, S is the measure of dispersal. The natural spontaneous direction of any process is toward greater dispersal of matter and of energy. Dispersal of matter: Thermodynamics We analyze the constraints on a system. The more the system is constrained, the less dispersed it is. Dispersal of energy: We analyze the flow of heat to measure how much energy is spread out in a particular process at a specific temperature. Change in, S 2, S measures the dispersal of matter S = k ln W where W is the number of ways of describing the system, and k is the Boltzmann constant (1.38 x 10-23 J K -1 ). The more ways that the system can reside, the greater the, S. S = k log W Note: If there is only one way to describe the system when W=1, the system is fully constrained. S=0 because ln (1)=0 The change in entropy is S. S = S 2 -S 1 where S 1 is the entropy of the initial state, and S 2 is the entropy of the final state. Like E and H, S is a state function. 3 Image from: http://en.citizendium.org/wiki/_%28thermodynamics%29 measures the dispersal of matter measures the dispersal of matter Example 1: Melting of ice H 2 O (s) H 2 O (l) Example 2: Rusting of iron 4 Fe (s) + 3 O 2 (g) + 3 H 2 O (l) 2 Fe 2 O 3 3H 2 O(s) S = S 2 -S 1 S >0 S < 0 In the ice crystal, all the molecules are constrained to fixed positions. This is state 1 corresponding to an entropy state, S 1. In liquid water, molecules are free to move around. This is state 2 corresponding to an entropy state, S 2. Image credit: http://www.visionlearning.com/library/module_viewer.php?mid=57 Image credit: http://www.splung.com/content/sid/3/page/batteries 5 6 1
measures the dispersal of matter measures the dispersal of matter Example 3: Combustion of methane CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (l) Example 1: Melting of water S > 0 Example 2: Iron rusting S < 0 Example 3: Combustion of methane S < 0 S < 0 All of the above processes are spontaneous at room temperature and atmospheric pressure. Conclusion: When a system changes spontaneously, there may be in increase of entropy ( S > 0), or a decrease in entropy ( S < 0). The above entropy changes are looking at the S of the system. But in the real world, there is always heat exchanges with the surroundings. Therefore, we must consider the entropy change of the universe. Usually when we write S, it implies S of the system ( S system ). 7 8 Change in of the Universe, S universe The Second Law of Thermodynamics: change measures the dispersal of energy How much energy is spread out in a particular process, or how widely spread out it becomes at a specific temperature. In any spontaneous process, there is always an increase in the entropy of the universe. S universe = S system + S surroundings > 0 Every process has a preferred direction natural spontaneity. For a spontaneous process, if there is a decrease in the S system, then the heat is absorbed by the surroundings must have caused an increase in the S surroundings such that overall there is an increase in the S universe (i.e. S universe > 0). Spontaneous processes: - Contraction of a stretched elastic band - Heat flows from a warm body to a cold body These processes increases random kinetic energy of the system. As a result, an increase of energy dispersion results. 9 10 change measures the dispersal of energy We analyze the flow of heat to measure how much energy is spread out in a particular process at a specific temperature. Every process has a preferred direction natural spontaneity. Let s look at what energy dispersal means on a microscopic scale with micro-states. S = k ln W is a measure of the available microstates Consider a chamber with 4 particles in it. Where the particle can only have energies 2, 4, 6, 8, 10, (i.e. the particles energy is quantized) The lowest energy state is when all 4 particles have energy = 2 or total energy is 8. There is only one way the particles can be arrange in this configuration. 4 particles with a total energy of 8 11 2
