AP Chemistry
Chapter 19 Chemical Thermodynamics
Chapter 19. Chemical Thermodynamics
Common Student Misconceptions
•
•
Students often believe that a spontaneous process should occur very quickly. They do not appreciate the
difference between kinetics and thermodynamics.
Students have a problem distinguishing between absolute thermodynamic quantities and the change in
thermodynamic quantities.
Lecture Outline
19.1 Spontaneous Processes
Chemical thermodynamics is concerned with energy relationships in chemical reactions,
i.e. Will a reaction occur?
Two components:
• enthalpy (H)
• entropy (S) (randomness or disorder): note that in nature things tend towards disorder, much easier to
go from low S  high S, but natural tendency to disorder can be overcome by enthalpy.
First law of thermodynamics: energy is conserved: ΔE= q + w
ΔE = change in internal energy,
q = heat absorbed by the system from the surroundings
w = work done.
Spontaneous process = any process that occurs without outside intervention.
• Two eggs are dropped  spontaneously break.
• Reverse reaction (two eggs leaping into your hand with their shells back intact) is not spontaneous.
 a spontaneous process has a direction.
•
A process that is spontaneous in one direction is not spontaneous in the opposite direction.
•
Temperature may also affect the spontaneity of a process.
Reversible and Irreversible Processes
•
A reversible process = one that can go back and forth between states along the same path.
•
Reverse process restores the system to its original state.
•
Path taken back to the original state is exactly the reverse of the forward process.
•
There is no net change in the system or the surroundings when this cycle is completed.
•
Completely reversible processes are too slow to be attained in practice.
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AP Chemistry
Chapter 19 Chemical Thermodynamics
e.g. the interconversion of ice and water at 1 atm, 0oC.
•
Ice and water are in equilibrium.
•
Add heat to the system from the surroundings.
1 mole of ice is melted to form 1 mole of liquid water.
q=ΔHfus
•
To return to the original state we reverse the procedure.
• We remove the same amount of heat from the system to the surroundings.
Sample Exercise 19.1 (p. 737)
Predict whether the following processes are spontaneous as described, are spontaneous in the reverse direction,
or are in equilibrium:
a)
When a piece of metal heated to 150oC is added to water at 40oC, the water gets hotter.
b)
Water at room temperature decomposes into H2(g) and O2(g).
c)
Benzene vapor, C6H6(g), at a pressure of 1 atm condenses to liquid benzene at the normal boiling point
of benzene, 80.1oC.
Practice Exercise 19.1
Under 1 atm pressure CO2(s) (“dry ice”) sublimes at -78oC. Is the transformation of CO2(s) to CO2(g) a
spontaneous process at -100oC and 1 atm pressure?
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AP Chemistry
Chapter 19 Chemical Thermodynamics
•
An irreversible process cannot be reversed to restore the system and surroundings back to their original
state.
• A different path (with different values of q and w) must be taken.
•
Consider a gas in a cylinder with a piston.
•
Remove the partition and the gas expands to fill the space.
•
No P-V work is done on the surroundings.
 w = 0.
•
Now use the piston to compress the gas back to the original state.
•
The surroundings must do work on the system.
 w > 0.
•
A different path is required to get the system back to its original state.
• Note that the surroundings are NOT returned to their original conditions.
•
For a system at equilibrium, reactants and products can interconvert reversibly.
•
For a spontaneous process, the path between reactants and products is irreversible.
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AP Chemistry
Chapter 19 Chemical Thermodynamics
19.2 Entropy and the Second Law of Thermodynamics
Why do spontaneous processes occur?
•
Consider the expansion of an ideal gas:
•
Consider an initial state: two 1-L flasks connected by a closed stopcock.
• One flask is evacuated and the other contains 1 atm of gas.
• We open the stopcock while maintaining the system at constant T.
• Initial state: an ideal gas confined to a cylinder kept at constant T in a water bath.
• The process is isothermal: at constant T
• ΔE = 0 for an isothermal process.
• Thus, q = –w.
•
Allow the gas to expand from V1 to V2.
•
Pressure ↓ from P1 to P2.
• The final state: two flasks connected by an open stopcock.
