CHEMISTRY Spring, 2018 QUIZ 1 February 16, NAME: HANS ZIMMER Score /20

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1 QUIZ 1 February 16, 018 NAME: HANS ZIMMER Score /0 [Numbers without decimal points are to be considered infinitely precise. Show reasonable significant figures and proper units. In particular, use generally accepted units for various quantities.] 1. (4 points) Without doing any calculations, for a pure gas that obeys the Boltzmann distribution, put the following parameters in order of increasing magnitude: <v>, vmp, and vrms. [Indicate the relative order with the less than sign (<).] vmp < <v> < vrms. (6 points) In pure water at 5 C, on average how far does a water molecule move (in three dimensions) in one millisecond by diffusion? From the Handbook, one may find that, in three dimensions, < rr > < xx > + < yy > + < zz > 6DDDD From this equation one obtains, from the data in the Handbook, the result: rr rrrrrr 6DDDD 6 ( mm ss 1 )( ss) mm cccc 3. (10 points) The sedimentation coefficient of lysozyme is determined by centrifugation at 55,000 rpm in water at 0 C. The boundary was monitored as a function of time of centrifugation. Determine the sedimentation coefficient of lysozyme in water at 0 C. [A plot may help.] Time (min) x b (cm) x b/x(0) ln(x b/x(0)) One starts by creating the proper function for a plot. The information is in the slope of the plot of ln (xb/x(0)) versus time. Slope min sec -1. This slope is ω s. So division of this by the square of the angular frequency gives the sedimentation coefficient. SS ssssssssss ssssss 1 ωω (ππ ssssss 1 ) sec 1.89 SSSS But, of course, one has to express this in the proper unit, the svedberg (10-13 sec), not in units of sec.

2 QUIZ February 3, 018 NAME: JOHNNY GREENWOOD Score /0 [Numbers without decimal points are to be considered infinitely precise. Show reasonable significant figures and proper units. In particular, use generally accepted units for various quantities.] 1. (6 points) For some unusual reactions, the rate law is of zero order in the reactant. (a) Write the rate equation for this situation. (b) Derive an equation for the time dependence of the reactant concentration, C R(t), as a function of time, assuming that the concentration at time t 0 is C R(0) and the rate constant is k 0. [Note: the result of (b) cannot be correct if the result of (a) is incorrect.] (a) The rate equation is: ddcc RR 0 kk dddd 0 CC RR [ kk 0 ] (b) Integrating this equation from t 0 to t t gives the result one seeks for the time-dependent concentration of the reactant: CC RR (tt) CC RR (0) kk 0 tt. (4 points) From data in the Handbook, estimate the second-order rate constant for the destruction of ozone (O 3) by nitric oxide (NO) at room temperature (98.15 K). The Arrhenius parameters for this reaction are found in Table 10. on page 10-. Using Arrhenius s equation, kk(tt) AA eeeeee EE RRRR kk(98.15kk) cccc 3 mmmmmm 1 ss JJ mmmmmm 1 eeeeee JJ mmmmmm 1 KK 1 (98.15KK) cccc 3 mmmmmm 1 ss 1 3. (8 points) One mechanism of ozone destruction involves reaction with HO radicals to create oxygen and a hydroxyl radical, by this gas-phase reaction HHOO (gg) + OO 33 (gg) OOOO (gg) + OO (gg). The initial rate of this reaction has been determined in several experiments, as shown in the table. (a) Give the initial orders of reaction with respect to the two reactants, and explain how you got those values. (b) What is the rate constant for this reaction under the conditions quoted? [HO ] (molecules cm -3 ) [O 3] (molecules cm -3 ) Rate (molecules cm -3 s -1 ) (a) One finds the ratios of rates compared to ratios of reactant concentrations, where n is the initial RR order by the ratio method: AA CC AA nn Using the first two rows of the table, one finds the order RR BB CC BB with respect to ozone: mmmmmmmmmmmmmmeess cccc 3 ss mmmmmmmmmmmmmmmmmm cccc 3 ss mmmmmmmmmmmmmmmmmm cccc mmmmmmmmmmmmmmmmmm cccc 3 nn, which gives nn. The order with respect to ozone is 1. A similar determination using the first and third lines (in which the concentration of HHHH is changed) gives the order with respect to it, which is 1, as well. Thus, the initial rate law looks like this: dd[oo 3] kk [OO dddd 3 ] [HHHH ] ) [OO 3 ][HHHH ] (b) From any of the results, on may find kk ( dd[oo 3 ] dddd. Substitution of the first row, for example, gives kk mmmmmmmmmmmmmmmmmm cccc 3 ss 1 ( mmmmmmmmmmmmmmmmmm cccc 3 )( mmmmmmmmmmmmmmmmmm cccc 3 ) cccc 3 mmmmmmmmmmmmmmmm ss 1

