1. 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.

2. Entropy, Free Energy, and Equilibrium18
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

3. Spontaneous Processes
18.1
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.

4. 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° = kJ/mol
The sign of ΔH alone is insufficient to predict spontaneity in every circumstance:
H2O(l) H2O(s) T > 0°C; ΔH° = kJ/mol

5. Entropy
18.2
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.

6. Entropy
Spontaneity is favored by an increase in entropy.
k is the Boltzmann constant (1.38 x 10–23 J/K)
W is the number of different arrangements
S = k ln W
The number of arrangements possible is given by:
X is the number of cells in a volume
N is the number of molecules
W = X N

7. Entropy

8. 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.

9. Entropy Changes in a System
18.3
The change in entropy for a system is the difference in entropy of the final state and the entropy of the initial state.
Alternatively:
ΔSsys = Sfinal – Sinitial

10. 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 = J/K∙mol, n = 1.0 mole, Vfinal = 5.0 L, and Vinitial = 2.5 L.
Solution
Vfinal
Vinitial
8.314 J
K ∙ mol
5.0 L
2.5 L
ΔSsys = nR ln
= 1.0 mol ×
×ln
= 5.8 J/K
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.

11. Worked Example 18.1 (cont.)
Solution These equilibrium concentrations are then substituted into the equilibrium expression to give
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 ( x) M ≈ M
Therefore, the equilibrium expression simplifies to
and x = 3.6×10-5 M. According to the equilibrium table, [H+] = x, so pH = –log(3.6×10-5) = 4.44.
(x)( x)
0.010 – x
1.8×10-5 =
(x)(0.050)
0.010
1.8×10-5 =

12. 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.

13. 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)

14. Entropy Changes in a System
In addition to translational motion, molecules exhibit vibrations and rotations.

15. Entropy Changes in a System
For a chemical reaction
aA + bB → cC + dD
Alternatively,
ΔS°rxn = [cS°(C) + dS°(D)] – [aS°(A) + bS°(B)]
ΔS°rxn = ΣnS°(products) – ΣmS°(reactants)

16. 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)] = J/K∙mol, S°[N2(g)] = J/K∙mol, S°[H2(g)] = J/K∙mol, S°[NH3(g)] = J/K∙mol, S°[Cl2(g)] = J/K∙mol, and S°[HCl(g)] = J/K∙mol.

17. 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)
= 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.

18. 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

19. 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

20. 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.
Think About It For reactions involving only liquids and solids, predicting the sign of ΔS° can be more difficult, but in many such cases an increase in the total number of molecules and/or ions is accompanied by an increase of entropy.
Strategy Consider the change in energy/mobility of atoms and the resulting change in number of possible positions that each particle can occupy in each case. An increase in the number of arrangements corresponds to an increase in entropy and therefore a positive ΔS.
Solution Increases in entropy generally accompany solid-to-liquid, liquid-to-gas, 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
(c) negative
(d) negative
(e) positive

21. Entropy Changes in the Universe
18.4
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.
A cup of hot water spontaneously cools to room temperature.
The entropy of both the system AND surroundings are important!
Perspective
Components ΔS
System
ice
positive
Surroundings
everything else
negative
Perspective
Components ΔS
System
hot water
negative
Surroundings
everything else
positive

22. Entropy Changes in the Universe
The change in entropy of the surroundings is directly proportional to the enthalpy of the system.
The second law of thermodynamics states that for a process to be spontaneous, ΔSuniverse must be positive.
ΔSuniverse = ΔSsys + ΔSsurr

23. Entropy Changes in the Universe
The second law of thermodynamics states that for a process to be spontaneous, ΔSuniverse must be positive.
ΔSuniverse > 0 for a spontaneous process
ΔSuniverse < 0 for a nonspontaneous process
ΔSuniverse = 0 for an equilibrium process
ΔSuniverse = ΔSsys + ΔSsurr

24. 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 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.

