Entropy Calculations in Chemistry: Second Law Applications - Prof. Stefan Franzen, Study notes of Physical Chemistry

Download Entropy Calculations in Chemistry: Second Law Applications - Prof. Stefan Franzen and more Study notes Physical Chemistry in PDF only on Docsity! 1 Chemistry 331 Lecture 15 Second Law Applications NC State University Summary of entropy calculations In the last lecture we derived formula for the calculation of the entropy change as a function of temperature and volume changes. These are summarized in the table below. Sometimes we only have pressure information and the entropy change can be rewritten as follows. P1V1 = P2V2 ⇒ P1 P2 = V2 V1 ∆S = nR ln V2 V1 = nR ln P1P2 Constant temperature ∆S = nRln(V2/V1) Constant volume ∆S = nCvln(T2/T1) Constant pressure ∆S = nCpln(T2/T1) Entropy of mixing When two substances can mix there is a spontaneous tendency for this occur. We quantify this using the entropy state function. If we consider two containers separated by a stopcock. It N2 gas is in one and Br2 gas is in the other we know from experience that the gases will mix once the stopcock is opened. N2 + Br2 N2 Br2 N2 + Br2 Entropy of mixing For each gas we can describe the mixing as a volume change. The N2 gas is originally on the right side contained in volume V at pressure P. After opening the stopcock, the available volume is 2V and the partial pressure of P1 = x1P. The same is true for Br2. Its initial pressure is P2 and final pressure is P2 = x2P. We treat the entropy as the sum of two expansions (i.e. an expansion for each gas). ∆mixS = ∆expS1 + ∆expS2 ∆mixS = – n1R ln P1 P – n2R ln P2 P ∆mixS = – n1R ln x1 – n2R ln x2 ∆mixS = – nR x1lnx1 + x2lnx2 Note that the mole fraction applies to the final composition Entropy of phase transition The entropy of phase transition can be calculated using the enthalpy and temperature of the transition. Example, using the data in the Table calculate the entropy of vaporization for the following compounds. ∆ fusS = qP,fus Tfus = ∆ fusH Tfus Compound ∆vapH kJ/mol Tb K C6H6 30.8 353.2 C2H6 14.7 184.6 CCl4 30.0 350.0 CH4 8.18 111.7 Br2 29.7 332.4 H2S 18.7 212.8 ∆vapS = ∆vapH Tvap Solution: Use the following relation Entropy of phase transition The entropy of phase transition can be calculated using the enthalpy and temperature of the transition. Example, using the data in the Table calculate the entropy of vaporization for the following compounds. ∆ fusS = qP,fus Tfus = ∆ fusH Tfus Note the similarity in the values for the entropy of vaporization. This is known as Trouton’s rule. Compound ∆vapH kJ/mol Tb K ∆vapS J/mol-K C6H6 30.8 353.2 87.2 C2H6 14.7 184.6 79.6 CCl4 30.0 350.0 85.7 CH4 8.18 111.7 73.2 Br2 29.5 332.4 88.5 H2S 18.7 212.8 87.7 2 Conformational entropy The entropy of a polymer or a protein depends on the number of possible conformations. This concept was realized first more than 100 years ago by Boltzmann. The entropy is proportional to the natural logarithm of the number of possible conformations, W. For a polymer W = MN where M is the number of possible conformations per monomer and N is the number of monomers. For a typical polypeptide chain in the unfolded state M could be a number like 6 where the conformations include different φ,ψ angles and side chain angles. On the other hand, when the protein is folded the conformational entropy is reduced to W = 1 in the theoretical limit of a uniquely folded structure. Thus, we can use statistical considerations to estimate the entropy barrier to protein folding. S = R ln W Problem solving We can identify the following main types of problems that involve entropy change: Isothermal expansion/compression (reversible/irreversible) Temperature change Equilibration Mixing Phase Transition Statistical or Conformational Entropy Adiabatic (trick question) if q = 0 then ∆S = 0. Whenever you are solving an entropy problem remember to consider both system and surroundings. The system is always calculated along a reversible path. Isothermal compression Calculate the entropy for an irreversible compression of oxygen gas. The initial pressure of the gas is 1 bar in a volume of 100 L. The final pressure of the of the gas is 10 bar and the temperature is 400 K. Solution: Note that you need to obtain the number of moles. The problem does not ask you for a molar entropy. Write down the expression for the entropy: We need either the ratio of volumes or the ratio of pressures. We are given the pressures so we can use those. P2/P1 = 10 bar/ 1 bar = 10. ∆S = nR ln V2 V1 = nR ln P1P2 Isothermal compression We obtain the number of moles using the ideal gas law: Now we can substitute into the entropy expression: n = P1V1RT = 105 Pa 0.1 m3 8.31 J/mol–K 400 K = 1 bar 100 L 0.0831 L–bar/mol–K 400 K = 3.00 moles ∆S = nR ln P1P2 = 3.0 mol 8.31 J/mol–K ln 10 = 249 J/K Equilibration We have seen a simple example where there are two metal blocks, both made of the same material. However, this need not be the case. For any arbitrary materials 1 and 2 in contact we need to know the initial temperatures and heat capacities to calculate the final temperature. Now we solve for the equilibrium temperature Teq. q1 = – q2 Cp,1 T1 – Teq = – Cp,2 T2 – Teq Cp,1T1 – Cp,1Teq = – Cp,2T2 + Cp,2Teq Cp,1T1 + Cp,2T2 = Cp,1 + Cp,2 Teq Teq = Cp,1T1 + Cp,2T2 Cp,1 + Cp,2 Equilibration: Calculating the entropy Once you have obtained the equilibrium temperature, the entropy is easily calculated from: In such problems, you are assuming that the two objects in thermal contact are a closed system. The overall entropy for heat flow should be positive since heat flow from a hotter to a colder body is a spontaneous process. Example: The coffee cup problem. A hiker uses an aluminum coffee cup. If the mass of the cup is 4 grams and the ambient temperature (i.e. the cup’s temperature) is -10 oC and the hiker pours 50 ml. of coffee with a temperature of 90 oC into the cup: A. Calculate the equilibrium temperature. B. Calculate the entropy change. ∆S1 = nCp,1 ln Teq T1 , ∆S2 = nCp,2 ln Teq T2