3 QUIZ 3 March, 018 NAME: ALEXANDRE DESPLAT Score /0 [Numbers without decimal points are to be considered infinitely precise. Show reasonable significant figures and proper units. In particular, use generally accepted units for various quantities.] 1. (10 points) The reaction NNNN + OO NNOO is second order with a temperature-dependent rate constant, k: T (K) k (cm 3 molecule -1 s -1 ) / Using these data only, determine H for this reaction over this range. [An appropriate plot makes the analysis easy.] -4.8 The appropriate equation to plot is the following: ln(k/t) /T ln kk TT ln kk bb SS hccθθ + RR HH 1-43 R² RR TT -43. The slope of this plot is related to the enthalpy of activation HH KK ( JJ KK 1 mmmmmmmm 1 ) 6.16 kkkk mmmmmmmm ln(k/t) [Note that the question specifically asked for the enthalpy of activation. It did NOT ask for the activation energy. The two parameters are not the same.] /T (1/K). (4 points) The decay of 60 Co to form 60 Ni by gamma ray emission has been used by physical chemists, such as former Delaware professor Conrad Trumbore, as a gamma-ray source in the study of radiation damage in biological systems. The half-life for the first-order decay of 60 Co is days. What is the first-order rate constant for this process? By rearrangement, one finds the relation: kk llll tt 1/ llll dddddd dddddd 1 This can be expressed in other time units as h min s (6 points) Consider a bimolecular solution-phase reaction in water, for which the diffusion coefficient at K is m s -1. [You may assume that all the reactants diffuse similarly to the water.] Assuming the reacting molecules are roughly 1.0 Angstrφm unit in diameter, estimate the maximum rate constant for the reaction under these conditions. The question is answered by calculating the maximum rate constant for diffusion control: kk DD 4ππNN 0 (rr AA + rr BB )(DD AA + DD BB ) 4ππ( mmmmmmmm 1 )( mm)( mm ss 1 ) mm dddd 3 mmmmmmmm ss mmmmmmmm ss

4 QUIZ 4 March 16, 018 NAME: CARTER BURWELL Score /0 [Numbers without decimal points are to be considered infinitely precise. Show reasonable significant figures and proper units. In particular, use generally accepted units for various quantities.] 1. (8 points) A strong absorption line of the Hg atom has a wavelength of nm. By what energy (in J/mol) are the two states involved in the transition separated? We use this version of Planck s relation: h cc EE ( JJ ss)( mm ss 1 1 ) λ mm JJ/mmmmmmmmmmmmmmmm But this is the energy per molecule. We must multiply this by the number of molecules in a mol to determine what is asked. mmmmmmmmmmmmmmmm 3 EE mmmmmm NN 0 EE mmmmmmmmmmmmmmmm JJ mmmmmm mmmmmmmmmmmmmmmm JJ mmmmmm. (6 points) For the following operators, determine the action on the operand: (Symbols other than x, y, and θ are to be considered constants.) Operator Operand Result dd AA ssssss nnnn ddxx xx ππ nn AA ssssss nnnn xx yy + dd dddd AA eeeeee(iiiiii) (yy + iiii) AA eeeeee(iiiiii) θθ + dd dddd AA eeeeee( θθ /) 0 3. (6 points) In quantum mechanics, the square of the wave function must represent the probability of finding the particle in the region around the point, x. To represent the probability, the wave function must be normalized. Normalize the wave function, Ψ (xx) AA ssssss ππππ over the region 00 xx. To normalize the wave function, the integral of its square over the range must be 1. Ψ (xx)ψ(xx)dddd Or AA ssiiii ππππ AA ssssss ππππ dddd AA ssssss ππππ dddd 0 0 Making the substitution at the limits gives 0 1. AA xx 0 AA 0 4ππ ssssss(ππ) + 4ππ ssssss(0) AA 1. The requirement for normalization is AA is real and positive. 4ππ ssssss ππππ 0 1. or AA, where the choice is made that the constant AA

