#### Transcript Ch 20

Chapter 20 Entropy and Free Energy - Spontaneity of Reaction Thermodynamics: Entropy, Free Energy Direction of Chemical Reactions 1. The Second Law of Thermodynamics: Predicting Spontaneous Change First Law and enthalpy, Second Law and entropy, Standard Entropy and Third Law 2. Calculating the Change in Entropy of a Reaction The standard entropy of reaction, entropy change and equilibrium 3. Entropy, Free Energy, and Work Free energy change and spontaneity, Standard free energy changes 4. Free Energy, Equilibrium, and Reaction Direction Important criterions Spontaneous reaction, in chemical reaction terms, is one which occurs with the system releasing free energy in some form (often, but not always, heat) and moving to a lower energy, hence more thermodynamically stable, state. Enthalpy, H, is the sum of the internal energy and the product of the pressure and volume of a thermodynamic system. Entropy, S, is a state function which measures the degree of disorder (randomness) of a system. The larger the disorder, the greater the value of S in units of J/K. Free Energy is the measure of the spontaneity of a process and of the useful energy available from it. The First Law of Thermodynamics indicates that to conserve energy, the change in energy must equal the heat transferred plus the work performed, E = q + w. The Second Law of Thermodynamics states that for a spontaneous process, ∆Suniverse = ∆Ssystem + ∆Ssurroundings > 0 (∆Suniverse positive) Spontaneous reaction and exothermic /endothermic Common mistake – exothermic is spontaneous. Figure 1 A spontaneous endothermic chemical reaction. water Ba(OH)2.8H2O(s) + 2NH4NO3(s) Ba2+(aq) + 2NO3-(aq) + 2NH3(aq) + 10H2O(l) H0rxn = + 62.3 kJ The Concept of Entropy (S) Entropy refers to the state of order, measures the degree of disorder of a system. . A change in order is a change in the number of ways of arranging the particles, and it is a key factor in determining the direction of a spontaneous process. more order solid more order crystal + liquid more order crystal + crystal less order liquid gas less order ions in solution less order gases + ions in solution Figure 2 Spontaneous expansion of a gas stopcock closed evacuated 1 atm In thermodynamic terms, a change in order is the number of ways of arranging the particles. The change in order is likely the key factor to determine the direction of a spontaneous reaction. The number of possible arrangement is increased. stopcock opened 0.5 atm 0.5 atm The number is of ways of arranging a system is related to the entropy. 1877 Ludwig Boltzman S = k ln W where S is entropy, W is the number of ways of arranging the components of a system, and k is a constant (the Boltzman constant), R/NA (R = universal gas constant, NA = Avogadro’s number. A system with relatively few equivalent ways to arrange its components (smaller W) has relatively less disorder and low entropy. A system with many equivalent ways to arrange its components (larger W) has relatively more disorder and high entropy. Suniverse = Ssystem + Ssurroundings > 0 This is the second law of thermodynamics. Figure 4 Random motion in a crystal The third law of thermodynamics. A perfect crystal has zero entropy at a temperature of absolute zero. Ssystem = 0 at 0 K Predicting Relative S0 Values of a System 1. Temperature changes S0 increases as the temperature rises. 2. Physical states and phase changes S0 increases as a more ordered phase changes to a less ordered phase. 3. Dissolution of a solid or liquid S0 of a dissolved solid or liquid is usually greater than the S0 of the pure solute. However, the extent depends upon the nature of the solute and solvent. 