Chapter Eighteen
ENTROPY, FREE ENERGY, AND EQUILIBRIUM
•
•
•
The Second Law of Thermodynamics (18.2 – 18.3)
Gibbs Free Energy (18.4)
Free Energy and Equilibrium (18.5)
THE SECOND LAW OF THERMODYNAMICS (18.2 – 18.3)
STUDY OBJECTIVES
1. Explain the meaning of the term "spontaneous process."
2. Predict, for a given process, whether the entropy of the system increases or decreases.
3. State the second law of thermodynamics.
4. Calculate the standard entropy change for a given reaction using a table of standard absolute entropies.
Spontaneous Processes. An important part of experimental chemistry deals with spontaneous reactions,
that is, reactions that take place without having to continually supply energy from outside the system. One goal
of thermodynamics is to be able to predict whether a reaction will take place when a given set of reactants is
brought together. In this chapter we discuss those properties of a system that can be used as criteria for
predicting spontaneous reactions. The release of heat by a reaction was once thought to be an indication that the
reaction was spontaneous. The sign of the enthalpy change, ∆H, by itself is not an adequate guide to spontaneity
because while some spontaneous reactions are known to be exothermic (∆H is –), many endothermic reactions
(∆H is +) are known to be spontaneous as well.
It is also important to remember that the term spontaneous doesn't necessarily mean a fast reaction rate. The
rate of a reaction is controlled by several factors as we discussed in Chapter 13: reactant concentrations ,
catalysts, and temperature. Thermodynamics only tells us whether or not the reaction will occur, but it tells us
nothing about how fast.
Entropy. In addition to the heat absorbed or released in a spontaneous process, another factor called
entropy must be considered. Entropy is a measure of the disorder or randomness of a system. The entropy (S)
is a state function that increases in value as the disorder or randomness of the system increases. Entropy has the
units J/K·mol. Intuitively we consider a system to be "ordered" if it is arranged according to some plan or
pattern. The system is "disordered" when its parts are helter-skelter and their arrangement is random.
Order and disorder in chemical systems are discernible at the molecular level. Crystalline solids are highly
ordered, with molecules or ions occupying fixed lattice sites, and with the unit cell repeated identically over and
over again. Liquids are less ordered than solids because the solid lattice has broken down, and molecules or ions
have kinetic energy and therefore are in random motion. The molecular motion in liquids increases the disorder
compared to that of solids.
Gases are more random than liquids. On vaporization, the molar volume increases almost 1000-fold, and it
is much more difficult to locate the position of any one molecule. The entropy of a substance increases as its
molecules are distributed over an increasingly greater volume. For a given substance, the molecular disorder
increases from the solid to the gaseous state.
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S solid < Sliquid < Sgas
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Entropy Changes. We can estimate whether the change in entropy in a process is positive or negative.
For chemical reactions in which solids or liquids are converted to gases, the entropy change ∆S is positive. It is
negative for the condensation of a gas or the freezing of a liquid. When a crystal of a salt dissolves in water, the
disorder increases due to the increase in freedom of motion of ions in the solution compared to the highly ordered
crystal. Heating also increases the entropy of a system. The higher the temperature the more molecular motion
and its accompanying disorder. Figure 18.2 in the text shows several processes that lead to an increase in
entropy.
Hints for estimating the sign of ∆S:
1. If a reaction produces an increase in the number of moles of gaseous compounds, ∆S is positive.
2. If the total number of moles of gaseous compounds is decreased, ∆S is negative.
3. If there is no net change in the total number of gas molecules, then ∆S is either a small positive or a small
negative number.
The Second Law. The second law of thermodynamics is concerned with predicting the direction of
spontaneous change. It states that the entropy of the universe increases in a spontaneous change. The term
universe used here refers to a system and all of it's surroundings. For an isolated system:
∆S univ = ∆Ssys + ∆Ssurr ≥ 0
where ∆Ssys stands for the entropy change of the system, and ∆Ssurr is the entropy change of the surroundings.
