che359_sp10_hw5_soln - due Thursday, March 4, 2010 at 2:30...

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Unformatted text preview: due Thursday, March 4, 2010 at 2:30 pm Solutions prepared by Chris Singer 1. (25 points) Proton NMR (nuclear magnetic resonance) works by looking at the state of the protons in the system when a magnetic field is applied. A proton can be either spin-down (ground state; ε = 0 ) or spin-up. The spin up state is higher in energy than the spin down state by Δε = gµB , where g = 2.79 is the value of a proportionality constant for protons, µ = 5.05 × 10−24 J/Tesla is the magnetic moment, and B = 7 Tesla is magnetic field in a 300 MHz NMR instrument. The € populations of protons occupying the two states€ follow a Boltzmann distribution. € a. (10 points) Write the probabilities p↑ and p↓ of finding the system in the € spin-up and spin-down states as a function of the partition function Q . € For spin up ε = gµB , and spin down ε = 0 . ChE 359 Homework #5 1 e − gµB / kT p↓ = p↑ = Q Q € € € € € € N↑ − N↓ b. (10 points) Calculate the relative population difference € T = 298.15 K . (N ↑ + N↓ ) at € € N i = Npi so the common factor N factors out and you get € N↑ − N↓ p↑ − p↓ 1 − e −ε / kT = = 1 + e −ε / kT N↑ + N↓ p↑ + p↓ € ε gµB (2.79)(5.05 x10 −24 J / Tesla)( 7Tesla) = = = 0.0240 kT kT (1.38 x10 −23 J / K )(298.15K ) ( )( ) € € N↑ − N↓ (N ↑ + N↓ ) = 1 − e −0.024 = 1.20 x10 −2 1 + e −0.024 € c. (5 points) Determine qualitatively how the relative population difference changes with increasing temperature. Also, determine how the relative population difference changes with decreasing temperature. (Hint: you can simplify your expression in part b. so that the temperature dependence is very clear!) Page 1 of 7 ChE 359 SP2010 - Homework #5 ε ε for small values of so the relative population difference kT kT ε / kT ε ≈ simplifies to . 2 − ε / kT 2 kT e −ε / kT ≈ 1 − € This result shows that a€the temperature increases, the relative population s difference approaches zero (the microstates are equally filled). As the temperature approaches zero the relative population difference will approach € 1 (everything is in the ground state). 2. (15 points) We derived the translational partition function for a particle in a three-dimensional cubic box of volume V . Let’s derive the heat capacity of an ideal monoatomic gas of N molecules (particles). a. (5 points) We saw that the partition function of a single particle is 3 ȹ 2πmkT ȹ 2 € qt = ȹ ȹ V . Use the partition function to derive an expression for the 2 ȹ h € Ⱥ internal energy U of the gas, and then derive an expression for the constant-volume heat capacity CV . € ∂ ln q U = kT 2 € ∂T € ȹ 3 ∂ ln q ∂ ȹ 3 ȹ 2πmkT ȹ = ȹ lnȹ ȹ + ln V ȹ = ∂T ∂T ȹ 2 ȹ h 2 Ⱥ Ⱥ 2T 3 U = kT 2 ȹ ∂U ȹ 3 CV = ȹ ȹ = k ȹ ∂T ȺV 2 b. (10 points) In our system, we have one mole of O2 molecules in the gas phase at T = 273.15 K confined in a volume V = 22.4 × 10−3 m . The molecular weight of O2 is 32 g/mol. Use the expressions derived in part a. to calculate the translational component of the internal energy [ J / mol ] and K the constant-volume heat capacity [ J / mol ⋅€ ]. € To get molar values, multiply the (molecular) results obtained in (a) by € Avogadro's number. Recall that R = kN A : € € € € € 3 U = RT = 3.4 kJ / mol 2 € ȹ ∂U ȹ 3 CV = ȹ ȹ = R = 12.5 J / mol ȹ ∂T ȺV 2 € 3. (20 points) Let’s now consider a protein in a box. The protein can be modeled as € a sphere of diameter 40 Å and is trapped in a cubic pore of a