This system is now heated so that the total energy of the particles is 10. There are 4 microstates (arrangements of the particles) that satisfy E t =10. This system is now heated so that the total energy of the particles is 12. There are 10 microstates (arrangements of the particles) that satisfy E t =12. 4 microstates with 3 particles with E =2 and 1 particle with E = 6 4 particles with a total energy of 10 3 particles with E =2 and 1 particle with E = 6 This system is now heated so that the total energy of the particles is 12. There are 10 microstates (arrangements of the particles) that satisfy E t =12. 6 microstates with 2 particles with E =2 and 2 particles with E = 4 S = k ln This system is now heated so that the total energy of the particles is 12. There are 10 mircostates (arrangements of the particles) that satisfy E t =12. 4 microstates with 3 particles with E =2 and 1 particles with E = 6 6 microstates with 2 particles with E =2 and 2 particles with E = 4 Total of 10 microstates with total energy = 12 2 particles with E = 2 and 2 particles with E = 4 S = k ln This system is now heated so that the total energy of the particles is 14. There are 20 mircostates (arrangements of the particles) that satisfy E t =14. 4 microstates with 3 particles with E =2 and 1 particle with E = 8 12 microstates with 2 particles with E =2 and 1 particles with E = 4 and 1 particle with E = 6 4 microstates with 1 particle with E =2 and 3 particles with E = 4 As the energy of the system increases the number of microstates increases. Thus the entropy of the system increases. For 4 particles with quantized energy levels of E = 2, 4, 6, E 8 1 10 4 12 10 14 20 16 35 18 56 20 84 3
Consider 2 of these 4 particles system, one with E t = 8 and one with E t = 20. The systems are placed in thermal contact with each other. Consider 2 of these 4 particles system, one with E t = 8 and one with E t = 20. Total Energy = 28 The systems are placed in thermal contact with each other. System 1 E o = 8 System 2 E o = 20 System 1 E = E 1 System 2 E = E 2 Number of microstates 1 x 84 = 84 If E = 2 units of energy are transferred from system 2 to system 1 System 1 E = 10 System 2 E= 18 Number of micro states 4 x 56 = 224 System 1 System 2 E 1 1 E 2 2 total 8 1 20 84 84 10 4 18 56 224 12 10 16 35 350 14 20 14 20 400 16 35 12 10 350 18 56 10 4 224 20 84 8 1 84 Consider 2 of these 4 particles system, one with E t = 8 and one with E t = 20. Total Energy = 28 The probability of a finding a given amount of energy in system 1 is show in the graph On a molecular scale The number of microstates and, therefore, the entropy tends to increase with increases in: Temperature Volume (gases) The number of independently moving molecules Notice that the most likely outcome is when systems 1 and 2 have equal energy 22, S, S is related to the various modes of motion in molecules. Translational: Movement of the entire molecule from one place to another. Vibrational: Periodic motion of atoms within a molecule. Rotational: Rotation of the molecule on about an axis or rotation about bonds. More particles lead to more energy states, and therefore, more entropy Higher temperature lead to more energy states, and therefore, more entropy Less structure (gas is less structured than liquid, which is less structured than solid) lead to more states, and therefore, more entropy increases with the freedom of motion of molecules. Therefore, S(g) > S(l) > S(s) Image credit: Chemistry, The Central Science, 10th edition Theodore L. Brown, H. Eugene LeMay, Jr., and Bruce E. Bursten 23 24 4
More spontaneous processes: - Free expansion of a gas S > 0 This process neither absorbs nor emits heat. - Combustion of methane S < 0 CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (l) The process produces heat (exothermic). - Melting of ice cubes at room temperature H 2 O (s) H 2 O (l) S > 0 This process requires heat (endothermic). More spontaneous processes: Photosynthesis Complex and highly-energetic compounds (compared to the starting materials, CO 2 and H 2 O) are formed, it appears that there a decrease in entropy in the process. Conclusion: Naturally spontaneous processes could be exothermic, endothermic, or not involving heat. 25 26 A good example is photosynthesis. S<0; It appears that there a decrease in entropy in this natural process. For any spontaneous process, S universe = S system + S surroundings > 0 Image credit: http://www.eoearth.org/article/photosynthesis When the system and surroundings are considered, there is a net increase in entropy as a result of photosynthesis! Photosynthesis in a plant does not consist of a isolated system of the plant alone. Experiments and calculations indicate that the maximum efficiency of photosynthesis in most plants is in the 30% range. This means: 30% of the sunlight that strikes the plant is absorbed by the plant in the photosynthesis of substances that decreases entropy in the plant. 70% of the sunlight that strikes the plant is dispersed to the environment (an entropy increase in slightly heating the leaf and the atmosphere. 