For any spontaneous change:
∆S sys + ∆Ssurr > 0
and for a reaction at equilibrium (no net change):
∆Ssys + ∆Ssurr = 0
Entropy Changes in Chemical Reactions. Not only can the sign ∆S for a chemical reaction be
estimated, but the actual value of ∆S can also be calculated. The absolute value of the entropy S of one mole of
an element or compound can be determined by very careful experimentation. Appendix 3 lists experimental
values of the absolute entropy of a number of substances in their standard states at 1 atm and 25°C. Recall that
the degree superscript " ° " refers to the standard state of the substance. These values are called absolute or
sometimes standard entropies.
One can confirm that the entropy of a gas is greater than that of a liquid by comparing the absolute entropy
of H2O in the liquid and gas phases.
S° of H2O(l) = 69.94 J/K·mol
S° of H2O(g) = 188.72 J/K·mol
For a reaction:
aA + bB → cC + dD
the standard entropy change of the reaction is given by:
∆Sorxn = ∑nS° (products) – ∑mS° (reactants)
where m and n are stoichiometric coefficients. When applied to the above reaction, we get
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∆Sorxn = [cS°(C) + dS°(D)] – [aS°(A) + bS°(B)]
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EXAMPLE 18.1 Changes in Entropy
Predict the sign of ∆Ssys for each of the following reactions using just the qualitative ideas discussed in above.
1
O (g)
2 2
a.
H2O2(l) → H2O(l) +
b.
c.
H+(aq) + OH–(aq) → H2O(l)
CaO(s) + CO2(g) → CaCO3(s)
•Method of Solution
a.
The number of moles of gaseous compounds in the products is greater than in the reactant. Entropy
increases during this reaction. Answer: The sign of ∆S is +.
b. Two reactants combine into one product in this reaction. Order is increased and so entropy decreases.
Answer: The sign of ∆S is –.
c. The number of gas molecules is decreased. Answer: The sign of ∆S is –.
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EXAMPLE 18.2 The Second Law
The solubility of silver chloride is so low that it precipitates spontaneously from many solutions. The entropy
change of the system is negative for this process.
Ag+(aq) + Cl–(aq) → AgCl(s)
∆H° = –65 kJ
Since ∆S decreases in this spontaneous reaction, shouldn't the reaction be nonspontaneous?
•Method of Solution
For a spontaneous reaction, the second law states that ∆Suniv > 0:
∆S sys + ∆Ssurr > 0
In order to predict whether a reaction is spontaneous, both ∆Ssys and ∆Ssurr must be considered, not just ∆Ssys.
Since the entropy change of the system is negative, the only way for the inequality to be true is if ∆Ssurr is
positive and greater in amount than ∆Ssys. In this case, this is a reasonable assumption because the reaction is
exothermic. This means that heat is released to the surroundings, which causes increased thermal motion and
disorder of molecules in the surroundings. Therefore, the sum of ∆Ssys and ∆Ssurr is greater than zero, even
though ∆Ssys is negative.
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EXAMPLE 18.3 Calculation of the Entropy Change for a Reaction
Use absolute entropies to calculate the standard entropy of reaction ∆Sorxn :
H2(g) +
1
O (g)
2 2
→ H2O(l)
•Method of Solution
The standard entropy change is given by:
1
S°(O2)]
2
∆Sorxn = S°(H2O(l)) – [S°(H2) +
Look up the standard entropy values in Appendix 3 of the text.
130.1 J 
1
205.0 J  
69.9 J 

∆Sorxn = 1 mol 
+
mol
– 1mol
 K ⋅ mol 
 K ⋅ mol 
 K ⋅ mol  
2

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∆Sorxn = 69.9 J/K – 233.5 J/K = –163.6 J/K
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•Comment
This reaction is known to be spontaneous. The value of ∆Sorxn applies only to the system. We are not
calculating ∆Suniv here.
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EXERCISES
1.
For each pair of substances, choose the one having the higher standard entropy value at 25°C.
a. CS2(s) or CS2(l) b. SO2(g) or CO2(g) c. BaSO4(s) or BaSO4(aq)
2.