• Each flask contains gas at 0.5 atm.
 the gas does no work and heat is not transferred.
•
Why does the gas expand?
• Why is the process spontaneous?
• Why is the reverse process nonspontaneous?
• When the gas molecules spread out into the 2 L system there is an ↑ in the randomness or disorder.
• Processes in which the disorder or entropy of the system ↑ tend to be spontaneous.
When there are many molecules, it is much more probable that the molecules will distribute
among to the two flasks than all remain in only one flask.
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AP Chemistry
Chapter 19 Chemical Thermodynamics
Entropy
•
Consider the melting of ice.
• In the ice, the molecules are held rigidly in a lattice.
• When it melts, the molecules have more freedom to move (↑ degrees of freedom).
• The molecules are more randomly distributed.
•
Consider a KCl crystal dissolving in water.
• The solid KCl has ions in a highly ordered arrangement.
• crystal dissolves  the ions have more freedom  more randomly distributed.
• However, now the water molecules are more ordered.
• Some must be used to hydrate the ions.
• Thus this example involves both ordering and disordering.
• The disordering usually predominates (for most salts).
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AP Chemistry
•
Chapter 19 Chemical Thermodynamics
Entropy, S, is a thermodynamic term that reflects the disorder or randomness of the system.
• ↑ disordered or random the system  ↑ S.
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AP Chemistry
Chapter 19 Chemical Thermodynamics
Sample Exercise 19.2 (p. 743)
By considering the disorder in the reactants and products, predict whether ΔS is positive or negative for each of
the following processes:
a)
H2O(l)  H2O(g)
b)
Ag+(aq) + Cl-(aq)  AgCl(s)
c)
4 Fe(s) + 3 O2(g)  2 Fe2O3(s)
Practice Exercise 19.2
Indicate whether each of the following reactions produces an increase or decrease in the entropy of the system:
a)
CO2(s)  CO2(g)
b)
CaO(s) + CO2(g)  CaCO3(s)
Relating Entropy to Heat Transfer and Temperature
•
Entropy is a state function.
•
independent of path.
•
For a system, ΔS = Sfinal – Sinitial.
•
If ΔS > 0 the randomness ↑, if ΔS < 0 the order ↑.
•
Suppose a system changes reversibly between state 1 and state 2.
qrev is given by:
T
•
Then, the change in entropy
•
Where qrev is the amount of heat added reversibly to the system.
• The subscript “rev” reminds us that the path between states is reversible.
• Example: A phase change occurs at constant T with the reversible addition of heat.
!S =
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AP Chemistry
Chapter 19 Chemical Thermodynamics
Sample Exercise 19.3 (p. 744)
The element mercury, Hg, is a silvery liquid at room temperature. The normal freezing point of mercury is 38.9oC, and its molar enthalpy of fusion is ΔHfus = 2.29 kJ/mol. What is the entropy change of the system when
50.0 g of Hg(l) freezes at the normal freezing point?
(ΔSsys = -2.44 J/K)
Practice Exercise 19.3
The normal boiling point of ethanol, C2H5OH, is 78.3oC (see Figure 11.12), and its molar enthalpy of
vaporization is 38.56 kJ/mol. What is the change in entropy when 68.3 g of C2H5OH(g) at 1 atm pressure
condenses to liquid at the normal boiling point?
(-163 J/K)
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AP Chemistry
Chapter 19 Chemical Thermodynamics
The Second Law of Thermodynamics
•
The second law of thermodynamics explains why spontaneous processes have a direction.
•
In any spontaneous process, the entropy of the universe ↑.
•
The change in entropy of the universe is the sum of the change in entropy of the system and the
change in entropy of the surroundings.
ΔSuniv = ΔSsys + ΔSsurr
•
For a reversible process:
ΔSuniv = ΔSsys + ΔSsurr = 0
•
For a spontaneous process (i.e., irreversible):
ΔSuniv = ΔSsys + ΔSsurr > 0
•
Entropy is not conserved: ΔSuniv is continually ↑.
•
•
Note: The second law states that the entropy of the universe must ↑ in a spontaneous process.
It is possible for the entropy of a system to ↓ as long as the entropy of the surroundings ↑.
