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Thermodynamics Entropy, Energy and equilibrium Chapter 19 1 19.1 Thermodynamics Thermo: heat Dynamics: power Study of energy flow and its transformations (heat and energy flow) Determines direction of reactions (spontaneous or nonspontaneous under given conditions) State Functions: considers only initial and final states Does not consider pathways or rate Organized into three laws: 1st Law (ΔU = qp + w = qp – PΔV ) 2nd Law 3rd Law 2 19.1 Basic Definitions universe System Surrounding system surroundings Open system Closed system Isolated system State of system: defined by values of composition, pressure, T, V. State Function: defined only by initial and final condition of the system (Enthalpy, Entropy, Gibbs Free Energy). Energy change signaled by: accomplishment of work and/or appearance or disappearance of heat. 3 19.1 1st Law of Thermodynamics 1. A gas does 135 J of work while expanding, and at the same time it absorbs 156 J of heat. What is the change in internal energy? (21 J) 2. The internal energy of a fixed quantity of ideal gas depends only on its temperature. If a sample of an ideal gas is allowed to expand against a constant pressure at a constant temperature, a) What is ΔU for the gas? b) Does the gas do work? c) If any heat exchanged with the surroundings? a) 0 b) w = -P ΔV c)no, only work done by gas is energy leaving the system. Internal energy should decrease, so the temp; but temp is const.; therefore the int energy does not change; gas has to absorb enough heat from surroundings to compensate for the work. Q = -w; ΔU = -w+w 4 19.1 First Law of Thermodynamics 1st Law: Energy can be neither created nor destroyed, but it can be converted from one form to another or transferred from a system to the surroundings or vice versa. Energy of the universe is constant Important concepts from thermochemistry Enthalpy Hess’s law Purpose of 1st Law Energy bookkeeping How much energy? Exothermic or endothermic? What type of energy? Δu = q + w q = heat; w = work system does on the surroundings (-PΔV) 5 19.2 Spontaneous Processes: Expansion Spontaneous processes are those that can proceed without any outside intervention. Product favored at equilibrium. Product-favored at equilibrium May be fast or slow May be influenced by temperature The gas in vessel B will spontaneously effuse into vessel A, but once the gas is in both vessels, it will not spontaneously separate 6 19.1 Spontaneous Processes Processes that are spontaneous in one direction are nonspontaneous in the reverse direction. Examples: rusting, neutralization reaction, dissolution of sugar in water, heat flow, expansion of gas, spontaneous combustion (CH4 + O2), reaction of sodium with water 7 19.1 Spontaneous Processes Processes that are spontaneous at one temperature may be nonspontaneous at other temperatures. Above 0C it is spontaneous for ice to melt. Below 0C the reverse process is spontaneous. 8 19.2 Spontaneous vs. Nonspontaneous 3. Determine if the following processes are spontaneous or not and if they are exothermic or endothermic. (a) Gases expand into larger volumes at constant temperature ________________ (b) H2O(s) melts above 0C ____________ (c) H2O(l) freezes below 0C ____________ (d) NH4NO3 dissolves spontaneously in H2O ___________ (e) Steel (iron) rusts in presence of O2 and H2O ________ (f) Wood burns to form CO2 and H2O __________ (g) CH4 gas burns to form CO2 and H2O __________ Evolution of Heat (Exothermicity) : not enough to predict spontaneity 9 19.2 Spontaneous vs Nonspontaneous Nonspontaneous process Does not occur unless there is outside assistance (energy?) Reactants-favored at equilibrium All processes which are spontaneous in one direction cannot be spontaneous in the reverse direction Spontaneous processes have a definite direction Spontaneous processes are irreversible. Can be reversed with considerable input of energy. 