Predict, using the intuitive ideas about entropy, whether ∆Srxn will be positive, negative, or essentially zero
for each of the following:
a. CuSO4(s) → Cu2+(aq) + SO 24 – (aq)
b. SO2(g) +
1
O (g)
2 2
→ SO3(g)
c. Ca(OH)2(s) + CO2(g) → CaCO3(s) + H2O(g)
d. Ag+(aq) + 2CN–(aq) → Ag(CN )2– (aq)
Using tabulated values, calculate ∆S° for the following reaction:
3.
N2(g) + 3H2(g) → 2NH3(g)
GIBBS FREE ENERGY (18.4)
STUDY OBJECTIVES
1.
2.
Calculate free energy changes for chemical reactions, given a table of standard free energies of formation.
Predict, using given ∆S° and ∆H° values, at what temperature a reaction will be spontaneous under standard
conditions.
3. Calculate ∆S for phase transitions.
Gibbs Free Energy. Quite often it is not possible to calculate ∆Ssurr. This makes the second law
difficult to apply when it is in the form given in the previous section. The Gibbs free energy, expressed in terms
of enthalpy and entropy, refers only to the system, yet can be used to predict spontaneity. The Gibbs free energy
change (∆G) for a reaction carried out at constant temperature and pressure is given by:
∆G = ∆H – T∆S
where both ∆H and ∆S refer to the system.
The Gibbs free energy change is equal to the maximum possible work (w) that can be obtained from a
process. Any process that occurs spontaneously can be utilized to perform useful work. The Gibbs free energy of
the system will decrease (∆G < 0) in a spontaneous process, and will increase (∆G > 0) in a nonspontaneous
process. The free energy criteria (at constant temperature and pressure) are summarized as follows:
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If ∆G < 0, the forward reaction is spontaneous.
If ∆G = 0, the reaction is at equilibrium.
If ∆G > 0, the forward reaction is nonspontaneous. The reverse reaction will have a negative ∆G and
will be spontaneous.
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The standard free-energy change of reaction, ∆G orxn , is the free-energy change for a reaction when
it occurs under standard state conditions, when reactants in their standard states are converted to products in their
standard states.
Calculation of ∆G°rxn. The free energy change for a reaction can be calculated in two ways. When both
∆H and ∆S are known, then ∆G = ∆H – T∆S will give the free energy change at the temperature T. ∆G orxn can
also be calculated from standard free energies of formation in a manner analogous to the calculation of ∆H° by
using enthalpies of formation of the reactants and products. The standard free energies of formation ( ∆G of ) of
selected compounds are tabulated in Appendix 3 of the text. Just as for the standard enthalpies of formation, the
free energies of formation of elements in their standard states are equal to zero.
For a general reaction
aA + bB → cC + dD
The standard free energy change is given by
∆G orxn = [c ∆G of (C) + d ∆G of (D)] – [a ∆G of (A) + b ∆G of (B)]
In general:
∆G orxn = ∑ n ∆G of (products) – ∑ m ∆G of (reactants)
where n and m are stoichiometric coefficients. Example 18.4 illustrates this type of calculation.
Temperature and the Free Energy Change. From the equation ∆G = ∆H – T∆S we can see that
temperature too will influence the spontaneity of reaction. If both ∆H and ∆S are positive then at low
temperature, as long as ∆H > T∆S, ∆G is positive and the process will be nonspontaneous. However as
temperature increases, the T∆S term increases and eventually ∆H = T∆S. At this point ∆G is zero. With further
T increase, T∆S > ∆H, making ∆G < 0, and the reaction becomes spontaneous. Table 18.1 summarizes the
four possible situations affecting the ∆G of a reaction.
Table 18.1 Enthalpy and Entropy Factors that the sign of ∆G orxn
∆H
∆S
∆G
+
+
∆G is positive at low temperatures
and negative at high temperatures.
