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Entropy, Free Energy, and Equilibrium

Entropy, Free Energy, and Equilibrium. Spontaneous Processes and Entropy.

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Entropy, Free Energy, and Equilibrium

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  1. Entropy, Free Energy, and Equilibrium Entropy 2009-2010

  2. Spontaneous Processes and Entropy One of the main objectives in studying thermodynamics, as far as chemists are concerned, is to be able to predict whether or not a reaction will occur when reactants are brought together under a special set of conditions (for example, at a certain temperature, pressure, and concentration). Entropy 2009-2010

  3. A reaction that does occur under the given set of conditions is called a spontaneous reaction. If a reaction does not occur under specified conditions, it is said to be nonspontaneous. Entropy 2009-2010

  4. We observe spontaneous physical and chemical processes every day, including many of the following examples: A waterfall runs downhill, but never up, spontaneously. A lump of sugar spontaneously dissolves in a cup of coffee, but dissolved sugar does not spontaneously reappear in its original form. Water spontaneously freezes below 0 oC, and ice melts spontaneously above 0 oC (at 1 atm). Iron exposed to water and oxygen spontaneously forms rust, but rust does not spontaneously change back to iron. Entropy 2009-2010

  5. These examples show that processes that occur spontaneously in one direction cannot, under the same conditions, also take place spontaneously in the opposite direction. Entropy 2009-2010

  6. CH4(g) + 2O2(g) CO2(g) + 2H2O(l) DH = -890.4 kJ If we assume that spontaneous processes occur so as to decrease the energy of a system, we can explain why a ball rolls downhill and why springs in a clock unwind. Similarly, a large number of exothermic reactions are spontaneous. An example is the combustion of methane Entropy 2009-2010

  7. But consider a solid-to-liquid phase transition such as this spontaneous process that occurs above 0 oC H2O(s) H2O(l) DH = 6.01 kJ In this case, the assumption that spontaneous process always decrease a system’s energy fails. Entropy 2009-2010

  8. H2O NH4NO3(s) NH4+(aq) + NO3-(aq) DH = 25 kJ Another example that contradicts our assumption is the dissolution of ammonium nitrate in water: This process is spontaneous, and yet it is also endothermic. Entropy 2009-2010

  9. From the study of the examples mentioned and many more cases, we come to the following conclusion: Exothermicity favors the spontaneity of a reaction but does not guarantee it. Just as it is possible for an endothermic reaction to be spontaneous, it is possible for an exothermic reaction to be nonspontaneous. In other words, we cannot decide whether or not a chemical reaction will occur spontaneously solely on the basis of energy changes in the system. To make this kind of prediction we need another thermodynamic quantity, which turns out to be entropy. Entropy 2009-2010

  10. Entropy In order to predict the spontaneity of a process, we need to know two things about the system. One is the change in enthalpy, DH. The other is change in entropy, (DS). Entropy is a measure of the randomness or disorder of a system. The greater the disorder of a system, the greater its entropy. Conversely, the more ordered a system, the smaller its entropy. Entropy 2009-2010

  11. For any substance, the particles in the solid state are more ordered than those in the liquid state, which in turn are more ordered than those in the gaseous state. So for the same molar amount of a substance, we can write Ssolid < Sliquid << Sgas In other words, entropy describes the extent to which atoms, molecules, or ions are distributed in a disorderly fashion in a given region of space. Entropy 2009-2010

  12. It is possible to determine the absolute entropy of a substance, something we cannot do for enthalpy. According to the third law of thermodynamics, the entropy of a perfect crystalline substance is zero at absolute zero (0 K). If the crystal is impure or if it has defects, then its entropy is greater than zero even at 0 K because it would not be perfectly ordered. Entropy 2009-2010

  13. The important point about the third law is that it allows us to determine the absolute entropies of substances. The standard entropies (So) that are listed on your reference sheets are the absolute entropies of substances at 1 atm and 25 oC. These are the values that are generally used in calculations. Entropy 2009-2010

  14. The units of entropy are J/K or J/K.mol for 1 mole of the substance. We use joules rather than kilojoules because entropy values are typically quite small. Entropy 2009-2010

