* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Document
Spinodal decomposition wikipedia , lookup
Self-assembled monolayer wikipedia , lookup
Chemical potential wikipedia , lookup
Computational chemistry wikipedia , lookup
Thermomechanical analysis wikipedia , lookup
Crystallization wikipedia , lookup
Rate equation wikipedia , lookup
Liquid–liquid extraction wikipedia , lookup
Lewis acid catalysis wikipedia , lookup
Physical organic chemistry wikipedia , lookup
Electrochemistry wikipedia , lookup
Marcus theory wikipedia , lookup
Acid dissociation constant wikipedia , lookup
Click chemistry wikipedia , lookup
Chemical reaction wikipedia , lookup
Process chemistry wikipedia , lookup
Electrolysis of water wikipedia , lookup
Determination of equilibrium constants wikipedia , lookup
Implicit solvation wikipedia , lookup
Stability constants of complexes wikipedia , lookup
Thermometric titration wikipedia , lookup
Thermodynamics wikipedia , lookup
Stoichiometry wikipedia , lookup
Equilibrium chemistry wikipedia , lookup
Transition state theory wikipedia , lookup
Organosulfur compounds wikipedia , lookup
Chemical equilibrium wikipedia , lookup
Chemical thermodynamics wikipedia , lookup
Chemistry: Atoms First Julia Burdge & Jason Overby Chapter 18 Entropy, Free Energy, and Equilibrium Kent L. McCorkle Cosumnes River College Sacramento, CA Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 18 Entropy, Free Energy, and Equilibrium 18.1 Spontaneous Processes 18.2 Entropy A Qualitative Description of Entropy A Quantitative Description of Entropy 18.3 Entropy Changes in a System Calculating ΔSsys Standard S° Qualitatively Predicting ΔS°sys 18.4 Entropy Changes in the Universe Calculating ΔSsurr The Second Law of Thermodynamics The Third Law of Thermodynamics 18.5 Predicting Spontaneity Gibbs Free-Energy Change, ΔG Standard Free-Energy Changes, ΔG° Using ΔG and ΔG° to Solve Problems 18.6 Free Energy and Chemical Equilibrium Relationship Between ΔG and ΔG° Relationship Between ΔG° and K 18.7 Thermodynamics in Living Systems 18.1 Spontaneous Processes A process that does occur under a specific set of conditions is called a spontaneous process. A process that does not occur under a specific set of conditions is called nonspontaneous. Spontaneous Processes A process that results in a decrease in the energy of a system often is spontaneous: CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ΔH° = -890.4 kJ/mol The sign of ΔH alone is insufficient to predict spontaneity in every circumstance: H2O(l) H2O(s) T > 0°C; ΔH° = -6.01 kJ/mol 18.2 Entropy To predict spontaneity, both the enthalpy and entropy must be known. Entropy (S) of a system is a measure of how spread out or how dispersed the system’s energy is. Entropy Spontaneity is favored by an increase in entropy. S = k ln W k is the Boltzmann constant (1.38 x 10–23 J/K) W is the number of different arrangements The number of arrangements possible is given by: W = XN X is the number of cells in a volume N is the number of molecules Entropy Entropy There are three possible states for this system: 1) One molecule on each side (eight possible arrangements) 2) Both molecules on the left (four possible arrangements) 3) Both molecules on the right (four possible arrangements) The most probable state has the largest number of arrangements. 