Physical Chemistry Test 2 Fall 2011, Prof. Shattuck Name 1 Constants: R = 8.3145 J K -1 mol -1 = 0.083145 L bar K -1 mol -1 R = 0.08206 L atm K -1 mol -1 1 F = 96485 C mol -1 1 atm = 1.01325 bar 1 bar = 1x10 5 Pa Part I. Answer four (4) of the following six questions. If you answer more than 4, cross out the problems you wish not to be graded. 8 points each. 1. Give a statement of the Second Law of Thermodynamics. 2. A solution is prepared by adding 0.100 mol of pure acetic acid to 1.00 L of water. The solution is neutralized by adding solid NaOH. List the chemical constituents and give the number of thermodynamic components. (You don t need to list any reactions or chemical constraints) 3. Of sulfur dioxide and carbon dioxide, which molecule is predicted to have the greater constant pressure heat capacity? Give estimates for the Cp values for both molecules (neglecting vibration). 4. When p-nitroanisole and pyridine are photolyzed in aqueous solution the reaction is: NO 2 N hν N + + + NO 2 - H 2 O O CH3 O CH3 The quantum yield for a solution containing 1.00x10-5 M p-nitroanisole and 0.0100 M pyridine in 1% acetonitrile is 4.65x10-3 at 366 nm. The molar absorption coefficient at 366 nm of p- nitroanisole is 1990 M -1 cm -1. Calculate the photochemical rate constant for an optically thin solution assuming the incident flux is 6.70x10-6 mol L -1 s -1 for a 10.00 cm path length.
5. Calculate the standard internal energy of formation at 298.2 K of liquid methyl acetate, C3H6O2, from its standard enthalpy of formation, which is -442.0 kj mol -1 at 298.2 K. 2 6. For a closed system and PV work only, describe why the second, lost-work term in the entropy production always gives the change in entropy production as greater than zero for an isothermal process. dstot dt = 1 T 1 đq Tsurr dt + P T Pext T dv dt
Part II. Answer two (2) of the following four questions. If you answer more than 2, cross out the problems you wish not to be graded. 16 points each. 7. Use a cyclic process in a closed system to argue that the internal energy is conserved. 3
8. Consider the transfer of heat between two blocks of metal in an isolated system. Block A is slightly hotter than block B so the transfer of heat is reversible. Assume that the volume change of the blocks is negligible so dva = 0 and dvb = 0, giving: 4 dutot = dua + dub = UA SA V dsa + UB SB V dsb = 0 (isolated, cst. VA&VB) If the entropy, S, shows the direction for energy dispersal, then the sum of the overall change in S should be positive for the spontaneous heat transfer: dsa + dsb > 0 (energy dispersal) Prove that UA > SA V UB. A summary of the relationships is diagrammed below. SB V hotter colder đq A,rev= đq B,rev dutot = dua + dub = 0 (isolated) dutot = UA SA V dsa + UB SB V dsb = 0 (isolated, cst. VA&VB) dstot = dsa + dsb > 0 (isolated, spontaneous)
9. Consider a spontaneous adiabatic expansion of an ideal gas. (a). Is the entropy change of the system zero, greater than zero, or less than zero? (b). Is the entropy change of the surroundings zero, greater than zero, or less than zero? (c). If the entropy change of this adiabatic process for the system is not zero, why isn t the entropy change zero? (Don t just give an equation or two; explain the reasoning behind your answers. Specifically address the issue of reversible heat transfer versus the actual heat transfer.) 5 10. Calculate q, w, U, H, and S for a reversible adiabatic expansion of 1.00 mol of a monatomic ideal gas. The gas is initially at 5.00 bar and 298.2 K and expands to a final pressure of 1.00 bar.
Part III. Answer two (2) of the following three questions. If you answer more than 2, cross out the problem you wish not to be graded. 18 points each. 11. Yeasts convert glucose to ethanol. (a). Calculate the change in enthalpy if one mole of glucose is converted to ethanol at 298.2 K: C6H12O6 (s) 2 CH3CH2OH (l) + 2 CO2 (g) (b). Thermophilic organisms live at 80.0 C. Calculate the change in enthalpy for the reaction at 80.0 C. Substance fh (kj mol -1 ) C p (J K -1 mol -1 ) glucose -1274. 209. ethanol -277.69 111.5 CO 2-393.51 37.11 6 12. The Joule Thompson coefficient is given by µjt = 1 CP H P T. For CO2, µjt = 1.11 K bar -1 and CP for CO2 is 37.11 J K -1 mol -1. Calculate the change in enthalpy per mole for an isothermal process for a change in pressure of 1 bar. Assume that both µjt and CP are constant over the pressure range.
7 13. Show that H T = CP + H V P T α κτ [Show all your work with reasons for each step. For example: since H is a state function, from the definition of heat capacity, or dividing by dp gives the desired result are typical statements. If you use a partial derivative relationship that has a name, if you choose not to derive the relationship, just state the corresponding relationship name. (e.g. from the chain rule or from the Euler chain rule )]