  15. Consider a certain process in which a system changes from some initial state to some final state. The entropy change for the process, DS, is DS = Sfinal - Sinitial If the change results in an increase in randomness or disorder, then the change in entropy is positive (+DS) Entropy 2009-2010

  16. At absolute zero, a substance has a zero entropy value (assuming that it is a perfect crystalline solid). As it is heated, its entropy increases gradually because of greater molecular motion. At the melting point, there is a sizable increase in entropy as the more random liquid state is formed. Further heating increases the entropy of the liquid again due to enhanced molecular motion. At the boiling point there is a large increase in entropy as a result of the liquid to gas transition. Beyond that temperature, the entropy of the gas continues to rise with increasing temperature. Entropy increase of a substance as the temperature rises from absolute zero. Entropy 2009-2010

  17. Processes that lead to an increase in the entropy of the system include • melting • vaporization • dissolving* • heating Entropy 2009-2010

  18. *Dissolving ionic compounds in water does not always result in an increase in entropy. For ionic compounds that contain Al+3 or Fe+3, hydration (the process of surrounding the ions by water molecules) can increase the order of the water molecules so much that the entropy change for the overall process can actually be negative (-DS). Entropy 2009-2010

  19. The Second Law of Thermodynamics The connection between entropy and the spontaneity of a reaction is expressed by the second law of thermodynamics: the entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process. DSuniverse = DSsystem + DSsurroundings > 0 DSuniverse = DSsystem + DSsurroundings = 0 Entropy 2009-2010

  20. For a spontaneous process, the second law says that DSuniv must be greater than zero, but it does not place a restriction on either DSsys or DSsurr. Thus it is possible for either DSsys or DSsurr to be negative, as long as the sum of these two quantities is greater than zero. • What if for some process we find that DSuniv is negative? The reaction is spontaneous in the opposite direction. Entropy 2009-2010

  21. To calculate DSuniv, we need to know both DSsys and DSsurr. Entropy 2009-2010

  22. Entropy Changes in the System Suppose that the system is represented by the following reaction: aA + bB cD + dD As in the case for enthalpy of a reaction, the standard entropy of reaction, DSorxn is given by DSorxn = [cSo(C) + dSo(D)] - [aSo(A) + bSo(B)] or, in general DSorxn = SnSo(products) - SnSo(reactants) Entropy 2009-2010

  23. N2(g) + 3H2(g) 2NH3(g) To calculate DSrxn (which is DSsys), look up the values on your reference sheets. Calculate the standard entropy for the formation of ammonia from nitrogen gas and hydrogen gas at 25 oC. DSorxn = SnSo(products) - SnSo(reactants) DSorxn = (2 mol)(193 J/K.mol) - [(1 mol)(192 J/K.mol) + (3 mol)(131 J/K.mol)] = -199 J/K Does this reaction result in an increase or decrease in order? Decrease because DS is negative Entropy 2009-2010

  24. Typically, if a reaction produces more gas molecules than it consumes, DSo is positive, likewise, if the total number of gas molecules diminishes, DSo is negative. If there is no change in the total number of gas molecules, then DSo may be positive or negative, but will be relatively small numerically. Entropy 2009-2010

  25. Predict whether the entropy change of the system in each of the following reactions is positive or negative. (a) 2H2(g) + O2(g) 2H2O(l) (b) NH4Cl(s) NH3(g) + HCl(g) (c) H2(g) + Br2(g) 2HBr(g) -DS +DS ?DS it will be small Entropy 2009-2010

  26. Entropy Changes in the Surroundings When an exothermic process takes place in the system, the heat transferred to the surroundings enhances motion of the molecules in the surroundings. Consequently, there is an increase in disorder of the surroundings at the molecular level, and the entropy of the surroundings increases. Conversely, an endothermic process in the system absorbs heat from the surroundings and so decreases the entropy of the surroundings because molecular motion decreases. Entropy 2009-2010

  27. For constant-pressure processes the heat change is equal to the enthalpy change of the system. Therefore, the change in entropy of the surroundings, DSsurr, is proportional to DHsys. DSsurra -DHsys The minus sign is used because if the process is exothermic, DHsys is negative and DSsurr is a positive quantity, indicating an increase in entropy. On the other hand, for an endothermic process, DHsys is positive and the negative sign ensures that the entropy of the surroundings decreases. Entropy 2009-2010