25. Worked Example 18.4 (cont.)
Solution (a) S°[H2(g)] = J/K∙mol, S°[I2(g)] = J/K∙mol, S°[HI(g)] = J/K∙mol; ΔHf°[H2(g)] = 0 kJ/mol, ΔHf°[I2(g)] = 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 J/K∙mol] = J/K∙mol
ΔH °rxn = [2 ΔH °f (HI)] – [ΔH °f (H2) + ΔH °f (I2)]
= (2)(25.9 kJ/mol) – [0 kJ/mol kJ/mol] = −10.5 kJ/mol
ΔSsurr = = = kJ/K∙mol = 38.5 J/K∙mol
ΔSuniverse = ΔSsys + ΔSsurr = J/K∙mol J/K∙mol = 59.5 J/K∙mol
ΔSuniverse is positive; therefore the reaction is spontaneous at 0°C.
−ΔHrxn T
−(−10.5 kJ/mol)
273 K

26. Worked Example 18.4 (cont.)
Solution (b) In Worked Example 18.2(a), we determined that for this reaction, ΔS °rxn = J/K∙mol; ΔH °f [CaCO3(s)] = − kJ/mol, ΔH °f [CaO(s)] = −635.6 kJ/mol, ΔH °f [CO2(g)] = −393.5 kJ/mol.
(b), (c)
ΔS °rxn = J/K∙mol
ΔH °rxn = [ΔH °f (CaO) + ΔH °f (CO2)] – [ΔH °f (CaCO3)]
= [ kJ/mol + (–393.5 kJ/mol)] – (– kJ/mol) = kJ/mol
(b) T = 200°C and
ΔSsurr = = = −0.376 kJ/K∙mol = −376 J/K∙mol
ΔSuniverse = ΔSsys + ΔSsurr = J/K∙mol + (−376 J/K∙mol) = − 216 J/K∙mol
Δsuniverse is negative, therefore the reaction is nonspontaneous at 200°C.
−ΔHsys T
−(177.8 kJ/mol)
473 K

27. Worked Example 18.4 (cont.) Solution (c) T = 1000°C and
ΔSsurr = = = − kJ/K∙mol = −139.7 J/K∙mol
ΔSuniverse = ΔSsys + ΔSsurr = 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.
−ΔHsys T
−(177.8 kJ/mol)
473 K

28. Worked Example 18.4 (cont.)
Solution (d) S°[Na(s)] = J/K∙mol, S°[Na(l)] = J/K∙mol; ΔHf°[Na(s)] = 0 kJ/mol, ΔHf°[Na(l)] = 2.41 kJ/mol.
ΔS °rxn = S°[Na(l)] – S°[Na(s)]
= J/K∙mol – J/K∙mol = 6.51 J/K∙mol
ΔH °rxn = ΔHf°[Na(l)] – ΔHf°[Na(s)]
= 2.41 kJ/mol – 0 kJ/mol = 2.41 kJ/mol
ΔSsurr = = = − kJ/K∙mol = −6.50 J/K∙mol
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.
Think About It Remember that standard enthalpies of formation have units of kJ/mol, whereas standard absolute entropies have units of J/K∙mol. Make sure that you convert kilojoules to joules, or vice versa, before combining the terms.
−ΔHrxn T
−(2.41 kJ/mol)
371 K

29. 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.

30. Predicting Spontaneity
18.5
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.
The change in free energy for a system is:
G = H – TS
ΔG = ΔH – TΔS

31. Predicting Spontaneity
Using the Gibbs free energy, it is possible to make predictions on spontaneity.
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
ΔG = ΔH – TΔS

32. 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 = 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 a release of energy (ΔH being negative) and by an increase in entropy (ΔS being positive). When both quantities are positive, as in this case, only the entropy change favor spontaneity. For an endothermic process such as this, which requires the input of heat, it should make sense that adding more heat by increasing the temperature will shift the equilibrium to the right, thus making it “more spontaneous.”
Strategy The temperature that divides high from low is the temperature at which ΔH = TΔS (ΔG = 0). Therefore, we use ΔG = ΔH – TΔS, substituting 0 for ΔG and solving for T to determine temperature in kelvins; we then convert to degrees Celsius.
Solution
ΔS =
476 J
K∙mol
= kJ/K∙mol
1 kJ
1000 J
T = ΔH ΔS
= 419 K
199.5 kJ/mol
0.476 kJ/K∙mol =
= (419 – 273) = 146°C

33. 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

34. Entropy Changes in a System
For a chemical reaction
aA + bB → cC + dD
Alternatively,
ΔG°f for any element in its most stable allotropic form at 1 atm is defined as zero.
ΔG°rxn = [cΔG°f (C) + dΔG°f (D)] – [aΔG°f (A) + bΔG°f (B)]
ΔG°rxn = ΣnΔG°f (products) – ΣmΔG°f (reactants)

35. 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

36. 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.