4. Dissolution of a gas A gas becomes more ordered when it dissolves in a liquid or solid. 5. Atomic size or molecular complexity In similar substances, increases in mass relate directly to entropy. In allotropic substances, increases in complexity (e.g. bond flexibility) relate directly to entropy. Thermodynamics: Entropy, Free Energy, and the Direction of Chemical Reactions 1 Three Laws of Thermodynamics: First Law : ∆Universe = 0, Second Law: ∆S>0, G<0, Third law : Ssolid=0 2 Calculating the Change in Entropy of a Reaction ∆S0 = mS0products - nS0reactants Combine standard molar entropies to find the standard entropy for a reaction 3 Entropy, Free Energy, and Work G = H-TS, ∆ G= ∆H- ∆TS, ∆G= ∆H- T∆S, ∆G0= ∆H0 - T∆S0, 4 Free Energy, Equilibrium, and Reaction Direction Lecture 2 Summary Spontaneous reaction happens at a specific condition without continuous input of energy. Neither first law of thermodynamic or the sign of ∆H predicts the direction of reaction. The spontaneous reaction involves a change in the universe from more to less order. Entropy is the measure of disorder and directly related to the number of ways of arrangement for components in the system. The second law of thermodynamic states that the entropy of universe increases in a spontaneous process. Entropy value are absolute since the pure single crystal has a zero entropy at absolute zero degree. Standard molar entropy is dependant on the temperature, phase change, dissolution, atomic size, and molecular complexity. Sample Problem Predicting Relative Entropy Values PROBLEM: Choose the member with the higher entropy in each of the following pairs, and justify your choice [assume constant temperature, except in part (e)]: (a) 1mol of SO2(g) or 1mol of SO3(g) (b) 1mol of CO2(s) or 1mol of CO2(g) (c) 3mol of oxygen gas (O2) or 2mol of ozone gas (O3) (d) 1mol of KBr(s) or 1mol of KBr(aq) (e) Seawater in midwinter at 20C or in midsummer at 230C (f) 1mol of CF4(g) or 1mol of CCl4(g) PLAN: In general less ordered systems have higher entropy than ordered systems and entropy increases with an increase in temperature. SOLUTION: (a) 1mol of SO3(g) - more atoms (d) 1mol of KBr(aq) - solution > solid (b) 1mol of CO2(g) - gas > solid (e) 230C - higher temperature (c) 3mol of O2(g) - larger #mols (f) CCl4 - larger mass Important criterions Spontaneous reaction, in chemical reaction terms, is one which occurs with the system releasing free energy in some form (often, but not always, heat) and moving to a lower energy, hence more thermodynamically stable, state. Entropy, S, measures the degree of disorder of a system. The larger the disorder, the greater the value of S in units of J/K. Free Energy, G, is the measure of the spontaneity of a process and of the useful energy available from it. When a reaction occurs, chemists expect to learn how to predict and calculate the change in entropy using the following equation: ∆S0 = mS0products - nS0reactants Sample Problem PROBLEM: Calculating the Standard Entropy of Reaction, S0rxn Calculate S0rxn for the combustion of 1mol of propane at 25 0C. C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) PLAN: Use summation equations. It is obvious that entropy is being lost because the reaction goes from 6 mols of gas to 3 mols of gas. SOLUTION: Find standard entropy values in the Appendix or other table. S = [(3 mol)(S0 CO2) + (4 mol)(S0 H2O)] - [(1 mol)(S0 C3H8) + (5 mol)(S0 O2)] S = [(3 mol)(213.7J/mol*K) + (4 mol)(69.9J/mol*K)] - [(1 mol)(269.9J/mol*K) + (5 mol)(205.0J/mol*K)] Lecture 2 S = - 374 J/K Determining