Ch17.ppt - CHAPTER 17 Entropy and Free Energy 11 Review...

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Unformatted text preview: CHAPTER 17 Entropy and Free Energy 11 Review: System and Surroundings • SYSTEM – The object under study • SURROUNDINGS – Everything outside the system Universe = System + Surroundings 22 Review: Review: FIRST FIRST LAW LAW OF OF THERMODYNAMICS THERMODYNAMICS heat energy transferred ∆E = q + w energy change work done by the system Energy is conserved! 33 44 Review: Review: ENTHALPY ENTHALPY ∆H = q where H = enthalpy q = Heat transferred at constant P If If ∆H ∆H > 0 process is > 0 process is ENDOTHERMIC ENDOTHERMIC heat transfers from SURROUNDINGS to SYSTEM. If If ∆H ∆H < 0 process is < 0 process is EXOTHERMIC EXOTHERMIC heat transfers from SYSTEM to SURROUNDINGS. Spontaneous processes • Spontaneous process is a process that occurs without an outside intervention. • Examples: a basketball rolling down the hill; transfer of heat from a hot object to a cold object • A spontaneous process leads the system towards equilibrium. “Spontaneous” does not imply anything about the time it takes for process to occur. 55 66 Spontaneous processes • Product-favored reactions proceed spontaneously from reactants to products. • Most of the product-favored reactions are exothermic. Fe2O3(s) + 2 Al(s) ---> 2 Fe(s) + Al2O3(s) ∆H = - 848 kJ Spontaneous Reactions But not all spontaneous reactions are exothermic; many spontaneous reactions or processes are endothermic or have ∆H = 0. NH4NO3(s) + heat NH4NO3(aq) ∆ H= +25.7 kJ/mol 77 Entropy • In all spontaneous processes, the final state is more DISORDERED (“RANDOM“) than the original state. Spontaneity is related to an increase in disorder. • ENTROPY (S) is a property that describes the degree of disorder. • Greater disorder = higher value of S. 88 Entropy • The concept of entropy is based on the idea that spontaneous change results in dispersal of matter, dispersal of energy, or of both. • This is a statistical effect. • A state in which energy/matter is more dispersed has a higher probability of occurring, that’s why a system spontaneously goes there. 99 10 10 Dispersal of Matter 11 11 Dispersal of Matter • Solubility is a result of a dispersal of matter. • A solid KMnO4 dissolves in water because the ions tend to disperse over the entire container. 12 12 13 13 Dispersal of Energy Another key contribution to entropy is dispersal of energy over as many different energy states as possible. 14 14 Dispersal of Energy in exothermic reactions Exothermic reactions involve a release of stored chemical potential energy to the surroundings. The stored potential energy starts out in a few molecules but is finally dispersed over very many molecules. The final state—with energy dispersed—is more probable; and so the reaction is spontaneous. 15 15 Boltzmann’s Equation for Entropy 16 16 S = k ln W where S is entropy, W is the number of microstates (number of different ways of arranging the energy of a system), k is a constant (the Boltzmann constant), which is equal to: k = R/NA (R = universal gas constant, NA = Avogadro’s number). •A system with relatively few ways to distribute its energy (smaller W) has relatively less disorder and low entropy. •A system with many ways to distribute its energy (larger W) has relatively more disorder and high entropy. •The largest possible W (and highest entropy) is achieved at equilibrium. 17 17 Spontaneous expansion of a gas stopcock closed 1 atm evacuated stopcock opened 0.5 atm 0.5 atm 18 18 Expansion of a gas and the increase in number of microstates. 