Physical chemistry I
• Thermodynamics
― Fundamental laws of thermodynamics
― Free energy and chemical equilibrium
― Solution and mixture
https://www.tesla.com/
• Chemical kinetics
― The rate of chemical reactions
― Catalysis
― Multiple reactions
• Relevant applications
― Electrochemistry and batteries
― Surface chemistry and colloids
― Semiconductors
https://www.forbes.com/
https://www.iea.org/
1
The postulates of physical laws
• Newton's first law
• The first law of thermodynamics
― The principle of inertia
• Newton’s second law
• The second law of thermodynamics
― 𝐹 = 𝑚𝑎
• Newton’s third law
• The third law of thermodynamics
― Conserva@on of momentum
2
Lecture 1
• Mechanical work
• Internal energy
• The first law of thermodynamics
• Reaction and bond enthalpy
• Heat capacity
• Reversible isothermal compression
• Reversible adiabatic compression
Reference book: Physical Chemistry, Keith J. Laidler, John H. Meiser, and Bryan C. Sanctuary, 5th edition
3
The first law of thermodynamics
Energy conservation
∆𝑈 = 𝑞 + 𝑤
∆𝑈 = 𝑈! − 𝑈"
𝑈 ∶ internal energy
(total kinetic and potential energy of molecules in the system)
https://afdc.energy.gov/
𝑞 : transferred heat
(system adsorbs heat 𝑞 > 0, system releases heat 𝑞 < 0)
𝑃!"
#
𝑤: (1) mechanical work 𝑤 = − ∫# " 𝑃$% 𝑑V
!
!"
!"
!!
!!
𝑤 = − $ 𝐹𝑑𝑧 = − $
""
𝐹
𝐴𝑑𝑧 = − $ 𝑃#$ 𝑑V
𝐴
"!
(work done on the system 𝑤 > 0,
work done by the system 𝑤 < 0)
(2) electrical work
𝐼
∆𝑞
4
The internal energy of gas
Translational energy
𝐸! =
=
1
1
1
𝑚𝑣"# + 𝑚𝑣"# + 𝑚𝑣"#
2
2
2
1
1
1
3
𝑅𝑇 + 𝑅𝑇 + 𝑅𝑇 = 𝑅𝑇
2
2
2
2
Molar internal energy
https://chem.libretexts.org/
Linear molecule
Nonlinear molecule
(per atom)
Rotational energy
3
𝑈$ = 𝑈$ 0 + 𝑅𝑇
2
Molar internal energy at T=0 K,
stemming from the internal structure of atom
Linear molecule (e.g. N2 and CO2), translation and rotation
5
𝑈$ = 𝑈$ 0 + 𝑅𝑇
2
2 rotation modes
Nonlinear molecule (e.g. CH4 and H2O), translation and rotation
𝑈$ = 𝑈$ 0 + 3𝑅𝑇
Potential energy
Atkins' physical chemistry. Oxford university press, 2014.
3 rotation modes
Interactions, no simple expression 5
Mechanical work
Case 2: Expansion against constant 𝑃$%
Case 3: Reversible expansion
(e.g. against atmosphere)
Atkins' physical chemistry. Oxford university press, 2014.
Case 1: Free expansion
(e.g. expansion against vacuum)
("
𝑃%" = 0 𝑤 = − 7 𝑃%" 𝑑V = 0
𝑑𝑤 = −𝑃%" 𝑑𝑉 ≠ −𝑃𝑑𝑉
𝑑𝑤 = −𝑃%" 𝑑𝑉 = −𝑃𝑑𝑉
("
($
(!
(#
(!