18.3 Entropy Changes in a System The change in entropy for a system is the difference in entropy of the final state and the entropy of the initial state. ΔSsys = Sfinal – Sinitial Alternatively: Ssys Vfinal nR ln Vinitial Worked Example 18.1 Determine the change in entropy for 1.0 mole of an ideal gas originally confined to one-half of a 5.0-L container when the gas is allowed to expand to fill the entire container at constant temperature. Strategy This is the isothermal expansion of an ideal gas. Because the molecules spread out to occupy a greater volume, we expect there to be an increase in the entropy of the system. Use ΔSsys = Sfinal – Sinitial to solve for ΔSsys. R = 8.314 J/K∙mol, n = 1.0 mole, Vfinal = 5.0 L, and Vinitial = 2.5 L. Solution ΔSsys = nR ln Vfinal 8.314 J 5.0 L = 5.8 J/K = 1.0 mol × ×ln Vinitial K ∙ mol 2.5 L Think About It Remember that for a process to be spontaneous, something must favor spontaneity. If the process is spontaneous but not exothermic (in this case, there is no enthalpy change), then we should expect ΔSsys to be positive. Worked Example 18.1 (cont.) Solution These equilibrium concentrations are then substituted into the equilibrium expression to give (x)(0.050 + x) -5 1.8×10 = 0.010 – x Because we expect x to be very small (even smaller than 1.34×10-3 M–see above), because the ionization of CH3COOH is suppressed by the presence of CH3COO-, we assume (0.10 – x) M ≈ 0.10 M and (0.050 + x) M ≈ 0.050 M Therefore, the equilibrium expression simplifies to (x)(0.050) -5 1.8×10 = 0.010 and x = 3.6×10-5 M. According to the equilibrium table, [H+] = x, so pH = –log(3.6×10-5) = 4.44. Entropy Changes in a System The standard entropy is the absolute entropy of a substance at 1 atm. Temperature is not part of the standard state definition and must be specified. Entropy Changes in a System There are several important trends in entropy: S°liquid > S°solid S°gas > S°liquid S° increases with molar mass S° increases with molecular complexity S° increases with the mobility of a phase (for an element with two or more allotropes) Entropy Changes in a System In addition to translational motion, molecules exhibit vibrations and rotations. Entropy Changes in a System For a chemical reaction aA + bB → cC + dD ΔS°rxn = [cS°(C) + dS°(D)] – [aS°(A) + bS°(B)] Alternatively, ΔS°rxn = ΣnS°(products) – ΣmS°(reactants) Worked Example 18.2 From the standard enthalpy values in Appendix 2, calculate the standard entropy changes for the following reactions at 25°C. (a) CaCO3(s) → CaO(s) + CO2(g) (b) N2(g) + 3H2(g) → 2NH3(g) (c) H2(g) + Cl2(g) → 2HCl(g) Strategy Look up standard enthalpy values and calculate ΔS°rxn. Just as we did when we calculated standard enthalpies of reaction, we consider stoichiometric coefficients to be dimensionless–giving ΔS°rxn units of J/K∙mol. From Appendix 2, S°[CaCO3(s)] = 92.9 J/K∙mol, S°[CaO(s)] = 39.8 J/K∙mol, S°[CO2(g)] = 213.6 J/K∙mol, S°[N2(g)] = 191.5 J/K∙mol, S°[H2(g)] = 131.0 J/K∙mol, S°[NH3(g)] = 193.0 J/K∙mol, S°[Cl2(g)] = 223.0 J/K∙mol, and S°[HCl(g)] = 187.0 J/K∙mol. Worked Example 18.2 (cont.) Solution (a) S°rxn = [S°(CaO) + S°(CO2)] – [S°(CaCO3)] = [(39.8 J/K∙mol) + (213.6 J/K∙mol)] – (92.9 J/K∙mol) = 160.5 J/K∙mol (b) S°rxn = [2S°(NH3)] – [S°(N2) + 3S°(H2)] =(2)(193.0 J/K∙mol) – [(191.5 J/K∙mol) + (3)(131.0 J/K∙mol)] = –198.5 J/K∙mol (c) S°rxn = [2S°(HCl)] – [S°(H2) + S°(Cl2)] = (2)(187.0 J/K∙mol) – [(131.0 J/K∙mol) + (223.0 J/K∙mol)] = 20.0 J/K∙mol Think About It Remember to multiply each standard entropy value by the correct stoichiometric coefficient. Like Equation 10.18, Equation 18.5 can only be used with a balanced chemical equation. Entropy Changes in a System Several processes that lead to an increase in entropy are: Melting Vaporization or sublimation Temperature increase Reaction resulting in a greater number of gas molecules Entropy Changes in a System The process of dissolving a substance can lead to either an increase or a decrease in entropy, depending on the nature of the solute. Molecular solutes (i.e. sugar): entropy increases Ionic compounds: entropy could decrease or increase Worked Example 18.3 For each process, determine the sign of ΔS for the system: (a) decomposition of CaCO3(s) to give CaO(s) and CO2(g), (b) heating bromine vapor from 45°C to 80°C, (c) condensation of water vapor on a cold surface, (d) reaction of NH3(g) and HCl(g) to give NH4Cl(s), and (e) dissolution of sugar in water. Strategy Consider the change in energy/mobility of atoms and the resulting changeThink in number of possible positionsinvolving that eachonly particle can and occupy in each case. About It For reactions liquids solids, An increase in thethe number arrangements corresponds to in anmany increase in entropy predicting sign ofofΔS° can be more difficult, but such and therefore positive ΔS. cases ana increase in the total number of molecules and/or ions is accompanied increase of entropy. Solution Increasesby in an entropy generally accompany solid-to-liquid, liquid-togas, and solid-to-gas transitions; the dissolving of one substance in another; a temperature increase; and reactions that increase the net number of moles of gas. ΔS is (a) positive (b) positive (d) negative (e) positive (c) negative 18.4 Entropy Changes in the Universe Correctly predicting the spontaneity of a process requires us to consider entropy changes in both the system and the surroundings. An ice cube spontaneously melts in a room at 25°C. Perspective Components ΔS System ice positive Surroundings everything else negative A cup of hot water spontaneously cools to room temperature. Perspective Components ΔS System hot water negative Surroundings everything else positive The entropy of both the system AND surroundings are important! Entropy Changes in the Universe The change in entropy of the surroundings is directly proportional to the enthalpy of the system. Ssurr Hsys T The second law of thermodynamics states that for a process to be spontaneous, ΔSuniverse must be positive. ΔSuniverse = ΔSsys + ΔSsurr Entropy Changes in the Universe The second law of thermodynamics states that for a process to be spontaneous, ΔSuniverse must be positive. ΔSuniverse = ΔSsys + ΔSsurr ΔSuniverse > 0 for a spontaneous process ΔSuniverse < 0 for a nonspontaneous process ΔSuniverse = 0 for an equilibrium process Worked Example 18.4 