Reaction Spontaneity In many spontaneous reaction, the system become more ordered at a given condition, ∆S0rxn<0. The Second Law of Thermodynamics indicated that the decrease in entropy of a system only can occur when the increase the entropy in surrounding overweigh them. The entropy change of the surrounding is related to an opposite change in the heat of the system. ∆S0surr - qsys The entropy change of the surrounding is also reversely proportional to the temperature of the heat transferred. ∆S0surr 1/T Combining these changes, we obtain the equation: ∆S0surr = ∆S0 Lecture 2 surr = - qsys T ∆ Hsys T When a pressure is constant, the heat is ∆H. ( ∆H = q + ∆ PV) Sample Problem PROBLEM: Determining Reaction Spontaneity At 298K, the formation of ammonia has a negative S0sys; N2(g) + 3H2(g) 2NH3(g) S0sys = -197 J/K Calculate S0rxn, and state whether the reaction occurs spontaneously at this temperature. PLAN: S0universe must be > 0 in order for this reaction to be spontaneous, so S0surroundings must be > 197 J/K. To find S0surr, first find Hsys; Hsys = Hrxn which can be calculated using H0f values from tables. S0universe = S0surr + S0sys. SOLUTION: H0rnx = [(2 mol)(H0fNH3)] - [(1 mol)(H0fN2) + (3 mol)(H0fH2)] H0rnx = -91.8 kJ S0surr = -H0sys/T = -(-91.8x103J/298K) = 308 J/K S0universe = S0surr + S0sys = 308 J/K + (-197 J/K) = 111 J/K S0universe > 0 so the reaction is spontaneous. Lecture 2 Gibbs Free Energy (G) G, the change in the free energy of a system, is a measure of the spontaneity of the process and of the useful energy available from it. G0system = H0system - TS0system G < 0 for a spontaneous process G > 0 for a nonspontaneous process G = 0 for a process at equilibrium G0rxn = S mG0products - S nG0reactants Lecture 2 Sample Problem PROBLEM: Calculating G0 from Enthalpy and Entropy Values Potassium chlorate, a common oxidizing agent in fireworks and matchheads, undergoes a solid-state disproportionation reaction when heated. Note that the oxidation number of Cl in the reactant is higher in one of the products and lower in the other (disproportionation). +5 4KClO3(s) +7 -1 3KClO4(s) + KCl(s) Use H0f and S0 values to calculate G0sys (G0rxn) at 250C for this reaction. PLAN: Use Appendix B values for thermodynamic entities; place them into the Gibbs Free Energy equation and solve. SOLUTION: H0rxn = S mH0products - S nH0reactants H0rxn = (3 mol)(-432.8 kJ/mol) + (1 mol)(-436.7 kJ/mol) (4 mol)(-397.7 kJ/mol) H0rxn = -144 kJ Lecture 2 Sample Problem Calculating G0 from Enthalpy and Entropy Values continued S0rxn = S mS0products - S nS0reactants S0rxn = (3 mol)(151 J/mol*K) + (1 mol)(82.6 J/mol*K) (4 mol)(143.1 J/mol*K) S0rxn = -36.8 J/K G0rxn = H0rxn - T S0rxn G0rxn = -144 kJ - (298K)(-36.8 J/K)(kJ/103 J) G0rxn = -133 kJ Lecture 2 Entropy, Free Energy, Work Free Gibbs energy is the measure of the spontaneity of a process. • ∆G0sys = ∆H0sys - T∆S0sys • T∆S0univ = - (∆H0sys - T∆S0sys) • The sign for ∆G0sys and ∆S0univ can tell us the spontaneity of a reaction. • ∆G0sys < 0 for a spontaneous reaction; • ∆G0sys = 0 for a reaction at equilibrium; • ∆G0sys > 0 for a nonspontaneous reaction; • ∆S0univ > 0 for a spontaneous reaction; • ∆S0univ = 0 for a reaction at equilibrium; • ∆S0univ < 0 for a nonspontaneous reaction. Lecture 2 The spontaneity of reaction and the sign of ∆H, ∆G, ∆S. ∆G0 = ∆H0 - T∆S0 ∆H0 ∆S0 -T∆S0 ∆G0 Description - + - - Spontaneous at all T + - + + Non-spontaneous at all T + + - + or - Spontaneous at high T Non-spontaneous at low T - - + + or - Spontaneous at low T Non-spontaneous at high T Determining the effect of temperature on G PROBLEM: An