𝑤 = −𝑃%" (𝑉& − 𝑉' )
𝑤 = − 7 𝑃%" 𝑑V = − 7 𝑃𝑑V
Zero work
Irreversible work
Reversible work (maximum6 work)
Mechanical work
Isotheral reversible expansion
(constant T)
#$
𝑤 = − 0 𝑃𝑑V
##
Equation of state: a thermodynamic equation relating state variables
Ideal gas
𝑃𝑉 = 𝑛R𝑇
Real gas
𝑍=
𝑃𝑉
𝑛R𝑇
𝑍 : compressibility
For idea gas, the work of isothermal reversible expansion is
#$
#$ 𝑛𝑅𝑇
##
##
𝑤 = − 0 𝑃𝑑V = − 0
#$ 1
𝑉'
𝑑V = −𝑛𝑅𝑇 0
𝑑V = −𝑛𝑅𝑇 ln
𝑉
𝑉
𝑉(
##
Reversible work7
Enthalpy
Processes at constant V
Constant V → 𝑑𝑉 = 0
𝑑𝑈 = 𝑑𝑞 − 𝑃𝑑𝑉
è 𝑑𝑈 = 𝑑𝑞
Processes at constant P
𝑑𝑈 = 𝑑𝑞 − 𝑃𝑑𝑉 or 𝑑𝑞 = 𝑑𝑈 + 𝑃𝑑𝑉
)$
*$
)#
##
𝑞 = 0 𝑑𝑈 + 0 𝑃𝑑𝑉 = 𝑈' − 𝑈( + 𝑃𝑉' − 𝑃𝑉( = 𝑈' + 𝑃𝑉' − 𝑈( + 𝑃𝑉(
Definition of enthalpy
𝐻 ≡ 𝑈 + 𝑃𝑉
𝑞+ = 𝐻' − 𝐻( = ∆𝐻
Endothermic: 𝑞+ and ∆𝐻 > 0
𝑞+ : 𝑞 at constant pressure
𝐻, 𝑈, 𝑃 𝑎𝑛𝑑 𝑉 are all state functions
Exothermic: 𝑞+ and ∆𝐻 < 0
8
Enthalpy of chemical reaction
Standard state of substance:
Usually 1 bar and 25oC
How to calculate the formation enthalpy of compounds?
C graphite + 2H# (g) → CH) (g)
CH) (g) + 2O# (g) → CH# (g) + H# O(g)
∆𝐻 * = −802.37 𝑘𝐽/𝑚𝑜𝑙
1
C graphite + O# (g) → CO# (g)
∆𝐻 * = −393.50 𝑘𝐽/𝑚𝑜𝑙
2
2H# (g) + O# (g) → 2H# O(g)
∆𝐻 * = 2(−241.83) 𝑘𝐽/𝑚𝑜𝑙
3
C graphite + 2H# (g) → CH) (g)
∆𝐻 * = −393.50 + 2 −241.83
− −802.37 = −74.80 𝑘𝐽/𝑚𝑜𝑙
Enthalpy of formation of organic compounds is commonly obtained from the enthalpy of combustion
9
Bond enthalpy
C graphite + 2H# (g) → CH) (g)
∆𝐻 * = −74.81 𝑘𝐽/𝑚𝑜𝑙
C graphite → C gaseous atoms
∆𝐻 * = 716.7 𝑘𝐽/𝑚𝑜𝑙
1
H (g) → H (gaseous atoms)
2 #
∆𝐻 * = 218.0 𝑘𝐽/𝑚𝑜𝑙
CH) g → C + 4H (gaseous atoms)
∆𝐻 * = 716.7 + 4×218.0 − −74.81 = 1663.5 𝑘𝐽/𝑚𝑜𝑙
Enthalpy of atomization of methane is 1663.5 𝑘𝐽/𝑚𝑜𝑙
C − H bond enthalpy (bond strength) is 1663.5 / 4 = 415.9 𝑘𝐽/𝑚𝑜𝑙
http://aschemistry.weebly.com/
10
Heat capacity
Heat capacity: the amount of heat required to change the temperature of any substance by 1 K.