Determine if each of the following is a spontaneous process, a nonspontaneous process, or an equilibrium process at the specified temperature: (a) H2(g) + I2(g) → 2HI(g) at 0°C, (b) CaCO3(s) → CaO(s) + CO2(g) at 200°C, (c) CaCO3(s) → CaO(s) + CO2(g) at 1000°C, (d) Na(s) → Na(l) at 98°C. (Assume that the thermodynamic data in Appendix 2 do not vary with temperature.) Strategy For each process, use ΔS°rxn = ΣnS°(products) – ΣmS°(reactants) to determine ΔS°sys; ΔSsurr = (–ΔHsys/T)and ΔH°sys = ΣnΔH °(products) – f ΣmΔH °(reactants) to determine ΔH°sys and ΔS°surr. At the specified temperature, the f process is spontaneous if ΔS°sys and ΔS°surr sum to a positive number, nonspontaneous is they sum to a negative number, and an equilibrium process if they sum to zero. Note that because the reaction is the system, ΔSrxn and ΔSsys are used interchangeably. Worked Example 18.4 (cont.) Solution (a) S°[H2(g)] = 131.0 J/K∙mol, S°[I2(g)] = 260.57 J/K∙mol, S°[HI(g)] = 206.3 J/K∙mol; ΔHf°[H2(g)] = 0 kJ/mol, ΔHf°[I2(g)] = 62.25 kJ/mol, ΔHf°[HI(g)] = 25.9 kJ/mol. ΔS°rxn = [2S°(HI)] – [S°(H2) + S°(I2)] = (2)(206.3 J/K∙mol) – [131.0 J/K∙mol + 160.57 J/K∙mol] = 21.03 J/K∙mol ΔH°rxn = [2 ΔH°f (HI)] – [ΔH°f (H2) + ΔH°f (I2)] = (2)(25.9 kJ/mol) – [0 kJ/mol + 62.26 kJ/mol] = −10.5 kJ/mol ΔSsurr = −ΔHrxn −(−10.5 kJ/mol) = = 0.0385 kJ/K∙mol = 38.5 J/K∙mol T 273 K ΔSuniverse = ΔSsys + ΔSsurr = 21.03 J/K∙mol + 38.5 J/K∙mol = 59.5 J/K∙mol ΔSuniverse is positive; therefore the reaction is spontaneous at 0°C. Worked Example 18.4 (cont.) Solution (b) In Worked Example 18.2(a), we determined that for this reaction, ΔS°rxn = 160.5 J/K∙mol; ΔH°f [CaCO3(s)] = −1206.9 kJ/mol, ΔH°f [CaO(s)] = −635.6 kJ/mol, ΔH°f [CO2(g)] = −393.5 kJ/mol. (b), (c) ΔS°rxn = 160.5 J/K∙mol ΔH°rxn = [ΔH°f (CaO) + ΔH°f (CO2)] – [ΔH°f (CaCO3)] = [-635.6 kJ/mol + (–393.5 kJ/mol)] – (–1206.9 kJ/mol) = 177.8 kJ/mol (b) T = 200°C and ΔSsurr = −ΔHsys −(177.8 kJ/mol) = = −0.376 kJ/K∙mol = −376 J/K∙mol T 473 K ΔSuniverse = ΔSsys + ΔSsurr = 160.5 J/K∙mol + (−376 J/K∙mol) = − 216 J/K∙mol Δsuniverse is negative, therefore the reaction is nonspontaneous at 200°C. Worked Example 18.4 (cont.) Solution (c) T = 1000°C and ΔSsurr = −ΔHsys −(177.8 kJ/mol) = = −0.1397 kJ/K∙mol = −139.7 J/K∙mol T 473 K ΔSuniverse = ΔSsys + ΔSsurr = 160.5 J/K∙mol + (−139.7 J/K∙mol) = 20.8 J/K∙mol In this case, ΔSuniverse is positive; therefore, the reaction is spontaneous at 1000°C. Worked Example 18.4 (cont.) Solution (d) S°[Na(s)] = 51.05 J/K∙mol, S°[Na(l)] = 57.56 J/K∙mol; ΔHf°[Na(s)] = 0 kJ/mol, ΔHf°[Na(l)] = 2.41 kJ/mol. ΔS°rxn = S°[Na(l)] – S°[Na(s)] = 57.56 J/K∙mol – 51.05 J/K∙mol = 6.51 J/K∙mol ΔH°rxn = ΔHf°[Na(l)] – ΔHf°[Na(s)] Think About It Remember that standard enthalpies of formation = 2.41 kJ/mol – 0 kJ/mol = 2.41 kJ/mol have units of kJ/mol, whereas standard absolute entropies have units −(2.41 kJ/mol) of −ΔH J/K∙mol. sure that you convert kilojoules to joules, or vice rxn Make ΔSsurr = = = −0.0650 kJ/K∙mol = −6.50 J/K∙mol 371 K the terms. versa,Tbefore combining ΔSuniverse = ΔSsys + ΔSsurr = 6.51 J/K∙mol − 6.50 J/K∙mol = 0.01 J/K∙mol ≈ 0 ΔSuniverse is zero; therefore, the reaction is an equilibrium process at 98°C. In fact, this is the melting point of