Lecture 4
Chapter 4
Internal energy and enthalpy changes
associate with the chemical reactions
H = qp; H = Hf-Hi; Cp = dH/dT
(slope of curve)


• A) phase transitions:
solid  liquid
liquid  solid
liquid  gas
gas  liquids
solid  gas
gas  solid
melting; fusion (fusH)
freezing
vaporization (vapH )
condensation
sublimation (subH )
deposition
Solid
liquid
gas
Water
fusH = 6.01 kJ vapH = 40.7 kJ
Methane fusH = 0.94 kJ vapH = 8.2 kJ
Standard molar
enthalpy Hm
H
Transition under
standard conditions : 1
Hforward
Hreverse
mol of pure substance at
1 bar (s )-> 1 mol of
pure substance (l) at 1
Hforward = - Hreverse
bar
Standard state of substance – pure
substance at 1 bar (temperature and
physical state has to be specified)
gas
liquid
solid
subH = fusH + vapH
At constant T
Enthalpy of ionization
Ionization:
H(g)  H+(g) + e- (g)
ionH = 1312 kJ energy (heat) that must be supplied to ionize
1 mol of H (g) at 1bar
The standard enthalpy of sublimation
for Mg (s) is 148 kJ mol-1 at 295 K.
How much energy (heat; at constant
T and P) must be supplied to 1 mol of
magnesium metal to produce Mg2+
and electrons
H = subH  + H (first ion)+ H
(second ion)
H = 2337 kJ
Mg 2+(g) + e- (g)
1451 kJ
Mg +(g) + e- (g)
H
738 kJ
148kJ
Electron gain: H+(g) + e- (g)  H(g)
Standard enthalpy of electron gain
egH
=
-ionH
Mg (g)
Mg (s)
Note: The enthalpy for first ionization is smaller than
enthalpy for second ionization
• Bond enthalpies
- Energy (heat) to be supply to dissociate a
covalent bond
Potential energy curve for
diatomic molecule
U
N2 -> 2 N H = 945 kJ mol-1
r
- Value depends on molecules
H-O-H (g) -> H-O (g) + H (g) H = 499 kJ mol-1
H-O (g) -> O (g) + H (g)
H = 428kJ mol-1
Accurate determination is complicated since it
has to be done for each specific molecules
Mean bond enthalpies HB - mean bond
enthalpy based on several similar compounds
Bond dissociation enthalpy – precisely measured
value
Bond energy
R – equilibrium distance
H = 436.3 kJ mol-1
U = 432 kJ mol-1
H = U + pV
Bond energy  Bond enthalpy
(work term (1 mol -> 2 mols) and
molecule has rotational and
vibration energy
Resonance energy
Molecules with conjugated double bonds are more stable,
the difference in stability is called resonance energy
fH (benzene,g) = 82.93 kJ as measured
fH (benzene) = 241 kJ based on bound enthalpies
Molecules with conjugated double bounds are more stable than those with
single/double bound only
• Problem:
Estimate the enthalpy of combustion for methane using the
bond enthalpy values
Chemical Change and Reaction enthalpy
CH2=CH2 + H2  CH3 –CH3 H = -137 kJ heat due
to the cleavage/formation of covalent bond
A) Combustion
CH4(g) + 2O2 (g)  CO2 (g)+ 2H2O (l ) cH = -890 kJ
standard enthalpy of combustion – change in standard enthalpy per mol of combustible
substance ; measured using bomb calorimetry
B) Standard enthalpy of formation
rH =  fH (products) -  fH (reactants)  -
stoch. coeffic.
fH - standard molar enthalpy of formation – enthalpy change when 1 mol of a
compound is formed from its constituent element at 1 bar and 298 K.
