1. + - = + - DSuniv Thermodynamics -DHsys T DSsys DSsurr
Chapter 16 – A.P. Chemistry
Thermodynamics
Because of:
Occurs only with outside intervention
The 2nd Law
Occurs w/o outside intervention
The entropy of the univ. is increasing
spontaneous
nonspontaneous +
DSuniv -
-DHsys
Which is equal to: 1. 2. 3.
phases T
# of gaseous particles
A measure of disorder
The change in the entropy of the universe = +
DSsys
DSsurr
rankings
complexity of particles
measurable because of or
3rd Law
The little superscript zeros mean:
standard
DSuniv = DSsys – DH/T
Changed to
DG° = DH° – TDS°
Change in free energy
Which can be changed to:
Which says:
-TDSuniv = DH – TDS
= DG
Standard ________ in ______________
Standard ________ in ___________
Standard ________ in __________
The entropy of a perfect crystal is zero
change free energy
change enthalpy
change entropy -
form that is usable
Which must be ___ for spontaneous processes
and for which info is easily found.
But what if we are not at standard conditions?
←Q is the reaction quotient…the mass action expression w/ init. conditions
DG = DG° + RTlnQ
DG° = -RTlnK
DG = 0
0 = DG° + RTlnK or
And if we are at equilibrium?

2. E = E° -RT lnQ E = E° -0.0592 logQ Ch. 17 E °= 0 Electrochem.
anode ion(s) ion(s) cathode
spontaneous
nonspontaneous
( Forcing a current through a cell or changing electrical energy to chemical energy)
line notation
Galvanic Cells
Electrolytic Cells
shorthand
Device to: ___________
____________________
change chem. energy to electrical energy
2H+(aq) + 2e-→ H2(g)
standard
Parts
The test will have a blend of chapter 16 and 17 concepts focusing on using thermodynamic concepts in a electrochemical system
E °= 0
(half-reactions)
Plating reactions
cathode
anode
(reduction)
(oxidation)
DG° = -nFE°
Faraday C
Info for these under ___________ conditions for____________ reactions
standard
reduction
E = E° -RT lnQ
______ half reactions to get:
____________________
add nF
Key:
standard cell pot. (EMF), E°
nonstandard conditions
Nernst Eq.
E = E° logQ n
(At 298 K)

3. Cathode ► DG°= - RTln K K = 2.41 x 1061 ◄electron flow
(because, for the electrode to gain mass, the Sn2+ must be reduced)
+0.60 V = V + Eº anode
Eºanode = 0.74 V
Eºreduction = V
Chromium, Cr
3Sn2+ (aq) + 2Cr (s) → 3Sn (s) + 2Cr3+ (aq)
Find the cell potential under these conditions
Ecell = 0.60 V – [(0.0592/6) log [(0.10)2/(0.50)3] = 0.61 V
Using the standard cell potential calculate DGº for this reaction .
Describe two other ways to find the standard change in free energy for this process.
Determine the equilibrium constant for this reaction.
DGº = - (6 mol) (96,500 C/mol) (0.60 J/V) = -350 kJ
DG° = DH° – TDS°
DG°= SnDG°f(prod.) - SnDG°f(react.)
DG°= - RTln K K = 2.41 x 1061