10 19.2 Factors That Favor Spontaneity Spontaneous Processes driven by Enthalpy, H (Joules) Many, but not all, spontaneous processes tend to be exothermic. Entropy, S (Joules/K) Measure of the disorder of a system Many, but not all, spontaneous processes tend to increase disorder of the system Exothermicity favors spontaneity, but does not guarantee it. 11 19.2 Factors That Favor Spontaneity: Enthalpy Example of spontaneous reaction that is not exothermic: NH4NO3(s) → NH4+ (aq) + NO3-(aq) ΔH = 25 kJ/mol Expansion of gas: energy neutral Phase changes: endothermic processes that occurs spontaneously (ice to water). Chemical system: H2(g) + I2(g) ↔ 2HI(g) Equilibrium can be approached from both sides (spontaneous both ways) even though the forward reaction is endothermic and the reverse is exothermic. 12 19.2 Spontaneity: Examples 4. Based on your experience, predict whether the following processes are spontaneous, are spontaneous in reverse direction, or are in equilibrium: (a) When a piece of metal heated to 150 ºC is added to water at 40 ºC, the water gets hotter. (b) Water at room temperature decomposes into hydrogen and oxygen gases (c) Benzene vapor, C6H6(g), at a pressure of 1 atm condenses to liquid benzene at the normal boiling point of benzene, 80.1 ºC. 13 19.2 Reversible Processes In a reversible process the system changes in such a way that the system and surroundings can be put back in their original states by exactly reversing the process. Example: melting ice at its melting point 14 19.2 Irreversible Processes Irreversible processes cannot be undone by exactly reversing the change to the system. Different path has to be used. Spontaneous processes are irreversible. Example: expansion a gas into vacuum. 15 19.2 Entropy Entropy (S) is a term coined by Rudolph Clausius in the 19th century. Clausius was convinced of the significance of the ratio of heat delivered and the temperature at which it is delivered, q T 16 19.2 Entropy Direct measure of the randomness or disorder of the system. Related to probability describes # of ways the particles in a system can be arranged in a given state (position and/or energy levels) The most likely state – the most random More possible arrangements, the higher disorder, higher entropy Ordered state – low probability of occurring Disordered state: high probability of occurring 17 19.2 Entropy on the Molecular Scale Ludwig Boltzmann described the concept of entropy on the molecular level. Temperature is a measure of the average kinetic energy of the molecules in a sample. 18 19.2 Entropy on the Molecular Scale Molecules exhibit several types of motion: 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. 19 19.2 Entropy on the Molecular Scale Boltzmann envisioned the motions of a sample of molecules at a particular instant in time. This would be akin to taking a snapshot of all the molecules. He referred to this sampling as a microstate of the thermodynamic system. 20 19.2 Entropy on the Molecular Scale Each thermodynamic state has a specific number of microstates, W, associated with it. Entropy is S = k lnW where k is the Boltzmann constant, 1.38 1023 J/K; W: number of microstates 21 19.2 Entropy on the Molecular Scale The change in entropy for a process, then, is S = k lnWfinal k lnWinitial lnWfinal S = k ln lnWinitial • Entropy increases with the number of microstates in the system. 