Thermodynamics
tutorhour 4
February 22nd 2017
Gibbs energy
Gibbs energy is defined as:
G  H  TS
dG  dH  dTS  dH  TdS  SdT
At constant T it becomes: dG  dH  TdS
At constant T,p it becomes: dG  dQ  Td S
Second Law: dS  dS  dS
tot
Clausius inequality:
sys
env
 sys dQ 
  dS 
0
T 

dQ  TdS sys  0
So when p,T constant for a
spontaneous process holds:
dG  0
Spontaneous process?
6 CO2 + 6 H2O →
C6H12O6 + 6 O2
ΔrG = + 28.3·105 Jmol-1 > 0
C6H12O6 + 6 O2 → 6 CO2 + 6 H2O
ΔrG = – 28.3·105 Jmol-1 < 0
Spontaneous process?
ΔrGbattery = – 17.4·105 Jmol-1 < 0
_
+
ΔrGsystem = + 2.1·105 J mol-1 > 0
Zn (s) + Cu2+ →
Cu (s) + Zn2+
ΔrGtotal
ΔrG==––15.3·10
2.1·105 Jmol
J mol-1 < 0
Zn2+ + Cu (s) →
Cu2+ + Zn (s)
ΔrG = + 2.1·105 Jmol-1 > 0
electrochemistry

G  wmax
mixing
lowering of freezing point
elevation of boiling point
osmotic pressure
ΔG
colligative
properties
Does this
reaction
occur?
Yes, if…
G  0
G  G θ  RT ln Q
at equilibrium
ΔG = 0 and Q = K
so:
G θ   RT ln K
Answers:
Question 1
a) at 298 K: K = 9.4·1011
b) at 400 K: K = 1.3·109
Question 2
a) 0
b) 0.0218
c) 31.8 kJ/mol
Question 3
a) –11.89 kJ/mol
Question 4
397 K