Experiments

1. Experiments

2. Experiment 3 Heat Capacity Ratio For Gases1 Summary - The purpose of this experiment is to determine the value for the heat capacity ratio for two gases, a monatomic gas and a diatomic gas. The adiabatic expansion method, a relatively crude method, will be employed. The results are compared to those predicted from simple theory. For the diatomic gas comparison is made to both the classical and quantum mechanical predictions.

3. Heat Capacity2 The heat capacity, C, of a substance is defined as C = lim T0 (q/T) = (dq/dT) Since q is not a state function heat capacity is also in general not a state function. However, if we restrict ourselves to certain processes we can make C a state function. In particular, we may show the following CV = (dq/dT)V = (U/T)Vconstant volume heat capacity Cp = (dq/dT)p = (H/T)pconstant pressure heat capacity Since the right-hand terms above are state functions both CV and Cp are also state functions.

4. General Relationships and Definitions Consider one mole of an ideal gas. Using a general relationship that applies to all substances3 we may show the following to be true Cp,m - CV,m = R ; or Cp,m = CV,m + R Cp,m and CV,m are the constant pressure and constant volume molar heat capacities, respectively, that is, the heat capacities per mole of ideal gas. Finally, we may define , the heat capacity ratio, as  = (Cp,m/CV,m) = (Cp/CV) Note that we can (but are not required to) use the molar heat capacities to define . The heat capacity ratio is a useful quantity in the discussion of the adiabatic reversible expansion of an ideal gas.4

5. Total Energy For One Mole of an Ideal Gas We may use the relationship CV,m = (U/T)V,m to find the constant volume molar heat capacity for an ideal gas. We start by saying that the total energy for one mole of an ideal gas is Utotal = Etrans + Erot + Evib where total = total energy trans = translational energy rot = rotational energy vib = vibrational energy In writing this we are assuming a sharp separation between rotational and vibrational energy and neglecting electronic energy. We now look at the contributions to the total energy according to classical mechanics. After this, we consider the effects of quantum mechanics on our results.

6. Classical Result For Total Energy The average translational, rotational, and vibrational energy of one mole of an ideal gas may be found using the equipartition theorem,5 a general result that can be derived from statistical mechanics. This theorem predicts the energy per translational, rotational, and vibrational degree of freedom. Etrans = 1/2 RT per translational degree of freedom Erot = 1/2 RT per rotational degree of freedom Evib = RT per vibrational degree of freedom For a gas molecule composed of N atoms, there are 3N degrees of freedom, which may be assigned as follows. translation = 3 rotation = 2 (linear molecule); 3 (non-linear molecule) vibration = 3N - 5 (linear molecule); 3N - 6 (non-linear molecule)

7. CV,m, Cp,m, and  For a monatomic ideal gas there is only translational energy, so Utotal = Etrans = 3/2 RT CV,m = (U/T)V,m = 3/2 R Cp,m = CV,m + R = 5/2 R  = Cp,m/CV,m = 5/3 = 1.66... For a diatomic ideal gas there is translational, rotational, and vibrational energy, so Utotal = Etrans + Erot + Evib = 3/2 RT + RT + RT = 7/2 RT CV,m = (U/T)V,m = 7/2 R Cp,m = CV,m + R = 9/2 R  = Cp,m/CV,m = 9/7 = 1.28...

8. Quantum Mechanics The above relationships were derived using the virial theorem, a classical result that assumes that translational, rotational, and vibrational energy are continuous (can take on any value within a particular range). However, a consequence of quantum mechanics is that these energies are quantized, that is, that they can only take on particular values. At room temperature this only affects the expression for vibrational energy. At low temperatures (relative to vib, the characteristic temperature for a particular vibration6), Evib  0. As temperature increases the molecule will gain vibrational energy, though the amount will in general be smaller than predicted by the virial theorem.