chromatography Page 2 of 7 ChE 359 SP2010 - Homework #5 column. The protein’s molecular weight is 104 g/mol. The sides of the pore are all 100 Å. Assume that the temperature T = 300 K and that even though the protein is big, the system can still be modeled using the particle-in-a-box. a. (5 points) Calculate the value of the translational partition function q t . If quantum effects are important only when q t < 10 , comment on whether the € system needs to be treated quantum mechanically or whether it can just be treated classically. € 3/2 3/2 € ȹ 2πMkT ȹ ȹ 2πmkT ȹ qt = ȹ V ȹ V = ȹ 2 ȹ ȹ h 2 Ⱥ ȹ N A h Ⱥ ȹ 2π (10 kg / mol)(1.38 x10 −23 J / K )( 300K ) ȹ 3 / 2 = ȹ (100 x10 −10 m) 3 = 9.76 x1011 23 −1 −34 2 ȹ ȹ (6.02 x10 mol )(6.626 x10 J ⋅ S ) Ⱥ € qt >> 10 , therefore the protein can be treated classically. b. (5 points) Calculate the mass [kg] of a particle that is treated quantum mechanically. What types of particles have masses of this magnitude? ȹ 2πMkT ȹ 3 / 2 ȹ 10 ȹ 2 / 3 ȹ h 2 ȹ −31 V < 10 ⇒ m < ȹ ȹ ȹ ȹ ȹ = 7.83 x10 kg / particle 2 ȹ ȹ V Ⱥ ȹ 2πkT Ⱥ ȹ N A h Ⱥ Electrons, m e = 9.109 x10 −31 kg , have a mass of this magnitude. € € € c. (5 points) Use the translational partition function to derive an expression for the Helmholtz free energy F . € ȹ 3 ȹ 2πmkT ȹ ȹ F = −kT ln qt = − kTȹ lnȹ ȹ + ln V ȹ 2 ȹ 2 €ȹ h Ⱥ Ⱥ d. (5 points) If the protein’s mass is increased by 10% (via deuterating all the hydrogens in the protein), calculate the change in Helmholtz free energy [ J / mol ]. Comment on whether the free energy increases or decreases, and why. ȹ m2 ȹ 3 / 2 qt 2 ΔF = − kT ln = − kT lnȹ ȹ = −356 J / mol qt1 ȹ m1 Ⱥ Free energy decreases due to structure being more stable in its detuerated state. € € € 4. (20 points) Polyenes are linear double-bonded polymer molecules −(C = C − C ) N − where N is the number of (C = C − C ) monomers. Let’s model the polyene chain as a box, in which π -electrons are particles that move freely in this box. In each box, each carbon atom is separated from another by the bond length d = 1.4 Å . If € box are a distance d 2 past the terminal C atoms in a chain, € both edges of the€ € € Page 3 of 7 € € ChE 359 SP2010 - Homework #5 then the length of the box is 2( N + 1) d . For instance, for the chain C = C − C = C − C = C the length of the box is 6 d . Each energy level, representing the two electrons in each bond, is occupied by two electrons (particles) with paired spins. Suppose the N lowest energy levels are occupied by electrons, so € the absorption of energy involves excitation from level N to level N + 1. € a. (10 points) If we only consider translational motion of the particles, derive an expression for the amount of energy that would be necessary to excite € the particles from level N to level N + 1. Use this expression to calculate € € this energy in [ J / mol ] from level 1 to level 2. The energy levels for the protein are: € €2 2 € n 2!2 h nh n 2h 2 ε= = = 8 mL2 8 m(2( N + 1) d ) 2 32 m( N + 1) 2 d 2 Note that the length is fixed at 2( N + 1) d . When determining the energy of a specific state only the value of n should change. € Δε = ε N +1 − ε N = N 2h 2 (N + 1) h 2 (2 N + 1) h 2 − = 2 2 2 € 32 m( N + 1) d 2 32 m( N + 1) d 2 32 m( N + 1) d 2 2 = € 2N + 1 5 2 × 4.628 x10 J / mol (N + 1) 3 × 4.628 x10 5 J / mol = 3.47 x10 5 J / mol 4 Δε 21 = € € b. (10 points) Plot the wavelength of absorbed radiation λ = € hc as a function Δε of N , where h = 6.626 × 10−34 J ⋅ s is Planck’s constant and c = 2.998 × 10 8 m s is the speed of