CHM 113 Enthalpy Entropy PDF

Chapter 14 (and 15.4): Entropy and Free Energy Key topics: Spontaneous processes Entropy S Gibbs free energy G Begin with a review of some key ideas from Chapter 10 Why do we “run” chemical reactions? o we want the products of the reaction o we want the energy from the reaction Thermodynamics...

Chapter 14 (and 15.4): Entropy and Free Energy Key topics: Spontaneous processes Entropy S Gibbs free energy G Begin with a review of some key ideas from Chapter 10 Why do we “run” chemical reactions? o we want the products of the reaction o we want the energy from the reaction Thermodynamics o study of the interconversion of different forms of energy o separate the universe into system and surroundings The system is the part of the universe that is of interest, including the substances involved in a chemical reaction. The surroundings are everything else. Exothermic reaction: Energy transferred from the system to the surroundings. 2H2(g) + O2(g) → 2H2O(l) + energy Endothermic reaction: Energy transferred from the surroundings to the system. energy + 2HgO(s) → 2Hg(l) + O2(g) First law of thermodynamics: conservation of energy Energy cannot be created or destroyed: ΔUuniverse = 0 ΔUuniverse = ΔUsys + ΔUsurr ⇒ ΔUsys = -ΔUsurr Any process that causes energy to be exchanged between the system and the surroundings can be classified as either heat or work. ΔU = q + w q is heat o q > 0, system absorbs heat o q < 0, system releases heat w is work o w > 0, work is done on the system o w < 0, work is done by the system Enthalpy, H = U + PV o thermodynamic state function o cannot be measured directly o at constant pressure, ΔH = q Enthalpy of reaction o ΔH = ΔHrxn = ΔHproducts - ΔHreactants o ΔH > 0, endothermic reaction o ΔH < 0, exothermic reaction Standard enthalpies of formation, ΔH°f o heat change that results when 1 mole of a compound is formed from its constituent elements in their standard states o standard state is the most stable state found at 1 atm o ΔH°f for elements in their most stable state are zero o ΔH°f for many substances are found in Appendix 2 Standard enthalpy of reaction, ΔH°rxn For the generic chemical reaction aA + bB ! cC + dD Hrxn = [c Hf (C) + d Hf (D)] [a Hf (A) + b Hf (B)] Spontaneous Processes Spontaneous process: does occur under a specific set of conditions Non-spontaneous process: does not occur under a specific set of conditions (note: a spontaneous process might occur quickly or slowly) In Chem 1311, we used enthalpy (ΔH) to predict spontaneity. o ΔH < 0 (exothermic): spontaneous o ΔH > 0 (endothermic): not spontaneous But ΔH is not always a good guide. o water evaporates (ΔH > 0 but spontaneous) o water does not freeze at room temperature (ΔH < 0 but not spontaneous) o ammonium nitrate dissolves in water [cold packs] (ΔH > 0 but spontaneous) Something else must matter… what else is changing? o water evaporates: the system’s energy is more spread out (in space) o water does not freeze at room temperature: the system’s energy would become less spread out because the solid (ice) has less freedom of movement than the liquid (water) o ammonium nitrate dissolves in water: the system’s energy is more spread out because the ionic compound (solid) breaks into soluble ions Entropy (S): a measure of how spread out or dispersed the system’s energy is More spread out or dispersed = increased entropy (ΔS > 0) Nature tends to move towards states corresponding to an increase in entropy. Quantitative definition of entropy (Boltzmann) S = k ln W o k = Boltzmann constant (k = R / NA = 1.38 x 10-23 J K-1) o W = number of energetically equivalent microstates Macrostate: thermodynamic state of the system (P, T, V, n, U) Microstates: o specific arrangement of particles o many microstates correspond to one macrostate o we need to “count” the number of microstates W that