+
–
∆G is positive at all temperatures
–
+
∆G is negative at all temperatures
–
–
∆G is negative at low temperatures
and positive at high temperatures
In the above situation, where both ∆H° and ∆S° are positive, the temperature at which ∆H° = T∆S°, and also
at which ∆G° = 0, can be calculated from the equation T = ∆H°/∆S°. Above this temperature, the reaction favors
the products at equilibrium. See Example 18.5 for an example.
Phase Transitions. For a phase transition, ∆G = 0 when the two phases coexist in equilibrium. For
example, at the boiling point (Tbp) the liquid and vapor phases are in equilibrium, and
∆Hvap – Tbp ∆Svap = 0
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rearranging gives ∆Svap =
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∆H vap
Tbp
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This equation allows the calculation of the entropy of vaporization from knowledge of the heat of vaporization
and the boiling point.
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EXAMPLE 18.4 Calculation of the Free Energy Change for a Reaction
Calculate ∆G orxn at 25°C for the following reaction using Appendix 3 and given:
∆G of (Fe2O3) = –741.0 kJ/mol
2Al(s) + Fe2O3(s) → Al2O3(s) + 2Fe(s)
•Method of Solution
∆G orxn = [ ∆G of (Al2O3)+ 2 ∆G of (Fe)] – [2 ∆G of (Al)+ ∆G of ( Fe2O3)]
= [1 mol (–1576.41 kJ mol–1) + 0] – [0 + 1 mol (–741.0 kJ mol–1)]
∆G orxn = –1576.41 kJ + 741.0 kJ = –835.4 kJ
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EXAMPLE 18.5 Effect of Temperature on ∆G
Hydrated lime Ca(OH)2 can be reformed into quicklime CaO by heating.
Ca(OH)2(s) → CaO(s) + H2O(g)
At what temperatures is this reaction spontaneous under standard conditions (that is, where H2O is formed at 1
atm pressure)? Given the following data on Ca(OH)2 that is not in the Appendix:
∆H of (Ca(OH)2) = –986.2 kJ/mol
S°(Ca(OH)2) = 83.4 J/K.mol
•Method of Solution
This reaction is nonspontaneous at room temperature. The temperature above which the reaction becomes
spontaneous under standard conditions corresponds to ∆G° = 0, and is given by
T=
∆H o
∆So
∆H° and ∆S° must be calculated separately.
•Calculation
∆H° = [ ∆H of (CaO) + ∆H of (H2O)] – [ ∆H of (Ca(OH)2)]
From Appendix 3 and the given data
∆H° = 1 mol(–635.55 kJ /mol) + 1 mol(–241.83 kJ/mol) – 1 mol(–986.2 kJ/mol)
= 108.82 kJ (or 1.088 × 105 J)
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∆S° = S°(CaO) + S°(H2O) – S°(Ca(OH)2)
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∆S° = 1 mol(39.8 J/K·mol) + 1 mol(188.7 J/K·mol) – 1 mol(83.4 J/K·mol)
= + 145.1 J/K
The temperature at which ∆G° is equal to zero is:
T=
∆H° 1.088 × 10 5 J
= 750 K
=
145.1 J/K
∆S°
At temperatures above 750 K the reaction is spontaneous.
•Comment
Recall that this is an approximate value because of the assumption that neither ∆H° nor ∆S° change appreciably
from there values calculated at 25°C.
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EXAMPLE 18.6 Entropy of Fusion
The heat of fusion of water (∆Hfus) at 0°C is 6.02 kJ/mol. What is ∆Sfus for 1 mole of H2O at the melting
point?
•Method of Solution
∆Sfus =
∆H fus
6.02 × 10 3 J/mol
=
273 K
Tmp
= +22.1 J/K.mol
•Comment
The increase in entropy upon melting of the solid corresponds to the higher degree of molecular disorder in the
liquid state as compared to the solid state.
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EXERCISES
4.
Calculate ∆G orxn for the following reaction
3NO2(g) + H2O(l) → 2HNO3(l) + NO(g)
Given the following free energies of formation:
∆G of (kJ/mol)
H2O
–237.2
HNO3
–79.9
NO
86.7
NO2
51.8
Calculate ∆G orxn for the following reaction at 298 K:
O3(g) → O2(g) + O(g)
Given ∆H° = 106.5 kJ and ∆S° = 127.3 J/K.
