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Thermochemistry. Second
law of thermodynamics
PhD. Halina Falfushynska
Thermochemical Equations
• A thermochemical equation is a
balanced equation that includes the
heat of reaction.
2H2O(l)  2H2(g) + O2(g)
ΔHrxn = 572 kJ
Rules for Manipulating
Thermochemical Equations
1) If a reaction is reversed, the sign of
ΔHrxn changes.
2) Changing the coefficients by a
constant factor also changes ΔHrxn by
the same factor.
3) The ΔHrxn can be used in a
stoichiometric ratio.
Constant P Calorimetry
• In calorimetry, we
run a controlled
reaction and monitor
the temp. change.
• The important things
to remember are:
qsystem= -qsurroundings
qrxn= -qsoln
qrxn = ΔHrxn
Application and use of calorimetry in
pharmaceutical development
• characterize the stability of pharmaceutical
• Analysed small amounts of sample for
screening relative reactivity
• Should yield approximate relative reactivity
rates under the various stressing conditions,
which can be useful for planning how high of
temperatures, and how much time would be
required before enough degradation takes
place to obtain reliable HPLC results
Hess’s Law
• Some reactions are impossible or very difficult to
carry out, yet we can still calculate what ΔH would
• Hess’s Law: the enthalpy change of an
overall process is the sum of the enthalpy
changes of its individual steps
• This means that you can break up a reaction
into a series of smaller reactions that add up to
the big reaction, and the ΔHrxn of the big
reaction will be the sum of all the small reaction
Subsequences of Hess’s law
Н298c = -Нf298
2. Нр-ції=ΣυНf298(prod.) -ΣυНf298(subst.)
3. Нр-ції=ΣυНс298(prod) -ΣυНс298(subst.)
Sample Problem
• Calculate ΔHrxn for the reaction
• 2NO2(g) + ½ O2(g)  N2O5(g) given the
information below.
N2O5(g)  2NO(g) + 3/2 O2(g)
NO(g) + 1/2 O2(g)  NO2(g)
ΔH = 223.7 kJ
ΔH = -57.1 kJ
Sample Problem
• Calculate ΔHrxn for the reaction Ca(s) +
½ O2(g) + CO2(g)  CaCO3(s) using the
information below.
Ca(s) + ½ O2(g)  CaO(s)
CaCO3(s)  CaO(s) + CO2(g)
ΔH = -635.1 kJ
ΔH = 178.3 kJ
Standard Heats of Reaction
• Standard conditions are defined at
• 1 atm (101.3 kPa) and 25 °C.
• Products and reactants at these
conditions are said to be in their standard
states. Note that solutions must be 1 M.
• The “not” symbol is used to indicate
standard conditions, e.g. ΔH°rxn.
Heat of Combustion
• The heat from the reaction that
completely burns 1 mole of a
• C + O2(g)  CO2(g) + 393.5 kJ, H = -393.5
Container of ethanol
vapour mixed with air,
undergoing rapid
Heat in Changes of State
1. Molar Heat of Fusion (Hfus.) = the heat
absorbed by one mole of a substance in
melting from a solid to a liquid
q = mol x Hfus.
(no temperature change)
2. Molar Heat of Solidification (Hsolid.) =
the heat lost when one mole of liquid
solidifies (or freezes)
to a solid
q = mol x Hsolid
(no temperature change)
Thermite reaction
Adiabatic conditions
Fe2O3 + 2 Al → 2 Fe + Al2O3
Thermite reaction
• Energy produced by the reaction itself
should be calculate as:
• subtracting the enthalpy of the reactants from
the enthalpy of the products;
• subtracting the energy consumed to heating
the products (from their specific heat, when
the materials only change their temperature),
their enthalpy of fusion and
eventually enthalpy of vaporization, when the
materials melt or boil).
Formation Reactions
• A formation reaction is one that creates
exactly 1 mole of a substance from its
• Of course, standard heats of formation
occur at 1 atm and 25 °C.
Na(s) + ½ Cl2(g)  NaCl(s)
ΔH°f = -411.1 kJ
Heats of Vaporization and Condensation
• When liquids absorb heat at their
boiling points, they become vapors.
Molar Heat of Vaporization (Hvap.) =
the amount of heat necessary to
vaporize one mole of a given liquid.
q = mol x Hvap. (no temperature change)
Heats of Vaporization and Condensation
• Condensation is the opposite of vaporization.
Molar Heat of Condensation (Hcond.) =
amount of heat released when one mole
of vapor condenses to a liquid
q = mol x Hcond.
• Hvap. = - Hcond.
(no temperature change)
Heats of Vaporization and Condensation
• The large values for water Hvap. and
Hcond. is the reason hot vapors such as
steam are very dangerous!
–You can receive a scalding burn from
steam when the heat of condensation
is released!
H20(g)  H20(l) Hcond. = - 40.7kJ/mol
Heat of Solution
• Heat changes can also occur when a solute
dissolves in a solvent.
Molar Heat of Solution (Hsoln.) = heat
change caused by dissolution of one mole of
substance; q = mol x Hsoln.
• Sodium hydroxide provides a good example of
an exothermic molar heat of solution (next
Heat of Solution
Na1+(aq) + OH1-(aq)
Hsoln. = - 445.1 kJ/mol
• The heat is released as the ions
separate (by dissolving) and
interact with water, releasing 445.1
kJ of heat as Hsoln.
–thus becoming so hot it steams!
Kirchhoff equation
• The Kirchhoff Law can only be applied to
small temperature changes, about less
than 100 Kelvins because over a larger
temperature change, the heat capacity is
not constant.
CV 
Cp 
Mayer equation ( for mol of ideal gas)
C p  CV 
Using ΔH°f’s
• Standard heats of formation are very
powerful; they allow calculation of
ΔH°rxn for anything using a Hess’s Law
type of calculation.
ΔH°rxn = (sum ΔH°f products) – (sum ΔH°f reactants)
The second law of
• states that the entropy of an isolated
system never decreases, because
isolated systems spontaneously evolve
towards thermodynamic equilibrium—the
state of maximum entropy.
Entropy Changes
• In general, entropy
increases when
– Gases are formed from
liquids and solids.
– Liquids or solutions are
formed from solids.
– The number of gas
molecules increases.
– The number of moles
Entropy Changes
Entropy changes for a reaction can be
calculated the same way we used for H:
S° for each component is found in a table.
Note for pure elements:
Calculation of entropy for the
phase passing from the ice
to the vapour state at heating.
T fusion
C p ' dT
H fusion
T fusion
Т boiling
Т fusion
С р ' ' dT
H boiling
Т boiling
 ...
Where Ср’ і Ср’’ – molar heat (capacity) of water
at the isobaric process