correspond to the macrostate (to get the entropy) Consider the arrangement of 4 gas particles into 2 bulbs. The distribution with the largest number of (energetically equivalent) microstates is the most probable. It also has the most dispersed energy state (highest entropy). Entropy is a state function just like enthalpy S = Sfinal Sinitial and processes that increase the number of microstates correspond to more probable states and are favorable (ΔS > 0) o increase in entropy is favorable (spontaneous) Standard Entropy (S°): the absolute entropy of a substance at 1 atm (and typically at 25°C) Note that the units are J K-1 mol-1 (vs. kJ mol-1 for enthalpies) Standard entropies are always positive (even for elements in their standard states, unlike ΔH°f) Important trends: 1. S° for the gas phase is greater than the liquid or solid phase of the same substance H2O(g) has greater S° than H2O(l) 2. More complex structures have greater S° (with a similar molar mass, same phase) because more types of motion are possible O3(g) greater than F2(g) C2H6(g) greater than CH4(g)  3. Allotropes: the more ordered forms have lower S° Diamond has lower S° than graphite The carbon atoms in diamond are in a rigid 3-dimensional structure whereas in graphite the 2-dimensional sheets are free to slide by one another. Therefore graphite has a higher standard entropy than diamond. 4. Monatomic species: the heavier atom has greater S° helium has lower S° than neon (due to quantization of translational energy; higher mass corresponds to more closely spaced energy levels) Entropy changes in a system o S°liquid > S°solid o S°vapor > S°liquid  o S°aqueous > S°pure solid Hydrated ions organize the water around them which leads to a decrease in solvent entropy. When the charges are small (±1) the solute entropy dominates (S°system > 0), but when the charges are large (e.g., Al3+, Fe3+) the solvent (water) entropy can dominate, leading to S°system < 0 e.g., Make a qualitative prediction of the sign of ΔH°soln for AlCl3(s) o S°higher temp > S°lower temp  o S°more moles gas > S°fewer moles gas o S°larger volume of gas > S°smaller volume of gas e.g., Determine the sign of ΔS for the following (qualitatively) 1. Liquid nitrogen evaporates. 2. Aqueous solutions of potassium iodide and lead nitrate are mixed together. 3. Water is heated from room temperature to 60°C.  Entropy Changes in the Universe and The Second and Third Laws of Thermodynamics The universe is composed of two parts: o the system o the surroundings (everything else) Both parts undergo changes in S during physical and chemical processes Second Law of Thermodynamics: the entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process. Spontaneous: ΔSuniverse = ΔSsystem + ΔSsurroundings > 0 ΔSsystem can be negative as long as ΔSsurroundings is more positive Equilibrium process: ΔSuniverse = 0 e.g., water freezing at 0°C; chemical reaction where Q = K. Entropy changes in the system: Entropy can be calculated from a table of standard values just as enthalpy changes were calculated. For the general reaction aA + bB ! cC + dD Srxn = [cS (C) + dS (D)] [aS (A) + bS (B)] Entropy changes in the surroundings: We know how to calculate Ssystem = Srxn What about ΔSsurroundings ? Consider liquid water freezing below 0°C spontaneous, ΔSuniverse > 0 liquid → solid, ΔSsystem < 0 Therefore |ΔSsurr| > |ΔSsys| How does water freezing increase Ssurr ? o water freezing is exothermic o energy is being dispersed into the surroundings ΔSsurr > 0 for exothermic processes ΔSsurr < 0 for endothermic processes Therefore Ssurroundings / Hsystem (/ q) But water freezing above 0°C is not spontaneous. The magnitude of ΔSsurr depends on the temperature o T < 0°C, |ΔSsurr| > |ΔSsys| o T > 0°C, |ΔSsurr| < |ΔSsys| o but ΔSsys is virtually independent of T o thus |ΔSsurr| depends on T o ΔSsurr is inversely proportional to T 1 Ssurr / T Combining these two relations gives Hsys Ssurr = T e.g., Predict if the synthesis of ammonia at 25°C is spontaneous or not. Given information: N2 (g) + 3H2 (g) ! 