Week 8 Monday November 12, 2012 page 1
System with 2 phases: phase δ with component j and phase β with component j.
µjβ = µjδ at equilibrium
If µjβ > µjδ then j flows spontaneously from β → δ
If δ has no j it still has a value for µj.
𝜇𝑗𝛿 = (
𝜕𝐺
𝑠ℎ𝑜𝑢𝑙𝑑 𝑏𝑒 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑖𝑓 𝑠𝑢𝑏𝑠𝑡𝑎𝑛𝑐𝑒 𝑗 𝑖𝑠 𝑛𝑜𝑡 𝑖𝑛 𝑝ℎ𝑎𝑠𝑒 𝛿?
)
𝜕𝑛𝑗 𝑇, 𝑃
Begin chapter 5
Standard thermodynamic functions of reactions
a. solids and liquids
standard state
pure substance
P = 1 bar = 750 torr = 0.987 atm
T = temperature of interest
V°m,200
1 bar goes with SI units
before 1982 it was 1 atm
tables typically have 25°C
° means 1 bar,
m means molar,
200 means 200K
b. gases
pure gas at P = 1 bar
assuming perfect (ideal gas)
T = temperature of interest
Standard enthalpy of reaction
ΔH°T
∆𝐻𝑇° = ∑ 𝜈𝑖 𝐻𝑚,𝑇,𝑖
𝜈𝑖 > 0 𝑓𝑜𝑟 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 𝜈𝑖 < 0 𝑓𝑜𝑟 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠
2C6H6 + 15O2 → 12CO2 + 6H2O
ΔH°T = 12H°m,T(CO2(g)) + 6H°m,T(H2O) – 2H°m,T(C6H6) – 15H°m,T(O2)
H2(g) + ½ O2(g) → H2O
ΔH°298 = -286 kJ/mol
2H2(g) + O2(g) → 2H2O
ΔH°298 = -572 kJ/mol
The per mole means per mole of reaction as it’s written.
Standard enthalpy of formation Δ°f,T,i
Def: The enthalpy of formation of an element in its reference form is the most stable form of that
element at temperature T.
ΔH°f,298 (O2(g)) = 0
arbitrarily decided
So ΔH°f,298 (O3(g))≠0 and ΔH°f,298 (O(g))≠0
ΔH°f,298(Cgraphite) = 0
so ΔH°f,298 (Cdiamond)≠0 = 1.897 kJ/mol
Example: ∆H°f,298 (formaldehyde) = -115.8 kJ/mol
H2(g) + ½ O2 + Cgraphite → CH2O
Assuming P° = 1 bar T = 298K for reactants
gasses ideal
°
∆𝐻𝑇° (𝑟𝑥𝑛) = ∑ 𝜈𝑖 ∆𝐻𝑓,𝑇,𝑖
°
∆𝐻𝑓,𝑇,𝑖
𝑐𝑜𝑚𝑒𝑠 𝑓𝑟𝑜𝑚 𝑡𝑎𝑏𝑙𝑒
𝑖
C2H6(g) + 7/2 O2(g) → CO2 + 3H2O(g)
∆H(rxn) = 2∆H°f (CO2(g)) + 3∆H°f(H2O(g)) – (∆H°f(C2H6(g)) + 7/2 ∆H°f(O2(g)))
∆H(rxn) = 2(-393.51 kJ/mol) + 3(-241.82 kJ/mol) – 84.68 kJ/mol + 0 (standard state already)
∆H(rxn) = -1427.8 kJ/mol
mol as equation is written
How do we find ∆H°f?
1. calorimetry
(constant V or constant P)
2. spectroscopically
(bond energies)
3. Gibbs energy and 3rd law
∆H - T∆S = ∆G
4. measure temperature variation of KC
KC is equilibrium constant
In bomb calorimeter
∆U = CV∆T
∆H = ∆U + ∆ngasRT
Relation between ∆H° and ∆U°
H = U + PV
∆H = ∆ + P∆V
constant P
∆H°T = ∆U°T + ∆ngasRT
R = 8.314 J/(mol K)
∆𝑛𝑔𝑎𝑠 = ∑ 𝑛𝑖 (𝑔𝑎𝑠𝑒𝑜𝑢𝑠 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠) − ∑ 𝑛𝑖 (𝑔𝑎𝑠𝑒𝑜𝑢𝑠 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠)
𝑖
𝑖
∆𝑛 ℎ𝑎𝑠 𝑛𝑜 𝑢𝑛𝑖𝑡𝑠 𝑠𝑖𝑛𝑐𝑒 𝑖𝑡 ′ 𝑠
∆𝑛𝑔𝑎𝑠
𝑚𝑜𝑙