C (s, graphite) + O2 (g)  CO2 (g) H 395 kJ mol-1
 rH = 1xfH (CO2) – [1xfH (C, graphite) + 1xfH (O2)]
fH (CO2) = 395 kJ
fH (C, graphite) = 0
fH (O2) = 0
By convention, fH are assigned to be
zero for elements in their most stable
allotropic form
 rH = 395 kJ mol-1 – (0 kJ mol-1 + 0 kJ mol-1) = 395 kJ mol-1
fH (C, diamond) = 1.90 kJ mol-1
fH (O3) = 147. 7 kJ mol-1
For compounds that can be synthesized from their elements, the reaction
enthalpy can be determined directly.
Problem: calculated the value of  rH for reacting 1 g of solid glycylclycine
with oxygen to form solid urea, CO2 and liquid water at 1 atm and 25 oC
using tabulated enthalpy of formation.
Indirect methods for determination of reaction enthalpy
- Most of the compound can not be synthesized from their elements,
thus the enthalpy of formation can not be measured directly
Hess law: when reactant are converted into products, the change in enthalpy is the
same whether the reaction takes place in one step or multiple steps
Find the value of  fH for carbon monoxide formation
C + 1/2O2  CO
1) C (graphite) + O2  CO2
2 CO(g) + 1/2 O2  CO2
 rH = -393.5 kJ mol-1
 rH = -283 kJ mol-1
 rH (C+ ½ O2) +  rH (CO+ ½ O2) =  rH (C+O2)
CO + ½ O2
 rH = -283 kJ mol-1
C + O2
 rH
CO2
= -393.5 kJ mol-1
 rH (C+ ½ O2) + -283 kJ mol-1 = -393.5 kJ mol-1
 rH (C+ ½ O2) = -110.5 kJ mol-1.
• Temperature dependence of enthalpy of
reaction
rH =  fH (products) -  fH (reactants)
( rH/ T)P = [H (products) /T]P - [H (reactants )/T]P
= Cp (products) - Cp (reactants)
= Cp different in the heat capacities of products and
reactants
∆𝐻 𝑇2
𝑇2
Thus:
𝑑∆𝐻 =
∆𝐻 𝑇1
∆𝐶𝑝 𝑑𝑇
𝑇1
∆𝐻 𝑇2 − ∆𝐻 𝑇1 = ∆𝐶𝑝(𝑇2 − 𝑇1)
H(T2) = H(T1) + Cp (T2-T1) Kirchhoff’s law
Problem: The standard enthalpy change for the reaction formation of ozone is
285.4 kJ mol-1 at 295 K. Calculate the value at 380 K. Assume that heat capacities
are temperature dependent.
• The energy change for the reaction (U)
U = H - (PV)
At constant pressure:
U = H - PV
For solids and liquids, V is small and thus U ~ H
For gases (using perfect gas approximation) and at constant temperature
U = H - RTn
where n corresponds to n(products) - n(reactants)
• calorimetry - Constant volume calorimetry
Bomb container + water are isolated
from surrounding => no heat exchange
adiabatic process
Steel container – constant volume
Considering bomb – system and water
bath surrounding
U =  U(system) +  U (surrounding) =
0
and  U(system) = -  U(surrounding)
Cv – determined independently
T is measured
And U = is then calculated
H =U +RTn
 U(system) = qv –p  V
 U = qv = -Cv (T2-T1)
Cv is heat capacity of the bomb
calorimeter
• Differential scanning calorimetry
Sample and the reference are slowly heated electrically
in such matter that the T of sample and reference
reminds the same.
H =  CpdT
Protein denaturation:
For 2 state unfolding:
N  U
Tm corresponds to the temperature when 50 %
protein is in folded state
H = 200 kJ mol-1 to 800 kJ mol-1
Homework 4
• Homework:
2.27; 2.28; 2.29; 2.30; 3.32 (3.35); 3.31 (3.34);
3.29 (3.32) 3.26 (3.29), 3.27 (3.30) 3.25
(3.28); 3.22 (3.26); 3.10 (3.14); 3.11 (3.15);
3.17 (3.21)