Problem 1 Birefringence of a uniaxial material Consider a uniaxial, birefringent material. The direction of propagation of unpolarized light is parallel to the optical axis. Draw a scheme of the experiment and discuss the behaviour of lightwaves with dierent directions of polarization. The discussion is analogous to that presented during the lectures for a typical uniaxial material with an optical axis. The gure below is for a general direction of the light. In case of light propagating parallel to the optical axis, since the material is uniaxial, all the dierent directions of polarization of light will behave in the same way.
Problem 2 Molar polarization vs. 1/T The molar polarization, P m, of pentachloroethane vapour (boiling point: 162.0 C) varies linearly with T 1, and is 46.79 cm 3 mol 1 at 176.1 C and 44.45 cm 3 mol 1 at 291.1 C. Calculate the polarizability, α, and the dipole moment, µ, of pentachloroethane molecules. We remember the relation between molar polarization, P m, the polarizability, α, and the dipole moment, µ: P m = N ) A (α + µ2 (1) 3ε 0 3kT This is also a linear eq. where P m depends on T 1. We can calculate the polarizability, α, and the dipole moment, µ, from the parameters of this linear equation. The general linear equation can be written as In this case we have y = b + mx P m = b + mt 1 where T 1 is the independent variable, m is the slope and b the intercept. By comparison with eq. 1, we notice that the slope, m, corresponds to and the intercept, b, corresponds to m = N Aµ 2 9ε 0 k b = N Aα 3ε 0 We notice that we can calculate µ from the slope and α from the intercept, as ( ) 1/2 9ε0 km µ = (2) N A α = 3ε 0b N A (3) The slope, m, can be calculated from the experimental data, considering the two equations P m1 = b + mt 1 1 by subtracting each member, we have P m2 = b + mt 1 2 P m1 P m2 = b + mt1 1 b mt2 1 = m ( ) T1 1 T2 1 and nally
By doing the calculation, we nally have polarizability α = 1.558 10 39 m 2 C 2 J 1, polarizability volume α = 14.0 10 24 cm 3, dipole moment: 3.069 10 30 C m = 0.92 D. m = P m1 P ( m2 ) T 1 1 T2 1
Problem 3 Estimation of the enthalpy of mixing of toluene and benzene The polarizability volume of toluene is 11.8 10 24 cm 3 and its ionization energy is 8.8276 ev. The polarizability volume of benzene is 10.0 10 24 cm 3 and its ionization energy is 9.24384 ev. The average intermolecular distances are, toluenetoluene: 6.1 Å, benzenebenzene: 5.0 Å, toluenebenzene: 5.66 Å. Estimate the molar enthalpy of mixing for an equimolar mixture of liquid toluene with liquid benzene. Discuss the results. How can we estimate the molar enthalpy of mixing? We assume that, to make an equimolar mixture, me need to break the toluenetoluene and benzene benzene molecular bonds, and reform toluenebenzene bonds. The molar enthalpy of mixing is the dierence between the energy needed to break the molecular bonds of the two components and the energy gained to reform new bonds in the mixture. What do we need to calculate the energy of the molecular bonds? We can use the London approximate formula: V = 3 2 α 1α 2 I 1 I 2 (I 1 + I 2 ) r 6. We remember that the ionization energy must be converted in joule 1 J = 6.24150974 10 18 ev. I toluene = 1.41 10 18 J I benzene = 1.48 10 18 J An example of the calculation for the toluenetoluene interaction is the following (we can calculate the values rst, then add the powers of ten) V = 3 2 11.82 1.41 2 (1.41 + 1.41) (6.1) 6 10 60 10 36 10 18 10 60. V = 0.00287 10 18 = 2.87 10 21 J. The molecular interaction energies calculated in this way are V toluene toluene = 2.87 10 21 J V benzene benzene = 7.11 10 21 J V toluene benzene = 3.89 10 21 J To estimate the molar enthalpy of the bonds, bonds H, we remember that the number of nearest-neighbour interactions, in a simplied treatment, tends to be 3 the total n. of molecules. vap H = 3N A V The calculation for the molar enthalpy of the toluenetoluene intermolecular bond is bonds H toluene toluene = 3 6.02 10 23 ( 2.87 10 21) = 5.179 kj mol 1 The molar enthalpies of the bonds calculated in this way are bonds Htoluene toluene = 5.179 kj mol 1 bonds Hbenzene benzene = 12.843 kj mol 1
bonds Htoluene benzene = 7.037 kj mol 1 Finally, to make one mole of the mixture we mix half mole of toluene with half of benzene. The molar enthalpy of mixing is mix H = 1 2 5.179 + 1 12.843 7.037 = 1.975 kj mol 1 2 The molar enthalpy of mixing is estimated to be small and positive, which means that the mixture is slightly energetically unfavoured. We know that the real situation is dierent since toluene and benzene are completely miscible at all concentrations. Discussion The molar enthalpy of mixing is quite small, indicating that the mixture is almost ideal. We should remember that the London approximate formula is appropriate for nonpolar molecules. In our case this is almost correct because benzene is nonpolar but toluene has a low dipole moment of 0.36 D. If the molecules are more polar, the estimation of the interaction will be less accurate.
