an adjacent one. Using the Create Names command, we assigned the names in the first column to the values in the second (to avoid confusion with row names, we have named T1 and T2 as Temp1 and Temp2 respectively; added an underscore to ‘a’, ‘b’, and ‘c’, so these constant appears as a_, etc. in our formula). In the column on the right, we have written the formula out in the second row, then evaluated it at T1 and T2 in the third and fourth rows respectively. The next row contains our answer, 15.2 kJ/mol, determined simply by subtracting ‘H1’ from ‘H2’ (and dividing by 1000). Hint: we need to keep track of units. Excel won't do this for us.
Isothermal enthalpy changes refer to those occurring at constant temperature, for example, changes in enthalpy due to isothermal pressure changes. Though pressure changes at constant temperature are relatively rare in nature, hypothetical isothermal paths are useful in calculating energy changes. Since enthalpy is a state property, the net change in the enthalpy of a system depends only on the starting and ending state: the enthalpy change is path independent. Imagine a system consisting of a quartz crystal that undergoes a change in state from 25°C and 1 atm to 500°C and 400 atm. How will the enthalpy of this system change? Though in actuality the pressure and temperature changes may have occurred simultaneously, because the enthalpy change is path independent, we can treat the problem as an isobaric temperature change followed by an isothermal temperature change, as illustrated in Figure 2.11. Knowing how to calculate isothermal enthalpy changes is useful for this reason.
Figure 2.11 Transformations on a temperature–pressure diagram. Changes in state variables such as entropy and enthalpy are path independent. For such variables, the transformation paths shown by the solid line and dashed line are equivalent.
We want to know how enthalpy changes as a function of pressure at constant temperature. We begin from eqn. 2.63, which expresses the enthalpy change as a function of volume and pressure:
(2.63)
By making appropriate substitutions for dU, we can derive the following of enthalpy on pressure:
(2.113)
If changes are large, α, β, and V must be considered functions of T and P and integration performed over the pressure change. The isothermal enthalpy change due to pressure change is thus given by:
(2.114)
2.10.2 Changes in enthalpy due to reactions and change of state
We cannot measure the absolute enthalpy of substances, but we can determine the enthalpy changes resulting from transformations of a system, and they are of great interest in thermodynamics. For this purpose, a system of relative enthalpies of substances has been established. Since enthalpy is a function of both temperature and pressure, the first problem is to establish standard conditions of temperature and pressure to which these enthalpies apply. These conditions, by convention, are 298.15 K and 0.1 MPa (25°C and 1 bar). Under these conditions the elements are assigned enthalpies of 0. Standard state enthalpy of formation, or heat of formation, from the elements, ΔH°, can then be determined for compounds by measuring the heat evolved in the reactions that form them from the elements (e.g., Example 2.2). For example, the heat of formation of water is determined from the energy released at constant pressure in the reaction: H2 + ½O2 → H2O, which yields a ΔH° of −285.83 kJ/mol, where water is in the liquid state. The minus sign indicates heat is liberated in the reaction, that is, the reaction is exothermic (a reaction that consumes heat is said to be endothermic).
Having established such a system, the enthalpy associated with a chemical reaction is easily calculated using Hess's law, which is:
(2.115)
where νi is the stoichiometric coefficient for the ith species. In other words, the enthalpy of reaction is just the total enthalpy of the products less the total enthalpy of the reactants. The use of Hess's law is illustrated in Example 2.5 below.
The heat of vaporization of a substance is the energy required to convert that substance from liquid to gas, i.e., to boil it. If the reaction H2 + ½O2 → H2O is run to produce water vapor, the ΔH° turns out to be −241.81 kJ/mol. The difference between the enthalpy of formation of water and vapor, 44.02 kJ/mol, is the heat consumed in going from liquid water to water vapor. This is exactly the amount of energy that would be required to boil 1 mole of water. Analogously, the heat of melting (or fusion) is the enthalpy change in the melting of a substance. Because reaction rates are often very slow, and some compounds are not stable at 298 K and 1 MPa, it is not possible to measure the enthalpy for every compound. However, the enthalpies of formation for these compounds can generally be calculated indirectly.
Example 2.5 Enthalpies (or heats) of reaction and Hess's law
What is the energy consumed or evolved in the hydration of corundum (Al2O3) to form gibbsite (Al(OH)3)? The reaction is:
Answer: We use Hess's law. To use Hess's law, we need the standard state enthalpies for water, corundum, and gibbsite. These are: Al2O3: −1675.70 kJ/mol, H2O: −285.83 and Al(OH)3: −1293.13. The enthalpy of reaction is
This is the enthalpy of reaction at 1 bar and 298 K. Suppose you were interested in this reaction under metamorphic conditions such as 300°C and 50 MPa. How would you calculate the enthalpy of reaction then?
2.10.3 Entropies of reaction
Since
(2.62)
and
(2.57)
then at constant pressure
(2.116)
Thus,
