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CHAPTER 18 (MOORE)
THERMODYNAMICS: SPONTANEITY, ENTROPY, AND FREE ENERGY
This chapter deals with the idea of spontaneity in chemical and physical processes. Spontaneous
means that a particular process can take place without outside help. We will introduce two new
thermodynamic variables: entropy and free energy.
In this chapter, we look at entropy and how it combines with enthalpy changes to influence free
energy changes … the “silver bullet” for spontaneity.
Recall that, in Chapter 6 (Section 6.9), you looked at heats of reaction, enthalpy changes (ΔH) and
Hess’s Law
You should carefully review Hill, pp. 233-240 and 246-253 to refresh yourself about heats of
reaction and how to calculate enthalpy changes using Hess’s Law ….
Thermodynamics tells us what processes are possible;
Kinetics tells us whether the process is practical.
Spontaneous and Nonspontaneous Changes
Spontaneous processes (changes) occur in systems left to themselves; no action from outside the
system is required to cause the change. Nonspontaneous processes cannot take place in systems left
to themselves.
If a given process (e.g. an explosion) is spontaneous, then the reverse process is not spontaneous,
and vice versa.
Example 18.1
Indicate whether each of the following processes is spontaneous or nonspontaneous.
(a) The boiling of water at normal atmospheric pressure and 65 °C.
Nonspontaneous below 100 °C at 1 atm!!
(b) The melting of an ice cube.
Spontaneous!! (at room temperature or at any temperature above 0 °C)
Spontaneous and Nonspontaneous Changes
Familiarity with a large number of chemical reactions has convinced chemists that spontaneous
chemical reactions occur only for exothermic processes, when systems lose heat energy and ΔH is
negative.
2 H2(g) + O2(g) 2 H2O(g) ΔH = − 483.6 kJ (spontaneous and exothermic!)
In fact, this is true for most chemical reactions, but not for all chemical reactions.
So how can we explain this apparent discrepancy?
Falling water (higher to lower potential energy) is a spontaneous process.
As shown previously, H2 and O2 combine spontaneously to form water (exothermic) BUT …
… liquid water vaporizes spontaneously at room temperature; an endothermic process.
Conclusion: enthalpy alone is not a sufficient criterion for prediction of spontaneity.
Entropy (S) and Entropy Change (ΔS)
Entropy is a mathematical concept used to describe the distribution of energy within a system.
Entropy is a measure of the relative order/disorder in a system. Two gases mix spontaneously when
the valve is opened, even though there is no significant enthalpy change … why?
Concept of Entropy
The total energy of the system remains unchanged in the mixing of the gases … but the number of
ways the energy can be distributed increases.
Mixing two different but chemically similar liquids … “likes dissolve likes” … Benzene and
toluene have similar intermolecular forces … essentially no enthalpy change when mixed. But the
entropy (translate as relative disorder) of the mixture is higher than the entropies of the two
substances separated, so mixing is spontaneous …
The difference in entropy (S) between two states, S1 and S2, is the entropy change (ΔS).
ΔS = S2 − S1 , where where S2 > S1 implies an increase in disorder.
Increase in Entropy When Water Vaporizes
(less “disorder”) H2O(l) ' H2O(g) (more “disorder”) S2 > S1
We know that vaporization is spontaneous! So, ΔS = + (the sign of ΔS is positive)
Some Important Thermodynamic Principles
Distribution of energy among different “states,” with an increase in entropy, correlates with more
physical disorder at the microscopic level … we use entropy as a kind of measure of relative
order/disorder.
Spontaneous processes: occur when systems achieve lower energy states (ΔH is negative) and when
system energy is distributed among more states (ΔS is positive). ΔH and ΔS may work together or
in opposition. The sign of each one may be either positive or negative. Spontaneity depends upon
which factor “wins.”
The more different “configurations” of atoms, ions and molecules that exist among energy levels in
a particular state of a system, the larger the entropy.
Entropy usually increases when:
1. Solids liquids
2. Solids/liquids gases
3. Solids/liquids dissolve in a solvent to form solutions
4. Chemical reaction increases number of gas molecules
5. Temperature of a substances increases
Example 18.2 Predict whether each of the following leads to an increase or decrease in the entropy
of a system. If in doubt, explain why.
(a) N2(g) + 3 H2(g) 2 NH3(g)
(b) C12H22O11(s) C12H22O11(aq)
(c) CO(NH2)2(aq) Æ CO(NH2)2(s)
decreases (fewer gas molecules on right than on left)
increases (solid to liquid)
decreases (aqueous to solid)
Entropy Change (ΔS) Is a State Function
ΔS = qrxn/T where qrxn is “reversible heat”
Reversible processes may be reversed by very small changes, as in the expansion of this gas.
Reversible processes are never very far from equilibrium.
Entropy Increases with Temperature
Entropy always increases with temperature …entropy increases dramatically during phase changes.
(s l g).
Standard Molar Entropies
By definition, the entropy of a pure, perfect crystal is zero at 0 K (Third Law of Thermodynamics)
What happens as the temperature is increased? Entropy increases!! (ΔS is positive)
standard molar entropy, S°: the entropy of 1 mole of a compound or an element in its standard
state (most stable form at 1 atm, 298 K)
Since entropy increases with temperature, the signs of standard molar entropies are all positive.