  28. The change in entropy for a given amount of heat also depends on the Kelvin temperature. If the temperature of the surroundings is high, the molecules are already quite energetic. Therefore, the absorption of heat from an exothermic process in the system will have relatively little impact on molecular motion and the resulting increase in entropy will be small. However, if the temperature of the surroundings is low, then the addition of the same amount of heat will cause a more drastic increase in molecular motion and hence a larger increase in entropy. Entropy 2009-2010

  29. From the inverse relationship between DSsurr and temperature (in Kelvins) we can rewrite the relationship between DH, T and DS as DSsurr = -DHsys T Entropy 2009-2010

  30. DSsurr = -DHsys T DSsurr = -(-46300 J) = 155 J/K 298 K DSuniv = DSsys + DSsurr Would you predict the synthesis of gaseous ammonia from nitrogen gas and hydrogen gas to be spontaneous at 25 oC? (Will it be +DSuniv ?) 1/2 N2(g) + 3/2 H2(g) NH3(g) DH = -46.3 kJ or -46300 J DSsys = (1)(193.0) - [(1/2)(191.5) + (3/2)(131.0)] = -99.3 J/K.mol DSuniv = DSsys + DSsurr DSuniv = -99.3 J/K.mol +155 J/K.mol = 55.7 J/K Because DSuniv is positive, we predict that the reaction is spontaneous at 25 oC. Entropy 2009-2010

  31. It is important to keep in mind that just because a reaction is spontaneous does not mean that it will occur at an observable rate. Thermodynamics can tell us whether a reaction will occur spontaneously under specific conditions, but it does not say how fast it will occur. Entropy 2009-2010

  32. Gibbs Free Energy The second law of thermodynamics tells us that a spontaneous reaction increases the entropy in the universe (+DS), but in order to determine the sign of DSuniv, we would need to calculate both DSsys and DSsurr. In order to express the spontaneity of a reaction more directly, we can use another thermodynamic function called Gibbs free energy (G). Entropy 2009-2010

  33. The change in free energy (DG) of a system for a constant-temperature process is DG = DHsys - TDSsys If a particular reaction is accompanied by a release of usable energy (-DG), the reaction will occur spontaneously, if DG is equal to zero, the system is at equilibrium. Entropy 2009-2010

  34. For the reaction C(s) + O2(g) CO2(g) the values of DH and DS are known to be -393.5 kJ and 3.05 J/K, respectively. Would this reaction be spontaneous at 25 oC? DG = DH - TDS = -393500 J - (298 K)(3.05 J/K) = -394,000 J or -394 kJ -DG, therefore the reaction would be spontaneous Entropy 2009-2010

  35. The standard free-energy of reaction (DGorxn) is the free-energy change for a reaction when it occurs under standard-state conditions, when reactants in their standard states are converted to products in their standard states. Entropy 2009-2010

  36. To calculate DGorxn we start with the equation aA + bB cC + dD the standard free-energy change for this reaction is given by the equation DGorxn = [cDGf(C) + dDGf(D)] - [aDGf(A) + bDGf(B)] or, in general, DGorxn = SnGf(products) - SnGf(reactants) Entropy 2009-2010

  37. Calculate the standard free-energy changes for the combustion of 1 mol of methane at 25 oC. DGorxn = [1DGf(CO2) + 2DGf(H2O)] - [1DGf(CH4) + 2DGf(O2)] DGorxn = [(1 mol)(-394.4 kJ/mol) + (2 mol)(-237.2 kJ/mol)] - [(1 mol)(-50.8 kJ/mol) + (2 mol)(0 kJ/mol)] DGorxn = -818.0 kJ Entropy 2009-2010

  38. Summarizing the conditions for spontaneity and equilibrium at constant temperature and pressure in terms of DG DG < 0 The reaction is spontaneous in the forward direction DG > 0 The reaction is nonspontaneous. The reaction is spontaneous in the opposite direction. DG = 0 The system is at equilibrium. There is no net change. Entropy 2009-2010