light. For the range 1 ≤ N ≤ 4 , determine the part of the electromagnetic spectrum in which these wavelengths fall. € € € € Page 4 of 7 ChE 359 SP2010 - Homework #5 wavelength vs quantum number 1500 wavelength (nm) 1200 900 600 300 0 0 1 2 3 4 5 N 6 7 8 9 10 λ 1 = 345 nm    ultraviolet λ 2 = 323 nm    visible, blue violet λ 3 = 453 nm    visible, yellow green λ 4 = 582 nm    visible, deep red € 5. (20 points) Carbon monoxide is a linear diatomic molecule. The atomic weight of C is 12 g/mol, and the atomic weight of O is 16 g/mol. The molecule is at T = 300 K and P = 1 atm . a. (5 points) Calculate the translational entropy of 1 mol of CO, and express S your answer t in units of [ J / mol ⋅ K ]. n € When considering translational contributions to the entropy the gas molecules are indistinguishable, which means that the 1/N! correction is € € needed for the translational partition function. The Sackur Tetrode equation applies: ȹ ȹ 2πmkT ȹ 3 / 2 V ȹ ȹ q e 5 / 2 ȹ St = Nk lnȹ t = Nk lnȹ ȹ e 5 / 2 ȹ ȹ ȹ 2 ȹ N Ⱥ ȹ ȹ h Ⱥ N Ⱥ For an ideal gas, V kT = . NP € € ȹ ȹ 2πmkT ȹ 3 / 2 kT ȹ ȹ ȹ 2πmkT ȹ 3 / 2 kT ȹ 5/2 St = Nk lnȹ ȹ e ȹ = nR lnȹ ȹ e 5 / 2 ȹ ȹ ȹ 2 2 ȹ ȹ h Ⱥ P Ⱥ ȹ ȹ h Ⱥ P Ⱥ € ȹ ȹ 2πmkT ȹ 3 / 2 kT ȹ S ⇒ t = R lnȹ ȹ e 5 / 2 ȹ ȹ 2 n ȹ ȹ h Ⱥ P Ⱥ € Page 5 of 7 ChE 359 SP2010 - Homework #5 € = 150 J / mol⋅ K b. (5 points) Calculate the rotational entropy of 1 mol of CO, given that the S C − O bond length is R = 1.128 × 10−10 m , and express your answer r in n units of J / mol ⋅ K . € For the rotational component of the entropy, the gas molecules are no longer € treated as indistinguishable because they have a particular alignment in space € to the chemical bond. due € ȹ ȹ ∂ ln q N ȹ ∂ ln q ȹ Sr = kȹ ln q N + T ȹ ȹ = Nkȹ ln q + T ȹ ∂T Ⱥ ∂T Ⱥ ȹ ȹ ȹ 8π 2 IkT ȹ ȹ ȹ ȹ 8π 2 µR 2 kT ȹ ȹ ȹ Sr ∂ ln q ȹ ⇒ = Rȹ ln q + T ȹ = Rȹ lnȹ ȹ + 1ȹ = Rȹ lnȹ ȹ + 1ȹ 2 2 ȹ n ∂T Ⱥ Ⱥ Ⱥ ȹ ȹ σh Ⱥ Ⱥ ȹ ȹ σh For CO, s = 1 (as for all heteronuclear diatomic molecules), and the reduced mass, masses, m for atomic masses. € € µ= MC MO mC mO 1 = = 1.146 x10 −26 kg MC + MO mC + mO N A , where M stands for molar Sr = 47.28 J / mol⋅ K €n € c. (5 points) Calculate the reduced mass µ in [ kg ] of CO, and use this to calculate the frequency ν in [ cm-1 ] of a classical harmonic oscillator with N spring constant k = 1903.17 m . Compare this to the spectroscopically measured value of ν = 2143.26 cm-1 . € € € € The reduced mass was calculated in (b). 1 ν= 2π € k € 1 1903.17 N / m = = 6.51x1013 s−1 µ 2π 1.146 x10 −26 kg To get this result in terms of wavenumber, divide by the speed of light: € This result is similar to the measured experimental value. 6.51x1013 s−1 = 2170cm −1 3.0 x1010 cm / s € d. (5 points) Calculate the vibrational entropy of 1 mol of CO, and express S your answer v in units of [ J / mol ⋅ K ]. n The localization of the vibrations means that the vibrational degrees of freedom are distinguishable just as the rotational ones were. € € Page 6 of 7 ChE 359 SP2010 - Homework #5 ȹ ∂ ln q ȹ Sv = Nkȹ ln q + T ȹ ȹ ∂T Ⱥ e − hν / 2 kT hν q= − ln(1 − e − hν / kT ) − hν / kT ⇒ ln q = − 1−e 2 kT € € € € ∂ ln q hν hν e − hν / kT = +2 ∂T 2 kT 2 kT 1 − e − hν / kT ȹ hν e − hν / kT ȹ S − hν / kT ∴ v = Rȹ )ȹ = 2.83x10−3 J / mol⋅ K − hν / kT − ln(1 − e n ȹ kT 1 − e Ⱥ ⇒ Of all the components of the entropy, the translational degrees of freedom make up the greatest contribution followed by the rotational and then vibrational degrees of freedom. Page 7 of 7 ...
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