5 QUIZ 5 March 3, 018 NAME: MARY J. BLIGE Score /0 [Numbers without decimal points are to be considered infinitely precise. Show reasonable significant figures and proper units. In particular, use generally accepted units for various quantities.] 1. (7 points) Evaluate the simplest form of the commutator of the two complex operators. (You are to assume that xx and yy are independent variables.) [HINT: Be sure to give a formula for the commutator s simplest form.] yy, xx yy xx To begin, apply this to a function, ff, of xx and yy: yy, xx yy xx xx yy Then, taking the derivative of the product in each term: yy, xx ff yy + yyyy ff xx xxxx ff The second and fourth terms cancel because derivatives with respect to independent variables are commutative, as is functional multiplication. Thus, the commutator s action comes to yy, xx ff yy xx yy xx ff Hence, the commutator can be simplified to the form: yy, xx yy xx. (8 points) A potassium surface has a work function of.40 ev. In a photoemission experiment the surface is exposed to radiation at a wave length of 35 nm. During this experiment, joules are absorbed. How many electrons are emitted in the experiment? JJ mmmmmm Let s convert everything to joules:.40 eeee 1 eeee JJ/eeeeeeeeeeeeeeee eeeeeeeeeeeeeeeeee/mmmmmm jjjjjjjjjj ss mm ss mm Similarly, the light photon is EE pphoooooooo hcc JJ λ The light photon s energy is greater than the work function, so photoemission will occur. Assuming all of the energy went into causing emission and none into kinetic energy of the electrons, one finds the maximum number of emissions by dividing the total absorbed energy by the work function: JJ NN mmmmmm eeeeeeeeeeeeeeeeee JJ/eeeeeeeeeeeeeeee Of course, the number could be less than that, depending on the amount of energy that went into the kinetic energy of the electrons. Assuming all of the electrons were at the same level (and hence had exactly the same energy and same kinetic energy upon leaving the surface), one finds NN JJ JJ/eeeeeeeeeeeeeeee eeeeeeeeeeeeeeeeee 3. (5 points) A particle is attached to a solid surface by a spring with an asymmetric spring constant. The result is a potential energy of the form: kk VV(xx) xx + AAxx 33 where xx rr rr eeee, and kk and AA are constants that describe the strength of the interaction. If the particle has a mass mm, write out a complete, correct expression for the Hamiltonian operator. hh HH TT + VV dd ddxx + kk xx + AAxx 33

6 QUIZ 6 April 6, 018 NAME: SUFJAN STEVENS Score /0 1. (6 points) A coin having a mass of grams is suspended by a rubber band. When it is stretched and released, the system vibrates with a frequency of 7.83 Hz. What is the force constant of the rubber band, in J/m? [HINT: Assume that the rubber band contributes essentially no mass to the system.] The relationship among the various parameters is given by the formula: ν 0 1 ππ kk mm, where k is the force constant. Rearranging this equation, kk 4ππ ν 0 mm. Substitution into this equation for this system gives kk 44ππ ( HHHH) ( gg) gg ss. 33 kkkk ss. 33 JJ mm. (7 points) Assuming that the molar mass of a proton is exactly 1 g/mol and that the molar mass of the major isotope of fluorine is exactly 19 g/mol, calculate the position of the center of mass of a HF molecule (relative to the F atom) if the bond length is pm. mm The position of the center of mass is given (for two particles) as: RR CCCC AA mm RR mm AA + mm AA + BB RR BB mm AA + mm BB. One can BB refer this to the position of one of these by making the position of that reference point the zero of the co-ordinates (and the second becomes the position of the second particle relative to the first). We choose the position of the fluorine atom to be zero. Hence, mm HH 1 gg/mmmmmm RR CCCC RR mm HH + mm HHHH FF 1 gg (91.68 pppp) pppp + 19 gg/mmmmmm mmmmmm (The masses of the isotopes are, of course, not exactly these values in integer grams, so, in the real world, the position will be somewhat different.) 3. (7 points) For a one-dimensional harmonic oscillator, the lowest-energy state is described by the wave function 1 xx Ψ(xx) 4 eeeeee ππ αα αα where α is a constant and x is the position relative to equilibrium position. Determine the expectation value of the square of the position. The expectation value of the square of the position is given by < xx > Ψ xx Ψ dddd 0 αα ππ αα ππ xx eeeeee αα xx eeeeee xx αα dddd yy eeeeee( yy αα + )dddd ππ yy eeeeee( yy )dddd 0 1 αα ππ αα ππ + xx eeeeee xx αα dddd 1 4 ππ αα Consider this: The result implies that the average potential energy of the particle while in this state is: VV kk kk kkαα xx xx 4

7 QUIZ 7 April 0, 018 NAME: ROBERT LOPEZ Score /0 1. (10 points) The principal line in the emission spectrum of the potassium atom is in the violet region of the visible spectrum. On careful examination, the line is actually seen to be two very closely spaced lines at wavelengths of nm and nm. These arise from the P 3/ and P 1/ components of the excited state that emit to go to the S 1/ ground state. From this information, calculate the spin-orbit coupling constant for the excited state, in cm -1. The difference in the energies of these two emission lines is due to the difference in energy of the two components of the excited state EE EE SSSS PP3 EE SSSS PP 1/ SS The formula for the spin-orbit energy is EE SSSS LL JJ JJ(JJ + 1) LL(LL + 1) SS(SS + 1) The energy difference is AA EE (1 + 1) (1 + 1) AA AA AA Then, the experimental value is EE Then AA AA 10 9 nnnn mm nnnn mm mm nnnn/mm nnnn mm cccc 1 EE cccc cccc 1. (10 points) For each of the following atoms, write the ground-state configuration and the lowest-energy term that arises for that configuration. [HINT: Neglect spin-orbit coupling.] Atom Configuration Lowest-energy Term H 1s 1 S He 1s 1 S N 1s s p 3 4 S F 1s s p 5 P Ne 1s s p 6 1 S