•
Consider an isolated system.
• The system does not exchange energy or matter with the surroundings.
• ΔSsys = 0 for a reversible process and ΔSsys > 0 for a spontaneous process.
Sample Exercise 19.4 (p. 746)
Consider the reversible melting of 1 mol of ice in a large, isothermal water bath at 0oC and 1 atm pressure. The
enthalpy of fusion of ice is 6.01 kJ/mol. Calculate the entropy change in the system and in the surroundings,
and the overall change in entropy of the universe for this process.
(22.0 J/mol-K, -22.0 J/mol-K, 0)
Practice Exercise 19.4
The molar enthalpy of vaporization for liquid bromine is 30.71 kJ/mol. Calculate the entropy change in the
system, the surroundings, and the universe for the reversible vaporization of 1 mol of liquid bromine (Br2) at its
normal boiling point (59oC).
(93 J/K-mol)
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AP Chemistry
Chapter 19 Chemical Thermodynamics
19.3 The Molecular Interpretation of Entropy
•
The entropy of a system indicates its disorder.
•
A gas is less ordered than a liquid which is less ordered than a solid.
•
Any process that ↑ the number of gas molecules  ↑ in entropy.
•
When NO(g) reacts with O2(g) to form NO2(g), the total number of gas molecules ↓.
 the entropy ↓.
•
•
2NO(g) + O2(g)  2NO2(g)
How can we relate changes in entropy to changes at the molecular level?
•
Formation of the new N-O bonds “ties up” more of the atoms in the products than in the reactants.
•
The degrees of freedom associated with the atoms have changed.
•
↑ freedom of movement and degrees of freedom  ↑ the entropy of the system.
Individual molecules have degrees of freedom associated with motions within the molecule.
• There are three atomic modes of motion:
•
Translational motion.
• The moving of a molecule from one point in space to another.
•
Vibrational motion.
• The shortening and lengthening of bonds, including the change in bond angles.
•
Rotational motion.
• The spinning of a molecule about some axis.
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AP Chemistry
•
Chapter 19 Chemical Thermodynamics
Energy is required to get a molecule to translate, vibrate or rotate.
•
These forms of motion are ways molecules can store energy.
•
↑ energy stored in translation, vibration and rotation  ↑ entropy.
•
In a perfect crystal at 0 K there is no translation, rotation or vibration of molecules = state of perfect order
•
Third law of thermodynamics: The entropy of a perfect pure crystal at 0 K = 0.
•
Entropy will ↑ as we ↑ the temperature of the perfect crystal.
•
Molecules gain vibrational motion  ↑ degrees of freedom.
•
As we heat a substance from 0 K  the entropy must ↑.
•
The entropy changes dramatically at a phase change.
• When a solid melts, the molecules and atoms have a large ↑ in freedom of movement.
• Boiling corresponds to a much greater change in entropy than melting.
•
In general, entropy will ↑ when liquids or solutions are formed from solids.
•
•
Gases are formed from solids or liquids.
The number of gas molecules increase.
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AP Chemistry
Chapter 19 Chemical Thermodynamics
Sample Exercise 19.5 (p. 752)
Choose the sample of matter that has greater entropy in each pair, and explain your choice:
a)
1 mol of NaCl(s) or 1 mol of HCl(g) at 25oC
b)
2 mol of HCl(g) or 1 mol of HCl(g) at 25oC
c)
1 mol of HCl(g) or 1 mol of Ar(g) at 25oC
d)
1 mol of N2(s) at 24 K or 1 mol of N2(g) at 298 K
Practice Exercise 19.5
Choose the substance with the greater entropy in each case:
a)
1 mol of H2(g) at STP or 1 mol H2(g) at 100oC and 0.5 atm
b)
1 mol of H2O(s) at 0oC or 1 mol of H2O(l) at 25oC
c)
1 mol of H2(g) at STP or 1 mol of SO2(g) at STP
d)
1 mol of N2O4(g) at STP or 2 mol of NO2(g) at STP
Sample Exercise 19.6 (p. 752)
Predict whether the entropy change of the system in each of the following isothermal reactions is positive or
negative.
