William M. White - Geochemistry

Answer : To solve this problem, we need to apply eqn. 2.110. We need only consider the octahedral site containing Fe and Mg, because O and Si are the only kinds of atoms occupying the tetrahedral and anion sites. The values for X for these two sites will therefore be 1, and ln(1) = 0, so there is no contribution to configurational entropy.

For the octahedral site, , , and . Therefore, the configurational entropy will be:

From eqn. 2.68, we can see that the temperature derivative of enthalpy is simply the isobaric heat capacity:

and hence:

(2.111)

Thus, the change in enthalpy over some temperature interval may be found as:

(2.112)

C pis often a complex function of temperature, so the integration is essential. Example 2.4illustrates how this is done.

How does the enthalpy of a 1 mol quartz crystal change if it is heated from 25°C to 300°C if the temperature dependence of heat capacity can be expressed as , and a = 46.94, b = 0.0343, and c = 1129680? Assume pressure is constant.

Answer: The first step is to convert temperature to kelvins: all thermodynamic formulae assume temperature is in kelvins. So and . To solve this problem, we need to use eqn. 2.112. Substituting the expression for heat capacity into eqn. 2.112, we have:

Performing the integral, we have:

Now that we have done the math, all that is left is arithmetic. This is most easily done using a spreadsheet. Among other things, it is much easier to avoid arithmetical errors. In addition, we have a permanent record of what we have done. We might set up a spreadsheet to calculate this problem as follows:

This example is from Microsoft Excel™. On the left, we have written down the names for the various constants in one column, and their values in 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 T 1and T 2as 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 T 1and T 2in 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 d U , 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)