important reaction in production of sulfuric acid is the oxidation of SO2 (g) to SO3(g), 2SO2(g) + O2(g) 2SO3(g) At 298K, ∆G0 = -141.6 kJ/mol, ∆H0 = -198.4 kJ/mol, ∆S0 = -187.9kJ/mol, (a). Use the data to predict if this reaction is spontaneous reaction at 25 C. (b). Predict how ∆G0 changes when temperature increases. (c). Assume ∆H0 and ∆S0 are constant with increasing T, is the reaction spontaneous at 900 C? PLAN: Note the sign of ∆G0, ∆H0, and T∆S0 Use the equations ∆G0 = ∆H0 - T∆S0 Determining the effect of temperature on G SOLUTION: (a) Noting G = -141.6 kJ, is negative, so the reaction at room temperature is spontaneous. (b) Noting G = H – TS = -198.4 - (- 0.1879T) = -198.4 + 0.1879T If T increases, the G is less negative. Therefore, the reaction is less spontaneous. (c) Calculating G = H – TS at 900 C = -198.4 - (- 0.1879T) = -198.4 + 0.1879(900+273.15) = 22.0 kJ At 900 C, G is positive. Therefore, the reaction is non- spontaneous. Free Energy, Equilibrium and Reaction Direction •If Q/K < 1, then ln Q/K < 0; the reaction proceeds to the right (G < 0) •If Q/K > 1, then ln Q/K > 0; the reaction proceeds to the left (G > 0) •If Q/K = 1, then ln Q/K = 0; the reaction is at equilibrium ( G = 0) G = RT ln Q/K = RT lnQ - RT lnK Under standard conditions (1M concentrations, 1atm for gases), Q = 1 and ln Q = 0 so G0 = - RT lnK Table The Relationship Between G0 and K at 250C G0(kJ) K 100 3x10-18 50 2x10-9 10 2x10-2 1 7x10-1 0 1 -1 1.5 -10 5x101 -50 6x108 -100 3x1017 -200 1x1035 Essentially no forward reaction; reverse reaction goes to completion Forward and reverse reactions proceed to same extent Forward reaction goes to completion; essentially no reverse reaction REVERSE REACTION 9x10-36 FORWARD REACTION 200 Significance Calculating G at Nonstandard Conditions PROBLEM: The oxidation of SO2, 2SO2(g) + O2(g) 2SO3(g) is too slow at 298K to be useful in the manufacture of sulfuric acid. To overcome this low rate, the process is conducted at an elevated temperature. (a) Calculate K at 298K and at 973K. (G0298 = -141.6kJ/mol of reaction as written using H0 and S0 values at 973K. G0973 = -12.12kJ/mol of reaction as written.) (b) In experiments to determine the effect of temperature on reaction spontaneity, two sealed containers are filled with 0.500atm of SO2, 0.0100atm of O2, and 0.100atm of SO3 and kept at 250C and at 700.0C. In which direction, if any, will the reaction proceed to reach equilibrium at each temperature? (c) Calculate G for the system in part (b) at each temperature. PLAN: Use the equations and conditions. G0 = -RTlnK Calculating G at Nonstandard Conditions SOLUTION: (a) Calculating K at the two temperatures: G0 = -RTlnK so (G / RT) Ke 0 (-141.6kJ/mol)(103J/kJ) At 298, the exponent is -G0/RT = - = 57.2 (8.314J/mol*K)(298K) (G / RT) Ke 0 = e57.2 = 7x1024 (-12.12kJ/mol)(103J/kJ) At 973, the exponent is -G0/RT = 1.50 (8.314J/mol*K)(973K) (G / RT) Ke 0 = e1.50 = 4.5 Calculating G at Nonstandard Conditions 2SO2(g) + O2(g) 2SO3(g) pSO32 (b) The value of Q = (pSO2)2(pO2) (0.100)2 = = 4.00 (0.500)2(0.0100) Since Q is < K at both temperatures the reaction will shift right; for 298K there will be a dramatic shift while at 973K the shift will be slight. (c) The nonstandard G is calculated using G = G0 + RTLnQ G298 = -141.6kJ/mol + (8.314J/mol*K)(kJ/103J)(298K)(Ln4.00) G298 = -138.2kJ/mol G973 = -12.12kJ/mol + (8.314J/mol*K)(kJ/103J)(973K)(Ln4.00) G298 = -0.9kJ/mol Thermodynamic Important equations • ∆G0sys = ∆H0sys - T∆S0sys • ∆S0univ = - qsys/T • ∆G0 = ∆H0 - T∆S0 • ∆S0 = mS0products - nS0reactants • ∆G0 = mG0products - nG0reactants • ∆G0 = - RT lnK • ∆G = ∆G0 - RT lnQ