27 Spontaneous process, favoured by a decrease in H (exothermic) favoured by an increase in S Nonspontaneous process, favoured by an increase in H (endothermic) favoured by an decrease in S 28 What is q rev? measures the dispersal of energy At a specific temperature, the change in entropy, S, is where q rev is the heat transferred at a constant temperature T in Kelvin. What is q rev? Consider a gas in a piston and cylinder configuration. The gas is held in place by a pile of sand. The gas undergoes isothermal expansion. Recall E = q+w Since E = 0 for an isothermal process, q is the same magnitude of w but opposite in sign. Let s take the PV curve of five different paths to take the gas from State I to State F. Work done by the gas (w<0) is the area under the PV curve. w rev w irrev This is the reversible path. The area under the PV curve is the greatest. This path yields the smallest w (most negative), and the largest q. We refer to the heat transfer of a reversible path reversible q, q rev. 29 Of all the different possible paths to take the gas from State I to F, q rev is the limit of the observed heat transferred. 30 5
, S What is q rev? Likewise, consider a gas in a piston and cylinder configuration. The gas is held in place by a pile of sand. The gas undergoes isothermal compression. Since E = 0 for an isothermal process, q is the same magnitude of w but opposite in sign. Let s take the PV curve of five different paths to take the gas from State I to State F. Work done on the gas (w>0) is the area under the PV curve. What is q rev? A reversible change is one that is carried out in such a way, such that, when the change is undone, both the system and the surroundings remain unchanged. This is achieved when the process proceeds in infinitesimal steps that would take infinitely long to occur. w rev w irrev Reversibility is an idealization that is unachievable in real process, except when the system is in equilibrium. This is the reversible path. The area under the PV curve is the smallest. This path yields the smallest w, and the largest q (least negative). We refer to the heat transfer of a reversible path reversible q, q rev. Of all the different possible paths to take the gas from State I to F, q rev is the limit of observed heat transferred. 31 A process that is carried out reversibly has the smallest w rev, and the largest q rev. 32 measures the dispersal of energy To express entropy is to describe the number of arrangements for positions and/or energy levels available to a system a measurement of dispersion. At a specific temperature, the change in entropy, S, is where q rev is the heat transferred at a constant temperature T in Kelvin. Although most real processes are irreversible, there are real reversible processes. Similarly, consider water vapourizing at 100 o C, where H vapourization = 40.7 kj/mole This is a reversible process. If the heat is absorbed in small enough increments such that it does not disturb the equilibrium, this is a reversible process. Consider ice melting at 0 o C, where H fusion = 79.71 cal/g = 6.003 kj/mole This is a reversible process. If the heat is absorbed in small enough increments such that it does not disturb the equilibrium, this is a reversible process. There is an increase in the entropy of water when ice melts to form liquid water. There is an increase in the entropy of water when water is converted to steam. Note: S for the vapourization of water is much higher than that of fusion. 33 34 Change in in the surroundings, S surroundings Expansion of an Ideal Gas Recall how we define a reversible process. For the expansion of a gas in a piston and cylinder configuration under constant temperature condition. Changes in Surroundings Since E = 0 for an isothermal process, q is the same magnitude of w but opposite in sign. Work done by the gas (w<0) is the area under the PV curve. Heat that flows into or out of the system also changes the entropy of the surroundings. For an isothermal process: It follows that At constant pressure, q sys is H for the system. w rev or increases with an increase in volume or a decrease in pressure. 35 36 6
