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Chemistry: Atoms First Julia Burdge & Jason Overby
Chapter 16 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.
18.1 Spontaneous Processes
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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
Spontaneous Processes
A process that does occur under a specific set of conditions is called a spontaneous process.
A process that results in a decrease in the energy of a system often is spontaneous:
A process that does not occur under a specific set of conditions is called nonspontaneous.
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)
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.
H2O(s)
T > 0°C; ΔH° = -6.01 kJ/mol
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
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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
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
Alternatively:
ΔSsys = nR ln
Ssys nR ln
Vfinal Vinitial
Worked Example 18.1 (cont.) Solution These equilibrium concentrations are then substituted into the equilibrium expression to give (x)(0.050 + x) 1.8×10-5 = 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
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.
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.
Therefore, the equilibrium expression simplifies to (x)(0.050) 1.8×10-5 = 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.
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Entropy Changes in a System
Entropy Changes in a System
There are several important trends in entropy:
In addition to translational motion, molecules exhibit vibrations and rotations.
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
Worked Example 18.2 From the standard enthalpy values in Appendix 2, calculate the standard entropy changes for the following reactions at 25°C.
For a chemical reaction aA + bB → cC + dD
(a) CaCO3(s) → CaO(s) + CO2(g)
(b) N2(g) + 3H2(g) → 2NH3(g)
(c) H2(g) + Cl2(g) → 2HCl(g) ΔS°rxn = [cS°(C) + dS°(D)] – [aS°(A) + bS°(B)]
Alternatively, ΔS°rxn = ΣnS°(products) – ΣmS°(reactants)
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
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.
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
(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.
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Entropy Changes in a System
Worked Example 18.3
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
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 NH 3(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 each only particle can and occupy in each case. About It For reactions liquids solids, An increase in thethe number arrangements corresponds in entropy predicting sign ofofΔS° can be more difficult, to butaninincrease many 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.
18.4
Entropy Changes in the Universe
ΔS is (a) positive
(b) positive
(d) negative
(e) positive
(c) negative
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.
The change in entropy of the surroundings is directly proportional to the enthalpy of the system.
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
Ssurr
Hsys T
The second law of thermodynamics states that for a process to be spontaneous, ΔSuniverse must be positive. ΔSuniverse = ΔSsys + ΔSsurr
The entropy of both the system AND surroundings are important!
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) H 2(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 f°(products) – ΣmΔH f°(reactants) to determine ΔH °sys and ΔS °surr. At the specified temperature, the 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.
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Worked Example 18.4 (cont.)
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.
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.
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
(b), (c) ΔS °rxn = 160.5 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
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
ΔSsurr =
−ΔHrxn −(−10.5 kJ/mol) = = 0.0385 kJ/K∙mol = 38.5 J/K∙mol T 273 K
(b) T = 200°C and −ΔHsys −(177.8 kJ/mol) = = −0.376 kJ/K∙mol = −376 J/K∙mol T 473 K
ΔSuniverse = ΔSsys + ΔSsurr = 21.03 J/K∙mol + 38.5 J/K∙mol = 59.5 J/K∙mol
ΔSsurr =
ΔSuniverse is positive; therefore the reaction is spontaneous at 0°C.
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
18.5
The third law of thermodynamics states that the entropy of a perfect crystalline substance is zero at absolute zero.
Measurements on the surroundings are seldom made, limiting the use of the second law of thermodynamics.
Entropy increases in a substance as temperature increases from absolute zero.
Gibbs free energy (G) or simply free energy can be used to express spontaneity more directly.
Predicting Spontaneity
G = H – TS The change in free energy for a system is: ΔG = ΔH – TΔS
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Predicting Spontaneity Using the Gibbs free energy, it is possible to make predictions on spontaneity. ΔG = ΔH – TΔS ΔG < 0 The reaction is spontaneous in the forward direction. ΔG > 0 The reaction is nonspontaneous in the forward direction. ΔG = 0 The 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
Entropy Changes in a System
The standard free energy of reaction (ΔG°rxn) is free-energy change for a reaction when it occurs under standard-state conditions.
For a chemical reaction
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
Worked Example 18.6
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 (cont.)
(b) 2MgO(s) → 2Mg(s) + O2(g)
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.
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.
(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
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.
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.
Calculate the standard free-energy changes for the following reactions at 25°C: (a) CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
(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
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18.6
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
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.
or 87.7 J/K∙mol
Free Energy and Chemical Equilibrium
Free Energy and Chemical Equilibrium
Consider the following 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 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
2
PH2
PI2
3.0 9.0 2.25 2.0 2.0 4.0
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
2
Then:
2.60 kJ 8.314 103 kJ G 298 K ln2.25 4.60 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.
PHI
2
1.0
2
P P 2.0 2.0 H2
I2
1.0 0.25 4.0
Then: G
2.60 kJ 8.314 103 kJ 298 K ln0.25 0.8 kJ/mol mol K·mol
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.
The reaction quotient expression is QP =
(PNO2)2 (0.122)2 = = 0.0329 PN2O4 0.453
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Free Energy and Chemical Equilibrium
Free Energy and Chemical Equilibrium
At equilibrium, ΔG = 0 and Q = K:
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
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 is° consistent with the large, positive value of ΔG°. We know ΔG ° which = (2ΔG f [H2(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.
18.7
Thermodynamics of Living Systems
Many biological reactions have positive ΔG° value, making the reaction nonspontaneous.
Strategy Use ΔG° = −RT lnK to calculate ΔG°.
Sone spontaneous reactions can be coupled with spontaneous reactions in order to drive a process forward:
Solution R = 8.314×10-3 kJ/K∙mol and T = (25 + 273) = 298 K.
alanine + glycine → alanylglycine
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.
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
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Thermodynamics of Living Systems Many biological reactions have positive ΔG° value, making the reaction nonspontaneous.
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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
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