For the case of constant volume,
𝜕𝑞
𝐶# ≡
𝜕𝑇 #
𝑑𝑈 = 𝑑𝑞 − 𝑃𝑑𝑉
𝐶# : heat capacity at constant volume
𝜕𝑞
𝜕𝑈
𝑪𝑽 =
=
𝜕𝑇 #
𝜕𝑇 #
𝑪𝑽,𝒎: molar heat capacity at constant V
For the case of constant pressure,
𝜕𝑞
𝐶- ≡
𝜕𝑇 -
𝐶- : heat capacity at constant pressure
𝑑𝐻 = 𝑑𝑈 + 𝑃𝑑𝑉 + 𝑉𝑑𝑝 = 𝑑𝑞 − 𝑃𝑑𝑉 + 𝑃𝑑𝑉 + 𝑉𝑑𝑝 = 𝑑𝑞 + 𝑉𝑑𝑝
𝑪𝑷 =
𝜕𝑞
𝜕𝐻
=
𝜕𝑇 𝜕𝑇 -
𝑪𝑷,𝒎: molar heat capacity at constant P
11
Heat capacity
Water has a very high heat capacity
(highest heat capacity of all liquids)
Why?
https://xaktly.com/HeatCapacity.html
12
Case 1: Reversible compression at constant pressure
Constant Pressure
From state A to state B
Volume changes from Vm,1 to Vm,2
(Temperature changes from T1 to T2)
(%,$
1$
The adsorbed heat
The reversible work
𝑤+%, = − 7
For ideal gas, 𝑃𝑉$ = 𝑅𝑇
𝑤+%, = 𝑃-
𝑞0,$ = 7 𝐶0,$ 𝑑T
1#
(%,#
𝑃- 𝑑V = 𝑃- (𝑉$,- − 𝑉$,# )
𝑅𝑇- 𝑅𝑇#
−
= 𝑅(𝑇- − 𝑇# )
𝑃𝑃-
For idea gas, 𝐶0,$ is independent of T
𝑞0,$ = 𝐶0,$ (𝑇# − 𝑇- )
Molar enthalpy change
∆𝐻$ = 𝐶0,$ (𝑇# − 𝑇- )
Molar internal energy change
∆𝑈$ = 𝑞 + 𝑤 = 𝐶0,$ 𝑇# − 𝑇- + 𝑃- 𝑉$,- − 𝑉$,# = 𝐶0,$ 𝑇# − 𝑇- + 𝑅 𝑇- − 𝑇#
= 𝐶0,$ − 𝑅 𝑇# − 𝑇- = 𝐶,,$ 𝑇# − 𝑇-
∆𝑈$ = 𝐶,,$ 𝑇# − 𝑇-
13
Case 2: Reversible compression at constant volume
Constant volume
From state A to state C
Pressure changes from P1 to P2
(Temperature changes from T1 to T2)
The reversible work
The adsorbed heat
(%,#
𝑤+%, = − 7
𝑃𝑑V = 0
(%,#
1$
𝑞,,$ = 7 𝐶,,$ 𝑑T = 𝐶,,$ (𝑇# − 𝑇- )
1#
(For idea gas, 𝐶,,$ is independent of T)
Molar internal energy change
Molar enthalpy change
∆𝑈$ = 𝐶,,$ (𝑇# − 𝑇- )
∆𝐻$ = ∆𝑈$ + ∆ 𝑃𝑉$ = ∆𝑈$ + ∆ 𝑅𝑇 = 𝐶,,$ 𝑇# − 𝑇- + 𝑅 𝑇# − 𝑇- = (𝐶,,$ + R) 𝑇# − 𝑇∆𝐻$ = 𝐶0,$ 𝑇# − 𝑇-
𝐶0,$
(The derivation of 𝐶0,$ = 𝐶,,$ + R will be introduced later)
14
Case 3: Reversible isothermal compression
Isothermal compression
Constant V
Constant P
+
P
T1
P
T1
=
P
T1
T2
T2
T2
V
V
∆𝑈$ = 𝐶,,$ 𝑇# − 𝑇-
∆𝑈$ = 𝐶,,$ (𝑇- − 𝑇# )
∆𝑈$ = 0
∆𝐻$ = 𝐶0,$ (𝑇# − 𝑇- )
∆𝐻$ = 𝐶0,$ 𝑇- − 𝑇#
∆𝐻$ = 0
(%,$
The reversible work
The adsorbed heat
𝑉$,𝑤+%, = − 7 𝑃𝑑V = 𝑅𝑇 ln
𝑉$,#
(%,#
𝑉$,𝑞+%, = ∆𝑈$ − 𝑤+%, = 0 − 𝑅𝑇 ln
𝑉$,#
V
𝑤+%, = 𝑅𝑇 ln
𝑈$ and 𝐻$ are constant under
isothermal conditions.
