CHEMISTRY 444.10 Spring, 2018 QUIZ 1 February 16, 2018 NAME: HANS ZIMMER Score ______/20

[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

2. (6 points) In pure water at 25°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, < 𝑟𝑟2 > = < 𝑥𝑥2 > + < 𝑦𝑦2 > + < 𝑧𝑧2 > = 6𝐷𝐷𝐷𝐷 From this equation one obtains, from the data in the Handbook, the result:

𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 = √6𝐷𝐷𝐷𝐷 = �6 (2.26 × 10−9 𝑚𝑚2 𝑠𝑠−1)(1 × 10−3 𝑠𝑠) = 3.68 × 10−6 𝑚𝑚 = 3.68 × 10−4 𝑐𝑐𝑚𝑚

3. (10 points) The sedimentation coefficient of lysozyme is determined by centrifugation at 55,000 rpm in water at 20°C. The boundary was monitored as a function of time of centrifugation. Determine the sedimentation coefficient of lysozyme in water at 20°C. [A plot may help.]

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 = 0.000377 min-1 = 6.282×10-6 sec-1. This slope is ω2s. So division of this by the square of the angular frequency gives the sedimentation coefficient.

𝑆𝑆 = 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝜔𝜔2 =

6.282 × 10−6𝑠𝑠𝑠𝑠𝑐𝑐−1

(2𝜋𝜋 × 916.67𝑠𝑠𝑠𝑠𝑐𝑐−1)2 = 1.89 × 10−13 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.

Time (min) xb (cm) xb/x(0) ln(xb/x(0)) 0 6.00 1.00 0.00 30 6.07 1.012 0.0119 60 6.14 1.023 0.0227 90 6.21 1.035 0.0344 120 6.28 1.047 0.0459 150 6.35 1.058 0.056

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, CR(t), as a function of time, assuming that the concentration at time t = 0 is CR(0) and the rate constant is k0. [Note: the result of (b) cannot be correct if the result of (a) is incorrect.] (a) The rate equation is: 𝑑𝑑𝐶𝐶𝑅𝑅

𝑑𝑑𝑑𝑑 = − 𝑘𝑘0 𝐶𝐶𝑅𝑅0 [= − 𝑘𝑘0]

(b) Integrating this equation from t = 0 to t = t gives the result one seeks for the time-dependent concentration of the reactant:

𝐶𝐶𝑅𝑅(𝑡𝑡) = 𝐶𝐶𝑅𝑅(0) − 𝑘𝑘0 𝑡𝑡

2. (4 points) From data in the Handbook, estimate the second-order rate constant for the destruction of ozone (O3) by nitric oxide (NO) at room temperature (298.15 K). The Arrhenius parameters for this reaction are found in Table 10.2 on page 10-2. Using Arrhenius’s equation,

𝑘𝑘(𝑇𝑇) = 𝐴𝐴 𝑒𝑒𝑒𝑒𝑒𝑒 �−𝐸𝐸𝑎𝑎𝑅𝑅𝑇𝑇

�

𝑘𝑘(298.15𝐾𝐾) = 7.9 × 1011𝑐𝑐𝑐𝑐3𝑐𝑐𝑚𝑚𝑚𝑚−1𝑠𝑠−1 𝑒𝑒𝑒𝑒𝑒𝑒�−10.5 × 103𝐽𝐽 𝑐𝑐𝑚𝑚𝑚𝑚−1

8.3144349 𝐽𝐽 𝑐𝑐𝑚𝑚𝑚𝑚−1𝐾𝐾−1(298.15𝐾𝐾)�

= 1.14 × 1010𝑐𝑐𝑐𝑐3𝑐𝑐𝑚𝑚𝑚𝑚−1𝑠𝑠−1

3. (8 points) One mechanism of ozone destruction involves reaction with HO2· radicals to create oxygen and a hydroxyl radical, by this gas-phase reaction 𝑯𝑯𝑶𝑶𝟐𝟐 ∙ (𝒈𝒈) + 𝑶𝑶𝟑𝟑 (𝒈𝒈) → 𝑶𝑶𝑯𝑯 ∙ (𝒈𝒈) + 𝟐𝟐 𝑶𝑶𝟐𝟐 (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?