9. Quantum Effects on  We now consider the effects of quantum mechanics on our theoretical value for . Monatomic ideal gas. Since there is only translational energy, including quantum mechanics has no effect on the values for CV,m, Cp,m, and . Diatomic ideal gas. Since a diatomic molecule has one vibration, quantum mechanics does have an effect on the values for CV,m, Cp,m, and . “low T” “high T” CV,m5/2 R 7/2 R Cp,m7/2 R 9/2 R  7/5 = 1.40... 9/7 = 1.28… In the above table “low T” corresponds to T << vib, and “high T” to T >> vib. For T ~ vib (“intermediate T”) CV,m, Cp,m, and  will fall between the “low T” and “high T” values.

10. Experimental You should do a minimum of five trials for both your monatomic gas (Ar) and your diatomic gas (N2). If you can do more trials you will get better statistics in your data analysis. You should compare your experimental results to the theoretical values expected for a monatomic gas and a diatomic gas. For the diatomic gas you should compare with both the “classical” and “quantum” theoretical result. You do not need to answer the questions posed in the discussion section of the lab text. You should, however, discuss possible random and systematic errors associated with carrying out the adiabatic expansion of the gas. If separate groups carry out the measurements for Ar and N2 you should share your data with one another, and carry out the data analysis for both gases.

11. References 1) Experiments in Physical Chemistry, 8th Edition, 2009. C. W. Garland, J. W. Nibler, D. P. Shoemaker. p. 106-114. 2) Physical Chemistry, 9th Edition, 2010. P. W. Atkins, J. de Paula. Heat capacity is discussed on p. 57-63. 3) Atkins, p. 78; 84-85 (Further information 2.2); 122-124. 4) Atkins, p. 63-64; 68. 5) Atkins, p. 9; 47. 6) Atkins, p. 599 for a definition of vib. A table of values for vib is given in the Appendix (Table 12.2, p. 933).

12. Experiment 6 Heat of Combustion1 Summary - The purpose of this experiment is to determine the value for the heat of combustion for naphthalene or another organic compound. The experimental method that will be used is bomb calorimetry. Data on the combustion of benzoic acid will be used to determine the heat capacity for the calorimeter. This will then be used to analyze the experimental data for naphthalene. The experimental value for the enthalpy of combustion for naphthalene will be compared to a literature value.

13. The Combustion Reaction2 By definition, combustion refers to the reaction of one mole of a single substance with oxygen (O2(g)) to form combustion products. The combustion products for a few common elements are as follows carbon (C) – CO2(g) nitrogen (N) – N2(g) hydrogen (H) – H2O() sulfur (S) – SO2(g) halogens (X) – X2 When a small sample of a compound is burned in a large excess of oxygen carbon and hydrogen will be essentially completely converted into CO2(g) and H2O(). For elements such as nitrogen, sulfur, and the halogens a complex mixture of products is likely obtained, and so the “combustion products” listed here are chosen for convenience in doing thermochemical calculations.

14. Examples 1) Combustion of n-octane (C8H18()). C8H18() + 25/2 O2(g)  8 CO2(g) + 9 H2O() 2) Combustion of aniline (C6H5NH2(s)). C6H5NH2(s) + 31/4 O2(g)  6 CO2(g) + 7/2 H2O() + ½ N2(g)

15. Enthalpy of Combustion By definition Hcomb, the enthalpy of combustion (sometimes also called the heat of combustion), is the enthalpy change when one mole of a single compound undergoes combustion for standard conditions. Standard conditions3 are usually taken to be p = 1.00 bar (approximately the same as 1 atm) and T = 25.0 C, though different standard temperatures are sometimes used. For example, for the combustion of n-octane C8H18() + 25/2 O2(g)  8 CO2(g) + 9 H2O() Hcomb(C8H18()) = - 5471. kJ/mol Note that the enthalpy of combustion is negative, meaning the reaction is exothermic. This is commonly the case for combustion reactions.

16. Bomb Calorimetry The most common experimental technique for determining the enthalpy of combustion for a substance is bomb calorimetry. A sample of the compound in question is burned under conditions where a large excess of oxygen is present and the heat that is evolved is measured. Bomb calorimetry is discussed in detail in the text. Since the combustion reaction is carried out at constant volume in bomb calorimetry it is Ucomb, the energy of combustion, that is measured, from which Hcomb can be found as discussed below.