Problem 4 Relative permittivity of acetic acid Acetic acid vapour contains a proportion of planar, hydrogen-bonded dimers. The relative permittivity of pure liquid acetic acid is 7.14 at 290 K and increases with increasing temperature. Suggest an interpretation of the latter observation. What eect should isothermal dilution have on the relative permittivity of solutions of acetic acid in benzene? For most liquids, the static relative dielectric permittivity is a decreasing function of temperature, because enhanced thermal motion reduces the ability of the molecular dipoles to orient under the eect of an external electric eld. The relative permittivity is larger if the molecules are more polar and can orient under the eect of an external electric eld. In pure acetic acid in vapour phase, part of the molecules form hydrogen-bonded dimers so the dimers are non-polar. In pure liquid acetic acid at room temperatures the molecules form both dimers, that are non-polar. and (chain-like) hydrogen-bonded structures, that are polar. By increasing the temperature the dimers break up forming longer chain structures. In this way the presence of polar species increases at increasing temperature. More polar species will increase the relative permittivity. Discussion Normally the relative permittivity, ε r, should decrease with increasing temperature. The relation between ε r and T is given by the Debye equation We rearrange Finally ε r 1 ε r + 2 = ρp m M = C ε r 1 = Cε r + 2C ε r Cε r = 2C + 1 ε r (1 C) = 2C + 1 ε r = 2C + 1 (1 C) Here we see that ε r is proportional to C which, in turn. is proportional to the molar polarization, P m, which is inversely proportional to T : P m = N ) A (α + µ2. 3ε 0 3kT So, as T increases, P m should decrease and ε r should decrease.
In fact, for most liquids, the static relative dielectric permittivity is a decreasing function of temperature, because enhanced thermal motion reduces the ability of the molecular dipoles to orient under the eect of an external electric eld. This is an example for water From: http://www.ncbi.nlm.nih.gov/pubmed/22383366 Chemphyschem. 2012 Apr 10;13(5):1182-90. doi: 10.1002/cphc.201100949. Epub 2012 Mar 1. Temperature dependence of the dielectric permittivity of acetic acid, propionic acid and their methyl esters: a molecular dynamics simulation study. Riniker S1, Horta BA, Thijssen B, Gupta S, van Gunsteren WF, Hünenberger PH. For most liquids, the static relative dielectric permittivity is a decreasing function of temperature, because enhanced thermal motion reduces the ability of the molecular dipoles to orient under the eect of an external electric eld. Monocarboxylic fatty acids ranging from acetic to octanoic acid represent an exception to this general rule. Close to room temperature, their dielectric permittivity increases slightly with increasing temperature. Herein, the causes for this anomaly are investigated based on molecular dynamics simulations of acetic and propionic acids at dierent temperatures in the interval 283-363 K, using the GROMOS 53A6(OXY) force eld. The corresponding methyl esters are also considered for comparison. The dielectric permittivity is calculated using either the box-dipole uctuation (BDF) or the external electric eld (EEF) methods. The normal and anomalous temperature dependences of the permittivity for the esters and acids, respectively, are reproduced. Furthermore, in the EEF approach, the response of the acids to an applied eld of increasing strength is found to present two successive linear regimes before reaching saturation. The low-eld permittivity ε, comparable to that obtained using the BDF approach, increases with increasing temperature. The higher-eld permittivity ε' is slightly larger, and decreases with increasing temperature. Further analyses of the simulations in terms of radial distribution functions, hydrogen-bonded structures, and diusion properties suggest that increasing the temperature or the applied eld strength both promote a relative population shift from cyclic (mainly dimeric) to extended (chain-like) hydrogen-bonded structures. The lower eective dipole moment associated with the former structures compared to the latter ones provides an explanation for the peculiar dielectric properties of the two acids compared to their methyl esters.
Problem 5 Anion in a water molecule electric eld Suppose an H 2 O molecule (µ = 1.85 D) approaches an anion. What is the favourable orientation of the molecule? Calculate the electric eld (in volts per metre) experienced by the anion when the water dipole is (a) 1.0 nm, (b) 0.3 nm, (c) 30 nm from the ion. The main point is to nd the correct equation and the correct geometry of the interaction between the water molecule and the anion. The problem asks us to calculate the electric eld generated by the water molecule. We need to nd the equation of this electric eld: What are the vectors µ and r? E = 1 [ ] 1 µ (µ r)3r 4πε 0 r3 r 5 µis the dipole moment of water, r is the position where we are evaluating the electric eld, considering the dipole at the origin. The orientation of the dipole moment in the water molecule is the following The dipole orientation goes from minus to plus. Now we need to know the geometry of the interaction so that we know the orientation of the vector r with respect to µ. What is the favourable orientation of water with respect to an anion?. The anion will be on the positive side of the water molecule. So, the two vectors are parallel. Note that they are parallel, not antiparallel. Since the interaction occurs along one coordinate, we can consider just the magnitudes of the vectors and simplify the equation. and nally E = 1 [ ] 1 4πε 0 r µ 3µr2 = 1 [ µ 3 r 5 4πε 0 r 3µ ] = 1 [ 2µ ] = 1 2µ 3 r 3 4πε 0 r 3 4πε 0 r 3 E = µ 2πε 0 r 3
Result (c) should be, more accurately, 4115.8 V m 1. Discussion We must always remember that the equation assumes point dipoles. At small distances this assumption becomes more critical. What the equation would have been if instead of an anion we had a cation?