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5.111 Principles of Chemical Science
Fall 2008
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5.111 Lecture Summary #17
Readings for today: Section 7.1 Spontaneous Change, Sections 7.2 and 7.8
Entropy, Sections 7.12, 7.13, and 7.15 Free Energy.
Read for Lecture #18: Section 7.16 Free-Energy Changes in Biological Systems.
(Assigned sections in chapter 7 are the same for the 3rd and the 4th ed.
__________________________________________________________________________
Topics: Thermodynamics
I. Enthalpies of reactions: calculating Hr (continued from Lecture #16)
II. Spontaneous change and free energy
III. Entropy
IV. Free energy of formation
________________________________________________________________________________
Clicker question: review of hybridization
Hybridization in complex molecules. Example: the morphine rule
Morphine, a potent and addictive analgesic (painkiller), and related molecules share a
characteristic set of four features referred to as the morphine rule:
H H
HO C C
H
C C CH HC CH
HC CH
C C HC CH
H2 4) _______ N C CH2 C OCH2CH3
HC C H2
Nb O H2
a C C CH2 C C
C C C
H C C H
1) phenyl ring H2 HC C N H 2C O
CH2
CH CH CH3 N
C
2) _______ C 3) CH2CH2 HO C
H H3C H2
Morphine Demerol
These molecules mimic the action of (and have structural similarities to) endorphins.
I. ENTHALPIES OF CHEMICAL REACTIONS (continued from Lecture #16)
USING HESSS LAW TO CALCULATE Hr
Enthalpy is a STATE FUNCTION, which means H is __________________ of path.
Hess's Law: If two or more chemical equations are added to give another chemical
equation, corresponding _________ must be added.
H for glucose oxidation (all values in kJ):
We have covered 3 methods (so far) to calculate Hr:
1) __________ enthalpies (H or HB)
Hr = HB(___________________) - HB(___________________)
2) Standard enthalpies of formation (______)
Hr = Hf(_______________) - Hf (________________)
3) Hesss law
*** END OF EXAM #2 MATERIAL ***
___________________________________________________________________________
II. SPONTANEOUS CHANGE AND FREE ENERGY
A spontaneous change is a process that, given enough time, occurs without the need
for outside intervention.
For example, the following reactions are spontaneous at constant pressure:
4Fe(s) + 3O2(g) 2Fe2O3(s) H = __________ kJ/mol
H3O+(aq) + OH(aq) 2H2O(l) H = __________ kJ/mol
2
hydrolysis of ATP: NH2
NH2
C
N
C C N
N
C N O O
HC
O O O
HC C CH
+ 2H2O(l) +HPO42(aq) + H3O+(aq)
C CH O P O P O CH2 N
O P O P O P O CH2 N N
N O
O
O O HC CH
O O O HC CH
HC CH
HC CH
OH OH
OH OH
__________ (aq) __________ (aq) H = ________ kJ/mol
But so are these
H2O(s) H2O(l) H = +6.95 kJ/mol
NH4NO3(s) NH4+(aq) + NO3-(aq) H = +28 kJ/mol
Is H the key to spontaneity? _______!
Condition for spontaneity under constant P + T involves GIBBS FREE ENERGY, G.
G = H TS
where T = temperature and S = change in entropy, a measure of disorder.
G < 0 _____________________ process
G > 0 _____________________ process
G = 0 equilibrium
Under constant pressure and temperature, a process is spontaneous when G < 0,
not necessarily when H < 0. Why?
Figuring this out was one of the towering achievements of thermodynamics!!
For example, NH4NO3(s) NH4+(aq) + NO3-(aq)
Hr = +28 kJ/mol Sr = +109 JK-1mol-1
G = H TS
at T = 298 K, G = _______________ ________(____________________)
G = __________ kJ/mol - __________ kJ/mol
G = __________ kJ/mol
3
G is negative, even though H is positive. The reaction is spontaneous.
Now consider glucose oxidation at room temperature:
C6H12O6 + 6O2 6CO2 + 6H2O
H = 2,816 kJ/mol, S = +233 J/Kmol
G = __________ 298(__________) = __________kJ/mol
G more negative than H. This reaction is spontaneous at ______________
temperatures.
III. ENTROPY
Entropy, S, is a measure of the ________________ of a system.
S = change in entropy. S is a state function.
S positive _______________ in disorder
S negative ______________ in disorder
Disorder of gas ____ liquid ____ solid
In solids, molecules cannot move around freely- they are locked in ordered locations.
Internal degrees of freedom e.g. multiple molecular configurations, many quantum
states that can be occupied - also contribute to entropy
Without calculations, we can predict the sign of S for many reactions. For example,
2H2O2(l) 2H2O(l) + O2(g) has a ________________ S.
Entropy for reactions, Sr
can be calculated from absolute entropies of products and reactants,
Sr = S(_______________) - S(________________)
where S is the absolute standard entropy.
Unlike E or H, where the zero can be selected, S has an absolute zero the perfect
crystal at T = 0 K (perfectly ordered, no disorder at all).
For example, consider the decomposition of hydrogen peroxide.
4
2H2O2(l) 2H2O(l) + O2(g)
Sr = S(products) - S(reactants)
S = ___S(___________) + S(________) - ___S(________)
S = 2(70. JK1mol1) + (205 JK1mol1) 2(110. JK1mol1)
S = __________ JK1mol1
Why is S positive? The reaction converts liquid liquid and gas.
G = H TS
= -196 kJ/mol 298.15 K (___________ kJ K1mol1)
= -233 kJ/mol (The reaction is _______________)
Now consider ice melting at 298.15 K. H2O(s) H2O(l)
S = S(__________) S(__________) = 69.91 41.32 S = 28.59 J K1 mol1
Why is S > 0? _____________________
G = 6.95 298.15K(2.859 x 102 kJ/K mol)
= -1.57 kJ/mol
Ice melting is spontaneous at room temperature even though H is positive.
IV. FREEE ENERGY OF FORMATION, Gf
analogous to Hf
Gf standard Gibbs free energy of formation
Gr for formation of 1 mol of compound from its elements in their
most stable form in the standard states at P = 1 bar and T = 298.15 K.
Tabulated for many compounds like Hf, but can also be calculated from
_______ = _______ TS
For example,
C(gr) + O2(g) CO2(g) G = 394.36 kJ/mol = Gf
Gf is important because it is a measure of a compounds stability relative to its elements.
5
If Gf < 0, a compound is thermodynamically __________________ relative to its elements.
If Gf > 0, a compound is thermodynamically __________________ relative to its elements.
6C(gr) + 3H2(g) C6H6(l) Gf = 124 kJ/mol
C6H6(l) 6C(gr) + 3H2(g) G = 124 kJ/mol
The reverse reaction is spontaneous, but very, very slow!
Free energy tells whether or not a reaction will happen spontaneously, but it tells us
________________ about the rate of the reaction (for rate information we need
chemical kinetics).
To calculate G for a reaction
Gr = Gf(products) - Gf(reactants)
OR Gr = Hr TSr