22 19.2 Spontaneous Processes: Dispersal of Matter Isothermal (constant temperature) expansion of gas Two molecules present: 25% probability After opening stopcock the molecules could be in any arrangement shown (4 arrangements) Probability for each arrangement = (1/2)2 S = k (ln(4) 23 19.2 Spontaneous Process: Isothermal Gas Expansion Consider why gases tend to isothermally (constant temp.) expand into larger volumes. Gas Container = two bulbed flask Ordered State Gas Molecules 24 19.2 Spontaneous Process: Isothermal Gas Expansion Gas Container Ordered State S = k ln (W) = k (ln 1) = (1.38 x 10-23 J/K)(0) = 0 J/K For 3 particles, probability = (1/2)3 For N particles, probability = (1/2)N 25 Disordered States More probable that the gas molecules will disperse between two halves than remain on one side 26 Disordered States Driving force for expansion is entropy (probability);27gas molecules have a tendency to spread out Disordered States S = k(ln 7) = (1.38 x 10-23 J/K)(1.95) = 2.7 x 10-23 J/K 28 Total Arrangements Stotal = k(ln 23) = k(ln 8) = (1.38 x 10-23 J/K)(1.79) = 2.9 x 10-23 J/K 29 19.2 Entropy on the Molecular Scale The number of microstates and, therefore, the entropy tends to increase with increases in Temperature. Volume. The number of independently moving molecules. 30 19.2 Entropy and Temperature (a) A substance at a higher temperature has greater molecular motion, more disorder, and greater entropy than (b) the same substance at a lower temperature. 31 19.2 Entropy and Physical States Entropy increases with the freedom of motion of molecules. Therefore, S(g) > S(l) > S(s) 32 19.2 Entropy Changes In general, entropy increases when Gases are formed from liquids and solids. Liquids or solutions are formed from solids. The number of gas molecules increases. The number of moles increases. 33 19.2 Entropy in Temperature 34 Standard entropy, S(J/K) 50 What kind of changes are represented here? 40 30 20 10 0 50 100 150 200 Temperature (K) 250 300 35 19.2 Solutions Generally, when a solid is dissolved in a solvent, entropy increases. 36 19.2 Patterns of Entropy Change 6. Describe in words the entropy of the system 37 19.2 Entropy Like total energy, E, and enthalpy, H, entropy is a state function. Therefore, S = Sfinal Sinitial S > 0 represents increased randomness or disorder Note: The magnitude of change in entropy depends on temperature. 38 19.2 Entropy For a process occurring at constant temperature (an isothermal process), the change in entropy is equal to the heat that would be transferred if the process were reversible divided by the temperature: qrev S = T Units: Joule/K 39 19.2 Entropy: Example 5. ΔS = q/T The element mercury, Hg, is a silvery liquid at room temperature. The normal freezing point of mercury is -38.9 ºC, and its molar enthalpy of fusion is ΔHfusion = 2.331 kJ/mol. What is the entropy change when 50.0 g of Hg(l) freezes at the normal freezing point? (-2.48 J/K) 40 19.2 Entropy: Examples 7. Predict if ΔS increases, decreases or does not change (a) Freezing liquid mercury (b) Condensing H2O(vapor) (c) Precipitating AgCl (d) Heating H2(g) from 60.0 ºC to 80 ºC (e) Subliming iodine crystals (f) Rusting iron nail 41 19.2 Entropy - Examples 8. Predict which substance has the higher entropy: a) NO2(g) or N2O4(g) b) I2(g) or l2(s) 9. Predict whether each of the following leads to increase or decrease in entropy of a system If in doubt, explain why. a) The synthesis of ammonia: N2(g) + 3H2(g) ↔ 2NH3(g) b) C12H22O11(s) → C12H22O11(aq) c) Evaporation to dryness of a solution of urea, CO(NH2)2 in vapor. CO(NH2)2(aq) → CO(NH2)2(s) 42 19.3 Second Law of Thermodynamics 43 19.3 Second Law of Thermodynamics: System 10. Predict the sign of ΔS0 for each of the following reactions: a) Ca+2(aq) + 2OH-(aq) → Ca(OH)2(s) b) MgCO3(s) → MgO(s) + CO2(g) d) H2(g) + Br2(g) → 2HBr(g) 44 Second Law of Thermodynamics The second law of thermodynamics states that the entropy of the universe increases for spontaneous processes, and the entropy of the universe does not change for reversible processes. In other words: For reversible processes: Suniv = Ssystem + Ssurroundings = 0 For irreversible processes: Suniv = Ssystem + Ssurroundings > 0 For nonspontaneous Process: ΔS univ= ΔS syst. + ΔS surr. <0 45 Second Law of Thermodynamics These last truths mean that as a result of all spontaneous processes the entropy of the universe increases. 