[HO2] (molecules cm-3) [O3] (molecules cm-3) Rate (molecules cm-3 s-1) 1.0 × 1011 1.0 × 1012 1.9 × 108 1.0 × 1011 5.0 × 1012 9.5 × 108 3.0 × 1011 1.0 × 1012 5.7 × 108

(a) One finds the ratios of rates compared to ratios of reactant concentrations, where n is the initial

order by the ratio method: 𝑅𝑅𝐴𝐴𝑅𝑅𝐵𝐵

= �𝐶𝐶𝐴𝐴𝐶𝐶𝐵𝐵�𝑛𝑛 Using the first two rows of the table, one finds the order

with respect to ozone: 1.9 ×108 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑠𝑠 𝑚𝑚𝑚𝑚−3𝑠𝑠−1

9.5 ×108 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑠𝑠 𝑚𝑚𝑚𝑚−3𝑠𝑠−1 = �1.0 ×1012𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑠𝑠 𝑚𝑚𝑚𝑚−3

5.0 ×1012 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑠𝑠 𝑚𝑚𝑚𝑚−3�𝑛𝑛, which gives 0.2 = 0.2𝑛𝑛.

The order with respect to ozone is 1. A similar determination using the first and third lines (in which the concentration of 𝐻𝐻𝐻𝐻2 ∙ is changed) gives the order with respect to it, which is 1, as well. Thus, the initial rate law looks like this:

−𝑑𝑑[𝐻𝐻3]𝑑𝑑𝑡𝑡

= 𝑘𝑘 [𝐻𝐻3] [𝐻𝐻𝐻𝐻2 ∙] (b) From any of the results, on may find 𝑘𝑘 = (−𝑑𝑑[𝑂𝑂3] 𝑑𝑑𝑑𝑑⁄ )

[𝑂𝑂3][𝐻𝐻𝑂𝑂2∙]. Substitution of the first row, for example, gives

𝑘𝑘 = 1.9 × 108𝑐𝑐𝑚𝑚𝑚𝑚𝑒𝑒𝑐𝑐𝑚𝑚𝑚𝑚𝑒𝑒𝑠𝑠 𝑐𝑐𝑐𝑐−3𝑠𝑠−1

(1.0 × 1012𝑐𝑐𝑚𝑚𝑚𝑚𝑒𝑒𝑐𝑐𝑚𝑚𝑚𝑚𝑒𝑒𝑠𝑠 𝑐𝑐𝑐𝑐−3)(1.0 × 1011𝑐𝑐𝑚𝑚𝑚𝑚𝑒𝑒𝑐𝑐𝑚𝑚𝑚𝑚𝑒𝑒𝑠𝑠 𝑐𝑐𝑐𝑐−3) = 1.9 × 10−15 𝑐𝑐𝑐𝑐3

𝑐𝑐𝑚𝑚𝑚𝑚𝑒𝑒𝑐𝑐𝑚𝑚𝑚𝑚𝑒𝑒 𝑠𝑠−1

CHEMISTRY 444.10 Spring, 2018 QUIZ 2 February 23, 2018 NAME: JOHNNY GREENWOOD Score ______/20

[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 𝟐𝟐 𝑵𝑵𝑵𝑵 + 𝑵𝑵𝟐𝟐 → 𝟐𝟐 𝑵𝑵𝑵𝑵𝟐𝟐 is second order with a temperature-dependent rate constant, k: T (K) 203.8 222.4 272.2 307.2 k (cm3molecule-1 s-1) / 10-15 2.67 4.17 11.5 20.9

Using these data only, determine ∆H≠ for this reaction over this range. [An appropriate plot makes the analysis easy.] The appropriate equation to plot is the following:

ln �𝑘𝑘𝑇𝑇� = ln �

𝑘𝑘𝑏𝑏ℎ𝑐𝑐𝜃𝜃

� + ∆𝑆𝑆≠

𝑅𝑅 −

∆𝐻𝐻≠

𝑅𝑅 1𝑇𝑇

The slope of this plot is related to the enthalpy of activation.