𝑉$,𝑉$,#
𝑉$,#
𝑞+%, = 𝑅𝑇 ln
𝑉$,-
𝑤+%, > 0
𝑤+%, + 𝑞+%, = 0
𝑞+%, < 0
15
Case 4: Reversible adiabatic compression
Adiabatic condition: no heat exchange
From A to B, compression can introduce energy into the system è T2 > T1
The adsorbed heat
𝑑𝑞 = 0
𝑑𝑈 = 𝑑𝑞 − 𝑝𝑑𝑉 = 0 − 𝑝𝑑𝑉
For ideal gas, 𝑑𝑈 is only dependent on 𝑛 and 𝑇
2
Note: 𝑈$ = 𝑈$ 0 + # 𝑅𝑇
𝑑𝑈 = 𝑛𝐶,,$ 𝑑𝑇
𝑛𝐶,,$ 𝑑𝑇 + 𝑃𝑑𝑉 = 0
1$
($
𝑇# 𝐶0,$ − 𝐶,,$ 𝑉#
𝑇#
𝑉#
𝑑𝑇
𝑑𝑉
𝑑𝑇
𝑑𝑉
ln = 0
𝐶,,$
+𝑅
= 0 è 𝐶,,$ 7
+𝑅7
= 0 è 𝐶,,$ ln + 𝑅 ln = 0 è ln +
𝑇
𝐶
𝑉𝑇
𝑉
𝑇
𝑉
,,$
1# 𝑇
(# 𝑉
3',%
Define: 𝛾 = 3
(,%
ln
𝑇#
𝑉#
𝑇#
𝑉+ (𝛾 − 1) ln = 0 è
= ( )45𝑇𝑉𝑇𝑉#
or
𝑃- (𝑉- )4 = 𝑃# (𝑉# )4
𝑃- 𝑉- = 𝑃# 𝑉#
Adiabatic
Isothermal
16
Case 4: Reversible adiabatic compression
Adiabatic condition: no heat exchange
From A to B, compression can introduce energy into the system è T2 > T1
The adsorbed heat
𝑃- (𝑉- )4 = 𝑃# (𝑉# )4
3',%
where 𝛾 = 3
(,%
𝑑𝑞 = 0
𝛾=5/3 for perfect monatomic gas
𝛾>1 for all gases
∆𝑈 = 𝐶, (𝑇# − 𝑇- )
∆𝐻 = 𝐶0 (𝑇# − 𝑇- )
∆𝑈 = 0 = ∆𝐻 along the isotherm curve, so ∆𝑈 and ∆𝐻 only depend on the const V process
𝑊 = ∆𝑈 − 𝑞 = 𝐶, 𝑇# − 𝑇- = 𝑛𝐶,,$ 𝑇# − 𝑇17
Reversible compression
18
Lecture 1
• Mechanical work
• Internal energy
• The first law of thermodynamics
• Reaction and bond enthalpy
• Heat capacity
• Reversible isothermal compression
• Reversible adiabatic compression
19
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