Calculation of ΔS Using Hess’s Law
ΔS = Σ S°(products) – Σ S°(reactants)
Example 18.3. Use data from Appendix C to calculate the standard entropy change at 25 °C for the
process in which:
4 HCl(g) + O2(g) 2 Cl2(g) + 2 H2O(g)
186.8 J/ molK 205.0 J/ molK 233.0 J/molK 188.7 J/molK
ΔS0 = [2 mol(233.0 J/mol K +2 mol(188.7 J/mol K)]−[4 mol(186.8 J/mol K)+1 mol(205.0 J/mol K)]
ΔS0 = − 128.8 J/K
The Second Law of Thermodynamics
All spontaneous processes increase the entropy of the universe (the system and its surroundings).
Entropy can be used as a sole criterion for spontaneous change …but the entropy change of both the
system and its surroundings must be considered.
ΔS(universe) = ΔS(system) + ΔS(surroundings but, for a spontaneous process, ΔS(universe) > 0
so, ΔS(surr) = q(surr)/T = − ΔH(sys)/T @ thus: ΔS(univ) = ΔS(sys) – ΔH(sys)/T
(ΔS(universe) = ΔS(system) + ΔS(surroundings)
or, multiplying by T, –TΔS(univ) = ΔH(sys) – TΔS(sys)
Now, let –TΔS(univ) = ΔG, where ΔG is called the change in Gibbs free energy (G)
Gibbs Free Energy and Free Energy Change
*Thus, for any process at constant T and P: ΔG(sys) = ΔH(sys) – TΔS(sys), or
ΔG = ΔH – TΔS
If ΔG < 0 (negative), a process is spontaneous as written.
If ΔG > 0 (positive), a process is nonspontaneous as written.
If ΔG = 0, neither the forward nor the reverse process is favored; there is no net change, and
the process is at equilibrium.
ΔH, ΔS, ΔG and Spontaneous Change
At high T, TΔS is large: | –TΔS | > | ΔH |
At low T, TΔS is small: | ΔH | > | –TΔS |
Free Energy (Gibbs Free Energy)
Free energy is a thermodynamic function that relates enthalpy and entropy to spontaneity, and can
also be related to equilibrium constants. ΔG represents the maximum work available from a system.
Standard Free Energy Change, ΔG°
free energy change when reactants and products are in their standard states (most stable form, 1
atm, 298 K)
standard free energy of formation, ΔGf°
free energy change for the formation of 1 mol of substance in its standard state from its elements in
their standard states
Calculation of ΔG0 Using the Gibbs Equation
Example 18.6 Calculate ΔG° at 298 K for the reaction
kJ/mol
4 HCl(g) + O2(g) 2 Cl2(g) + 2 H2O(g)
186.8
205.0
223.0
188.7
ΔH° = –114.4 kJ
(a) using the Gibbs equation, ΔG0 = ΔH0 – TΔS0
Calculate ΔS0 from S0 values using Hess’s Law: get S0 values from Table C.1 (Appendix C)
ΔS0 = [2(223.0) + 2(188.7)] – [4(186.8) + 1(205.0)] = – 128.8 J/K = – 0.1288 kJ/K
ΔG0 = ΔH0 – TΔS0 = –114.4 kJ – (298 K)(– 0.1288 kJ/K) = – 76.0 kJ
Calculation of ΔG0 Using Hess’s Law: ΔG° = Σ ΔGf°(products) – Σ ΔGf°(reactants)
Example 18.6 (see Appendix J, pp. A.38-A.44, for ΔG0 values)
Calculate ΔG° at 298 K for the reaction, 4 HCl(g) + O2(g) 2 Cl2(g) + 2 H2O(g)
(b) from standard free energies of formation (ΔGf0), in kJ/mol
ΔG0 = [2(– 228.6) + 2(0)] – [4(– 95.30) + 1(0)] = – 76.0 kJ
Free Energy Change and Equilibrium
At equilibrium, ΔG = 0 and reaction is neither product-favored nor reactant favored.
or, ΔG = 0 = ΔH – TΔS at a temperature, T, where the process is at equilibrium
Thus, at the equilibrium temperature, ΔH = TΔS and ΔS = ΔH / T
ΔSvap = ΔHvap /Tbp (Trouton’s rule) or, Tbp ≅ ΔHvap / ΔSvap
Works best with nonpolar substances but not for liquids with lots of hydrogen bonding.
Estimating Normal Boiling Point – Trouton’s Rule
Example 18.7 At its normal boiling point, the enthalpy of vaporization of pentadecane,
CH3(CH2)13CH3, is 49.45 kJ/mol. What should its approximate normal boiling point temperature
be?
Tbp ≈ ΔHvap / ΔSvap = 49.45 kJ/mol / 0.087 kJ/mol K) ≈ 570 K
ΔG° and the Equilibrium Constant, KC
When ΔG = 0, then ΔG° = − RT ln KC @ ln KC = − ΔG0 / (RT)
Calculating Equilibrium Constants
ln KC = − ΔG0 / (RT)
(R = 8.3143 J/mol K)
Example 18.9
Determine the value of KC at 25 °C for the reaction (Hint: first convert ΔG0 to J/mol)
2 NO2(g) ' N2O4(g) ΔG0 = − 4.78 kJ = − 4.78 x 103 J
ln KC = − ΔG0 / (RT) = − ( − 4.78 x 103 J)/{(8.3143 J/mol K)(298 K)}
ln KC = 1.93
KC = exp(1.93) = 6.9
Sign and Magnitude of ΔG°
Large, negative ΔG°; equilibrium lies far to right.
Large, positive ΔG°; equilibrium lies far to left.
Intermediate ΔG°; KC ≈ 1 ??
Summary:
Thermodynamics determines the equilibrium state of any system and can predict the relative
proportions of products and reactants at equilibrium.
Since kinetics determines the pathway by which equilibrium is reached, high activation
energies can effectively block reactions that are “thermodynamically favored.”