ENTROPY AND THE SECOND LAW
5.4 THE THIRD LAW
The third law of thermodynamics states that the entropy, unlike the energy and enthalpy, has a natural zero point. The entropy of a perfect crystal is zero at 0 K.
3The product of two negative numbers.
THE THIRD LAW 79 Because of the third law, it is possible to obtain a standard molar entropy (often called the “absolute” entropy) of any pure substance at any temperature. The task is simple but not easy. One must determine the molar heat capacity at constant pressure Cpfor the crystal at many temperatures until it undergoes the first phase transition.
By the methods shown in Section 5.2.1, the integral taken down to low T S=
T1 0
Cp
T d T
gives the standard entropy at the transition temperature T1. Since Cp is a molar heat capacity, S is an standard molar entropy. Normally one wants the entropy at some higher temperature, say 298 K, and often the phase transition takes place at a temperature T1 lower than 298 K. Therefore we must add the entropy contribution from the phase transition to the value of S that we already have to obtain the standard molar entropy after the phase change at T1. The new phase then is heated over a temperature interval T1 →T2 where the new temperature may be 298 K or may be the temperature of a new phase change. Melting and vaporization are handled in the same way as crystalline phase changes. Eventually, over a few or many phase changes, one arrives at the desired temperature. The standard entropy is the summation of all the contributions along the way:
S = T1
0
Cp
T d T +Htrans
Ttrans + Thigher
Tlower
Cp
T d T 5.4.1 Chemical Reactions (Again)
The change in entropy of a chemical reaction can be determined by carrying out a determination of the standard molar entropies of all of the reactants and all of the products as just described and taking the sum
Sr =
S(products)−
S(reactants)
The entropy of ordering or disordering that occurs when, for example, the product state is in the gaseous phase and the reactants are in a condensed phase (liquid or solid) is included in this sum because terms likeHvap/Tbare included in
S(products) but not in
S(reactants). It would be attractive to adopt the simplistic attitude that all spontaneous chemical and physical reactions produce an entropy increase for the reacting system, but, once again, things are more complicated than that. A spontaneous change is driven both by the tendency of a system to reduce its energy and enthalpy and by the tendency of a system to increase its disorder. A composite function is needed that includes both the enthalpy and entropy, and this is the function found and described in mathematical detail by the great American thermodynamicist J. Willard Gibbs. The composite function the Gibbs free energy or, more simply, the Gibbs function G=H−T S now bears his name. There is a comparable function
involving the energy and the entropy used more by engineers than by chemists; this is called the Helmholtz free energy, A=E−T S.
PROBLEMS AND EXAMPLE
Example 5.1 The Standard Entropy of Silver The Cp/T vs. T data set for silver from 15 to 300 K is
T Cp Cp/T ln T
15.0000 0.6700 2.7100 0.0447
30.0000 4.7700 3.4000 0.1590
50.0000 11.6500 3.9100 0.2330
70.0000 16.3300 4.2500 0.2333
90.0000 19.1300 4.5000 0.2126
110.0000 20.9600 4.7000 0.1905
130.0000 22.1300 4.8700 0.1702
150.0000 22.9700 5.0100 0.1531
170.0000 23.6100 5.1400 0.1389
190.0000 24.0900 5.2500 0.1268
210.0000 24.4200 5.3500 0.1163
230.0000 24.7300 5.4400 0.1075
250.0000 25.0300 5.5200 0.1001
270.0000 25.3100 5.6000 0.0937
290.0000 25.4400 5.6700 0.0877
300.0000 25.5000 5.7000 0.0850
The Cp/T vs. T curve for silver is given as Fig. 5.2. Select from packaged software (for example, SigmaPlotC) or write a short program of your own that will enable you to integrate the data set for silver to find the standard entropy S at 300 K. The problem is simplified by the lack of phase transitions in solid Ag over the temperature range, including the melting and boiling points, which are well above 298 K. The CRC Handbook of Chemistry and Physics, 2008–2009 (89th ed.) value for the standard S298Ag at 298 is 42.67 J K−1mol−1.