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6.
The following reaction is nonspontaneous at 25°C.
Cu2O(s) → 2Cu(s) +
1
O (g)
2 2
∆G° = 141 kJ
If ∆S° = 75.8 J/K, above what temperature will the reaction become spontaneous?
7.
The enthalpy of vaporization of mercury is 58.5 kJ/mol and the normal boiling point is 630 K. What is the
entropy of vaporization of mercury?
8.
What is the sign of ∆G for the melting of ice at 5°C?
FREE ENERGY AND EQUILIBRIUM (18.5)
STUDY OBJECTIVES
1. Calculate ∆G, the free energy change under nonstandard state conditions.
2. Calculate an equilibrium constant from a knowledge of ∆G°, and vice versa.
∆G and ∆G°. Recall that ∆G° refers to the standard free energy change. All the values we have calculated
so far relate to processes in which the reactants are present in their standard states and are converted to products
in their standard states. However, in many cases neither the reactants nor the products are present at standard
concentration (1 M) and standard pressure (1 atm). Under nonstandard state conditions, we use the symbol ∆G.
The relationship between ∆G and ∆G° is:
∆G = ∆G° + RT ln Q
where R is the gas constant (8.314 J/K·mole), T is the absolute temperature, and Q is the reaction quotient. For
a certain reaction at a given temperature the value of ∆G° is constant, but the value of Q depends on the
composition of the reacting mixture; therefore ∆G will depend on Q. To calculate ∆G, first find ∆G°, then
calculate Q from the given concentrations of reactants and products, and substitute into the preceding equation.
Under special conditions this equation reduces to an extremely important relationship. At equilibrium,
Q = K, and therefore ∆G = 0. The equation then becomes:
0 = ∆G° + RT ln K
or
∆G° = – RT ln K
This equation relates the equilibrium constant of a reaction to its standard free energy change. Thus, if ∆G° can
be calculated, K can be determined, and vice versa. In the equation Kp is used for gases and Kc for reactions in
solution.
Three possible relationships exist between ∆G° and K, because ∆G° can be negative, positive, or zero.
When ∆G° is negative, ln K is positive and K > 1. The products are favored over reactants at equilibrium.
The extent of reaction is large.
2. When ∆G° is positive, ln K is negative and K < 1. The reactants are favored over products at equilibrium.
The extent of reaction is small.
3. When ∆G° = 0, ln K is zero and K = 1. The reactants and products are equally favored at equilibrium.
1.
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When calculating ∆G and ∆G° using the two equations above, we need to express both free energy changes in
units of kJ/mol. The "per mole" refers to "a mole of reaction" which is the reaction of exactly the molar
amounts written in the balanced equation.
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EXAMPLE 18.7 Calculating the Equilibrium Constant
The standard free energy change for the reaction
1
N (g)
2 2
+
3
H (g)
2 2
NH3(g)
is ∆G orxn = 26.9 kJ/mol at 700 K. Calculate the equilibrium constant at this temperature.
•Method of Solution
The equilibrium constant is related to the standard free energy change by the equation:
∆G orxn = – RT ln Kp
•Calculation
Since the gas constant R has units involving joules and the free energy change has units involving kilojoules,
we must be careful to use consistent units. In terms of joules, we get
26.9 × 103 J/mol = – (8.31 J/mol.K)(700 K) ln Kp
– 4.62 = ln Kp
Taking the antilog of both sides:
Kp = e– 4.62
Use of a calculator with an ex key yields:
K p = 9.8 × 10–3
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EXAMPLE 18.8 ∆G at Nonstandand State Conditions
Using data given in the preceding example, calculate ∆G at 700 K if the reaction mixture consists of 30.0 atm
of H2, 20.0 atm of N2, and 0.500 atm of NH3.
