Chem/Biochem 471 Exam 3 12/19/07 Page 1 of 6 Name:____________________________ (Please print clearly on all pages) KEY 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 all 5 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 20 points. The total exam grade will be normalized so that the maximum number of course points for this exam will be 20. For example, getting 80 points on 4 questions equals 100 points on 5 questions equals 20 points toward the final grade. Getting 80 points on 5 questions would be worth 80% 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 “DO NOT GRADE THIS PAGE ” 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 at least 4 problems (of your choosing) or all 5, as you prefer. Answers without explanations (where indicated) are not complete. For grading purposes: question 1 2 3 4 5 Tot scoreChem/Biochem 471 Exam 3 12/19/07 Page 2 of 6 Name:____________________________ (Please print clearly on all pages) KEY 1. In the technique of nuclear magnetic resonance spectroscopy, the nucleus of a hydrogen atom (a proton) is subjected to a strong external magnetic field. This field causes the magnetic moment of the nucleus to align parallel or antiparallel to the magnetic field. The energy of the nucleus with its magnetic moment aligned in the direction of the applied field decreases by -μB while the energy if the magnetic moment aligns antiparallel to the field increases by +μB. Here μ is the magnetic moment and B is the magnetic field strength. a) In the following incomplete diagram, give the value of the high and low energy levels in the presence of the magnetic field (middle) and circle the direction of the nuclear magnetic moment corresponding to each energy level (right). (The energy answers are formulas, not numbers.) b) When a magnetic field B is applied, what is the probability that the nuclear magnetic moment points in the same direction as the magnetic field? (The answer is a formula, not a number.) c) When the temperature of the nucleus is very high (approaching infinity), what is probability that the nucleus will be in the high energy state above? (The answer is a number.) d) When the temperature of the nucleus is 100 K, you find that 49% of the nuclei are in the high energy state (and thus 51% are in the low energy state). What is the value of the energy gap Ehigh- Elow? DO NOT GRADE THIS PAGE  Eground energy Ehigh = Elow = Magnetic moment or or B = 0 B > 0, pointing up ↑ {{Eground + μB Eground - μB This is a situation in which energies of the possible states determine the probability of being in the state according to the Boltzmann distribution. For the low energy state, ()() ()TkBTkBTkBTkBETkBETkBEMiTkEiTkElowlowBBBBgroundBgroundBgroundBiBloweeeeeeegegPμμμμμμ−+++−−−−−=−−+=+==∑1. The probability of the high energy state is TkBTkBTkBhighBBBeeePμμμ−+−+=. As T → ∞, the exponential terms approach 1, so the probability approaches 0.5. We know that TkBTkBTkEhighBBBhigheeePμμ−+−+= and TkBTkBTkElowBBBloweeePμμ−+−+= so ()TkEETkETkEhighlowBlowhighBhighBloweeePP−+−−== . Plugging in the values we have, ()()(){}KKJEElowhighe1001038.12349.051.0−×−+=, or ()(){}JKKJEElowhigh23231052.549.051.0ln1001038.1−−×=×=−. You can also calculate the value of μB and note that the energy gap is twice that value.Chem/Biochem 471 Exam 3 12/19/07 Page 3 of 6 Name:____________________________ (Please print clearly on all pages) KEY OCl- + H2O HOCl + OH-, fast I- + HOCl → HOI + Cl-, slow HOl + OH- H2O + OI-, fast K3 K1 k2 2. The initial velocity of the reaction I- + OCl- → OI- + Cl- in basic solution was studied at different solution concentrations as shown at right. a) What is the order of this reaction for [I-]? b) Write down the rate law for this reaction. (This is a formula.) c) In the formula of part b), what is the value of the rate constant? (This is a number.) d) Shown at right is a set of elementary reactions that make up the proposed mechanism for this reaction. What is the value of the rate constant in part b) in terms of the equilibrium and rate constants in the elementary reactions? (This is a formula.) DO NOT GRADE THIS PAGE  [I-]0 (mM) [OCl-]0 (mM) [OH-]0 (M) v0 (mM/s of chloride)2 1.5 1 0.18 4 1.5 1 0.36 2 3 2 0.18 4 3 1 0.72 The first and second lines of the table vary [I-]0, holding other concentrations fixed. When [I-]0 doubles, v0 doubles. Thus, the velocity depends on [I-]0 to the first power. The reaction is first order in [I-]0. Doing the same thing for [OCl-]0 and [OH-]0, we find that the reaction is first order in [OCl-]0 and negative one order in [OH-]0. Thus, the rate law is ][]][[0−−−=OHOClIkv. Given that the rate law is ][]][[0−−−=OHOClIkv, we can pick any line of data in the table to get k. Using the first line, ()()MmMmMkmM15.12sec18.0 = or ()( )sec160sec106.05.121sec18.0 ===mMMmMmMMmMk For the top reaction we have an equilibrium, so ][]][[1−−=OClHOClOHK and for the middle reaction we have ]][[20HOClIkv−=. (In the top equilibrium reaction, water is solvent, so its activity is 1.) Rearranging the equilibrium relationship, we have that ][][][1−−=OHOClKHOCl. Substituting this into the second velocity equation leaves us with ][]][[120−−−=OHOClIKkv. Comparing to the answer in b), we see that k = k2K1.Chem/Biochem 471 Exam 3 12/19/07 Page 4 of 6 Name:____________________________ (Please print clearly on all pages) KEY 3. Chlorophyll absorbs photons strongly in two bands, one between 340 and 450 nm and the other between 650 and 680 nm. Chlorophyll collects and transfers the photon energy to fix CO2 into carbohydrate CH2O units according to CO2 + H2O (CH2O) + O2 where the (CH2O) indicates one carbon unit of the