19 19 Expansion of a gas and the increase in number of microstates. Entropy 20 20 ENTROPY is the thermodynamic property that describes the degree of disorder. Like energy (E) and enthalpy (H), ENTROPY (S) is a state function. “State function” means: change in entropy ( S) for any process depends only on the initial and final states of the system, but not on the pathway by which the process occurs. In a spontaneous process, the entropy increases (the 2nd law of thermodynamics) 21 21 The entropy of liquid water is greater than the entropy of solid water (ice). S (H2O liq) > S (H2O solid) Entropy and States of Matter Entropy of a gas (vapor) is greater than entropy of a liquid. 22 22 Standard Molar Entropies 23 23 24 24 Entropy SSoo (J/K•mol) (J/K•mol) H H22O(liq) O(liq) 69.95 69.95 H H22O(gas) O(gas) 188.8 188.8 S (gases) > S (liquids) > S (solids) • Units of entropy are J/K•mol • Entropy depends on temperature. • So is the standard entropy of a substance (entropy at 298K) Entropy Entropy of a substance increases with temperature. Molecular motions of heptane, C7H16 Molecular motions of heptane at different temps. 25 25 Entropy Entropy and and Temperature Temperature S increases slightly with T S increases a large amount with phase changes 26 26 Entropy Increase in molecular complexity generally leads to increase in S. 27 27 28 28 Entropy and vibrational motion. NO NO2 N2O4 29 29 Entropy Entropies of ionic solids depend on coulombic attractions. SSoo (J/K•mol) (J/K•mol) Mg2+ & O2- Na+ & F- MgO MgO 26.9 26.9 NaF NaF 51.5 51.5 Entropy Entropy Change Change Entropy increases as a more ordered state changes to a more disordered state. solid more order crystal + liquid more order liquid gas less order ions in solution less order S > 0 S > 0 30 30 Entropy Entropy Change Change Entropy usually increases when a pure liquid or solid dissolves in a solvent. 31 31 32 32 The entropy change accompanying the dissolution of a salt. pure solid MIX pure liquid solution 33 33 The small increase in entropy when ethanol dissolves in water. Ethanol Water Solution of ethanol and water 34 34 The large decrease in entropy when a gas dissolves in a liquid O2 gas O2 gas in H2O 35 35 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 S 0 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, increase in molar mass relates directly to entropy. Increases in complexity (e.g. bond flexibility) increase the entropy. 36 36 Sample Problem 1 Entropy Comparisons PROBLEM: Choose the member with the higher entropy in each of the following pairs, and justify your choice: (a) 1mol of O2(g) or 1mol of O3(g) (b) 1mol of CF4(g) or 1mol of CCl4(g) (c) 3mol of oxygen gas (O2) or 2mol of ozone gas (O3) PLAN: In general less ordered systems have higher entropy than ordered systems. SOLUTION: (a) 1mol of O3(g) - more atoms, larger molecular complexity. (b) CCl4 - larger molar mass (c) 3mol of O2(g) - larger #mols, more randomness in arranging molecules. 37 37 Sample Problem 2 Entropy Comparisons PROBLEM: Determine the sign of S for each of the following processes: (a) CO2(s) ---> CO2(g) (b) KBr(s) ---> KBr(aq) (c) 3H2(g) + N2(g) ---> 2NH3(g) PLAN: Entropy is higher when a system is more disordered; entropy depends on state of matter. SOLUTION: (a) S is positive because gas is more disordered than solid; entropy increases. (b) S is positive because aqueous solution is more disordered than solid crystal; entropy increases. (c) S is negative because we went from 4 moles of gas (add coefficients: 3+1=4) down to 2 moles of gas; some entropy is lost. Entropy changes in chemical reactions The standard entropy change during a reaction (∆Srxno) is the entropy change that occurs when all reactants and products are in their standard states. oo oo oo ∆∆SSrxn = S (products) S (reactants) = S (products) S (reactants) rxn ∆Srxno = the sum of standard entropies (So) of the products minus the sum of standard entropies of reactants. Take values of So for each involved substance from reference table (appendix L). 