sodium. Entropy Changes in the Universe The third law of thermodynamics states that the entropy of a perfect crystalline substance is zero at absolute zero. Entropy increases in a substance as temperature increases from absolute zero. 18.5 Predicting Spontaneity Measurements on the surroundings are seldom made, limiting the use of the second law of thermodynamics. Gibbs free energy (G) or simply free energy can be used to express spontaneity more directly. G = H – TS The change in free energy for a system is: ΔG = ΔH – TΔS Predicting Spontaneity Using the Gibbs free energy, it is possible to make predictions on spontaneity. ΔG = ΔH – TΔS ΔG < 0The reaction is spontaneous in the forward direction. ΔG > 0The reaction is nonspontaneous in the forward direction. ΔG = 0The system is at equilibrium Worked Example 18.5 According to Table 18.4, a reaction will be spontaneous only at high temperatures if both ΔH and ΔS are positive. For a reaction in which ΔH = 199.5 kJ/mol and ΔS = 476 J/K∙mol, determine the temperature (in °C) above which the reaction is spontaneous. Think About It Spontaneity is favored by alow release energy (ΔHat Strategy The temperature that divides high from is theoftemperature being negative) an increase entropy (ΔS–being which ΔH = TΔS (ΔG =and 0).by Therefore, we in use ΔG = ΔH TΔS,positive). substituting 0 for both are positive, as in this case, only entropy ΔG andWhen solving forquantities T to determine temperature in kelvins; wethe then convert to favor spontaneity. For an endothermic process such as this, degreeschange Celsius. which requires the input of heat, it should make sense that adding Solution more heat by increasing the temperature will shift the equilibrium to 476 J 1 kJ the right, thusΔS making it “more spontaneous.” = = 0.476 kJ/K∙mol K∙mol 1000 J T= ΔH 199.5 kJ/mol = = 419 ΔS 0.476 kJ/K∙mol K = (419 – 273) = 146°C Predicting Spontaneity The standard free energy of reaction (ΔG°rxn) is free-energy change for a reaction when it occurs under standard-state conditions. The following conditions define the standard states of pure substances and solutions are: Gases 1 atm pressure Liquids pure liquid Solids pure solid Elements the most stable allotropic form at 1 atm and 25°C Solutions 1 molar concentration Entropy Changes in a System For a chemical reaction aA + bB → cC + dD ΔG°rxn = [cΔG°f (C) + dΔG°f (D)] – [aΔG°f (A) + bΔG°f (B)] Alternatively, ΔG°rxn = ΣnΔG°f (products) – ΣmΔG°f (reactants) ΔG°f for any element in its most stable allotropic form at 1 atm is defined as zero. Worked Example 18.6 Calculate the standard free-energy changes for the following reactions at 25°C: (a) CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) (b) 2MgO(s) → 2Mg(s) + O2(g) Strategy Look up the ΔG°f values for the reactants and products in each equation, and use ΔG°rxn = ΣnΔG°f (products) – ΣmΔG°f (reactants) to solve for ΔG°rxn. Solution From Appendix 2, we have the following values: ΔG°f [CH4(g)] = −50.8 kJ/mol, ΔG°f [CO2(g)] = −394.4 kJ/mol, ΔG°f [H2O(l)] = −237.2 kJ/mol, and ΔG°f [MgO(s)] = −569.6 kJ/mol. All the other substances are elements in their standard states and have, by definition, ΔG°f = 0. (a) ΔG°rxn = (ΔG°f [CO2(g)] + 2ΔG°f [H2O(l)]) – (ΔG°f [CH4(g)] + ΔG°f [O2(g)]) = [(−394.4 kJ/mol) + (2)(−237.2 kJ/mol)] − [(−50.8 kJ/mol) + (2)(0 kJ/mol)] = −818.0 kJ/mol Worked Example 18.6 (cont.) Solution From Appendix 2, we have the following values: ΔG°f [CH4(g)] = −50.8 kJ/mol, ΔG°f [CO2(g)] = −394.4 kJ/mol, ΔG°f [H2O(l)] = −237.2 kJ/mol, and ΔG°f [MgO(s)] = −596.6 kJ/mol. All the other substances are elements in their standard states and have, by definition, ΔG°f = 0. (b) ΔG°rxn = (2ΔG°f [Mg(s)] + ΔG°f [O2(g)]) – (2ΔG°f [MgO(s)]) = [(2)(0 kJ/mol) + (0 kJ/mol)] − [(2)(−569.6 kJ/mol)] = 1139 kJ/mol Think About It Note that, like standard enthalpies of formation (ΔH°f ), standard free energies of formation (ΔG°f ) depend on the state of matter. Using water as an example, ΔG°f [H2O(l)] = −237.2 kJ/mol and ΔG°f [H2O(g)] = −228.6 kJ/mol. Always double-check to make sure you have selected the right value from the table. Worked Example 18.7 The molar heats of fusion and vaporization of benzene are 10.9 and 31.0 kJ/mol, respectively. Calculate the entropy changes for the solid-to-liquid and liquid-tovapor transitions for benzene. At 1 atm pressure, benzene melts at 5.5°C and boils at 80.1°C. Strategy The solid-liquid transition at the melting point and the liquid-vapor transition at the boiling points are equilibrium processes. Therefore, because ΔG About Itin For same is zeroThink at equilibrium, eachthe case we substance, can use ΔGΔS=vap ΔHis–always TΔS, substituting 0 for larger ΔSfus. the Theentropy changechange in number of ΔG andsignificantly solving for ΔS, to than determine associated with the arrangements is always bigger in a liquid-to-gas transition than in a process. solid-to-liquid transition. ΔHfus 10.9 kJ/mol Solution ΔSfus = = Tmelting 278.7 K = 0.0391 kJ/K∙mol ΔSvap = or 39.1 J/K∙mol ΔHvap 31.0 kJ/mol = Tboiling 353.3 K = 0.0877 kJ/K∙mol or 87.7 J/K∙mol 18.6 Free Energy and Chemical Equilibrium It is the sign of ΔG (not ΔG°) that determines spontaneity. The relationship between ΔG and ΔG° is: ΔG = ΔG° + RT lnQ R is the gas constant (8.314 J/K·mol). T is the kelvin temperature. Q is the reaction quotient. Free Energy and Chemical Equilibrium Consider the following equilibrium: H2(g) + I2(g) ⇌ 2HI(g) ΔG° at 25°C = 2.60 kJ/mol ΔG depends on the partial pressures of each chemical species. If PH2 = 2.0 atm; PI2 = 2.0 atm; and PHI = 3.0 atm: QP PHI P P H2 Then: 2 I2 3.0 9.0 2.25 2.0 2.0 4.0 2 2.60 kJ 8.314 103 kJ G 298 K ln2.25 4.60 kJ/mol mol K mol Free Energy and Chemical Equilibrium The spontaneity can be manipulated by changing the partial pressures of the reaction components: H2(g) + I2(g) ⇌ 2HI(g) ΔG° at 25°C = 2.60 kJ/mol If PH2 = 2.0 atm; PI2 = 2.0 atm; and PHI = 1.0 atm: QP PHI 2 1.0 2 P P 2.0 2.0 H2 I2 1.0 0.25 4.0 Then: 2.60 kJ 8.314 103 kJ G 298 K ln0.25 0.8 kJ/mol mol K·mol Worked Example 18.8 The equilibrium constant, KP, for the reaction N2O4(g) ⇌ 2NO2(g) is 0.113 at 298 K, which corresponds to a standard free-energy change of 5.4 kJ/mol. In a certain experiment, the initial pressures are PN2O4 = 0.453 atm and PNO2 = 0.122 atm. Calculate ΔG for the reaction at these pressures, and predict the direction in which the reaction will proceed spontaneously to establish equilibrium. Strategy Use