CHM 101 – Chapter Nineteen

CHM 101 - ReevesDepartment of Chemistry and Biochemistry

CHM 101 – Chapter Nineteen

• Spontaneous Processes

• Entropy & the Second Law of Thermodynamics

• The Molecular Interpretation of Entropy

• Entropy Changes in Chemical Reactions

• Gibbs Free Energy

• Free Energy and Temperature


Molecular Interpretation of Entropy

The Third Law of Thermodynamics defines zero entropy: The entropy of a perfectly ordered crystalline solid at 0K is 0. Under all other circumstances, absolute entropies are positive.


Molecular Interpretation of Entropy

Absolute entropies have been measured for many substances. Appendix C provides a comprehensive list.

Arrange the following in order of increasing entropy (S)

c lic k he re

C (graphite) C (diamond) C (g) Kmol

JS 0

CH3CH2CH3(g) CH4(g) CH3CH2OH(l) Kmol

JS 0


The Second Law of ThermodynamicsA reversible change is one for which a very slight (infinitesimal) change in condition reverses the direction of the change. Consider melting ice.

CT osys 0

H2O(s) H2O(l) H = 6 kJ


The Second Law of Thermodynamics

The entropy change (S) for any process is defined as:

The Second Law of Thermodynamics states that in any spontaneous process, the entropy of the Universe always increases. Thus:

In the case of melting one mole of ice at the infinitesimal temperature difference described above :



Most changes are irreversible, And a slight change does not change the direction of the process.

CT osys 0



In the case of a finite difference where Tsurr>Tsys

If the temperature of the surroundings is less than that of the system (say Tsurr = -1oC), then the heat

flows in the opposite direction and S is still positive.


Entropy Changes in Chemical Reactions

• Although absolute entropies (S) are always positive, entropy changes (S) for chemical reactions can be either positive or negative.

• Since the entropies of gases are so much larger than entropies of solids or liquids, the sign of S will depend on whether there are more gaseous moles of reactants or products.

• If there are more moles of gaseous product, S will usually be positive. Conversely, more moles of gaseous reactants indicates a negative S.


CHM 101 – Chapter NineteenEntropy changes in Chemical Reactions

Calculate the entropy change (S) the reaction of gaseous hydrogen peroxide (H2O2) with hydrogen to form liquid water.

Com pound So (J/mol-K)

H2O(l) 69.93232.9

130.6H 2(g)

H 2O 2(g)

Appendix C: Thermodynamic Quantities for Selected Compounds at 298K


Entropy Changes in Chemical Reactions

• An exothermic reaction (Hrxn< 0) releases heat, dispersing energy that had been localized in the chemical bonds of the reactants.

• As a result, the surroundings experience a positive entropy change:

• An increase in the entropy of the system (Srxn>0) disperses the reactant atoms into products that can be arranged in many more configurations.


The Gibbs Free Energy

Recall that according to the Second Law:

J Willard Gibbs summarized this result by defining the Free Energy (G) as G = H - TS, or at constant T,

Thus for any spontaneous process at const T & P,


Because G = H - TS, the sign of the Free Energy change (G) depends on the signs of the

enthalpy (H) and entropy (S) changes.

SHSpontaneous

(G < 0)+ ++ -- -- +




Calculate the standard free energy change (G0) associated with boiling water at 25oC and 1 atm

Compound So

(J/mol-K)

H2O(l) 69.93

188.8H 2O(g)

Ho

(kJ/mol)-285.8

-241.8

f



Estimate the temperature at which liquid water is in equilibrium with its vapor at 1 atm. pressure.

0HH2O(l) H2O(g) kJ44 0S K

J119

CHM 101 – Chapter Nineteen

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