46 19.3 Second Law: Entropy Changes EQUILIBRIUM PROCESSES (reversible) ♦ ΔS universe= ΔS syst. + ΔS surr. =0 ♦ ΔS syst = ΔS surr ♦ ΔS syst = ΔSº 0 ΔS final initial 47 19.3 Entropy Changes in a System (Reactions) Entropy changes in a system aA + bB → cC + dD Standard entropy change ΔSº (25 ºC, 1atm). Only changes in entropy can be measured. Each element has an entropy value (compare to enthalpy). Absolute value for each substance can be determined. For a chemical system: S° = nS°(products) - mS°(reactants) where n and m are the coefficients in the balanced chemical equation. Standard Molar entropy, S0, is the entropy of one mole of a substance in its standard state (298 K) 48 19.2 Standard Entropies These are molar entropy values of substances in their standard states. Standard entropies tend to increase with increasing molar mass. 49 19.2 Standard Entropies Larger and more complex molecules have greater entropies. 50 19.3 Second Law: Example 11. Using standard molar entropies, calculate S°rxn for the following reaction at 25°C: 2SO2(g) + O2(g) → 2SO3(g) S° = 248.1 205.1 256.6 (J · K-1mol-1) (Ans.: -187.9 J/K) 51 19.3 Entropy of Reactions (System) 12. Using thermodynamic tables, calculate the standard entropy changes for the following reactions at 25 ºC a) Evaporation of 1.00 mol of liquid ethanol to ethanol vapor. b) The oxidation of one mole pf ethanol vapor (combustion reaction) c) Are the reactions spontaneous under the given conditions. Answers: a) 122.0 J/K b) 96.09 J/K 52 19.3 Entropy Changes in the System In a reaction More gas molecules produced: entropy increases Less gas molecules produced: entropy decreases No net change of # of gas molecules produced: entropy changes, but slightly Liquid, solid products: hard to estimateneeds calculations 53 19.3 Entropy of Surrounding (a) When an exothermic reaction occurs in the system (ΔH < 0), the surroundings gain heat and their entropy increases (Δ Ssurr > 0). (b) When an endothermic reaction occurs in the system (Δ H > 0), the surroundings lose heat and their 54 entropy decreases (Δ Ssurr < 0). 19.3 Entropy Changes in Surroundings Heat that flows into or out of the system changes the entropy of the surroundings. For an isothermal process: Ssurr = qsys T • At constant pressure, qsys is simply H for the system. (negative sign needed to make entropy positive in an exothermic process) 55 19.3 Second Law of Thermodynamics (System) 13. Examples: H2O(s) melts above 0C (endothermic). What about entropy? Steel (iron) rusts in presence of O2 and H2O (exothermic). What about entropy? 4Fe(s) + 3O2(g) 2Fe2O3(s) CH4 gas burns to form CO2 and H2O (exothermic). What about entropy? CH4(g) + 2O2(g) CO2(g) + 2H2O(g) Each process increases entropy of the universe. 