Entropy 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 Entropy, 2 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 = 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 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, is a state function. 3 Image from: http://en.citizendium.org/wiki/entropy_%28thermodynamics%29 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 2 -S 1 >0 < 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
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 > 0 Example 2: Iron rusting < 0 Example 3: Combustion of methane < 0 < 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 ( > 0), or a decrease in entropy ( < 0). The above entropy changes are looking at the 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, it implies of the system ( system ). 7 8 Change in Entropy of the Universe, universe The Second Law of Thermodynamics: Entropy change measures the dispersal of energy In any spontaneous process, there is always an increase in the entropy of the universe. How much energy is spread out in a particular process, or how widely spread out it becomes at a specific temperature. universe = system + surroundings > 0 Every process has a preferred direction natural spontaneity. For a spontaneous process, if there is a decrease in the system, then the heat is absorbed by the surroundings must have caused an increase in the surroundings such that overall there is an increase in the universe (i.e. 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 Entropy 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. Entropy is a measure of available micro-states. Chamber 1 contains 4 oxygen atoms Consider two chambers containing a few oxygen atoms. w x y z b c a Chamber 2 contains 3 oxygen atoms Define a system such that each atom can reside in 5 quantized states. The energy of small particles are quantized. 11 12 2
Micro-states and Entropy Let s heat chamber 2. Heating causes the atoms move faster. After heating, there are 10 microstates with energy = 12 q b c a Heat chamber 2 Recall the Second Law of Thermodynamics: There is an inherent direction in which any system which is not in equilibrium moves. In any spontaneous process, the entropy of the universe will increase. In terms of micro-states, it means that all spontaneous process must increase the number of available micro-states. (i.e. The number of available micro-states after the event/reaction will always be greater than the number of available micro-states before the event/reaction). What happens if we bring the heated Chamber 2 in contact with Chamber 1? After heating, the oxygen atoms have more entropy. The number of available micro-states increases.(10 micro-states from 1 micro-state) 13 14 Chamber 1 contains 4 oxygen atoms Energy = 8 1 micro-state w x y z q b a c q Let s look at what an increase of energy dispersion mean for the process of heat spontaneous flow from a warm body to a cold body. Chamber 2 contains 3 oxygen atoms Energy =12 10 micro-states Chamber 1 contains 4 oxygen atoms Energy = 10 4 micro-states w x y z b c a Let s look at what an increase of energy dispersion mean for the process of heat spontaneous flow from a warm body to a cold body. Chamber 2 contains 3 oxygen atoms Energy =10 6 micro-states The total number of micro-states for this system in the initial state is 10 micro-states (1x10). The total number of micro-states for this system in the final state is 24 micro-states (4x6). Micro-state 1: Micro-state 1: Micro-state 4: Micro-state 2: Micro-state 3 etc: 15 Micro-state 2: Micro-state 3: Micro-state 5: Micro-state 6 etc: Nature spontaneously proceeds toward the states that have the highest probabilities of existing. 16 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 Entropy 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. Image credit: Chemistry, The Central Science, 10th edition Theodore L. Brown, H. Eugene LeMay, Jr., and Bruce E. Bursten 17 18 3
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 Entropy increases with the freedom of motion of molecules. Therefore, S(g) > S(l) > S(s) More spontaneous processes: - Free expansion of a gas > 0 This process neither absorbs nor emits heat. - Combustion of methane < 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) > 0 This process requires heat (endothermic). Conclusion: Naturally spontaneous processes could be exothermic, endothermic, or not involving heat. 19 20 More spontaneous processes: Photosynthesis A good example is photosynthesis. <0; It appears that there a decrease in entropy in this natural process. Photosynthesis in a plant does not consist of a isolated system of the plant alone. 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. 21 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! 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. 22 For any spontaneous process, Entropy measures the dispersal of energy universe = system + surroundings > 0 At a specific temperature, the change in entropy,, is Spontaneous process, favoured by a decrease in H (exothermic) favoured by an increase in where q rev is the heat transferred at a constant temperature T in Kelvin. Nonspontaneous process, favoured by an increase in H (endothermic) favoured by an decrease in What is q rev? 23 24 4
What is q rev? Entropy, What is q rev? Consider a gas in a piston and cylinder configuration. The gas is held in place by a pile of wheat grains. 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. Likewise, consider a gas in a piston and cylinder configuration. The gas is held in place by a pile of wheat grains. 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. w rev 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. w irrev 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 the observed heat transferred. 25 Of all the different possible paths to take the gas from State I to F, q rev is the limit of observed heat transferred. 26 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. Entropy 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,, 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. Consider ice melting at 0 o C, where H fusion = 79.71 cal/g = 6.003 kj/mole This is a reversible process. Reversibility is an idealization that is unachievable in real process, except when the system is in equilibrium. A process that is carried out reversibly has the smallest w rev, and the largest q rev. 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. 27 28 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. 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. 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. It follows that There is an increase in the entropy of water when water is converted to steam. Note: for the vapourization of water is much higher than that of fusion. w rev or 29 Entropy increases with an increase in volume or a decrease in pressure. 30 5