  39. Temperature and Chemical Reactions The temperature at which a reaction occurs (becomes spontaneous) is important to the practical chemist. We can make a reasonable estimate of that temperature using the data on your reference sheets. Entropy 2009-2010

  40. Calcium oxide (CaO) also called quicklime, is an extremely valuable inorganic substance. It is prepared by decomposing limestone (CaCO3) in a kiln at a high temperature according to the following reaction CaCO3(s) CaO(s) + CO2(g) Estimate the temperature at which decomposition becomes spontaneous. DG = DH - TDS DH = [(1 mol)(-635.6 kJ/mol) + (1 mol)(-393.5 kJ/mol)] - [(1 mol)(-1206.9 kJ/mol)] DH = 177.8 kJ DS = [(1 mol)(39.8 J/K.mol) + (1 mol)(213.6 J/K.mol)] - [(1 mol)(92.9 J/K.mol)] DS = 130.0 J/K.mol Entropy 2009-2010

  41. Since DG is a large positive quantity, we conclude that the reaction is not spontaneous at 25 oC. In order to make DG negative, we first have to find the temperature at which DG is zero, the point at which the system is at equilibrium. DG = DH - TDS DG = 177.8kJ - (298 K)(.1605 kJ/K) DG = 130.0 kJ At 835 oC the system is at equilibrium. At temperatures higher than 835 oC, DG becomes negative, indicating that the decomposition is spontaneous. Reactions with +DH and +DS are spontaneous at high temps! 0 = DH - TDS -DH = -TDS DH = TDS DH = T DS T = 177800 J = 1108 K or 835 oC 160.5 J/K Entropy 2009-2010

  42. Two things need to be kept in mind regarding the previous calculation. First, we used DH and DS values at 25 oC to calculate changes that occur at a much higher temperature, so the value of DG will just be close to the actual value. Second, some decomposition will occur at below 835 oC (just like some water will evaporate at temps below 100 oC. Entropy 2009-2010

  43. Phase Changes Phase changes will occur when a system is at equilibrium (DG = 0). At approximately what temperature are the liquid and gaseous bromine at equilibrium? (Or in other words, estimate the normal boiling point of liquid Br2.) Br2(l)  Br2(g) DH = 31.0 kJ/mol DS = 93.0 J/K.mol Entropy 2009-2010

  44. DG = DH – TDS 0 = DH – TDS = 3.10 x 104 J/mol = 333 K 93.0 J/K . mol DH = TDS T = DH DS So this means that at any temp ABOVE 333 K, the above process will be spontaneous, therefore, 333 K is Bromine’s boiling point. Entropy 2009-2010

  45. Free Energy and the Equilibrium Constant The temperature at which a reaction occurs (becomes spontaneous) is important to the practical chemist. We can make a reasonable estimate of that temperature using the data on your reference sheets. Entropy 2009-2010

  46. We can use the value of DGo to calculate the value of DG under nonstandard conditions. DG=DGo + RT lnQ In this equation, R is the ideal gas constant, 8.314 J/mol K, T is the absolute temperature, and Q is the reactant quotient. Entropy 2009-2010

  47. Under standard conditions, all the reactants and products are equal to 1. Thus, under standard conditions, Q = 1 and therefore, lnQ = 0, and DG=DGo, as it should under standard conditions. When the concentrations of reactants and products are nonstandard, we must calculate the value of Q to determine DG Entropy 2009-2010

  48. Calculate DG for the formation of ammonia at 298 for a reaction mixture that consists of 1.0 atm N2, 3.0 atm H2, and 0.50 atm NH3. Entropy 2009-2010

  49. (PNH3)2 (PN2)(PH2)3 (.50)2 (1.0)(3.0)3 Calculate DG at 298 for a reaction mixture that consists of 1.0 atm N2, 3.0 atm H2, and 0.50 atm NH3, DGo = -33.3 kJ 3H2 + N2 2NH3 = 9.3 x 10-3 Q = = Entropy 2009-2010

  50. DG=DGo + RT lnQ DG= -33.3 kJ/mol + (.008314 kJ/mol K)(298 K)ln(9.3 x 10-3) DG= -33.3 kJ/mol + (-11.6 kJ/mol) = -44.9 kJ/mol Entropy 2009-2010

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