56 19.3 Second Law of Thermodynamics Consider transferring the same amount of heat to two different systems, one at 298K and another at 500K. Q 298K q rev S T Q 500K q q 298K 500 K More entropy (disorder) is created in the system at lower temp. At high temperatures the system is already disordered 59 19.3 Second Law of Thermodynamics To determine Suniv for a process, both Ssystem and Ssurroundings need to be known: Ssystem related to matter dispersal in system Ssurroundings determined by heat exchange between system and surroundings and T at which it occurs Sign of Ssurr depends on whether process in system is endothermic (Ssurr< 0) or exothermic (Ssurr >0) Magnitude of Ssurr depends on T Ssurr = -Hsystem/T 61 19.3 Second Law of Thermodynamics 14. Reaction: N2(g) + 3H2(g) → 2NH3(g) ΔHº = -92.6kJ ΔSsys = -199 J/K at 25 ºC ΔSsurr = -(-92.6 x1000)J/298K = 311 J/K ΔSuniv = -199 J/K + 311 J/K = 112 J/K Reaction spontaneous at 25 ºC 62 19.3 Conditions for Spontaneity ΔSsystem ΔS surr ΔSuniv Spontaneity + + + Yes - - - + - ? No, reaction towards reactants Yes, if ΔSsys> ΔSsurroun - + ? Yes, if ΔSsurr> ΔSsyst 63 19.3 Entropy and Spontaneity 15. Consider the vaporization of liquid water to steam at a pressure of 1 atm. a) Is this process endothermic or exothermic, explain. b) In what temperature range is the process spontaneous? c) In what temperature range is it a nonspontaneous process? d) At what temperature are the two phases at equilibrium? 64 19.3 Third Law of Thermodynamics The entropy of a pure crystalline substance at absolute zero is 0. Perfect crystal: its internal arrangement is absolutely regular. Nothing is in motion (vibrations, rotations and translations) 65 19.3 Entropy and Third Law Third Law of Thermo Gives us a starting point, S at 0K is equal to zero. All others must be >0. Standard Entropies Sº ( at 298 K and 1 atm) of substances are listed. Products - reactants to find Sº (a state function) More complex molecules higher Sº. 66 19.4 Gibbs Free Energy Heat that flows into or out of the system changes the entropy of the surroundings. For an isothermal process: Ssurr = qsys T • At constant pressure, qsys is simply H for the system. 67 19.4 Gibbs Free Energy The universe is composed of the system and the surroundings. Therefore, Suniverse = Ssystem + Ssurroundings For spontaneous processes Suniverse > 0 68 19.4 Gibb’s Free Energy This becomes: Suniverse = Ssystem + Hsystem T Multiplying both sides by T, TSuniverse = Hsystem TSsystem TΔS universe is defined as the Gibbs free energy, G. When Suniverse is positive, G is negative. Therefore, when G is negative, a process is spontaneous. 69 19.4 Gibb's Free Energy at any Conditions G=H-TS Never used this way. At constant temperature G=Hsys –TSsys G function eliminates the need to deal with entropy of the surroundings If G is negative at constant T and P, the process is spontaneous. We deal only with the SYSTEM. 70 19.4 Gibbs Free Energy The Gibbs Free Energy is a measure of the maximum amount of work, at a given temperature and pressure, that can be done on the surroundings by a system. Never really achieved because some of the free energy is changed to heat during a change, so it can’t be used to do work. Wmax = G For a spontaneous process: Maximum amount of energy released by the system that can do useful work on the surroundings Energy available from spontaneous process that can be used to drive non-spontaneous process. For a nonspontaneous process: Minimum amount of work that must be done to force the process to occur 71 19.4 Gibbs Free Energy Summary of Conditions for Spontaneity G < 0 reaction is spontaneous in the forward direction (Suniv > 0) G > 0 reaction is nonspontaneous in the forward direction (Suniv < 0) G = 0 SYSTEM IS AT EQUILIBRIUM (Suniv = 0) Remember- Spontaneity tells us nothing about rate. 72 19.4 Gibbs Standard Energy Free Energy The standard free energies of formation, G°, are the free energy values for the formation of a substance under standard conditions. The standard free energy of formation for any element in its standard state is zero. (Like enthalpy, but not entropy) Compare: G = Hsys - TSsys (any conditions) Gº = H0sys - TS0sys (standard conditions) 73 19.4 Convention for Standard States State of Matter Standard State Gas 1 atm pressure Liquid Pure liquid Solid Pure solid Elements Gºf = 0 Solution 1 molar 74 19.4 Free Energy in Reactions Gº = standard free energy change. Free energy change that will occur if reactants in their standard state turn to products in their standard state. Can’t be measured directly, can be calculated from other measurements. The reaction: aA + bB → cC + dD or Gºrxn = ΣnGº (products) - ΣmGº (reactants) Gº=Hº-TSº Use adapted Hess’s Law with known reactions. 