38 38 Calculating ∆S for a Reaction o oo oo oo ∆∆SSrxn = S (products) S (reactants) = S (products) S (reactants) rxn Ex. Calculate ∆Srxno for reaction: 2 H2(g) + O2(g) 2 H2O(l) ∆So = 2So(H2O) - [2So(H2) + So(O2)] ∆So = 2 mol (69.9 J/K•mol) [2 mol (130.7 J/K•mol) + 1 mol (205.3 J/K•mol)] ∆So = -326.9 J/K Note that there is a decrease in S because 3 mol of gas give 2 mol of liquid. 39 39 40 40 Sample Problem 3 PROBLEM: Calculating the Standard Entropy of Reaction, S0rxn Calculate S0rxn for the combustion of 1mol of propane at 250C. C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) PLAN: Use summation equations. 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 4. 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)] S = - 374 J/K Entropy Change Change of entropy during a reversible process is: ∆S = qrev/T where qrev = heat absorbed in a reversible process T = temperature (in K) at which the process occurs 41 41 Entropy Change ∆S = qrev/T What is a reversible process? • At every step along a reversible pathway between the two states, the system remains at equilibrium. • Spontaneous processes are not reversible, because they proceed in one direction and their pathways involve non-equilibrium conditions. • Adding heat slowly and in very small increments approximates a reversible process. 42 42 Entropy Change ∆S = qrev/T Examples of reversible processes: 1) melting of ice; 2) boiling of water. For H2O (l) H2O(g) q = ∆H = +40,700 J/mol qq 40, 700 J/mol 40, 700 J/mol = + 109 J/K • mol S = = S = = = + 109 J/K • mol TT 373.15 K 373.15 K 43 43 Entropy Changes and Spontaneity We mentioned that spontaneity is related to increase of entropy. More specifically, In any spontaneous process, the entropy of the universe increases. This is the 2nd law of thermodynamics. 44 44 2nd Law of Thermodynamics A reaction is spontaneous if ∆S for the universe is positive. ∆Suniverse > 0 for spontaneous process If ∆Suniverse < 0, the process cannot be spontaneous; If ∆Suniverse = 0, the system is at equilibrium. ∆Suniverse = ∆Ssystem + ∆Ssurroundings 45 45 2nd Law of Thermodynamics ∆Suniverse = ∆Ssystem + ∆Ssurroundings We will see that ∆Ssystem is change of entropy created by dispersal of matter ∆Ssurround is change of entropy created by dispersal of energy 46 46 Entropy of Surroundings ∆Suniverse = ∆Ssystem + ∆Ssurroundings During a reaction, the change in entropy of the system (∆Ssys ) is: ∆Ssys = ∆Srxn ∆∆SSsys = S (products) - S (reactants) sys = S (products) - S (reactants) But how do we get the change of entropy of surroundings? Ssurr qΔH surr = T surr T Ssurroundings = - = ΔHsys Hsystem T T 47 47 Entropy of Surroundings ∆Suniverse = ∆Ssystem + ∆Ssurroundings Ssurroundings = - Hsystem T The change in entropy of the surroundings is directly related to a change in the enthalpy of the system with reversed sign. For exothermic reactions Hsystem < 0, so Ssurroundings > 0 For endothermic reactions Hsystem > 0, so Ssurroundings < 0 48 48 2nd Law of Thermodynamics = ∆S + ∆S ∆Suniverse system surroundings Dissolving NH4NO3 in water—an entropy driven process. Occurs because of dispersion of matter. 49 49 2nd Law of Thermodynamics 2 H2(g) + O2(g) ---> 2 H2O(l) ∆Sosystem = -326.9 J/K qqsurroundings ­H ­Hsystem surroundings system == == TT TT oo surroundings surroundings S S Calculate ∆Hosystem = ∆Horxn = -571.7 kJ ­ (­571.7 kJ)(1000 J/kJ) == ­ (­571.7 kJ)(1000 J/kJ) 298.15 K 298.15 K oo surroundings surroundings S S ∆Sosurroundings = +1917 J/K 50 50 2nd Law of Thermodynamics 2 H2(g) + O2(g) ---> 2 H2O(l) ∆Sosystem = -326.9 J/K ∆Sosurroundings = +1917 J/K ∆Souniverse = ∆Sosystem + ∆Sosurroundings =+1590 J/K Is the reaction spontaneous in the forward direction? • The entropy of the universe is increasing, so the reaction is spontaneous (product-favored). 