process, the entropy of the universe (i.e., the system plus its surroundings) increases. More simply: The entropy of the universe is increasing. Note that the entropy of a system can decrease if that of the surroundings increases. Suniv = Ssys + Ssurr Because S = qrev / T and qrev > qirrev it follows that S > qirrev / T Combining these two equations yields: S q / T where the equality applies to a reversible process. In an isolated system there is no transfer of heat into or out of the system, that is, q=0. In this case S 0. The universe itself (the system plus its surroundings) is an isolated system. Thus, it follows that: 1) In a reversible process the total entropy of a system plus its surroundings is unchanged, Suniv = 0. 2) In an irreversible process the total entropy of a system plus its surroundings must increase, Suniv > 0. 3) A process for which Stot < 0 is impossible. (The process is spontaneous in the opposite direction.) Suniv > 0 Spontaneous process in direction written. Suniv < 0 Impossible (the process is spontaneous in opposite direction). Suniv = 0 System at equilibrium. (Reversible) Definition of Gibbs free Energy: G = H TS It can be shown that Suniv = Gsys/T and thus, G = H TS at constant T at constant P and T, which implies the following: Gsys and the spontaneity of a given process carried out at constant P and constant T Gsys < 0 Gsys > 0 Gsys = 0 for spontaneous processes for nonspontaneous processes when the system is at equilibrium (reversible processes) Note: Because G = H TS, an endothermic reaction (H > 0) can be spontaneous if S > 0 but only at sufficiently high temperatures such that H < TS, H TS < 0 and thus G < 0, indicating the reaction is spontaneous. Note: When subscripts are not included, it is assumed that all quantities refer to the system. See Table 10.6 pg 424 of your text. The Third Law of Thermodynamics: The entropy of a perfect crystal at 0 K is zero. Thus, the absolute entropies, So, for all substances at a given temperature greater than 0 K will be great than zero. Values of So at 25oC are tabulated in Appendix 4. So increases as the complexity of a molecule increases. For the reaction: a A + b B --> c C + d D So = [c So (C) + d So (D)] [a So (A) + b So (B)] Gibbs Free Energy and Equilibrium: where Q is the reaction quotient and Go = G = Go + RT ln Q At equilibrium, G = 0 and Q = K. Thus, Go = RT ln K and Ho TSo ln K = G = H + S RT RT R o o o A graph of ln K versus 1/T yields a straight line with a slope = Ho/R and intercept = So/R van't Hoff equation ln K2 K1 = o Hvap Vapor Pressure ln P 2 P1 = Hvap R o R 1 1 T2 T1 1 1 T2 T1 Given Ho and K at one temperature, the van't Hoff equation can be used to calculate K at another temperature, assuming Ho and So are constant over the temperature range T1 to T2. For the equilibrium, H2O(l) H2O(g), the equilibrium constant K = PH2O(g), the van't Hoff equation is written in terms of vapor pressures P1 and P2 at temperatures T1 and T2 . If P1 = 1 atm then T1 = Tb , the normal boiling point.
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