Solution 5.1 SigmaPlotC contains a macro that carries out integration under curves that are displayed as a smooth function. First plot your function, then execute Tools → macro → run → compute. Be sure to designate your plot be- low the macro. The SigmaPlot output for this integration is 42.2076 over the interval from 15 K to 298 K.
The result is pretty close to the standard entropy in the handbook, but it lacks a contribution below 15 K. This problem is usually handled by the Debye method (Problem 5.7), which assumes a third-power equation leading to
Cp =AT3=0.67 J K−1mol−1
PROBLEMS AND EXAMPLE 81 Cp/ T vs. T Silver(s)
T,K
0 50 100 150 200 250 300 350
Cp/T, J mol–1
0.00 0.05 0.10 0.15 0.20 0.25
FIGURE 5.2 Cp/T vs. T for metallic silver Ag(s). There are no phase transitions for solid Ag over this temperature range.
at 15 K and leads to
S015 =Cp
3 = 0.67
3 =0.22 J K−1mol−1
This small addition yields S0298=42.43 J K−1mol−1, which is within 0.6% of the handbook value.
Problem 5.1
Hexa-1,3,5-triene has a boiling point of 355 K under atmospheric pressure. Estimate the enthalpy of vaporization of hexa-1,3,5-triene.
Problem 5.2
(a) What is the entropy change brought about by heating 2.5 mol of helium from 300 to 400 K at constant volume?
(b) What is the entropy change brought about if the same heating process takes place at constant pressure?
Problem 5.3
What is the entropy of isothermal mixing of 1 mol of helium with 1 mol of argon if the two gases start out in separate chambers, each at 1 bar pressure, and they produce 2 mol of mixed gases also at 1 bar pressure?
Problem 5.4
Industrial production of ammonia NH3is carried out by combination of the elements at about T =650 K and p=400 bar (Metiu, 2006):
3H2+N2 →2NH3
What is the enthalpy change for this reaction at this temperature and pressure? For simplicity, assume ideal behavior of all three gases.
Problem 5.5
What is the enthalpy change for the pressure change from 1.0 bar to 40 MPa?
Problem 5.6
Sulfur dioxide has a heat of fusion of 7.41 kJ mol−1 at its melting point of 200 K.
Find the entropy change for the melting process SO2(s)→SO2(l)
How does this compare with the entropy change for melting ice which has Hfusion= 333.6 J g−1 (CRC Handbook of Chemistry and Physics, 2008–2009, 89th ed.)
Problem 5.7
On theoretical grounds, Peter Debye proposed what is known as the Debye third- power law for the entropy of perfectly crystalline solids near absolute zero K.
Cp=AT3, T <15 K
Solid chlorine Cl2(s) has a heat capacity of Cp =3.72 J K−1mol−1. What is the entropy of Cl2(s) at T =15 K?
PROBLEMS AND EXAMPLE 83 Problem 5.8
The Cp/T vs. T data set for solid lead Pb(s) is written in BASIC as follows:
DATA 0,0,5,.061,10,.28,15,.4666,20,.54,25,.564,30,.55,50,.428, 70,.333,100,.245,150,.169 200,.129,250,.105,298,.089
N = 14
The data are in 14 pairs. The first number of the pair is T and the second is Cp. Devise a program of your own or use a canned program to estimate the standard entropy of Pb(s).
Problem 5.9
Phase changes occur reversibly. For example, the transition from solid ice to liquid water, which occurs at a temperature that is infinitesimally above the melting point of 273.15 K, can be reversed by lowering the ambient temperature to slightly less than 273.15 K.
(a) The standard enthalpy of fusionfusH◦is 6.01 kJ mol−1. What is the entropy of fusion,fusS◦?
(b) The standard enthalpy of vaporization of water isvapH◦=40.7 kJ mol−1. What is the entropy of vaporization of water, l →v?
(c) Why are the results so different?