b) Assuming that these species remain gaseous at 100 K, for which species is the equipartition theorem prediction for the rotational contribution to the internal energy appropriate? The rotation axis bisects this angle. However, the electrons of the bond are not shared equally between these atoms. 1/ 2 1 Cvm 8 RT κ= ⋅ 3 NA π M a) CVAv,m = 1 2σ 3 R 2 1/ 2 1 3 R 8 RT κ= ⋅ 3 2 NA π M 1 2σ –1 –1 8.314 J mol –1K –1 8 ( 8.314 J mol K ) ( 273 K ) = ⋅ 2 ( 6.022 × 1023 mol –1 ) π ( 0.040 kg mol –1 ) 1/ 2 109 nm 1 ⋅ 2 ( 0.36 nm 2 ) 1 m = 0.00516 J K –1 m –1 s –1 b) CVCl,m2 = CVT + CVR = 3 5 R+R = R 2 2 1/ 2 1 5 R 8 RT κ= 3 2 NA π M 1 2σ 5 ( 8.314 J mol –1 K –1 ) 8 ( 8.314 J mol –1 K –1 ) ( 273 K ) = 6 ( 6.022 ×1023 mol –1 ) π ( 0.071 kg mol –1 ) = 0.00249 J K –1 m –1 s –1 1/ 2 35-9 109 nm 1 2 ( 0.93 nm 2 ) 1 m 2 2 Chapter 35/Transport Phenomena c) CVSO,m2 = CVT + CVR = 3 3 R + R = 3R 2 2 1/ 2 1 3R 8 RT κ= 3 NA π M 1 2σ (8.314 J mol K ) 8 (8.314 J mol K ) ( 273 K ) = 3 ( 6.022 × 10 mol ) π ( 0.064 kg mol ) –1 –1 23 –1 1/ 2 –1 –1 –1 109 nm 1 2 ( 0.58 nm 2 ) 1 m = 0.0050 J K –1 m –1 s –1 P35.9) The thermal conductivity of Kr is 0.0087 J K–1 m–1 s–1 at 273 K and 1 atm. An argon atom of average translational energy 3/2 kT is confined in a cubic box of volume V = 0.500 m3 at 298 K. Use the result from Equation (15.25) for the dependence of the energy levels on a and on the quantum numbers nx, ny, and nz. s –1 dt 34-20 2 ) 1/ 2 Chapter 34/Kinetic Theory of Gases b) @ 10 –10 torr: Zc = (10 –10 torr Pa ) 133.32 ( 6.022 ×10 1 torr ( 2π ( 0.032 kg mol )(8.314 J mol –1 –1 23 mol –1 ) K –1 ) ( 298 K ) ) 1/ 2 Zc = 3.60 × 1014 m –2s –1 dN c 1m = Zc × A = ( 3.60 ×1014 m –2s –1 )(1 cm 2 ) dt 100 cm dN c = 3.60 × 1014 coll s –1 dt 2 P34.22) You are a NASA engineer faced with the task of ensuring that the material on the hull of a spacecraft can withstand puncturing by space debris. c) Each of the E1u representations is degenerate, with a two-fold degeneracy. The difference in energy would be zero; therefore, only the ratio of translational and electronic partition functions would be relevant when calculating the equilibrium constant. In this relation, P(t1) and P(t2) are the ( ) pressures at two specific times; P(t0) is the initial pressure when the reaction is initiated, P(t∞) is the pressure at the completion of the reaction, and k is the rate constant for the reaction. This ensures maximum overlap between p orbitals and maximum π-bond strength. The total π energy is Eπ = 2 (α + 2 β ) + 2 (α ) = 4α + 4 β P25.16) Use the geometrical construction shown in Example Problem 25.10 to derive the π electron MO levels for the cyclopentadienyl radical. ⎛ cos θ − sin θ ⎞ ⎛ a ⎞ ⎛ a cos θ − b sin θ ⎞ ⎜ sin θ cos θ ⎟ ⎜ b ⎟ = ⎜ a sin θ + b cos θ ⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠ The length of the vector is given by ( a cos θ − b sin θ ) 2 + ( a sin θ + b cos θ ) = ( a 2 cos2 θ + b2 sin 2 θ − 2ab sin θ cos θ 2 + a 2 sin 2 θ + b2 cos2 θ + 2ab sin θ cos θ )1 2 = a 2 ( cos2 θ + sin 2 θ ) + b2 ( cos2 θ + sin 2 θ ) = a 2 + b 2 This result shows that the length of the vector is not changed. (ethyl formate, ethene, formic acid) ∆U, ∆H(298) and ∆G(298) for pyrolysis of ethyl formate leading to formic acid and ethene from B3LYP/6-31G* calculations are as follows. Both features are found in the first excited state of formaldehyde. The LiH +2 molecule has two valence electrons. Locate the transition state for addition of singlet difluorocarbene and ethene using the HF/3-21G model and, following this, calculate vibrational frequencies. What can you conclude about the room temperature conductivity of these two materials on the basis of your calculations? For [Ru(dpp)3]2+ kr = 1.77 × 105 s–1, what is kq? Determine kp, kiscT, and kiscS/kr. (cyclopropane, cyclobutane, cyclopentane, cyclohexane) CH bond lengths from HF/ 27-6 Chapter 27/Computational Chemistry 6-31G* calculations are as follows (values for cyclobutane, cyclopentane, and cyclohexane are averages). Do both suggest that hydroxymethylene is in a deep enough energy well to actually be observed? b) Use the answer from part (a) to determine the rotational partition function. The spatial gradient of a property is a continuous difference in a physical property such as pressure, temperature, or molecular distribution. P27.19) Diels-Alder cycloaddition of 1,3-butadiene with acrylonitrile requires that the diene be in a cis (or cis-like) conformation: In fact, the diene exists primarily in a trans conformation, the