8 QUIZ 8 April 7, 018 NAME: DIANE WARREN Score /0 1. (10 points) The equilibrium bond length of 7 Li 1 H has been found to be pm. Using data in the Handbook, find the value of B e for 7 Li 1 H, as precisely as you can. Report your answer in cm -1. The formula for Be in the Handbook gives the parameter in Hz. To convert to cm -1, one must divide by the speed h of light. So, the appropriate equation is: BB ee 8ππ ccccrr ee. So, we need to calculate the reduced mass, using the nuclear mases given in Table 1.. mm LLLL mm HH gg μμ mmmmmm gg mmmmmm mm LLLL + mm HH gg mmmmmm gg gg kkkk/mmmmmmmmmmmmmmmm mmmmmm mmmmmm Then, by substitution JJJJ BB ee 8ππ ( mm ss 1 )( kkkk mmmmmmmmmmmmmmmm 1 )( mm 1 mm) cccc 1. (10 points) In the table, match the phrase in column A with the best phrase in column B by placing the number in blank beside the entry in column A. There is only one correct answer for each. Column A Column B 9 Anharmonic potential 1. Overtone spectroscopy. 4 Beer-Lambert law. P, Q, R, S, and T branches. 6 Carbon dioxide 3 Dipole approximation _10 Raman effect 3. Theoretical method of specifying selection rules. 4. Strength of absorption of light with sample size. 5. Of the terms derived from a configuration, the term with highest spin multiplicity is of lowest energy. 6. Has two degenerate modes of vibration. 7. Multiplex advantage. 8. Nonlinear molecule. 9. Treated by introducing the Morse potential. 10. Scattering of light with frequency change.

9 QUIZ 9 May 4, 018 NAME: BENJ APSEK Score /0 1. (10 points) The hydrogen molecule ion (with nuclear centers A and B) serves as the exemplary diatomic molecule. For each one-electron LCAO molecule orbital listed in the table, indicate the appropriate term symbol. (N is an appropriate normalization constant, in each case. You do not have to indicate whether the orbital is bonding or antibonding. ) Molecular Orbital NN Ψ 1ss,AA + Ψ 1ss,BB Symmetry Symbol σσ gg NN Ψ 1ss,AA Ψ 1ss,BB σσ uu NN Ψ ppzz,aa Ψ ppzz,bb σσ gg NN Ψ pp1,aa + Ψ pp1,bb ππ uu NN Ψ pp1,aa Ψ pp1,bb ππ gg. (10 points) In the table below, match the phrase in column A with the phrase in column B that best describes it. Column A Column B c AO e Basis functions a Bond order h Born-Oppenheimer approximation (a) [bonding electrons antibonding electrons]/ (b) A measure of the degree to which AOs have nonzero values in the same region. (c) Atomic orbital. (d) For system subject to a Coulomb potential, the average kinetic and potential energies are related. f Delocalization g Molecular configuration b Overlap integral i Secular determinant j Symmetry designation d Virial theorem (e) The set from which one forms MOs. (f) Extension of an MO over the whole molecule. (g) Obtained by putting electrons into MOs, in order of energy and obeying Hund s rule. (h) The procedure that assumes the positions of nuclear centers are fixed. (i) Expression for finding the eigenvalues of the Hamiltonian matrix. (j) Classification of a molecular orbital according to the physical layout of the function.

10 QUIZ 10 May 11, 018 NAME: LONNIE R. LYNN Score /0 1. (10 points) For each object below, give its point group. [Please use the Schoenfliess symbol, not the Hermann-Mauguin symbol.] Object Point Group HH DD h HHHHHH CC vv SSSSSSSSSSSS bbbbbbbb II h CCHH 44 TT dd CC HH 44 DD h. (6 points) In group theory, the action of two sequential operations (indicated as a product) is also an operation of the group. For C v symmetry, determine the operation that is equivalent to the product in Column A and put that in column B. Column A CC (zz)cc (zz) Column B EE CC (zz)ee CC (zz) CC (zz)σσ(xxxx) σσ(yyyy) 3. (4 points) For water (with the molecular plane being the xz plane), consider the following possible MOs. What is the symmetry label of each one-electron MO in C v symmetry? [C and α are constants in each case.] Possible Molecular Orbital Representation CC(1111 HHHH HHHH + αα OO ) 1 CC(1111 HHHH 1111 HHHH ) bb 1