•Method of Solution
Under nonstandard conditions, ∆G is related to the reaction quotient Q by the equation
∆G = ∆G° + RT ln Qp
where
Qp =
PNH 3
1/2 3/2
PN 2 PH 2
=
(0.500)
(20.0)1 / 2 (30.0)3 / 2
Q p = 6.80 × 10–4
From Example 18.7, ∆G° = 26.9 kJ/mol. Substitution yields:
∆G = 26.9 kJ/mol + (8.31 J/K.mol)(700 K) ln (6.80 × 10–4)
= 26.9 kJ/mol – (42,400 J/mol ×
1 kJ
) = 26.9 kJ/mol – 42.4 kJ/mol
10 3 J
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= –15.5 kJ/mol
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•Comment
By making the partial pressures of N2 and H2 high and that of NH3 low, the reaction is spontaneous in the
forward reaction. This condition corresponds to Qp < Kp, and so the reaction proceeds in the forward direction
until Qp = Kp.
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EXERCISES
9. Explain the difference between ∆G and ∆G°.
10. Hydrogen peroxide (H2O2) decomposes according to the equation:
H2O2(l) → H2O(l) + 12 O2(g)
a. Is this reaction spontaneous at 25°C?
b. From the following data calculate the value of Kp for this reaction at 25°C.
∆H° = –98.2 kJ
∆S° = +70.1 J/K
11. The autoionization of water at 25°C has the equilibrium constant
2H2O(l)
H3O+(aq) + OH–(aq) Kc = 1.0 × 10–14
Calculate the value of ∆G° for this reaction.
12. The equilibrium constant for the reaction:
AgBr(s)
Ag+(aq) + Br–(aq)
is the solubility product constant, Ksp = 7.7 × 10–13 at 25°C. Calculate ∆G for the reaction when [Ag+] =
1.0 × 10–2 M and [Br–] = 1.0 × 10–3 M. Is the reaction spontaneous or nonspontaneous at these
concentrations?
13. Calculate ∆G for the following reaction at 25°C when the pressure of CO2 is 0.001 atm.
CaCO3(s) → CaO(s) + CO2(g)
Given ∆H° = 177.8 kJ and ∆S° = 160.5 J/K.
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CONCEPTUAL QUESTION
When the environment is contaminated by a toxic chemical spill or an oil spill, the substance tends to
disperse. How is this consistent with the second law of thermodynamics. In the same regard, which requires
less work: cleaning the environment after a spill, or keeping the substance contained before a spill.
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PRACTICE TEST
1.
Which of the following processes are spontaneous?
a. melting of ice at –10°C and 1 atm pressure
b. evaporation of water at 30°C when the relative humidity is less than 100 percent
c. Water + NaCl(s) → salt solution
2.
From each pair of substances, choose the one having the larger standard entropy at 25°C.
a. H2O(l) or H2O(g) b. SiO2(s) or CO2(g) c. Ag+(g) or Ag+(aq)
d. F2(g) or Cl2(g)
3.
e. 2Cl(g) or Cl2(g)
Predict, using the intuitive ideas about entropy, whether ∆Ssys will be positive, negative, or essentially zero
for each of the following:
a. Ca(OH)2(s) + CO2(g) → CaCO3(s) + H2O(g)
b. CuSO4(s) → Cu2+(aq) + SO 24 – (aq)
c. 2HCl(g) + Br2(l) → 2HBr(g) + Cl2(g)
d. SO2(g) +
1
O (g)
2 2
→ SO3(g)
e. Cu2+(aq) + 4NH3(aq) → Cu( NH 3 )24 + (aq)
4.
At the boiling point, 35°C, the heat of vaporization of MoF6 is 25 kJ/mol. Calculate ∆S for the
vaporization of MoF6.
5.
Calculate ∆G orxn for the following reaction at 298 K:
2H2(g) + CO(g)
CH3OH(g)
given that ∆H° = –90.7 kJ and ∆S° = –221.5 J/K for this process.
6.
For the reaction at 298 K
1
O (g)
2 2
Mg(s) +
→ MgO(s)
∆H° = –602 kJ and ∆G° = –569 kJ. Calculate ∆S°.