51 51 52 52 Spontaneous or Not? Remember is proportional to –∆H˚ sys Remember that that ∆S˚ ∆S˚surr surris proportional to –∆H˚sys An > 0. An exothermic exothermic process process has has ∆S˚ ∆S˚surr surr > 0. Components of S0universe for spontaneous reactions 53 53 ∆Suniverse = ∆Ssystem + ∆Ssurroundings exothermic endothermic system becomes more disordered exothermic system becomes more disordered system becomes more ordered 54 54 Sample Problem 4 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 S0universe, and state whether the reaction occurs spontaneously at this temperature. PLAN: S0universe = S0sys + S0surr To find S0surr, first find Hsys; Hsys= Hrxn which can be calculated using H0f values from tables in Appendix 4. S0universe must be > 0 in order for this reaction to be spontaneous. SOLUTION: H0rx = [(2 mol)(H0fNH3)] - [(1 mol)(H0fN2) + (3 mol)(H0fH2)] H0rx = -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. 3rd Law of Thermodynamics Where is a reference point for entropy? The third law of thermodynamics says: Ssystem = 0 at 0 K A perfect crystal at a temperature of absolute zero has no disorder (entropy = 0) So 0 for elements in standard states 55 55 Gibbs Free Energy 56 56 ∆Suniv = ∆Ssurr + ∆Ssys It would be more convenient to deal with system only, not with surroundings. Suniv = Hsys T + Ssys Multiply through by –T : -T∆Suniv = ∆Hsys - T∆Ssys J. Willard Gibbs 1839-1903 Gibbs defined the free energy G for the system as G = H - TS Free Energy 57 57 -T∆Suniv = ∆Hsys - T∆Ssys Since G = H – TS, then: ∆Gsys = ∆Hsys - T∆Ssys It follows that ∆Gsystem = -T∆Suniv Change of free energy of the system is the criterion of spontaneity For any spontaneous process,∆Gsystem< 0 Under standard conditions — o ∆Gosys = ∆Hohere, sys - T∆S sys T is temperature in Kelvin 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 58 58 Free Energy Gibbs free 59 59 o ∆Go = ∆Ho -T∆S energy change = total energy change for system (∆ Ho) - energy lost in disordering the system ( T∆So) 1) If reaction is • exothermic (negative ∆ Ho) (energy dispersed) • and entropy increases (positive ∆So) (matter dispersed) • then ∆Go is always NEGATIVE • reaction is spontaneous (and product-favored). Free Energy 60 60 o ∆Go = ∆Ho -T∆S 2) If reaction is • endothermic (positive ∆Ho) • and entropy decreases (negative ∆So) • then ∆Go must be POSITIVE • reaction is not spontaneous favored). (reactant- 3) If signs of ∆Ho and ∆So are the same, reaction may be spontaneous depending on temperature 61 61 o Free Energy ∆Go = ∆Ho -T∆S ∆Ho ∆So ∆Go Reaction exo(–) increase(+) – Spontaneous endo(+) decrease(-) + not spontaneous exo(–) decrease(-) ? T dependent endo(+) increase(+) ? T dependent The effect of entropy change (sign of ∆So) becomes a dominant factor at higher temperatures. Free Energy H0 62 62 o ∆Go = ∆Ho -T∆S Description S0 G0 + + + or - Spontaneous at higher T; nonspontaneous at lower T - - + or - Spontaneous at lower T; nonspontaneous at higher T Calculating ∆Gorxn 63 63 Combustion of acetylene at 298 K C2H2(g) + 5/2 O2(g) --> 2 CO2(g) + H2O(g) Use enthalpies of formation to calculate ∆Horxn = -1238 kJ Use standard molar entropies to calculate ∆Sorxn = -97.4 J/K or -0.0974 kJ/K ∆Gorxn = -1238 kJ - (298 K)(-0.0974 kJ/K) = -1209 kJ Reaction is spontaneous in spite of negative ∆Sorxn. Reaction is “enthalpy driven” Calculating ∆Gorxn NH4NO3(s) + heat ---> NH4NO3(aq) Is the dissolution of ammonium nitrate productfavored? If so, is it enthalpy- or entropy-driven? 