a)
CaCO3(s)  CaO(s) + CO2(g)
b)
N2(g) + 3 H2(g)  2 NH3(g)
c)
N2(g) + O2(g)  2 NO(g)
Practice Exercise 19.6
Predict whether ΔS is positive or negative in each of the following processes:
a)
HCl(g) + NH3(g)  NH4Cl(s)
b)
2 SO2(g) + O2(g)  2 SO3(g)
c)
Cooling nitrogen gas from 20oC to -50oC
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AP Chemistry
Chapter 19 Chemical Thermodynamics
19.4 Entropy Changes in Chemical Reactions
•
Absolute entropy can be determined from complicated measurements.
• Values are based on a reference point of zero for a perfect crystalline solid at 0 K (the 3rd law).
•
Standard molar entropy, So: molar entropy of a substance in its standard state.
• Similar in concept to ΔHo.
• Units: J/mol-K.
• Note: the units of ΔH are kJ/mol.
•
Some observations about So values:
• Standard molar entropies of elements ≠ zero.
•
•
Sogas > Soliquid or Sosolid.
•
So tends to ↑ with ↑ molar mass of the substance.
•
So tends to ↑ with the number of atoms in the formula of the substance.
For a chemical reaction which produces n products from m reactants:
#S ° =
•
! nS °(products)" ! mS °(reactants)
Example: Consider the reaction:
N2(g) + 3H2(g)  2NH3(g)
ΔSo = {2So(NH3) – [So(N2) + 3So(H2)]}
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AP Chemistry
Chapter 19 Chemical Thermodynamics
Sample Exercise 19.7 (p. 753)
Calculate ΔSo for the synthesis of ammonia from N2(g) and H2(g) at 298 K.
N2(g) + 3 H2(g)  2 NH3(g)
(-198.3 J/K)
Practice Exercise 19.7
Using the standard entropies in Appendix C, calculate the standard entropy change,
reaction at 298 K:
So for the following
Al2O3(s) + 3 H2(g)  2 Al(s) + 3 H2O(g)
(180.39 J/K)
Entropy Changes in the Surroundings
•
For an isothermal process,
• ΔSsurr = –qsys / T
•
For a reaction at constant pressure,
• qsys = ΔH
•
Example: consider the reaction:
N2(g) + 3H2(g)  2NH3(g)
•
The entropy gained by the surroundings is > the entropy lost by the system.
•
This is the sign of a spontaneous reaction: the overall entropy change of the universe is positive.
•
ΔSuniv > 0
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AP Chemistry
Chapter 19 Chemical Thermodynamics
19.5 Gibbs Free Energy
•
For a spontaneous reaction the entropy of the universe must ↑.
•
Reactions with large negative ΔH values tend to be spontaneous.
•
How can we use ΔS and ΔH to predict whether a reaction is spontaneous?
•
The Gibbs free energy, (free energy), G, of a state is:
G = H–TS
• Free energy is a state function.
• For a process occurring at constant temperature, the free energy change is: ΔG = ΔH–TΔS
•
Recall:
• ΔSuniv = ΔSsys + ΔSsurr = ΔSsys + [–ΔHsys / T]
•
•

–TΔSuniv = ΔHsys – TΔSsys
The sign of ΔG is important in predicting the spontaneity of the reaction.
•
If ΔG < 0 then the forward reaction is spontaneous.
•
If ΔG = 0 then the reaction is at equilibrium and no net reaction will occur.
•
If ΔG > 0 then the forward reaction is not spontaneous.
• However, the reverse reaction is spontaneous.
• If ΔG > 0, work must be supplied from the surroundings to drive the reaction.
The equilibrium position in a spontaneous process is given by the minimum free energy available to the
system.
•
The free energy ↓ until it reaches this minimum value.
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AP Chemistry
Chapter 19 Chemical Thermodynamics
Standard Free-Energy Changes
•
We can tabulate standard free energies of formation, ΔGof .
•
Standard states are: pure solid, pure liquid, 1 atm (gas), 1 M concentration (solution), and ΔGo = 0 for
elements.
• We most often use 25oC (or 298 K) as the temperature.
