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Entropy 2009-20101 Entropy, Free Energy, and Equilibrium.

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Presentation on theme: "Entropy 2009-20101 Entropy, Free Energy, and Equilibrium."— Presentation transcript:

1 Entropy 2009-20101 Entropy, Free Energy, and Equilibrium

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).

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.

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.

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.

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

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 system’s energy fails.

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.

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.

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.

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 << S gas 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.

12 Entropy 2009-201012 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.

13 Entropy 2009-201013 The important point about the third law is that it allows us to determine the absolute entropies of substances. The standard entropies (S o ) that are listed on your reference sheets are the absolute entropies of substances at 1 atm and 25 o C. These are the values that are generally used in calculations.

14 Entropy 2009-201014 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.

15 Entropy 2009-201015 Consider a certain process in which a system changes from some initial state to some final state. The entropy change for the process,  S, is  S = S final - S initial If the change results in an increase in randomness or disorder, then the change in entropy is positive (+  S)

16 Entropy 2009-201016 Entropy increase of a substance as the temperature rises from absolute zero. 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.

17 Entropy 2009-201017 Processes that lead to an increase in the entropy of the system include melting vaporization dissolving* heating

18 Entropy 2009-201018 *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 (-  S).

19 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

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.

21 Entropy 2009-201021 To calculate  S univ, we need to know both  S sys and  S surr.

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)

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

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.

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

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.

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.

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.

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

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

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.

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).

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 system for a constant-temperature process is  G =  H sys - T  S sys

34 Entropy 2009-201034 For the reaction C(s) + O 2 (g) CO 2 (g) the values of  H and  S are known to be -393.5 kJ and 3.05 J/K, respectively. Would this reaction be spontaneous at 25 o C?  G =  H - T  S = -393500 J - (298 K)(3.05 J/K) = -394,000 J or -394 kJ -  G, therefore the reaction would be spontaneous

35 Entropy 2009-201035 The standard free-energy of reaction (  G o rxn ) 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.

36 Entropy 2009-201036 To calculate  G o rxn we start with the equation aA + bB cC + dD the standard free-energy change for this reaction is given by the equation  G o rxn = [c  G f (C) + d  G f (D)] - [a  G f (A) + b  G f (B)] or, in general,  G o rxn =  nG f (products) -  nG f (reactants)

37 Entropy 2009-201037 Calculate the standard free-energy changes for the combustion of 1 mol of methane at 25 o C.  G o rxn = [1  G f (CO 2 ) + 2  G f (H 2 O)] - [1  G f (CH 4 ) + 2  G f (O 2 )]  G o rxn = [(1 mol)(-394.4 kJ/mol) + (2 mol)(-237.2 kJ/mol)] - [(1 mol)(-50.8 kJ/mol) + (2 mol)(0 kJ/mol)]  G o rxn = -818.0 kJ

38 Entropy 2009-201038 Summarizing the conditions for spontaneity and equilibrium at constant temperature and pressure in terms of  G  G < 0The reaction is spontaneous in the forward direction  G > 0 The reaction is nonspontaneous. The reaction is spontaneous in the opposite direction.  G = 0The system is at equilibrium. There is no net change.

39 Entropy 2009-201039 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.

40 Entropy 2009-201040 Calcium oxide (CaO) also called quicklime, is an extremely valuable inorganic substance. It is prepared by decomposing limestone (CaCO 3 ) in a kiln at a high temperature according to the following reaction CaCO 3 (s) CaO(s) + CO 2 (g) Estimate the temperature at which decomposition becomes spontaneous.  G =  H - T  S  H = [(1 mol)(-635.6 kJ/mol) + (1 mol)(-393.5 kJ/mol)] - [(1 mol)(-1206.9 kJ/mol)]  H = 177.8 kJ  S = [(1 mol)(39.8 J/K. mol) + (1 mol)(213.6 J/K. mol)] - [(1 mol)(92.9 J/K. mol)]  S = 130.0 J/K. mol

41 Entropy 2009-201041  G =  H - T  S  G = 177.8kJ - (298 K)(.1605 kJ/K)  G = 130.0 kJ Since  G is a large positive quantity, we conclude that the reaction is not spontaneous at 25 o C. In order to make  G negative, we first have to find the temperature at which  G is zero, the point at which the system is at equilibrium. 0 =  H - T  S -  H = -T  S  H = T  S  H = T  S T = 177800 J= 1108 K or 835 o C 160.5 J/K At 835 o C the system is at equilibrium. At temperatures higher than 835 o C,  G becomes negative, indicating that the decomposition is spontaneous. Reactions with +  H and +  S are spontaneous at high temps!

42 Entropy 2009-201042 Two things need to be kept in mind regarding the previous calculation. First, we used  H and  S values at 25 o C to calculate changes that occur at a much higher temperature, so the value of  G will just be close to the actual value. Second, some decomposition will occur at below 835 o C (just like some water will evaporate at temps below 100 o C.

43 Entropy 2009-201043 At approximately what temperature are the liquid and gaseous bromine at equilibrium? (Or in other words, estimate the normal boiling point of liquid Br 2.) Br 2 (l)  Br 2 (g)  H = 31.0 kJ/mol  S = 93.0 J/K. mol Phase changes will occur when a system is at equilibrium (  G = 0). Phase Changes

44 Entropy 2009-201044 = 3.10 x 10 4 J/mol = 333 K 93.0 J/K. mol  G =  H – T  S 0 =  H – T  S  H = T  S T =  H  S So this means that at any temp ABOVE 333 K, the above process will be spontaneous, therefore, 333 K is Bromine’s boiling point.

45 Entropy 2009-201045 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.

46 Entropy 2009-201046 We can use the value of  G o to calculate the value of  G under nonstandard conditions.  G =  G o + 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.

47 Entropy 2009-201047 Under standard conditions, all the reactants and products are equal to 1. Thus, under standard conditions, Q = 1 and therefore, lnQ = 0, and  G =  G o, as it should under standard conditions. When the concentrations of reactants and products are nonstandard, we must calculate the value of Q to determine  G

48 Entropy 2009-201048 Calculate  G for the formation of ammonia at 298 for a reaction mixture that consists of 1.0 atm N 2, 3.0 atm H 2, and 0.50 atm NH 3.

49 Entropy 2009-201049 Calculate  G at 298 for a reaction mixture that consists of 1.0 atm N 2, 3.0 atm H 2, and 0.50 atm NH 3,  G o = -33.3 kJ 3H 2 + N 2  2NH 3 Q = (P NH 3 ) 2 (P N 2 )(P H 2 ) 3 (.50) 2 (1.0)(3.0) 3 = 9.3 x 10 -3 =

50 Entropy 2009-201050  G =  G o + RT lnQ  G = -33.3 kJ/mol + (.008314 kJ/mol K)(298 K)ln(9.3 x 10 -3 )  G = -33.3 kJ/mol + (-11.6 kJ/mol) = -44.9 kJ/mol

51 Entropy 2009-201051 We can use the previous equation to derive the relationship between  G o and the equilibrium constant, k, for the reaction.  G =  G o + RT lnQ 0 =  G o + RT ln k  G o = - RT ln k at equilibrium,  G = 0, and Q = k

52 Entropy 2009-201052 Given the value of  G o from the previous calculation, solve for the equilibrium constant for the formation of ammonia at 298 K.  G o = - RT ln k -33,300 J/mol = - (8.314 J/mol K)(298 K) ln k 13.4 = ln k Taking the inverse ln of both sides… 6.60x 10 5 = k Which makes sense, a –  G should equal a k > 1!

53 Entropy 2009-201053 The End


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