2C Thermochemistry
thermochemistry:
the study of the energy transferred as heat during the course of
chemical reactions
Use calorimetry to measure the energy supplied or discarded as heat by a reaction,
identify
ΔU if V = constant; ΔU = qV
ΔH if p = constant; ΔH = qp
Endothermic process
- absorbs energy; cool the surroundings
- ∆H > 0 at constant p
Exothermic process
- releases energy; heat the surroundings
- ∆H < 0 at constant p
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Standard enthalpy changes
Changes in enthalpy normally reported for processes taking place
under a set of standard conditions.
Standard enthalpy change, ∆Hө:
the change in enthalpy for a process in which the initial and
final substances are in their standard states.
or
the difference between the enthalpy of products in their standard
states
and the reactants in their standard states, all at the same
specified temperature. (Convention T = 298.15 K = 25.00 °C)
standard state:
of a substance at a specified temperature is its pure form, x = 1,2 1
bar.
Example of a standard enthalpy change:
The standard enthalpy of vaporization, ∆vapHө, is the enthalpy change per mol
when a pure liquid at 1 bar vaporizes to a gas at 1 bar, as in
Here, the symbol ө refers only to
the pressure and the mol fraction
x = 1 as standard
A note on good practice
The attachment of the name of the transition to the symbol ∆:
∆vapH: modern convention.
∆Hvap: older convention, but still widely used.
The new convention is more logical because the subscript identifies the type of
change, not the physical observable related to the change.
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Enthalpies of physical change
standard enthalpy of transition, ∆trsHө:
the standard enthalpy change that accompanies a change of
physical state
Because enthalpy is a state function:
- a change in enthalpy is independent of the path between the
two states:
- same value of ∆Hө will be obtained however the change is
brought about between the same initial and final states.
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Example:
Conversion of a solid to a vapour either as occurring by sublimation,
or as occurring in two steps, first fusion (melting) and then vaporization of the resulting
liquid:
Because the overall result of the indirect path is the same as that of the direct path, the
overall enthalpy change is the same in each case, for processes occurring at the same
temperature:
The standard “concentration” for ice and for liquid water is x = 1.
The standard “concentration” for water vapour is p = 1 bar.
The difference between the standard states is contained
in the tabulated thermodynamic constants.
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Because H is a state function; the standard enthalpy changes of
a forward process and its reverse differ in sign:
Example:
The enthalpy of vaporization of water is +44 kJ mol−1 at
298 K, its enthalpy of condensation at that temperature is −44 kJ
mol−1.
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To summarise:
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Enthalpies of chemical change
How do we report the change in enthalpy that
accompanies a chemical reaction?
We write a thermochemical equation, a combination of a chemical equation
and the corresponding change in standard enthalpy:
∆Hө applies for the reaction as written (not per mol)
∆Hө: the change in enthalpy when reactants in their standard states change
quantitatively (to 100%) to products in their standard states:
Pure, separate reactants in their standard states →
pure, separate products in their standard states
Standard state: gases at 1 bar, H2O(l) at x = 1 (pure water)
Note:
Except in the case of ionic reactions in solution, the enthalpy changes
accompanying mixing and separation are insignificant in comparison with the
contribution from the reaction itself (mixH 0)
(Later: This does not apply to mixG and mixS)
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For the reaction
the standard reaction enthalpy is
where H өm(J) is the standard molar enthalpy of species J at the temperature of
interest, x = 1 (liquid, solid) and a pressure of 1 bar (gas).
In general,
where in each case the molar enthalpies of the species are multiplied by their
stoichiometric coefficients, ν.
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Hess’s law
Hess’s law:
The standard enthalpy of an overall reaction is the sum of the
standard enthalpies of the individual reactions into which a
reaction may be divided (some terms of the sum may be
negative!).
- The individual steps need not be realizable in practice: they may be
hypothetical reactions, the only requirement being that their chemical
equations should balance.
- The thermodynamic basis of the law is the path-independence of the value
of ∆rHө
- Information about a reaction of interest, which may be difficult to
determine directly, can be assembled from information on other reactions.
(Self-study: Do Example 2C.1)
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Standard enthalpies of formation
Standard enthalpy of formation, ∆fHө
the standard reaction enthalpy for the formation of a compound from its
elements in their reference states.
Reference state:
the most stable state of an element at the specified temperature
(usually 298 K) and 1 bar.
Standard enthalpies of formation of elements in their reference states
are zero at all temperatures
∆rHө = ∆fHө = +49.0 kJ mol−1
for the formation of 1 mol of benzene
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The reaction enthalpy in terms of enthalpies
of formation (Hess’s law)
Regard rxn as proceeding by decomposition of
reactants into their elements; then forming
those elements into the products.
∆rHө for the overall reaction is the sum of these
‘unforming’ and forming enthalpies.
‘unforming’ is reverse of forming, enthalpy of an
unforming step is the negative of the enthalpy
of formation
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Illustration 2C.3 Using standard enthalpies of formation
The standard reaction enthalpy of
2 HN3(ℓ) + 2 NO(g) → H2O2(l) + 4 N2(g)
is calculated as follows:
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Temperature dependence of rxn enthalpies
Kirchhoff’s law:
(assume no phase transition takes place in
the temperature range of interest.)
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Example 2C.2 Using Kirchhoff’s law
The standard enthalpy of formation of gaseous H2O at 298 K is −241.82 kJ mol−1.
Estimate its value at 100°C given the following values of the molar heat capacities
at constant pressure: H2O(g): 33.58 J K−1 mol−1; H2(g): 28.84 J K−1 mol−1; O2(g):
29.37 J K−1 mol−1. Assume that the heat capacities are independent of
temperature.
Answer
When ∆Cpө is independent of temperature in the range T1 to T2,the integral
evaluates to (T2 − T1)∆r Cpө. Therefore,
The reaction is H2(g) + ½O2(g) → H2O(g), so
It then follows that
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