75 19.4 Free Energy Changes At temperatures other than 25°C, G ° = H TS How does G change with temperature? There are two parts to the free energy equation: H — the enthalpy term TS — the entropy term The temperature dependence of free energy, then comes from the entropy term. 76 19.4 Gibbs free Energy: Example 16. For a particular reaction, Hrxn = 53 kJ and Srxn = 115 J/K. Is this process spontaneous a) at 25°C, and b) at 250°C? (c) At what temperature does Grxn = 0? Look at the equation G=H-TS Spontaneity can be predicted from the sign of H and S. (ans.: a) G = 18.7 kJ, nonspontaneous; b) –7.1 kJ, spontaneous; c) 460.9 K or 188C) 77 19.4 Gibbs Free Energy and Spontaneity Effect of Temperature on the Spontaneity of Reactions ΔH ΔS -|TΔS| 1 2 3 4 + - + + - - + ΔG = ΔH-TΔS Reaction Characteristics Example Always negative Spontaneous at all temperatures 2O3(g) →3O2(g) + Always positive Nonspontaneous at 3O2(g) all temperatures →2O3(g) + Negative at low T; Spontaneous at low H2O(l) positive at high T T; nonspontaneous →H2O(s)+ at high T. Enthalpy driven - - Positive at low T; negative at high T Nonspontaneous at H2O(s) → low T; becomes H2O(l) spontaneous at high T. Entropy driven. 78 19.4 Gibbs Free Energy and Spontaneity Free energy change as a function of temperature. 79 19.4 Gibbs Free Energy and Temperature Spontaneous Processes H2O(s) melts above 0C (endothermic) H > 0 (endo process), S > 0 ENTROPY DRIVEN Steel (iron) rusts in presence of O2 and H2O at 25 C (exothermic) 4Fe(s) + 3O2(g) 2Fe2O3(s) H < 0 (Exo) , S < 0 (entropy decreases) ENTHALPY DRIVEN G < 0 for each process T determines sign of G: G = H -TS 80 19.4 Gibbs Free Energy: Example (p. 748) 17. Predict which of the four cases in Table (slide #78) you expect to apply to the following reactions: a) C6H12O6 (s) + 6O2(g) → 6CO2(g) + 6H2O(g) ΔH = -2540 kJ b) Cl2(g) → 2Cl(g) 18. Calculate ΔG0 at 298 K for the reaction 4HCl(g) + O2(g) → 2Cl2(g) + 2H2O(g) ΔH0 = 114.4 kJ a) using Gibbs free Energy equation b) from standard free energies of formation. ( -76.0 kJ) 81 19.4 Gibbs free Energy: Example 18. For the reaction SO2(g) + 2H2S(g) →3S(s) + 2H2O(g) Calculate the temperature at which ΔG0 = 0 Values of ΔH0, kJ/mol ΔS0 (kJ/(K mol) SO2 -296.8 0.2481 H2S -20.6 0.2057 S 0.0 0.0318 H2O -241.8 +0.1887 Answer: 780K 82 19.4 Temperature and Chemical Reactions – Problem Set 19. Calculate the temperature at which the reaction: CaCO3(s) → CaO(s) + CO2(g) becomes spontaneous. ΔS º = 160.5 J/K; ΔH º = 177.8 kJ Answer: 835 ºC 20. At its normal boiling point, the enthalpy of vaporization of pentadecane, CH3(CH2)13CH3, is 49.45 kJ/mol. What should be its approximate normal boiling point temperature be? ΔS0 of vaporization is 87 J mol-1 K-1 (570 K) 83 19.4 Gibbs Free Energy 21. Consider the reaction C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) ΔH0 = -2220 kJ (a) Without using data from the Thermodynamic Tables, predict whether ΔG0 for this reaction is more negative, or less negative than ΔH0. (b) Use data from the Tables to calculate ΔG0 at 298 K. Is your prediction correct? Values of ΔG0 (in kJ/mol): C3H8(g), -23.5; O2(g), 0; CO2(g) -394.4; H2O(l) -237.2 (a) less negative; (b) -2108 kJ 84 19.5 Gibbs