17. Calculations There are two key equations for q, the value for heat for the combustion reaction q = mC Ucomb(C) + mW Ucomb(W) (6.1) q = - Cbomb T (6.2) In the above equations q is the value for heat for the combustion reaction. C and W refer to the compound (C) and wire (W) (a small amount of wire, used to initiate combustion, reacts with oxygen and must be accounted for in the equations), which have energies of combustion Ucomb(C) and Ucomb(W), respectively (these are usually given in units of J/g) and masses mC and mW. Cbomb is the heat capacity of the bomb, and T is the experimental change in temperature observed for the combustion reaction.

18. Determination of Cbomb A convenient method for finding the value for Cbomb, the heat capacity of the calorimeter, is to burn a known amount of a compound for which Ucomb is known. Benzoic acid (C6H5COOH(s)) is a convenient compound to use because it has an accurately known value for the energy of combustion4, can be obtained at high purity, can be easily pressed into pellets, and burns completely to form combustion products. If we burn a known amount of benzoic acid (and wire)4 and measure the temperature change associated with combustion, we may use equation (6.1) above to find q. Equation (6.2) can then be used to find Cbomb. Assuming that the value for Cbomb is insensitive to the compound being burned, this value can be used to determine q for the combustion of a known mass of a different compound, such as naphthalene, using equation (6.2). Substituting this value for q into equation (6.1) gives a relationship where the only unknown is Ucomb for the compound, which can then be found from the equation.

19. Conversion of Ucomb To Hcomb The experimental data in this experiment are sufficient to determine Ucomb, the energy of combustion. However, literature values are usually given as Hcomb, the enthalpy of combustion. These two quantities are related by the approximate expression5 Hcomb  Ucomb + ngRT where Ucomb and Hcomb are given on a per mole basis, and  ng is the change in the number of moles of gas per mole of reaction (which would, for example, be equal to – 9/2 for the combustion of n-octane). The expression approximate because it is derived assuming all gas reactants and products behave ideally and that the volume of solid and liquid reactants and products is negligible.

20. Experimental You should do a minimum of three trials with benzoic acid to determine the heat capacity of the calorimeter, and three trials with naphthalene to determine the enthalpy of combustion. For each trial there are only three final experimental results that are obtained and used in the calculations; mC, the mass of compound burned, mW, the mass of wire burned, and T, the temperature change for combustion. There are several methods that can be used to determine T, and so you should clearly indicate the method you have used. After finding the Ucomb for naphthalene you will need to convert this to Hcomb by the method indicated above. This will require the balanced combustion reaction for naphthalene, which should be given in the lab report. You should compare your value for Hcomb for napthalene to a literature value (available in Atkins) and discuss what you find. You should think about, but do not need to answer, the questions in the discussion section of your lab manual.

21. References 1) Experiments in Physical Chemistry, 8th Edition, 2009. C. W. Garland, J. W. Nibler, D. P. Shoemaker. p. 145-158. 2) A good general discussion of thermochemistry, including combustion, is given in Physical Chemistry, 9th Edition, 2010. P. W. Atkins, J. de Paula, p. 65-72. 3) Atkins, p. 5; 65. 4) Ucomb(benzoic acid) = - 26.41 kJ/g; Ucomb(wire) = - 6.68 kJ/g, from Garland. 5) Atkins, p. 57-59.

22. Experiment 13 Vapor Pressure of a Pure Liquid1 Summary - The purpose of this experiment is to determine the value for the enthalpy of vaporization for water. The boiling point method, in which the pressure at which a liquid boils is measured as a function of temperature, will be used. Two independent trials will be carried out and compared to each other and to the literature value for the enthalpy of vaporization.

23. Vaporization Vaporization refers to the process M()  M(g) where M is a pure chemical substance. The amount of heat that must be added to convert one mole of a liquid in equilibrium with its vapor into gas is called Hvap, the enthalpy of vaporization. When vaporization occurs at a pressure p = 1.00 atm, then the temperature is called the normal boiling point for the liquid, and the enthalpy change is called Hvap, the standard enthalpy of vaporization.