2NH3 (g) Hrxn = 92.6 kJ/mol 1 1 S [N2 (g)] = 191.5 J K mol S [H2 (g)] = 131.0 J K 1 mol 1 S [NH3 (g)] = 193.0 J K 1 mol 1 Solution: ΔSsys = ΔSrxn = 2(193.0) – (191.5) – 3(131.0) = -198.5 J K–1 mol–1 ΔSsurr = - (-92.6 kJ mol–1) / 298 K = 311 J K–1 mol–1 ΔSuniverse = ΔSsys + ΔSsurr = 112 J K–1 mol–1 spontaneous e.g., Calculate ΔSuniverse and determine the spontaneity of the reaction H2O2(l) → H2O2(g) at 163°C. Given information (from Appendix 2): 1 1 S [H2 O2 (l)] = 109.6 J K 1 mol Hf [H2 O2 (l)] = 187.6 kJ mol 1 1 S [H2 O2 (g)] = 232.9 J K 1 mol Hf [H2 O2 (g)] = 136.1 kJ mol Solution: ΔHrxn = -136.1 – (-187.6) = 51.5 kJ mol–1 ΔSsurr = - (51.5 kJ mol–1) / 436 K = -118 J K–1 mol–1 ΔSsys = ΔSrxn = 232.9 – 109.6 = 123.3 J K–1 mol–1 ΔSuniverse = ΔSsys + ΔSsurr = 5.3 J K–1 mol–1 spontaneous e.g., The reaction NH3(g) + HCl(g) → NH4Cl(s) is spontaneous at room temperature. Determine the temperature at which it is no longer spontaneous. Given information (from Appendix 2): 1 1 1 S [NH3 (g)] = 193.0 J K mol Hf [NH3 (g)] = 46.3 kJ mol 1 1 1 S [HCl(g)] = 187.0 J K mol Hf [HCl(g)] = 92.3 kJ mol 1 1 1 S [NH4 Cl(s)] = 94.56 J K mol Hf [NH4 Cl(s)] = 315.39 kJ mol Solution: ΔHrxn = -315.4 – (-46.3 – 92.3) = -176.8 kJ mol–1 ΔSsys = ΔSrxn = 94.56 – (193.0 + 187.0) = -285.44 J K–1 mol–1 ΔSuniverse = ΔSsys + ΔSsurr = -0.28544 +(176.8 / T) > 0 This inequality holds as long as T < 176.8 / 0.28544 = 619 K e.g., Determine the boiling point of Br2. Given information (from Appendix 2): S [Br2 (l)] = 152.3 J K 1 mol 1 Hf [Br2 (l)] = 0 kJ mol 1 1 1 1 S [Br2 (g)] = 245.13 J K mol Hf [Br2 (g)] = 30.7 kJ mol Solution: For the equilibrium Br2(l) Br2(g) –1 ΔHrxn = 30.7 kJ mol ΔSsys = ΔSrxn = 245.13 – 152.3 = 92.8 J K–1 mol–1 ΔSuniverse = ΔSsys + ΔSsurr = 0 ⇒ T = 331 K = 58°C Third law of thermodynamics The entropy of a perfect crystalline substance at T = 0 K is zero. S = k ln W = k ln 1 = 0 The importance of this law is that is allows us to calculate absolute entropy values (S°) Gibbs Free Energy Recall that the 2nd Law of Thermodynamics says that a process is spontaneous if ΔSuniverse = ΔSsystem + ΔSsurroundings > 0 It would be convenient if we could predict spontaneity based only on what happens to the system. We can do this using ΔSsurr = –ΔHsys / T because then ΔSuniverse = ΔSsys – ΔHsys / T > 0, or TΔSsys – ΔHsys > 0, or ΔHsys – TΔSsys < 0 Based on this inequality, we define the Gibbs Free Energy as G = H – TS At constant temperature and pressure (ΔH = q), we have (where all terms are for the system): G= H T S G < 0 ) spontaneous in the forwards direction G > 0 ) not spontaneous in the forwards direction G = 0 ) the system is at equilibrium Case 1: ΔH < 0 and ΔS > 0 ΔG = ΔH – TΔS = (–) – T(+) < 0 always e.g., 2H2O2(aq) → 2H2O(l) + O2(g) Case 2: ΔH > 0 and ΔS < 0 ΔG = ΔH – TΔS = (+) – T(–) > 0 always e.g., 3O2(g) → 2O3(g) Case 3: ΔH > 0 and ΔS > 0 ΔG = ΔH – TΔS = (+) – T(+) < 0 at high T e.g., ice melting; water evaporating Case 4: ΔH < 0 and ΔS < 0 ΔG = ΔH – TΔS = (–) – T(–) < 0 at low T e.g., water freezing; water vapor condensing (from http://ch301.cm.utexas.edu) Determining T required for spontaneity: What does “high T” or “low T” mean? For water, we know the answer: o water freezing: H2O(l) → H2O(s) spontaneous at low T (below 0°C) o water evaporating: H2O(l) → H2O(g) spontaneous at high T (above 100°C) In general, we set ΔG = 0 (equilibrium condition): 0 = ΔH – TΔS ⇒ T = (ΔH) / (ΔS) T in Kelvin Note: for phase changes this relation also allows us to compute the entropy change ΔS = (ΔH) / (T) o H2O(s) H2O(l) ΔH = ΔHfusion; T = melting point o H2O(l) H2O(g) ΔH = ΔHvap; T = boiling point e.g., The reaction 4Fe(s) + 3O2(g) + 6H2O(l) 4Fe(OH)3(s) is