Chemistry 471 Final exam 12/18/06 Page 1 of 6 Name:

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1 Chemistry 47 Final exam /8/6 Page of 6 Please leave the exam pages stapled together. The formulas are on a separate sheet. This exam has 5 questions. You must answer at least 4 of the questions. You may answer more questions if you wish. Answering 5 questions can be an advantage if you are unsure of some of your answers (this will distribute the risk ). Answering 4 questions is advantageous if you are very sure of your answers. Each page is worth points. The total exam grade will be normalized so that the maximum number of course points for this exam will be 5. For example, getting 8 points on 4 questions equals points on 5 questions equals 5 points toward the final grade. Getting 8 points on 5 questions would be worth 8% of the maximum grade. If you leave a page blank, it will not be included in the grading. If you work on a page and then decide that you do not want it to be graded, be sure to mark the box at the bottom of the page. If you work on the page and fail to mark the box, the page will be graded. Work either 4 or 5 problems, as you prefer. For grading purposes: question Tot score

2 Chemistry 47 Final exam /8/6 Page of 6. Bacteriophage T7 has a standard sedimentation coefficient of s,w = 453 S, a diffusion constant of D,w = 6-8 cm /sec, and partial specific volume v =.64 cm 3 /g. (Note: S = -3 sec, the density of water at C is.998 g/cm 3 and the viscosity of water at C is. -3 N sec/m.) a) Calculate the weight per mole of the bacteriophage from these data at T = C. RTs The formula relating molar mass and sedimentation coefficient is M = D We know all the values on the right, so plugging them into the formula gives M = (8.345 J/K.mol)(93.5 K)(453-3 sec)/[(6 - m /sec){ (.64 cm 3 /g)(.998 g/cm 3 )}] = kg/mol = g/mol ( vρ ). b) What is the friction coefficient f for bacteriophage T7 in water at C? The formula relating sedimentation coefficient and f is ( vρ ) s f rearranged to ( vρ ) m m = which can be f =. We have the mass per bacteriophage T7 from part a): m = s M/N A = (5.9 4 kg/mol)/(6. 3 /mol) = kg. Plugging in this plus the values we already know gives f = { (.64 cm 3 /g)(.998 g/cm 3 )}( kg)/ (453-3 sec) = kg/sec. c) If bacteriophage T7 can be assumed to be a spherical particle, what is its radius? The formula relating f and r is f = 6πηr. Thus, r = f/6πη = ( kg/sec)/{6π (. -3 N sec/m )} = m = 34 Å. d) Suppose bacteriophage T7 was centrifuged for 3 minutes at 3, rpm. It moves a distance d in the centrifuge tube. If a different bacteriophage with identical partial specific volume (and density) but of half the mass of bacteriophage T7, would the new bacteriophage move more, less or the same distance compared to bacteriophage T7? (You may assume that both bacteriophages are spherical in shape.) Explain. The new bacteriophage would move a lesser distance as compared to the bacteriophage T7. This is because the sedimentation constant is directly proportional to the mass of the particle that is sedimenting. It is also proportional to D, but D depends on particle size according to D = k B T/6πηr, and r is smaller by a factor (M new /M T7 ) /3 because volume is proportional to M for both (they have the same density). Thus, s will decrease according to (M new /M T7 ) /3. NOTE: One could argue that this occurs in a very small centrifuge tube, so that the centrifugation is at equilibrium. Although it seems that both proteins would be at the bottom of the tube in this case, the heavier particle will be, on average, slightly closer to the tube bottom due to its lower diffusion constant.