Change in in the surroundings, S surroundings A phase change is isothermal (no change in T). For H 2 O: H fusion = 6.003 kj/mole H vapourization = 40.7 kj/mole Example: 1.00 mole of a monatomic ideal gas at 273 K is expanded via two paths: (a) isothermally and reversibly from 22.4 L to 44.8 L. Find w, q, E, H, S system, S surroundings, and S universe. q (q > 0) External pressure, P ext (w < 0) Under isothermal condition, T=0, therefore E = H = 0 Since E depends only on temperature, E = 0 and H = E + (PV) = E + (nrt) = 0 This is because work of expansion done by the gas equals to the heat flow into the system. Net overall change in internal energy and enthalpy is zero. If this is done reversibly: S surr = S sys This is S system. What is S surroundings? What is S universe? q rev = S T = 5.764 273 = 1.574 kj Since E = 0, w = -q = -1.574 kj 37 Since q surr = -q sys, q surr = -1.574 kj This is true for a reversible process. The system is at equilibrium. 38 Example: 1.00 mole of a monatomic ideal gas at 273 K is expanded via two paths: (b) via free expansion from 22.4 L to 44.8 L. Find w, q, E, H, and S system, S surroundings, and S universe. This is an irreversible process with no change in temperature, T=0. As a result, E = H = 0 w = 0 because P ext = 0 w = -q, q =0 This is S system. It has the same disorder as calculated in Part (a). What is S surroundings? What is S universe? Since the no heat flowed into the surroundings, q surr = 0, Since E depends only on temperature, E = 0 and H = E + (PV) = E + (nrt) = 0 To calculate S, we need to find a reversible path that will take the gas from the same initial state to the same final state. Part (a) is such a path. This is a spontaneous process. S universe > 0 associated with temperature change At a specific temperature T Recall, q = n C T where n = number of moles C = molar heat capacity T = temperature difference For a change in temperature of n moles of substance from T 1 to T 2 where the heat capacity is constant over the temperature range. Under constant volume condition Under constant pressure condition 39 40 associated with temperature change For a change in temperature of n moles of substance from T 1 to T 2 and the heat capacity is expressed as a power series, a + bt + ct 2 + T in Kelvin At a specific temperature T Calculate S for the change in state S Increase the volume of the container. The amount of gaseous H 2 O present is governed by the vapour pressure of water at 25 o C. Given C p (H 2 O,liq) = 18 cal mole -1 K -1 C p (H 2 O, g) = 9.0 cal mole -1 K -1 H vap = 9713 cal/mole (at 373K) Let s devise a reversible process to calculate S. S S H Warming liquid water, S 1 Cooling gaseous water, S 3 S 2 Compare 41 S = S 1 + S 2 + S 3 42 7
Calculate S for the change in state S > 0 S Given C p (H 2 O,liq) = 18 cal mole -1 K -1 C p (H 2 O, g) = 9.0 cal mole -1 K -1 H vap = 9713 cal/mole (at 373K) S S < 0 Trouton s Rule A useful generalization that works for many liquids at their normal boiling point (i.e. boiling point at 1 atm) is that, the standard molar entropy of vapourization has a value of about 22 cal mole -1 K -1 or 88 J mole -1 K -1 Since The idea is, if the degree of disorder associated in the phase transformation of 1 mole of liquid to 1 mole of vapour at 1 atm is the same, then the S o vapourization of different liquids should be similar. Warming liquid water, S 1 Cooling gaseous water, S 3 S o vapourization = 22 cal mole -1 K -1 or 88 J mole -1 K -1 Trouton s rule is useful method for estimating the enthalpy of vapourization of a liquid if its normal boiling point is known. S > 0 S 2 S = S 1 + S 2 + S 3 = 4.04 + 26.04-2.02 = 28.1 cal K -1 Example: We can estimate the enthalpy of vapourization of Hg using Trouton s rule from the normal boiling point of Hg. The normal boiling point of Hg is 357 o C. H vapourization = 630 K 88 J mole -1 K -1 = 55 kj mole -1 Compare: Heat of vaporization of mercury: 59.11 kj mol -1 Cited value: Wikipedia http://en.wikipedia.org/wiki/mercury_%28element%29 43 44, S Trouton s Rule S o vapourization = 22 cal mole -1 K -1 or 88 J mole -1 K -1 The Third Law of Thermodynamics: Substance H vap (kcal mole -1 ) T b ( o C) S vap (cal mole -1 K -1 ) O 2 1.630-182.97 18.07 The entropy of a pure perfect crystal is zero when the crystal is at zero Kelvin (0 K) CH 4 1.955-161.49 17.51 H 2O 9.717 100.0 26.04 NH 3 5.581-33.43 23.28 C 2H 5OH 9.23 78.3 26.27 A pure perfect crystal is one in which every molecule is identical, and the molecular alignment is perfectly even throughout the substance. CHCl 3 7.02 61.2 20.8 C 6H 6 7.36 80.1 20.83 (C 2H 5) 2O ether 6.21 34.5 20.18 For non-pure crystals, or those with less-than perfect alignment, there will be some dispersal of the imperfection, so the entropy cannot be zero. (CH 3) 2CO acetone 7.22 56.2 21.90 In liquid state, hydrogen bonding between molecules produces a greater degree of order than expected. As a result, the degree of disorder produced in the vapourization process is generally greater than the nominal 22 cal mole -1. 