Change in Entropy in the surroundings, surroundings Change in Entropy in the surroundings, surroundings Entropy Changes in Surroundings Heat that flows into or out of the system also changes the entropy of the surroundings. For an isothermal process: 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 At constant pressure, q sys is H for the system. If this is done reversibly: surr = sys 31 32 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, system, surroundings, and universe. 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 system, surroundings, and 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. This is an irreversible process with no change in temperature, T=0. As a result, E = H = 0 Since E depends only on temperature, E = 0 and H = E + (PV) = E + (nrt) = 0 w = 0 because P ext = 0 w = -q, q =0 To calculate, 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 system. What is surroundings? What is universe? Since q surr = -q sys, q surr = -1.574 kj q rev = T = 5.764 273 = 1.574 kj Since E = 0, w = -q = -1.574 kj This is system. It has the same disorder as calculated in Part (a). What is surroundings? What is universe? This is a spontaneous process. universe > 0 Since the no heat flowed into the surroundings, q surr = 0, This is true for a reversible process. The system is at equilibrium. 33 34 Entropy 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 Entropy 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 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 H Compare 35 36 6
Calculate for the change in state Calculate for the change in state 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. > 0 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) Warming liquid water, 1 Cooling gaseous water, 3 2 < 0 Warming liquid water, 1 Cooling gaseous water, 3 > 0 2 = 1 + 2 + 3 = 4.04 + 26.04-2.02 = 28.1 cal K -1 = 1 + 2 + 3 37 38 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 o vapourization = 22 cal mole -1 K -1 or 88 J mole -1 K -1 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 o vapourization of different liquids should be similar. Trouton s rule is useful method for estimating the enthalpy of vapourization of a liquid if its normal boiling point is known. 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 Trouton s Rule Substance H vap (kcal mole -1 ) T b ( o C) vap (cal mole -1 K -1 ) O 2 1.630-182.97 18.07 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 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 (CH 3) 2CO acetone o vapourization = 22 cal mole -1 K -1 or 88 J mole -1 K -1 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. 39 40 The Third Law of Thermodynamics: The entropy of a pure perfect crystal is zero when the crystal is at zero Kelvin (0 K) A pure perfect crystal is one in which every molecule is identical, and the molecular alignment is perfectly even throughout the substance. 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. 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. 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. Image credit: Chemistry, The Central Science, 10th edition Theodore L. Brown, H. Eugene LeMay, Jr., and Bruce E. Bursten 41 42 7
Standard Molar o Standard Molar o Standard entropies (measured at 298K) have been carefully measured for many substances. Standard entropies (measured at 298 K) 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 Silver 42.6 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. 43 44 Standard Molar o Standard Molar 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 diatomic gases. Standard entropies (measured at 298 K) have been carefully measured for many substances. For example, the standard molar entropies (standard entropy per mole) for more 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. CH 4 (g) 186 C 2 H 2 (g) 201 C 2 H 4 (g) 221 C 2 H 6 (g) 229.5 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. Image credit: Chemistry, The Central Science, 10th edition Theodore L. Brown, H. Eugene LeMay, Jr., and Bruce E. Bursten 45 46 Standard Molar o Standard Entropy for Aqueous Ions For example, the standard molar entropies (standard entropy per mole) for compounds. H 2 O (l) 69.9 H 2 O (g) 188.7 H 2 O 2 (l) 110.0 CO (g) 197.8 CH 3 OH(l) 126.9 CO 2 (g) 213.6 NO (g) 210.6 CH 3 Cl(l) 145.3 NO 2 (g) 240.4 CHCl 3 (l) 202.9 N 2 O 4 (g) 304.3 SO 2 (g) 248.4 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. 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 A NO molecule has only one type of vibrational molecule A NO 2 molecule has three types of vibration. H + (aq) + OH - (aq) H 2 O (l) o = S o (H 2 O (l)) (S o (H + (aq)) - S o (OH - (aq)) o = 70.0 (0-10.7) o = 80.7 J K -1 47 48 8
reaction o Calculate the change in Entropy of a reaction from Standard Molar Entropy. 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) =? 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. 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))) o = 2(213.6) + 3 (188.7) (229.5 + 7 / 2 (205.1)) o = 46.3 J K -1 49 50 Summary: The Second Law of Thermodynamics: In any spontaneous process, there is always an increase in the entropy of the universe. universe = system + surroundings > 0 Change in entropy,, at a specific temperature, T. Change in entropy,, 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. 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, o can be calculated from the reactants and products standard molar entropies. 51 9