spontaneous below 1950°C. Estimate ΔS° for Fe(OH)3(s). Given information (from Appendix 2): 1 1 S [Fe(s)] = 27.2 J K 1 mol Hf [Fe(s)] = 0 kJ mol 1 1 S [O2 (g)] = 205.0 J K 1 mol Hf [O2 (g)] = 0 kJ mol 1 1 1 S [H2 O(l)] = 69.9 J K mol Hf [H2 O(l)] = 285.8 kJ mol 1 1 1 S [Fe(OH)3 (s)] = x J K mol Hf [Fe(OH)3 (s)] = 824.25 kJ mol Solution: From ΔG = ΔH – TΔS, at T = 1950°C we have ΔS = (ΔH) / (T). We must convert to Kelvin and solve for the missing data. ΔS = 4x -4(27.2) -3(205.0) -6(69.9) = 4x – 1143.2 ΔH = 4(-824.25) – 6(-285.8) = -1582.2 4x 1143.2 1582.2 1 1 = ) x = 107.9 J K mol 1000 1950 + 273 Standard free energy changes, ΔG° The standard free energy change of reaction ΔG°rxn is the ΔG value under standard state conditions, which are o Gases 1 atm pressure o Liquids pure liquid o Solids pure solid o Elements most stable allotrope at 1 atm, 25°C o Solutions 1 molar concentration ΔG° can be calculated from a table of standard values (ΔG°f) just like enthalpy and entropy values aA + bB ! cC + dD Grxn = [c Gf (C) + d Gf (D)] [a Gf (A) + b Gf (B)] Here ΔG°f is the standard free energy of formation of a compound, namely the free energy change that occurs when 1 mole of the compound is synthesized from its consistuent elements, each in their standard state. Therefore, ΔG°f = 0 for elements in their standard states. ΔG° values at different temperatures: In the equation ΔG° = ΔH° – TΔS° the ΔH° and ΔS° terms are somewhat insensitive to changes in temperature, so to get a rough estimate of ΔG° we can do: GT = H298 T S298 Chemical Equilibrium and Free Energy Relationship between ΔG and ΔG° We can get ΔG° from tables (Appendix 2) but we need ΔG to predict spontaneity. For ΔG, there is a relationship with Q vs K Notice that ln(Q/K) has the same sign as ΔG In fact, they are proportional to each other G / ln(Q/K) G = RT ln(Q/K) G = RT ln(Q) RT ln(K) Now, if we put everything at standard conditions, then o ΔG = ΔG° o Q = 1 since all concentrations 1 M, all pressures 1 atm Therefore we have the relation G = RT ln K K is the equilibrium constant, which we can now calculate if we have the available thermodynamic data. e.g., Calculate ΔG° for BaF2(s) Ba2+(aq) + 2F–(aq) given Ksp = 1.7 x 10–6 for BaF2(s) at 25°C. Solution: ΔG° = –RT ln K = -(8.314 x 10–3 kJ mol–1K–1)(298 K)(-13.285) = 32.9 kJ/mol ΔG° o is constant for a specific reaction at a specific T o contains the same information as the magnitude of K o if we change T, K and ΔG° change: ΔG°T = –RT ln KT Note: this gives us a new way to predict how temperature affects equilibrium (previously, we used Le Chatelier’s Principle based on the sign of ΔH) ΔG o contains the same information as Q vs K comparison o equal to the maximum work that can be done (expansion work, electrical work, or chemical work) o changes as the reaction proceeds (→ zero at equilibrium) Since ΔG° is related to the logarithm of K, small changes in ΔG° lead to large changes in K. We can use this to determine very large or very small K values from thermodynamic data. Put everything together: ( G = RT ln(Q) RT ln(K) G} + RT ln Q G = | {z | {z } G = RT ln K fixed varies e.g., H2(g) + I2(s) 2HI(g) ΔG° = 2.60 kJ/mol at 25°C Calculate ΔG with initial values PH2 = 5.25 atm; PHI = 1.75 atm Solution: The reaction quotient is Q = (1.75)2 / (5.25) = 0.583 ΔG = 2.60 kJ/mol + (8.314 x 10–3 kJ mol–1K–1)(298 K)(-0.539) = 1.26 kJ/mol. The reaction with thus run backwards spontaneously. Free energy diagram (free energy vs reaction coordinate) G = Gproducts Greactants - from tables - changes with T Lowest point is equilibrium ΔG is the slope at any point Can approach Text equilibrium from either side (Q vs K) ΔG tells us the direction the reaction needs to proceed to achieve equilibrium ΔG° tells us the position of equilibrium (K) Large |ΔG°| means the equilibrium lies further to the reactant or product side