45 Image credit: Chemistry, The Central Science, 10th edition Theodore L. Brown, H. Eugene LeMay, Jr., and Bruce E. Bursten 46, S Standard Molar, S o The entropy of the crystal gradually increases with temperature as the average kinetic energy of the particles increases. As the crystal warms to temperatures above 0 K, the particles in the crystal start to move, generating some disorder. The entropy of the liquid gradually increases as the liquid becomes warmer because of the increase in the vibrational, rotational, and translational motion of the particles. Standard entropies (measured at 298K) have been carefully measured for many substances. For example, the standard molar entropy of some solids: C(diamond) 2.38 C (graphite) 5.74 Sodium 51.3 Phosphorus (white) 41.1 Sulfur (rhombic) 31.8 At the melting point, T m, the entropy of the system increases in entropy without a temperature change as the compound is transformed into a liquid, which is not as well ordered as the solid. At the boiling point, there is another increase in the entropy of the substance without a temperature change as the compound is transformed into a gas. Silver 42.6 47 48 8
Standard Molar, S o Standard Molar, S o Standard entropies (measured at 298 K) have been carefully measured for many substances. For example, the standard molar entropies (standard entropy per mole) for gases are usually higher because heat of melting and heat of vapourization must be included. The standard molar entropies for noble gases are: He 126.0 Ne 146.2 Ar 154.7 Kr 164.0 Xe 169.6 For liquids and gases S o are usually higher because heat of melting and heat of vaporization must be included. Standard entropies (measured at 298 K) have been carefully measured for many substances. For example, the standard molar entropies (standard entropy per mole) for diatomic gases. H 2 130.6 N 2 191.5 O 2 205.1 Halogens: F 2 203.7 Cl 2 222.9 Br 2 245.4 I 2 260.6 For diatomic gases, they are usually higher than those of monatomic noble gases as there are more degree of freedom of motion. 49 50 Standard Molar, S o Standard Molar, S o Standard entropies (measured at 298 K) have been carefully measured for many substances. For example, the standard molar entropies (standard entropy per mole) for compounds. For example, the standard molar entropies (standard entropy per mole) for more gases. CH 4 (g) 186.3 C 2 H 2 (g) 200.8 C 2 H 4 (g) 219.5 C 2 H 6 (g) 229.6 C 3 H 8 (g) 270.3 In general, the more complicate the molecule and the heavier the molar mass, the higher the standard entropy. H 2 O (l) 69.9 H 2 O 2 (l) 110.0 CH 3 OH(l) 126.9 CH 3 Cl(l) 145.3 CHCl 3 (l) 202.9 A NO molecule has only one type of vibrational molecule H 2 O (g) 188.7 CO (g) 197.8 CO 2 (g) 213.6 NO (g) 210.6 NO 2 (g) 240.4 N 2 O 4 (g) 304.3 SO 2 (g) 248.4 A NO 2 molecule has three types of vibration. Image credit: Chemistry, The Central Science, 10th edition Theodore L. Brown, H. Eugene LeMay, Jr., and Bruce E. Bursten 51 52 Standard for Aqueous Ions S reaction o These are not third law entropies, presumably because the task of working up from absolute zero is too complex a task for aqueous ions. The hydrogen ion, is arbitrarily chosen to have an entropy of zero and all other ions are compared with it. Calculate the change in of a reaction from Standard Molar. Some of these other ions have greater entropy than the hydrogen ions, some lower, hence the negative values. H + (aq) OH (aq) H 2 O(l) S o (J K -1 mole -1 ) 0-10.7 70.0 H + (aq) + OH - (aq) H 2 O (l) S o = S o (H 2 O (l)) (S o (H + (aq)) - S o (OH - (aq)) S o = 70.0 (0-10.7) S o = 80.7 J K -1 53 54 9
S reaction o Calculate the standard entropy change accompanying the burning of ethane, C 2 H 6. C 2 H 6 (g) + 7 / 2 O 2 (g) 2 CO 2 (g) + 3 H 2 O (g) S =? Summary: The Second Law of Thermodynamics: In any spontaneous process, there is always an increase in the entropy of the universe. S universe = S system + S surroundings > 0 Change in entropy, S, at a specific temperature, T. C 2 H 6 (g) 229.5 O 2 (g) 205.1 CO 2 (g) 213.6 H 2 O (g) 188.7 In general, the more complicate the molecule and the heavier the molar mass, the higher the standard entropy. Change in entropy, S, measured over a temperature range from T 1 to T 2. Assuming that C v and C p are constant from T 1 to T 2. For an isothermal process expansion or compression of a gas. S o = 2S o (CO 2 (g)) + 3 S o (H 2 O (g)) (S o (C 2 H 6 (g)) + 7 / 2 S o (O 2 (g))) S o = 2(213.6) + 3 (188.7) (229.5 + 7 / 2 (205.1)) S o = 46.3 J K -1 The Third Law says the entropy of a pure perfect crystal is zero when the crystal is at zero Kelvin (0 K). For a chemical reaction, S o can be calculated from the reactants and products standard molar entropies. 55 56 10