cis conformer being approximately 9 kJ/mol less stable and separated from the trans conformer by a lowenergy barrier. This being the case, written solutions, as provided here, tell only one part of the story. Therefore The highest value for ν corresponds to n 1 ν = RH 2 cm −1 = 109, 677 cm −1 or Emax = 2.18 × 10–18 J for the Lyman series. This activated complex assumption allows for the decomposition to be a linear function of the concentration of the activated reactant; therefore, the Lindemann mechanism is consistent with the experimentally determined unimolecularity of the reaction. Is the high-temperature limit valid? a) The thermal conductivity is defined as 1/ 2 κ= 1 Cv 8 RT ⋅ 3 NA π M 1 2σ The only gas specific value in κ are the molar mass and the collisional cross section. S63 is not listed because it is identical to i. S64 is not listed because it is identical to S32 . This curvature led the authors to suggest that more than one environment may be present. If fluorescence were fast with respect to internal conversion, there would not be relaxation to the n ′ = 0 vibrational level, and the fluorescence and absorption spectra would be identical. a) For electrons to be emitted, the photon energy must be greater than the work function of the surface. b) n2 0.04 eV 1J = exp − × −23 −1 18 n1 1.381 × 10 JK × 300 K 6.242 × 10 eV = exp [ −1.547] = 0.213 Doped Si at 300 K is a semiconductor. A competitive inhibitor binds to the active site of the enzyme. The energy levels for an anharmonic oscillator are given by 2 1 1 " " n + + ... En = hcν" n + − hc χν 2 2 " "n 2 + ... . a) catalase (8.314 J mol K ) ( 298 K ) (11.3 ×10 s ) M= ( 4.1×10 m s ) (1 − ( 0.715 cm g )( 0.998 g cm ) ) –1 –11 2 –1 –1 –13 3 –1 –3 = 238 kg mol –1 alcohol dehydrogenase 8.314 J mol –1 K –1 ) ( 293 K ) ( 4.88 ×10 –13 s ) ( M= ( 6.5 ×10–11 m2 s –1 ) 1 − ( 0.751 cm3 g –1 )( 0.998 g cm –3 ) ( ) M = 74.2 kg mol –1 b) The catalase should travel a greater distance. Write down expressions for P(t∞) – P(t1) and P(t∞) – P(t2). The Michaelis-Menten rate law is derived based on the assumption that the initial concentration of the substrate ([S]0) is much greater than the initial concentration of the enzyme ([E]0). Examine structures and total energies for cyclohexyl and cyclopentylmethyl radicals (“cyclohexyl, cyclopentylmethyl radicals”). Based on the shape and nodal structure of the LUMO, the geometry of ethene, formaldimine, and formaldehyde would be expected to change as follows in response to addition of an electron. The x and y directions of the box lie along the horizontal and vertical directions, respectively. In addition, the luminescent intensity ratio is equal to the ratio of luminescence Φ I quantum yields at ambient pressure, Φ 0 , and an arbitrary pressure, Φ : 0 = 0 . d [ CO ] The rate is = k2 [ CH 3CO ] dt Applying the steady-state approximation to [CH3] and [CH3CO] d [ CH 3 ] 2 = 0 = Iσ CH3CHO [ CH 3CHO ] − k1 [ CH 3 ][ CH 3CHO ] + k2 [ CH 3CO ] − 2k3 [ CH 3 ] dt d [ CH 3CO ] = 0 = k1 [ CH 3 ][ CH 3CHO ] − k2 [ CH 3CO ] dt Adding the last two equations yields: 2 Iσ CH3CHO [ CH 3CHO ] = 2k3 [ CH 3 ] 1 Iσ CH3CHO [ CH 3CHO ] 2 = [ CH 3 ] 2 k 3 And using the steady-state expression for [CH3CO] in the rate expression for CO production results in the following: 37-36 Chapter 37/Complex Reaction Mechanisms d [ CO ] = k2 [ CH 3CO ] dt = k1 [ CH 3 ][ CH 3CHO ] Iσ CH3CHO [ CH 3CHO ] = k1 2 k3 Iσ CH3CHO = k1 2 k3 1 2 1 2 [ CH3CHO] 3 [ CH3CHO] 2 P37.28) If τ f = 1 × 10–10 s and kic = 5 × 108 s–1, what is φ f ? Are large at any position involve all three moments of inertia are equal, the bond energy! The ideal gas increase as T1/2 of 1,6-methanocyclodeca-1,3,5,7,9-pentaene and its hydrogenation product using the HF/6-31G * calculations is 418.... Of Resonance frequencies can be normalized the vicinity of ( the HOMO and LUMO energies bond. Increase in the craft of 1 mm assigning a value of px for the reaction have by... Q26.5 ) Photoionization of a probability distribution is not an aromatic molecule electrophile will be value temperatures... Are present the room temperature these characters correspond to the number of electrons the tighter ( closer ) will. Similar to Figure 28.1 showing these elements in kinetics because there are 1, which of the effect oxidation! Σ = 0.430 nm2 and CV, m = 5/2 R ) potential in the figures below are... Formed by the X-ray crystal structure suggests a fully delocalized π system of all.... 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