7. Using Appendix 3 of the text calculate ∆G° values for the following reactions:
a. 3CaO(s) + 2Al(s) → 3Ca(s) + Al2O3(s)
1
b. ZnO(s) → Zn(s) + O2(g)
2
8.
Consider the following three reactions. Which one will have the greatest equilibrium constant?
a. N2 + O2
b. N2 + 2O2
1
c. N2 + O2
2
2NO
N2O4
N2O
Given: ∆G of (NO) = +86.7 kJ/mol
∆G of (N2O4) = +98.3 kJ/mol
∆G of (N2O) = +103.6 kJ/mol
Given the equilibrium constant at 400°C for the reaction:
H2(g) + I2(g)
2HI(g)
Kp = 64
calculate the value of ∆G orxn at this temperature.
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9.
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Entropy, Free Energy, and Equilibrium / 377
10. Calculate ∆G orxn and Kp at 25°C for the following reaction:
1
O (g)
2 2
NO(g) +
→ NO2(g)
11. The synthesis of O2(g) is often carried out in chemistry lab by the decomposition of KClO3:
KClO3(s) → KCl(s) +
1
O (g)
2 2
for which ∆H° = –44.7 kJ and ∆S° = +247.2 J/K. Is this reaction spontaneous at 25°C under standard
conditions?
12. For the reaction:
N2 + O2 → 2NO
the following are given: ∆H° = 180.7 kJ and ∆S° = 24.7 J/K.
a. Is this reaction spontaneous at 25°C?
b. Above what temperature will this reaction become spontaneous under standard conditions?
13. For the reaction 2SO2(g) + O2(g)
2SO3(g), Kp = 7.4 × 104 at 700 K. If, in a reaction vessel at 700 K,
we have the following partial pressures, what is ∆G?
PSO2 = 1.2 atm
PO2 = 0.5 atm
PSO3 = 50 atm
Predict the direction of reaction.
ANSWERS
Exercises
1.
2.
3.
4.
5.
6.
7.
8.
9.
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11.
12.
13.
a. CS2(l) b. SO2(g)
c. BaSO4(aq)
a. + b. – c. cannot predict the sign, ∆S will be essentially zero d. –
–198.5 J/K
8.7 kJ
∆G orxn = 68.6 kJ
2160 K
∆Svap = 92.6 J/K·mol
negative
∆G° refers to the free energy change when the reactants and products are both present in their standard states.
Their concentrations are all 1 atm or 1 molar. ∆G refers to the free energy change when the reactants and
products are present at concentrations other than those for the standard state.
a. yes b. 7.7 × 1020
∆G orxn = 79.8 kJ
∆G° = +69.1 kJ and ∆G = +40.6 kJ. Therefore the reaction is nonspontaneous.
∆G = 112.9 kJ
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378 / Entropy, Free Energy, and Equilibrium
Conceptual Question
1.
Think about the system. Let it be the oil or toxic substance? When an oil spill spreads out it is becoming
more disordered as its molecules have a much larger area to occupy. Any process that gives rise to a more
random distribution of the particles of the substance in space gives rise to an increase in entropy of the
substance. In the reverse process, that of cleaning up a dispersed substance, work must be done to bring the
substance together into a more concentrated space. This reduces the disorder of the system. This is a
nonspontaneous process requiring a continual work input. Keeping the substance contained in the first place
takes less work.
Practice Test
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
b and c
a. H2O(g) b. CO2(g) c. Ag+(g) d. Cl2(g) e. 2Cl(g)
a. Essentially zero b. Positive c. Positive d. Negative
∆Svap = 81 J/mol·K
∆G orxn = –24,700 J
∆S° = – 110 J/K
a. ∆G° = 236 kJ b. ∆G° = 318.2 kJ
b.
∆G orxn = –23.2 kJ
∆G orxn = –34.85 kJ; Kp = 1.29 × 106
Yes, ∆G° < 0
a. No, ∆G° > 0 b. T = 7,320 K
∆G = –17,800 J. Spontaneous in the forward direction.
e. Negative
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