Entropy 2009-20101 Entropy, Free Energy, and Equilibrium

  • View
    219

  • Download
    2

Embed Size (px)

Text of Entropy 2009-20101 Entropy, Free Energy, and Equilibrium

  • Slide 1
  • Entropy 2009-20101 Entropy, Free Energy, and Equilibrium
  • Slide 2
  • Entropy 2009-20102 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).
  • Slide 3
  • Entropy 2009-20103 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.
  • Slide 4
  • Entropy 2009-20104 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 o C, and ice melts spontaneously above 0 o C (at 1 atm). Iron exposed to water and oxygen spontaneously forms rust, but rust does not spontaneously change back to iron.
  • Slide 5
  • Entropy 2009-20105 These examples show that processes that occur spontaneously in one direction cannot, under the same conditions, also take place spontaneously in the opposite direction.
  • Slide 6
  • Entropy 2009-20106 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 CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(l) H = -890.4 kJ
  • Slide 7
  • Entropy 2009-20107 But consider a solid-to-liquid phase transition such as this spontaneous process that occurs above 0 o C H 2 O(s) H 2 O(l) H = 6.01 kJ In this case, the assumption that spontaneous process always decrease a systems energy fails.
  • Slide 8
  • Entropy 2009-20108 Another example that contradicts our assumption is the dissolution of ammonium nitrate in water: NH 4 NO 3 (s) NH 4 + (aq) + NO 3 - (aq) H = 25 kJ H2OH2O This process is spontaneous, and yet it is also endothermic.
  • Slide 9
  • Entropy 2009-20109 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.
  • Slide 10
  • Entropy 2009-201010 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, H. The other is change in entropy, ( S). 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.
  • Slide 11
  • Entropy 2009-201011 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 S solid < S liquid
  • Entropy 2009-201019 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. S universe = S system + S surroundings > 0 S universe = S system + S surroundings = 0
  • Slide 20
  • Entropy 2009-201020 For a spontaneous process, the second law says that S univ must be greater than zero, but it does not place a restriction on either S sys or S surr. Thus it is possible for either S sys or S surr to be negative, as long as the sum of these two quantities is greater than zero. What if for some process we find that S univ is negative? The reaction is spontaneous in the opposite direction.
  • Slide 21
  • Entropy 2009-201021 To calculate S univ, we need to know both S sys and S surr.
  • Slide 22
  • Entropy 2009-201022 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, S o rxn is given by S o rxn = [cS o (C) + dS o (D)] - [aS o (A) + bS o (B)] or, in general S o rxn = nS o (products) - nS o (reactants)
  • Slide 23
  • Entropy 2009-201023 To calculate S rxn (which is S sys ), 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 o C. N 2 (g) + 3H 2 (g) 2NH 3 (g) S o rxn = nS o (products) - nS o (reactants) S o rxn = (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 S is negative
  • Slide 24
  • Entropy 2009-201024 Typically, if a reaction produces more gas molecules than it consumes, S o is positive, likewise, if the total number of gas molecules diminishes, S o is negative. If there is no change in the total number of gas molecules, then S o may be positive or negative, but will be relatively small numerically.
  • Slide 25
  • Entropy 2009-201025 Predict whether the entropy change of the system in each of the following reactions is positive or negative. (a) 2H 2 (g) + O 2 (g) 2H 2 O(l) (b) NH 4 Cl(s) NH 3 (g) + HCl(g) (c) H 2 (g) + Br 2 (g) 2HBr(g) -S-S +S+S ? S it will be small
  • Slide 26
  • Entropy 2009-201026 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.
  • Slide 27
  • Entropy 2009-201027 For constant-pressure processes the heat change is equal to the enthalpy change of the system. Therefore, the change in entropy of the surroundings, S surr, is proportional to H sys. S surr - H sys The minus sign is used because if the process is exothermic, H sys is negative and S surr is a positive quantity, indicating an increase in entropy. On the other hand, for an endothermic process, H sys is positive and the negative sign ensures that the entropy of the surroundings decreases.
  • Slide 28
  • Entropy 2009-201028 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.
  • Slide 29
  • Entropy 2009-201029 From the inverse relationship between S surr and temperature (in Kelvins) we can rewrite the relationship between H, T and S as S surr = - H sys T
  • Slide 30
  • Entropy 2009-201030 Would you predict the synthesis of gaseous ammonia from nitrogen gas and hydrogen gas to be spontaneous at 25 o C? (Will it be + S univ ?) 1/2 N 2 (g) + 3/2 H 2 (g) NH 3 (g) H = -46.3 kJ or -46300 J S sys = (1)(193.0) - [(1/2)(191.5) + (3/2)(131.0)] = -99.3 J/K. mol S surr = - H sys T S surr = -(-46300 J) = 155 J/K 298 K S univ = S sys + S surr S univ = -99.3 J/K. mol + 155 J/K. mol = 55.7 J/K Because S univ is positive, we predict that the reaction is spontaneous at 25 o C. S univ = S sys + S surr
  • Slide 31
  • Entropy 2009-201031 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.
  • Slide 32
  • Entropy 2009-201032 Gibbs Free Energy The second law of thermodynamics tells us that a spontaneous reaction increases the entropy in the universe (+ S), but in order to determine the sign of S univ, we would need to calculate both S sys and S surr. In order to express the spontaneity of a reaction more directly, we can use another thermodynamic function called Gibbs free energy (G).
  • Slide 33
  • Entropy 2009-201033 If a particular reaction is accompanied by a release of usable energy (- G), the reaction will occur spontaneously, if G is equal to zero, the system is at equilibrium. The change in free energy ( G) of a s

Recommended

View more >