64 64 Calculating ∆Gorxn NH4NO3(s) + heat ---> NH4NO3(aq) From tables of thermodynamic data we find ∆Horxn = +25.7 kJ ∆Sorxn = +108.7 J/K or +0.1087 kJ/K ∆Gorxn = +25.7 kJ - (298 K)(+0.1087 kJ/K) = -6.7 kJ Reaction is spontaneous in spite of positive ∆Horxn. Reaction is “entropy driven” 65 65 Sample Problem 5 PROBLEM: 66 66 Calculating G0 from Enthalpy and Entropy Values Potassium chlorate undergoes a disproportionation reaction when heated: 4KClO3(s) 3KClO4(s) + KCl(s) Use H0f and S0 values to calculate G0sys (G0rxn) at 250C for this reaction. PLAN: Use Appendix L to find thermodynamic values (H0f and S0); place them into the Gibbs Free Energy equation and solve. Make sure H0 and TS0 are both in the same units. SOLUTION: H0rxn = H0products - H0reactants H0rxn = (3 mol)(-432.8 kJ/mol) + (1 mol)(-436.7 kJ/mol) (4 mol)(-397.7 kJ/mol) H0rxn = -144 kJ Sample Problem 5 67 67 Calculating G0 from Enthalpy and Entropy Values continued S0rxn = S0products - S0reactants 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 H0rxn = -144 kJ Make sure units are consistent !! S0rxn = -36.8 J/K = (-36.8 J/K) x (1kJ/1000J) = -0.0368 kJ/K G0rxn = -144 kJ - (298K)(-0.0368 kJ/K) = -144 kJ + 11.0 kJ G0rxn = -133 kJ Free Energy, G 68 68 Two methods of calculating ∆Go a) Determine ∆Horxn and ∆Sorxn and use Gibbs equation (∆Go = ∆Ho - T∆So) b) Use tabulated values of free energies of formation, ∆Gfo. oo oo ∆∆G = ∆G (products) ∆G Goorxn = ∆G (products) ∆G (reactants) ff f f (reactants) rxn The method b) works only at 298 K. 69 69 Free Energy of Formation The standard free energy of formation of a compound, (∆G˚f ) is the free energy change when forming 1 mole of the compound from the component elements, with all involved substances in their standard states. Note that ∆G˚f for an element = 0 70 70 oo ∆∆G G rxn rxn Calculating ∆Gorxn == ∆G ∆G (products) (products) -- ∆G ∆G (reactants) (reactants) oo ff oo ff Combustion of carbon C(graphite) + O2(g) --> CO2(g) ∆Gorxn = ∆Gfo(CO2) - [∆Gfo(graph) + ∆Gfo(O2)] ∆Gorxn = -394.4 kJ - [ 0 + 0] Remember that free energy of formation of an element in its standard state is 0. ∆Gorxn = -394.4 kJ Reaction is spontaneous as expected. 71 71 Sample Problem 6 Calculating G0rxn from G0f Values PROBLEM: Use G0f values to calculate Grxn for the reaction of disproportionation of potassium chlorate: 4KClO3(s) 3KClO4(s) + KCl(s) PLAN: Use the G summation equation. Remember to multiply values of G0f from appendix 4 by corresponding number of moles in the chemical equation. SOLUTION: G0rxn = G0products - G0reactants G0f values: of KCl (s) = -408.8 kJ/mol of KClO3 (s) = -296.3 kJ/mol of KClO4 (s) = -303.2 kJ/mol G0rxn = (3mol)(-303.2kJ/mol) + (1mol)(-408.8kJ/mol) (4mol)(-296.3kJ/mol) G0rxn = -134kJ Free Energy and Temperature 2 Fe2O3(s) + 3 C(s) ---> 4 Fe(s) + 3 CO2(g) ∆Horxn = +467.9 kJ ∆Sorxn = +560.3 J/K ∆Gorxn = +300.8 kJ at 298 K Reaction is reactant-favored at 298 K At what T does ∆Gorxn just change from being (+) to being (-)? When ∆Gorxn = 0 = ∆Horxn - T∆Sorxn H 467.9 kJ H 467.9 kJ = 835.1 K rxn rxn T = = T = = = 835.1 K S 0.5603 kJ/K Srxn 0.5603 kJ/K rxn 72 72 The effect of temperature on reaction spontaneity. 73 73 74 74 Sample Problem 7 PROBLEM: Determining the Effect of Temperature on G0 For the reaction of oxidation of SO2(g) to SO3(g): 2SO2(g) + O2(g) 2SO3(g) At 298K, G0 = -141.6kJ; H0 = -198.4kJ; and S0 = -187.9J/K (a) Use the data to decide if this reaction is spontaneous at 250C, and predict how G0 will change with increasing T. (b) Assuming H0 and S0 are constant with increasing T, is the reaction spontaneous at 9000C? PLAN: The sign of G0 tells us whether the reaction is spontaneous and the signs of H0 and S0 will be indicative of the T effect. Use the Gibbs free energy equation for part (b). SOLUTION: (a) The reaction is spontaneous at 250C because G0 is (-). Since H0 is (-) but S0 is also (-), G0 will become less spontaneous as the temperature increases. (b) G0rxn = H0rxn - T S0rxn G0rxn ...
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