the partial pressures of N2O4 and NO2 to calculate the reaction quotient QP, and then use ΔG = ΔG° + RT lnQ to calculate ΔG. The reaction quotient expression is (PNO2)2 (0.122)2 QP = = = 0.0329 PN2O4 0.453 Worked Example 18.8 (cont.) Solution ΔG = ΔG° + RT lnQ 5.4 kJ 8.314×10-3 kJ = + (298 K)(ln 0.0329) mol K∙mol = 5.4 kJ/mol – 8.46 kJ/mol = –3.1 kJ/mol Because ΔG is negative, the reaction proceeds spontaneously from left to right to reach equilibrium. Think About It Remember, a reaction with a positive ΔG° value can be spontaneous if the starting concentrations of reactants and products are such that Q < K. Free Energy and Chemical Equilibrium At equilibrium, ΔG = 0 and Q = K: 0 = ΔG° + RT ln K ΔG° = –RT ln K Free Energy and Chemical Equilibrium At equilibrium, ΔG = 0 and Q = K: 0 = ΔG° + RT ln K ΔG° = –RT ln K Free Energy and Chemical Equilibrium At equilibrium, ΔG = 0 and Q = K: 0 = ΔG° + RT ln K ΔG° = –RT ln K Worked Example 18.9 Using data from Appendix 2, calculate the equilibrium constant, KP, for the following reaction at 25°C: 2H2O(l) ⇌ 2H2(g) + O2(g) Strategy Use data from Appendix 2 and ΔG°rxn = ΣnΔG°f (products) – ΣmΔG°f (reactants) to calculate ΔG° for the reaction. Then use ΔG° = −RT lnK to solve for KP. Think About It This is an extremely small equilibrium constant, Solution with the large, positive value of ΔG°. We know ΔG°= which (2ΔG°f is [Hconsistent 2(g)] + ΔG° f [O2(g)]) – (2ΔG° f [H2O(l)]) everyday+ experience water doeskJ/mol)] not decompose =from [2(0 kJ/mol) (0 kJ/mol)]that − [(2)(−237.2 =spontaneously 474.4 kJ/mol into its constituent elements at 25°C. ΔG° = −RT ln KP 474.4 kJ 8.314×10-3 kJ = (298 K) ln KP mol K∙mol −191.5 = ln KP KP = e−191.5 = 7×10-84 Worked Example 18.10 The equilibrium constant, Ksp, for the dissolution of silver chloride in water at 25°C: AgCl(s) ⇌ Ag+(aq) + Cl-(aq) is 1.6×10-10. Calculate ΔG° for the process. Strategy Use ΔG° = −RT lnK to calculate ΔG°. Solution R = 8.314×10-3 kJ/K∙mol and T = (25 + 273) = 298 K. ΔG° = −RT ln Ksp 8.314×10-3 kJ = (298 K) ln (1.6×10-10) K∙mol = 55.9 kJ/mol Think About It The relatively large, positive ΔG°, like the very small K value, corresponds to a process that lies very far to the left. Note that the K in ΔG° = −RT lnK can be any type of Kc (Ka, Kb, Ksp, etc.) or KP. 18.7 Thermodynamics of Living Systems Many biological reactions have positive ΔG° value, making the reaction nonspontaneous. None spontaneous reactions can be coupled with spontaneous reactions in order to drive a process forward: alanine + glycine → alanylglycine ATP + H2O → ADP + H3PO4 ΔG° = 29 kJ/mol ΔG° = –31 kJ/mol ATP + H2O + alanine + glycine → ADP + H3PO4 + alanylglycine ΔG° = 29 kJ/mol + –31 kJ/mol = –2 kJ/mol Thermodynamics of Living Systems Many biological reactions have positive ΔG° value, making the reaction nonspontaneous. 18 Key Concepts A Qualitative Description of Entropy A Quantitative Description of Entropy Calculating Δssys Standard S° Qualitatively Predicting ΔS°sys Calculating Δssurr The Second Law of Thermodynamics The Third Law of Thermodynamics Gibbs Free-Energy Change, ΔG Standard Free-Energy Changes, ΔG° Using ΔG and ΔG° to Solve Problems Relationship Between ΔG and ΔG° Relationship Between ΔG° and K