•
The standard free-energy change for a process is given by:
"G ° =
•
! n"G° (products)# ! m"G° (reactants)
f
f
The quantity ΔGo for a reaction tells us whether a mixture of substances will spontaneously react
to produce more reactants (ΔGo > 0) or products (ΔGo < 0).
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AP Chemistry
Chapter 19 Chemical Thermodynamics
Sample Exercise 19.8 (p. 758)
a)
By using data from Appendix C, calculate the standard free-energy change for the following reaction at
298 K:
P4(g) + 6 Cl2(g)  4 PCl3(g)
(-1102.8 kJ)
b)
What is ΔGo for the reverse of the above reaction?
(+1102.8 kJ)
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AP Chemistry
Chapter 19 Chemical Thermodynamics
Practice Exercise 19.8
By using the data from Appendix C, calculate ΔGo at 298 K for the combustion of methane:
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
(-800.7 kJ)
Sample Exercise 19.9 (p. 758)
In Section 5.7 we used Hess’s law to calculate ΔHo for the combustion of propane gas at 298 K (see Figure
5.22):
C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l)
ΔHo = -2220 kJ
a)
Without using data from Appendix C, predict whether ΔGo for this reaction is more negative or less
negative than ΔHo.
b)
Use data from Appendix C to calculate the standard free-energy change for the reaction at 298 K. Is
your prediction from part (a) correct?
(-2108 kJ)
Practice Exercise 19.9
Consider the combustion of propane to form CO2(g) and H2O(g) at
298 K:
C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(g)
ΔHo = -2220 kJ
Would you expect ΔGo to be more negative or less negative than ΔHo?
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AP Chemistry
Chapter 19 Chemical Thermodynamics
19.6 Free Energy and Temperature
•
The sign of ΔG tells us if the reaction is spontaneous.
•
Focus on ΔG = ΔH – TΔS.
•
•
If ΔH <0 and –TΔS <0:
• ΔG will always be <0.
• Thus the reaction will be spontaneous.
•
If ΔH >0 and –TΔS >0:
• ΔG will always be >0.
• Thus the reaction will not be spontaneous.
•
If ΔH and –TΔS have different signs:
• The sign of ΔG will depend on the sign and magnitudes of the other terms.
• Temperature will be an important factor.
•
For example, consider the following reaction:
H2O(s)  H2O(l)
ΔH >0, ΔS > 0
• At a T < 0oC:
• ΔH > TΔS
• ΔG> 0
• The melting of ice is not spontaneous when the temperature is < 0oC.
•
At a T > 0oC:
• ΔH < TΔS
• ΔG < 0
• The melting of ice is spontaneous when the temperature is > 0oC.
•
At 0oC:
• ΔH = TΔS
• ΔG = 0
• Ice and water are in equilibrium at 0oC.
Even though a reaction has a negative ΔG it may occur too slowly to be observed.
• Thermodynamics gives us the direction of a spontaneous process; it does not give us the rate of the
process.
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AP Chemistry
Chapter 19 Chemical Thermodynamics
Sample Exercise 19.10 (p. 760)
The Haber process for the production of ammonia involves the following equilibrium:
N2(g) + 3 H2(g)
2 NH3(g)
Assume that ΔHo and ΔSo for this reaction do not change with temperature.
a)
b)
Predict the direction in which ΔGo for this reaction changes with increasing temperature.
Calculate the values of ΔGo for the reaction at 25oC and 500oC.
(-33.3 kJ, 61 kJ)
Practice Exercise 19.10
a)
Using standard enthalpies of formation and standard entropies in Appendix C, calculate ΔHo and ΔSo at
298 K for the following reaction:
2 SO2(g) + O2(g)  2 SO3(g).
(ΔHo = -196.6 kJ, ΔSo = -189.6 J/K)
b)
Using the values obtained in part (a), estimate ΔGo at 400 K.
(ΔGo = -120.8 kJ)
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AP Chemistry
Chapter 19 Chemical Thermodynamics
19.7 Free Energy and the Equilibrium Constant
•
Recall that ΔGo and Keq (equilibrium constant) apply to standard conditions.
•
Recall that ΔG and Q (equilibrium quotient) apply to any conditions.