37. 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-to- vapor transitions for benzene. At 1 atm pressure, benzene melts at 5.5°C and boils at 80.1°C.
Think About It For the same substance, ΔSvap is always significantly larger than ΔSfus. The change in number of arrangements is always bigger in a liquid-to-gas transition than in a solid-to-liquid transition.
Strategy The solid-liquid transition at the melting point and the liquid-vapor transition at the boiling points are equilibrium processes. Therefore, because ΔG is zero at equilibrium, in each case we can use ΔG = ΔH – TΔS, substituting 0 for ΔG and solving for ΔS, to determine the entropy change associated with the process.
ΔSfus =
10.9 kJ/mol
278.7 K
ΔHfus
Tmelting =
Solution
= kJ/K∙mol or 39.1 J/K∙mol
ΔSvap =
31.0 kJ/mol
353.3 K
ΔHvap
Tboiling =
= kJ/K∙mol or 87.7 J/K∙mol

38. Free Energy and Chemical Equilibrium
18.6
It is the sign of ΔG (not ΔG°) that determines spontaneity.
The relationship between ΔG and ΔG° is:
R is the gas constant (8.314 J/K·mol).
T is the kelvin temperature.
Q is the reaction quotient.
ΔG = ΔG° + RT lnQ

39. 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:
Then:

40. 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:
Then:

41. Worked Example 18.8 The equilibrium constant, KP, for the reaction
N2O4(g) ⇌ 2NO2(g)
is 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 = atm and PNO2 = 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
QP =
(0.122)2
0.453
(PNO2)2
PN2O4 = =

42. Worked Example 18.8 (cont.) Solution
Because ΔG is negative, the reaction proceeds spontaneously from left to right to reach equilibrium.
ΔG = ΔG° + RT lnQ
(298 K)(ln ) =
8.314×10-3 kJ
K∙mol
5.4 kJ
mol +
= 5.4 kJ/mol – 8.46 kJ/mol
= –3.1 kJ/mol
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.

43. Free Energy and Chemical Equilibrium
At equilibrium, ΔG = 0 and Q = K:
0 = ΔG° + RT ln K
ΔG° = –RT ln K

44. Free Energy and Chemical Equilibrium
At equilibrium, ΔG = 0 and Q = K:
0 = ΔG° + RT ln K
ΔG° = –RT ln K

45. Free Energy and Chemical Equilibrium
At equilibrium, ΔG = 0 and Q = K:
0 = ΔG° + RT ln K
ΔG° = –RT ln K

46. 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)
Think About It This is an extremely small equilibrium constant, which is consistent with the large, positive value of ΔG°. We know from everyday experience that water does not decompose spontaneously into its constituent elements at 25°C.
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.
Solution
ΔG ° = (2ΔG °f [H2(g)] + ΔG °f [O2(g)]) – (2ΔG °f [H2O(l)])
= [2(0 kJ/mol) + (0 kJ/mol)] − [(2)(−237.2 kJ/mol)]
= kJ/mol
ΔG° = −RT ln KP
474.4 kJ
mol
(298 K) ln KP =
8.314×10-3 kJ
K∙mol
−191.5 = ln KP
KP = e−191.5 = 7×10-84

47. AgCl(s) ⇌ Ag+(aq) + Cl-(aq)
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× Calculate ΔG° for the process.
Strategy Use ΔG° = −RT lnK to calculate ΔG°.
Solution
R = 8.314×10-3 kJ/K∙mol and T = ( ) = 298 K.
ΔG° = −RT ln Ksp
(298 K) ln (1.6×10-10) =
8.314×10-3 kJ
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.

48. Thermodynamics of Living Systems
18.7
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 ΔG° = 29 kJ/mol
ATP + H2O → ADP + H3PO4 ΔG° = –31 kJ/mol
ATP + H2O + alanine + glycine → ADP + H3PO4 + alanylglycine
ΔG° = 29 kJ/mol + –31 kJ/mol = –2 kJ/mol

49. Thermodynamics of Living Systems
Many biological reactions have positive ΔG° value, making the reaction nonspontaneous.

50. 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