24. Thermodynamics of Vaporization For general phase transitions occurring among the solid, liquid, or gas phases of a pure substance the Clapeyron equation2 applies dp = Spt = Hpt (13.1) dT Vpt T Vpt where “pt” stands for “phase transition” (fusion, vaporization, sublimation, or a solid to solid phase transition). The above equation can be solved for Hpt and therefore used to find the enthalpy change for a phase transition.

25. The Clausius-Clapeyron Equation For vaporization (or sublimation) where the final phase is the vapor phase, the Clapeyron equation can be rewritten based on the following assumptions: 1) The enthalpy of vaporization is approximately independent of temperature. 2) The vapor phase can be described by the ideal gas law. 3) The volume of the liquid phase is negligible. Based on these assumptions one obtains the Clausius-Clapeyron equation3 d ln p = - Hvap (13.2) d (1/T) R This suggests that the value for the enthalpy of vaporization can be found by finding the slope of a plot of ln p vs 1/T.

26. Effect of Nonideal Behavior of Water Vapor The Clausius-Clapeyron equation assumes that water vapor behaves as an ideal gas. However, since water is a polar molecule, nonideal behavior is expected even at low temperature. If we define the compressibility factor, Z, as Z = pVm (13.3) RT then the Clausius-Clapeyron equation may be rewritten4 as d ln p = - Hvap (13.4) d (1/T) RZ where Z is the compressibility factor for water vapor. At the normal boiling point, T = 100.0 C, the experimental value for the compressibility factor for water vapor is Z = 0.986.5

27. Temperature Dependence of Hvap The Clausius-Clapeyron equation also assumes that the value for the enthalpy of vaporization, Hvap, is independent of temperature. In fact the enthalpy of vaporization is temperature dependent, as is true for enthalpy changes for other processes.6 This leads to curvature in a plot of ln p vs 1/T. In the present experiment the amount of curvature expected is small compared to the experimental error in the measurements, and so likely will not be noticed. The enthalpy of vaporization measured in the present experiment corresponds to the value at the average temperature of your data. It can be adjusted to the value expected at T = 100.0 C as discussed below.

28. Accounting For the Dependence of Hvap on T The general expression for the temperature dependence of the enthalpy of reaction,6 applied to the process of vaporization, gives Hvap(T) = H(Tave) + TaveT (Cp,m(g) - Cp,m()) dT where Hvap(T) is the enthalpy of vaporization at the normal boiling point temperature T, H(Tave) is the experimental enthalpy of vaporization, corresponding to the value at the average temperature of the data, and Cp,m() and Cp,m(g) are the constant pressure molar heat capacities of the liquid and gas.

29. Accounting For the Dependence of Hvap on T If we assume that Cp,m() and Cp,m(g) are independent of temperature (true if T - Tave is small) then Hvap(T)  H(Tave) + (Cp,m(g) - Cp,m()) (T - Tave) You should use the above equation to adjust your value for the enthalpy of vaporization to the value expected at T = 100.0 C, the normal boiling point for water. Use Tave = 90.0 C, Cp,m() = 75.29 J/mol.K, and Cp,m(g) = 33.58 J/mol.K in doing this correction.

30. Experimental Two independent trials should be carried out for water. Each trial should be analyzed independently, and a value for Hvap, along with the confidence limits for this value, should be found from a plot of ln p vs 1/T, and using Z = 0.986 for the compressibility factor for water vapor. The experimental values should then be compared to each other, and to a literature value for the enthalpy of vaporization (available in Atkins). The pressure data do not need to be corrected for the small change in the density of mercury with temperature.7

31. References 1) Experiments in Physical Chemistry, 8th Edition, 2008. C. W. Garland, J. W. Nibler, D. P. Shoemaker. p. 199-207. 2) Physical Chemistry, 9th Edition, 2010. P. W. Atkins, J. de Paula, p. 146-147. 3) Atkins, p. 148-149. 4) Garland, p. 200. 5) Garland, p. 201. 6) Atkins, p. 73-74; Garland, p 201. 7) These corrections are discussed in Garland. The correction for the difference in the density of mercury at room temperature (about 20. C) and 0. C is small (~ 0.4 %), and to a first approximation cancel out.