•
It is useful to determine whether substances will react under specific conditions:
ΔG = ΔGo + RTlnQ
Sample Exercise 19.11 (p. 762)
As we saw in Section 11.5, the normal boiling point is the temperature at which a pure liquid is in equilibrium
with its vapor at a pressure of 1 atm.
a)
Write the chemical equation that defines the normal boiling point of liquid carbon tetrachloride, CCl4(l).
b)
What is the value of ΔGo for the equilibrium in part (a)?
c)
Use thermodynamic data in Appendix C and Equation 19.20 to estimate the normal boiling point of
CCl4. (70oC)
Practice Exercise 19.11
Use data in Appendix C to estimate the normal boiling point, in K, for elemental bromine, Br2(l).
(The experimental value is given in Table 11.3). (330 K)
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AP Chemistry
Chapter 19 Chemical Thermodynamics
Sample Exercise 19.12 (p. 763)
We will continue to explore the Haber process for the synthesis of ammonia:
N2(g) + 3 H2(g)
2 NH3(g)
Calculate ΔG at 298 K for a reaction mixture that consists of 1.0 atm N2, 3.0 atm H2, and 0.50 atm NH3.
(-44.9 kJ/mol)
Practice Exercise 19.12
Calculate ΔG at 298 K for the reaction of nitrogen and hydrogen to form ammonia if the reaction mixture
consists of 0.50 atm N2, 0.75 atm H2, and 2.0 atm NH3.
(-26.0 kJ/mol)
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AP Chemistry
Chapter 19 Chemical Thermodynamics
• At equilibrium, Q = Keq and ΔG = 0, so:
ΔG = ΔGo + RTlnQ
0 = ΔGo + RTlnKeq
∴ ΔGo = – RTlnKeq
•
From the above we can conclude:
•
If ΔGo < 0, then Keq > 1.
•
If ΔGo = 0, then Keq = 1.
•
If ΔGo > 0, then Keq < 1.
Sample Exercise 19.13 (p. 764)
Use standard free energies of formation to calculate the equilibrium constant Keq at 25oC for the reaction
involved in the Haber process:
N2(g) + 3 H2(g)
2 NH3(g)
(7 x 105)
Practice Exercise 19.13
Use data from Appendix C to calculate the standard free-energy change, ΔGo, and the equilibrium constant, Keq,
at 298 K for the following reaction:
H2(g) + Br2(l)
2 HBr(g)
(-106.4 kJ/mol; 5 x 1018)
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AP Chemistry
Chapter 19 Chemical Thermodynamics
Driving Nonspontaneous Reactions
•
If ΔG > 0, work must be supplied from the surroundings to drive the reaction.
•
Biological systems often use one spontaneous reaction to drive another nonspontaneous reaction.
• These reactions are coupled reactions.
•
The energy required to drive most nonspontaneous reactions comes from the metabolism of foods.
•
Example: Consider the oxidation of glucose:
C6H12O6(s) + 6O2(g)  6CO2(g) + 6H2O(l)
•
The free energy released by glucose oxidation is used to convert low energy adenosine diphosphate
(ADP) and inorganic phosphate into high energy adenosine triphosphate (ATP).
•
When ATP is converted back to ADP the energy released may be used to “drive” other reactions.
ΔGo = –2880 kJ.
Sample Integrative Exercise 19 (p. 766)
Consider the simple salts NaCl(s) and AgCl(s). We will examine the equilibria in which these salts dissolve in
water to form aqueous solutions of ions:
NaCl(s)
AgCl(s)
Na+(aq) + Cl-(aq)
Ag+(aq) + Cl-(aq)
a)
Calculate the value of ΔGo at 298 K for each of the preceding reactions.
b)
The two values from part (a) are very different. Is this difference primarily due to the enthalpy term or
the entropy term of the standard free-energy change?
c)
Use the values of ΔGo to calculate Ksp values for the two salts at 298 K.
d)
Sodium chloride is considered a soluble salt, whereas silver chloride is considered insoluble. Are these
descriptions consistent with the answers to part (c)?
e)
How will ΔGo for the solution process of these salts change with increasing T? What effect should this
change have on the solubility of the salts?
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