3 Chemistry 47 Final exam /8/6 Page 3 of 6 Initial [A] Initial [B] Initial rate. The stoichiometric equation for a reaction is A + B C + D... (M sec - ) The initial rate of formation of C follows the values shown in the table. a) What is the order of the reaction with respect to A? From the first and second lines, it must be second order in A because as [A] doubles, the velocity goes up by a factor of 4. b) What is the order of the reaction with respect to B? From the first and third lines, it must be zero order in B because as [B] doubles, there is no effect on velocity. c) In the following differential equation, indicate the values shown by the letters a, b and c: From the first and third lines, it must be zero a = order in B because as [B] doubles, there is no d[ C] a b c = k[ A] [ B] [ C] b = effect on velocity. There is no reverse reaction dt here (no back arrow, so this reaction is zero c = order in C and D. The powers of the concentrations are just the orders for each. d) Calculate the value of the rate constant k in part c. This is a second-order reaction. Thus, v = k[a]. From the first line we have the values of v and [A], so just plug them in: (. -3 M sec - ) = k( M), so k = (. -3 M sec - )/ ( M) =. -3 M - sec -. The same value can be obtained using any row of data in the table. e) How long would it take for the concentration of C to reach.5 M for the three initial conditions shown in the table above? (You may assume that there is no back reaction.) Initial [A] Initial [B] Time to.5 M (sec) For a second order reaction, concentration and time are related by k [ A] [ ] [ kt A ] + Similar values can be calculated for other lines. Note that the first and third lines can use the formula for half life, t / =, because.5 M is half of. M. k[ A ] [ A] =, which can be rewritten as t =. For the first line of data, = sec A t =. 3.5M M M sec

4 Chemistry 47 Final exam /8/6 Page 4 of 6 3. The I H stretching vibration of HI can be treated as a quantum mechanical harmonic oscillator. The potential energy of stretching for a harmonic oscillator is U = ½kx where k is the force constant for the bond and x is the deviation of the bond length from its equilibrium value. a) The mass of a hydrogen atom is.67-7 kg and that of an iodine atom is. -5 kg. Thus, the iodine atom can be considered to be immobile while the hydrogen vibrates in space. Write the Schrödinger equation for the stretching of the I H bond. Indicate what mass is included in the equation. Because the mass of the hydrogen is much smaller than that of the iodine, the mass included in the Schrödinger equation is only that of the hydrogen. (Basically, /m H is much larger than /m I.) The general Schrödinger equation is h. Plugging in the value for U from above Ψ + UΨ = EΨ 8π i mi and noting that this is a -D oscillator, we have that h d Ψ + kx Ψ = EΨ for this case. 8π m dx b) The solutions to the Schrödinger equation in part a) have energy level values equal to E n = n + hν where n is an integer (,,, ) and ν k =, k being the force π m constant and m the mass from part a). Measurement of the HI bond vibration spectral absorption shows a series of absorption lines. The longest wavelength absorption in the series is 4.8 μm. What is the value of ν? The longest wavelength absorption is the lowest energy absorption, which corresponds to the smallest energy gap. Note that all energy levels are equally spaced. Thus the gap between any two adjacent levels equals this smallest energy gap. So, choosing n = and n = as the two adjacent energy levels, the energy gap is 3 Δ = =. This equals the energy of the absorbed photon, which is E = hν = hc/λ = E E E hν hν = hν ( J sec)(3 8 m/sec)/(4.8-6 m) = J. Thus, ν = ΔE/h = ( J)/( J sec) = sec -. c) What is the wavelength of the next shortest absorption line in the HI vibration spectrum? The next largest energy gap is between energy levels where the intervening level is skipped. Since energy levels are evenly spaced, this will gap will exist between levels n and n+. Let s pick and : 5. The value is twice that from part b), or J. This gap will equal Δ E = E = E = hν hν hν the energy of the absorbed photon, which is E = hν = hc/λ, thus λ = hc/δe= ( J sec)(3 8 m/sec)/(9.7 - J) =.5-6 m =.5 μm. It is shorter than the longest wavelength absorption by a factor of ½. d) If the absorption spectrum of DI was measured, how would your answers to b) and c) change? If a value changes, state by what amount (relative or absolute) the value changes. (Note: D is deuterium. Its mass is kg.) The energy levels of the harmonic oscillator are proportional to ν, which itself is proportional to. m Thus, if m doubles, ν decreases by a factor of to sec -. The wavelength of the absorption in part c) is proportional to /E or /ν, so the wavelength will increase by a factor of to.9 μm.