free Energy and Equilibrium 85 19.5 Free Energy and Equilibrium ΔG and ΔGº are not the same. ΔG = ΔH – T ΔS ΔGº = ΔH º – T ΔS º ΔGº is only at standard conditions (values obtained from Tables) ΔGº = Gº(products) - ΔGº (reactants) ΔG any conditions (no Tables of values available) 86 19.5 Free Energy and Equilibrium Predicament: we start a reaction with all reactants in standard state (1 atm, 25ºC, 1M solution). Is the standard state preserved as the reaction progresses? It can be shown mathematically that at nonstandard conditions: ΔG = ΔGº +RTlnQ [products] Q, reaction quotient [reac tan ts] 87 K<Q K>Q K=Q If ΔG is negative, the forward reaction is spontaneous. If Δ G is 0, the system is at equilibrium. If G is positive, the reaction is spontaneous in the reverse direction. 88 19.5 Free Energy and the Equilibrium Constant At equilibrium: Useful equation to determine small Keq 0 G RT ln K eq G RT ln K eq K = e-(ΔGº /RT) From the above we can conclude: If G < 0, then K > 1 Product favored If G = 0, then K = 1 Equilibrium If G > 0, then K < 1 Reactants favored Kc used for solutions and molarities, Kp used for gases 89 19.519.5 FreeTemperature Energy and Dependence Chemical Equilibrium of K Gº= -RTlnK = Hº - TSº ln(K) = Hº/R(1/T)+ Sº/R A straight line of lnK vs 1/T Slope? Y-intercept? 90 19.5 Reaction Path and ΔGº ΔGº = -RTln K Value of ΔGº Sign of K Path of reaction Negative, <0 K>0 Products are favored Positive, >0 K<0 Reactants are favored zero, =0 K=1 Equilibrium 91 19.5 Keq: Problem Set – 19.6 (p. 755) 22. Determine the value of Keq at 25 ºC for the reaction: 2NO2(g) ↔ N2O4(g). Note: calculate ΔG0 from tables Use ΔG0 = -RTln Kp ( 6.9) 92 19.5 Keq: Problem Set 19.7 (p. 755) 23. Using the solubility product of of silver iodide at 25 ºC (8.5 x 10-17), calculate ΔGº for the process: AgI(s) ↔ Ag+(aq) + I-(aq) Answer: 24. Estimate the value of ΔS0298 for the dissociation of copper (II) oxide. CuO(s) ↔ 2Cu2O(s) + O2(g) ΔH0298 = 283 kJ ( 0.203 kJ K-1 93 19.5 Gibbs and Equilibrium: Example 25. Calculate Grxn for the reaction below: 2A(aq) + B(aq) C(aq) + D(g) if G°rxn = 9.9 x 103 J/mol and 150 °C (a) [A] = 0.8 M, [B] = 0.5 M, [C] = 0.05 M, and PD = 0.05 atm, and (b) (b) [A] = 0.1 M, [B] = 1 M, [C] = 0.5 M, and PD = 0.5 atm. (c) Is the reaction spontaneous under these conditions? 94 19.5 Gibbs and Equilibrium: Example 26. Calculate G°rxn for the ionization of acetic acid, HC2H3O2 (Ka = 1.8 x 10-5) at 25°C. Is this reaction spontaneous under standard state conditions? (ans.: 27 kJ) 27. Calculate G° for the neutralization of a strong acid with a strong base at 25°C. Is this process spontaneous under these conditions? For the reaction below, K = 1.0 x 1014. H+ + OH- H2O (ans.: -80 kJ) 95 19.5 Gibbs and Equilibrium: Example 28. Calculate Keq for a reaction at (a) 25°C and (b) 250°C if H°rxn = 42.0 kJ and S°rxn = 125 J/K. At which temperature is this process product favored? (ans.: a) 0.15; b) 216; 250C) 96 19.5 Summary First Law of Thermodynamics: Second Law: ΔU = Δq + Δw ΔSuniv = ΔS sys + ΔSsurr ΔS sys= ΔH/T ΔG = ΔH – T ΔS ΔG = ΔGº +RTlnK ΔGº = -RTlnK 97 19.5 Gibbs Free Energy Many biological reactions essential for life are nonspontaneous Spontaneous reactions used to “drive” the nonspontaneous biological reactions Example: photosynthesis 6CO2 + 6H2O C6H12O6 + 6O2 G > 0 What spontaneous reactions drive photosynthesis? 98 19.5 Entropy and Life Processes If the 2nd law is valid, how is the existence of highly-ordered, sophisticated life forms possible? growth of a complex life form represents an increase in order (less randomness) lower entropy 99 19.5 Entropy and Life Processes Organisms “pay” for their increased order by increasing Ssurr. heat CO2 Over lifetime, Suniv > 0. H2O 100 102