5 Chemistry 47 Final exam /8/6 Page 5 of 6 4. The molar absorptivity of benzene equals M - cm - at 6 nm. (You may assume that the solvent in which benzene is diluted has no absorbance itself and has no effect on benzene's molar absorptivity.) a) What concentration of benzene in a cm path length cell would give an absorbance of.4 for 6 nm radiation? Absorbance, molar absorptivity, concentration and path length are related by A = εcl, so c = A/ εl = (.4)/( M - cm - )( cm) =.4 - M = 4 mm. b) If two cells as in part a) were placed back-to-back and 6 nm radiation of intensity I = intensity units was passed through both of them, what would be the intensity of the 6 nm radiation exiting from the second cell? This can be done two ways. From εcl I = I ( cm )(.4 M )(cm) (.8) ( i. u. ) = ( i. u. ) the calculation is direct: M I = =.58. The second way is to note that absorbance is additive, so twice the benzene will double A to.8. From A = -log(t) the value of T can be gotten as T = Since T = I/I, it is easy to calculate I =.58. c) A solution of benzene of the same concentration as in part a) is made. Some ATP is added to this solution (you may assume that the volume of the solution does not change). The absorbance of this new ATP-containing solution in a cm path length cell is measured to be.9 at 6 nm. The molar absorptivity of ATP at 6 nm is.6 μm - cm -. What is the concentration of ATP in the solution? Absorbance is additive. The total absorbance is that due to benzene (equal to.4) and that due to ATP. Thus, A ATP =.5. Now the problem is just like part a): c = A/ εl = (.5)/(.6 μm - cm - )( cm) = 3.3 μm.

6 Chemistry 47 Final exam /8/6 Page 6 of 6 5. A collection of 3 electrons is exposed to a magnetic field of magnitude B =.3 T. (A Tesla, T, is the SI unit of magnetic field strength.) The magnetic moment of an electron is J/T. a) Suppose all three electrons are arranged so that their magnetic moments are oriented all in the same direction, all pointing in the same direction as the applied magnetic field. For convenience let s set the energy reference so that we say this arrangement has zero energy. What would be the energy of the arrangement when one of the electron spins is flipped to point in the opposite direction as the applied magnetic field? The difference in energy is equal to E d E u for one spin where u signifies the direction of the magnetic field and d is the opposite. Since E u = -μb = -(9.8-4 J/T)(.3 T) = J and E d = -μb = -( J/T)(.3 T) =.78-4 J, the energy difference is = J. This is the value of the second highest energy level (if we set the lowest level at zero). b) Let s call E the amount of energy calculated in part a). So the lowest energy state has energy equal to and the next state has energy equal to E. There are two more energy states, one with two spins down and one up and one with all three spins down. Show the spin directions and energies on the following diagram. Energy = 3E Each flip of a spin will add.78-4 J to the energy of the system. Thus, if spin down has energy E, then two spins down has energy E and three spins down has energy 3E. Energy = E Energy = E Energy = c) Degeneracy is the number of different ways that a state can exist. What is the degeneracy g i of each of these energy levels? g 3E = There is only one way to arrange the 3 spins for the energy = and g E = 3 energy = 3E states (uuu and ddd, respectively). The energy = E state g E = 3 can come about three different ways (uud, udu, duu). Likewise, the energy = E state can come about three ways (udd, dud, ddu). g = d) What is the probability that the energy = E state will be occupied by this collection of 3 electrons if T = K? E k T The probability for each of the states follows the Boltzmann distribution: gie i B Pi = The g i values for each energy level are known from part c) and the N E j k g je energies from part b). The denominator in the expression evaluates to j= ()exp(-/k B T) + (3)exp{-( J)/[(.38-3 J/K)( K)]} + (3)exp{-(. -3 J)/[(.38-3 J/K)( K)]} + ()exp{-(.67-3 J)/[(.38-3 J/K)( K)]} = = The probability for the energy = E state is.88/7.54 =.38. (For the other 3 states, the probabilities are P =.33, P E =.367 and P 3E =.8. BT.