∆𝐻𝐻≠ = 747.6 𝐾𝐾 (8.3144349 𝐽𝐽 𝐾𝐾−1 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚−1) = 6.216 𝑘𝑘𝐽𝐽 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚−1

[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.] 2. (4 points) The decay of 60Co to form 60Ni 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 60Co is 1.9 × 103 days. What is the first-order rate constant for this process? By rearrangement, one finds the relation: 𝑘𝑘 = 𝑙𝑙𝑙𝑙2

𝑡𝑡1/2 = 𝑙𝑙𝑙𝑙2

1.9 ×103 𝑑𝑑𝑑𝑑𝑑𝑑 = 3.6 × 10−4 𝑑𝑑𝑑𝑑𝑑𝑑−1

This can be expressed in other time units as 1.5 × 10-5 h-1 = 2.5 × 10-7 min-1 = 4.2 × 10-9 s-1. 3. (6 points) Consider a bimolecular solution-phase reaction in water, for which the diffusion coefficient at 298.15 K is 2.25×10-9 m2 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:

𝑘𝑘𝐷𝐷 = 4𝜋𝜋𝑁𝑁0(𝑟𝑟𝐴𝐴 + 𝑟𝑟𝐵𝐵)(𝐷𝐷𝐴𝐴 + 𝐷𝐷𝐵𝐵) = 4𝜋𝜋(6.02211415 × 1023𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚−1)(1 × 10−10 𝑚𝑚)(2 × 2.25 × 10−9𝑚𝑚2𝑠𝑠−1)

= 3.41 × 106 𝑚𝑚3

𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 ∗ 𝑠𝑠 = 3.41 × 109

𝑑𝑑𝑚𝑚3

𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 ∗ 𝑠𝑠

CHEMISTRY 444.10 Spring, 2018 QUIZ 3 March 2, 2018 NAME: ____ALEXANDRE DESPLAT__________ Score ______/20 [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.]

ln(k/T) = -747.6/T - 40.54R² = 0.9984

-44.4

-44.2

-44

-43.8

-43.6

-43.4

-43.2

-43

-42.8

0.003 0.0035 0.004 0.0045 0.005

ln(k

/T)

1/T (1/K)

1. (8 points) A strong absorption line of the Hg atom has a wavelength of 184.957 nm. By what energy (in J/mol) are the two states involved in the transition separated? We use this version of Planck’s relation:

∆𝐸𝐸 = ℎ 𝑐𝑐λ

= (6.6260693 × 10−34 𝐽𝐽 𝑠𝑠)(2.99792458 × 108 𝑚𝑚 𝑠𝑠−1) �1

184.957 × 10−9𝑚𝑚�

= 1.07400 × 10−18 𝐽𝐽/𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑐𝑐𝑚𝑚𝑚𝑚𝑚𝑚 But this is the energy per molecule. We must multiply this by the number of molecules in a mol to determine what is asked.

∆𝐸𝐸𝑚𝑚𝑚𝑚𝑚𝑚 = 𝑁𝑁0 ∆𝐸𝐸𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 = 6.02211415 × 1023𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑐𝑐𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚

�1.07400 × 10−18𝐽𝐽

𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑐𝑐𝑚𝑚𝑚𝑚𝑚𝑚�

= 6.46777 × 105𝐽𝐽

𝑚𝑚𝑚𝑚𝑚𝑚

2. (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 𝑑𝑑2

𝑑𝑑𝑥𝑥2 𝐴𝐴 𝑠𝑠𝑠𝑠𝑠𝑠 �

𝑠𝑠𝑛𝑛𝑎𝑎𝑥𝑥� −�

𝑠𝑠2𝑛𝑛2

𝑎𝑎2� 𝐴𝐴 𝑠𝑠𝑠𝑠𝑠𝑠 �

𝑠𝑠𝑛𝑛𝑎𝑎𝑥𝑥�

𝑦𝑦 + 𝑑𝑑𝑑𝑑𝑦𝑦

𝐴𝐴 𝑚𝑚𝑥𝑥𝑒𝑒(𝑠𝑠𝑖𝑖𝑦𝑦) (𝑦𝑦 + 𝑠𝑠𝑖𝑖) 𝐴𝐴 𝑚𝑚𝑥𝑥𝑒𝑒(𝑠𝑠𝑖𝑖𝑦𝑦)

�𝜃𝜃 + 𝑑𝑑𝑑𝑑𝜃𝜃� 𝐴𝐴 𝑚𝑚𝑥𝑥𝑒𝑒(− 𝜃𝜃2/2) 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 �𝝅𝝅𝒙𝒙

𝒂𝒂� over the region 𝟎𝟎 ≤ 𝒙𝒙 ≤ 𝒂𝒂.

To normalize the wave function, the integral of its square over the range must be 1.

� Ψ∗(𝑥𝑥)Ψ(𝑥𝑥)𝑑𝑑𝑥𝑥𝑎𝑎

0 = 1.

Or ∫ 𝐴𝐴 𝑠𝑠𝑠𝑠𝑠𝑠 �𝜋𝜋𝜋𝜋𝑎𝑎�𝐴𝐴 𝑠𝑠𝑠𝑠𝑠𝑠 �𝜋𝜋𝜋𝜋

𝑎𝑎� 𝑑𝑑𝑥𝑥𝑎𝑎

0 = 𝐴𝐴2 ∫ 𝑠𝑠𝑠𝑠𝑠𝑠2 �𝜋𝜋𝜋𝜋𝑎𝑎� 𝑑𝑑𝑥𝑥𝑎𝑎

0 = 𝐴𝐴2 �𝜋𝜋2

|0𝑎𝑎 − 𝑎𝑎4𝜋𝜋

𝑠𝑠𝑠𝑠𝑠𝑠 �2𝜋𝜋𝜋𝜋𝑎𝑎� |0𝑎𝑎� = 1.

Making the substitution at the limits gives

𝐴𝐴2 �𝑎𝑎2

− 02

− 𝑎𝑎

4𝑛𝑛𝑠𝑠𝑠𝑠𝑠𝑠(2𝑛𝑛) +

𝑎𝑎4𝑛𝑛

𝑠𝑠𝑠𝑠𝑠𝑠(0)� = 𝐴𝐴2 �𝑎𝑎2� = 1.

The requirement for normalization is 𝐴𝐴2 = 2𝑎𝑎 or 𝐴𝐴 = �2

𝑎𝑎, where the choice is made that the constant 𝐴𝐴

is real and positive.

CHEMISTRY 444.10 Spring, 2018 QUIZ 4 March 16, 2018 NAME: CARTER BURWELL Score ______/20 [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

,𝒙𝒙 𝝏𝝏𝝏𝝏𝒚𝒚� = 𝒚𝒚� 𝝏𝝏

𝝏𝝏𝒚𝒚� − 𝒙𝒙 � 𝝏𝝏

𝝏𝝏𝒙𝒙�

To begin, apply this to a function, 𝑓𝑓, of 𝑥𝑥 and 𝑦𝑦: �𝑦𝑦 𝜕𝜕

𝜕𝜕𝜕𝜕, 𝑥𝑥 𝜕𝜕

𝜕𝜕𝜕𝜕� 𝑓𝑓 = 𝑦𝑦 𝜕𝜕

𝜕𝜕𝜕𝜕�𝑥𝑥 𝜕𝜕𝜕𝜕

𝜕𝜕𝜕𝜕� − 𝑥𝑥 𝜕𝜕

𝜕𝜕𝜕𝜕�𝑦𝑦 𝜕𝜕𝜕𝜕

𝜕𝜕𝜕𝜕�

Then, taking the derivative of the product in each term:

�𝑦𝑦𝜕𝜕𝜕𝜕𝑥𝑥

, 𝑥𝑥𝜕𝜕𝜕𝜕𝑦𝑦� 𝑓𝑓 = 𝑦𝑦

𝜕𝜕𝑥𝑥𝜕𝜕𝑥𝑥

�𝜕𝜕𝑓𝑓𝜕𝜕𝑦𝑦� + 𝑦𝑦𝑥𝑥 �

𝜕𝜕2𝑓𝑓𝜕𝜕𝑥𝑥𝜕𝜕𝑦𝑦

� − 𝑥𝑥𝜕𝜕𝑦𝑦𝜕𝜕𝑦𝑦

�𝜕𝜕𝑓𝑓𝜕𝜕𝑥𝑥� − 𝑥𝑥𝑦𝑦 �

𝜕𝜕2𝑓𝑓𝜕𝜕𝑦𝑦𝜕𝜕𝑥𝑥

�

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

, 𝑥𝑥𝜕𝜕𝜕𝜕𝑦𝑦� 𝑓𝑓 = 𝑦𝑦 �

𝜕𝜕𝑓𝑓𝜕𝜕𝑦𝑦� − 𝑥𝑥 �

𝜕𝜕𝑓𝑓𝜕𝜕𝑥𝑥� = � 𝑦𝑦 �

𝜕𝜕𝜕𝜕𝑦𝑦� − 𝑥𝑥 �

𝜕𝜕𝜕𝜕𝑥𝑥�� 𝑓𝑓

Hence, the commutator can be simplified to the form:

�𝑦𝑦𝜕𝜕𝜕𝜕𝑥𝑥

, 𝑥𝑥𝜕𝜕𝜕𝜕𝑦𝑦� = 𝑦𝑦 �

𝜕𝜕𝜕𝜕𝑦𝑦� − 𝑥𝑥 �

𝜕𝜕𝜕𝜕𝑥𝑥�

2. (8 points) A potassium surface has a work function of 2.40 eV. In a photoemission experiment the surface is exposed to radiation at a wave length of 325 nm. During this experiment, 5.00 × 10-3 joules are absorbed. How many electrons are emitted in the experiment? Let’s convert everything to joules: 2.40 𝑒𝑒𝑒𝑒 = 2.40�96485.3 𝐽𝐽 𝑚𝑚𝑚𝑚𝑚𝑚−1𝑒𝑒𝑒𝑒−1�

6.02211415×1023𝑒𝑒𝑚𝑚𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑚𝑚𝑒𝑒𝑒𝑒/𝑚𝑚𝑚𝑚𝑚𝑚 = 3.845 × 10−19 𝐽𝐽/𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒

Similarly, the light photon is 𝐸𝐸𝑝𝑝ℎ𝑚𝑚𝑒𝑒𝑚𝑚𝑒𝑒 = ℎ𝑒𝑒 λ

= �6.6260693×10−34𝑗𝑗𝑚𝑚𝑗𝑗𝑚𝑚𝑒𝑒−𝑒𝑒��2.99792458×108𝑚𝑚 𝑒𝑒−1�325×10−9𝑚𝑚

= 6.112 × 10−19 𝐽𝐽 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:

𝑁𝑁𝑚𝑚𝑚𝑚𝜕𝜕 = 5.00 × 10−3𝐽𝐽

3.845 × 10−19𝐽𝐽/𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 = 1.30 × 1016 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒

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 = 5.00 × 10−3 𝐽𝐽

6.112 × 10−19𝐽𝐽/𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 = 8.18 × 1015 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:

𝑽𝑽(𝒙𝒙) = 𝒌𝒌𝟐𝟐𝒙𝒙𝟐𝟐 + 𝑨𝑨𝒙𝒙𝟑𝟑

where 𝒙𝒙 = 𝒓𝒓 − 𝒓𝒓𝒆𝒆𝒆𝒆, 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� = 𝑻𝑻 + 𝑽𝑽 = −� 𝒉𝒉𝟐𝟐𝟐𝟐�

𝟐𝟐

𝟐𝟐𝒎𝒎 𝒅𝒅𝟐𝟐

𝒅𝒅𝒙𝒙𝟐𝟐 +

𝒌𝒌𝟐𝟐𝒙𝒙𝟐𝟐 + 𝑨𝑨𝒙𝒙𝟑𝟑

CHEMISTRY 444.10 Spring, 2018 QUIZ 5 March 23, 2018 NAME: MARY J. BLIGE Score ______/20 [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) A coin having a mass of 8.400 grams is suspended by a rubber band. When it is stretched and released, the system vibrates with a frequency of 7.823 Hz. What is the force constant of the rubber band, in J/m2? [HINT: Assume that the rubber band contributes essentially no mass to the system.] The relationship among the various parameters is given by the formula: ν0 = 1

2𝜋𝜋�𝑘𝑘𝑚𝑚

, where k is the force

constant. Rearranging this equation, 𝑘𝑘 = 4𝜋𝜋2ν02𝑚𝑚. Substitution into this equation for this system gives

𝒌𝒌 = 𝟒𝟒𝝅𝝅𝟐𝟐(𝟕𝟕.𝟖𝟖𝟐𝟐𝟖𝟖 𝑯𝑯𝑯𝑯)𝟐𝟐 (𝟖𝟖.𝟒𝟒𝟒𝟒𝟒𝟒 𝒈𝒈) = 𝟐𝟐.𝟒𝟒𝟖𝟖 × 𝟏𝟏𝟒𝟒𝟒𝟒 𝒈𝒈 𝒔𝒔−𝟐𝟐 = 𝟐𝟐𝟒𝟒.𝟖𝟖 𝒌𝒌𝒈𝒈 𝒔𝒔−𝟐𝟐 = 𝟐𝟐𝟒𝟒.𝟖𝟖 𝑱𝑱 𝒎𝒎−𝟐𝟐 2. (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 91.68 pm. The position of the center of mass is given (for two particles) as: 𝑹𝑹𝐶𝐶𝐶𝐶 = 𝑚𝑚𝐴𝐴

𝑚𝑚𝐴𝐴 + 𝑚𝑚𝐵𝐵𝑹𝑹𝐴𝐴 + 𝑚𝑚𝐵𝐵

𝑚𝑚𝐴𝐴 + 𝑚𝑚𝐵𝐵𝑹𝑹𝐵𝐵. One can

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,

𝑹𝑹𝐶𝐶𝐶𝐶 = 𝑚𝑚𝐻𝐻

𝑚𝑚𝐻𝐻 + 𝑚𝑚𝐹𝐹𝑹𝑹𝐻𝐻𝐹𝐹 =

1 𝑔𝑔/𝑚𝑚𝑚𝑚𝑚𝑚

1 𝑔𝑔𝑚𝑚𝑚𝑚𝑚𝑚 + 19 𝑔𝑔/𝑚𝑚𝑚𝑚𝑚𝑚

(91.68 𝑝𝑝𝑚𝑚) = 4.584 𝑝𝑝𝑚𝑚

(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

Ψ(𝑥𝑥) = 1

√𝜋𝜋4 √𝛼𝛼 𝑒𝑒𝑥𝑥𝑝𝑝 �−

𝑥𝑥2

2𝛼𝛼2�

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

< 𝑥𝑥2 > = � Ψ∗𝑥𝑥2Ψ 𝑑𝑑𝑥𝑥𝑎𝑎

0 =

1𝛼𝛼√𝜋𝜋

� 𝑒𝑒𝑥𝑥𝑝𝑝 �−𝑥𝑥2

2𝛼𝛼2� 𝑥𝑥2𝑒𝑒𝑥𝑥𝑝𝑝 �−

𝑥𝑥2

2𝛼𝛼2�𝑑𝑑𝑥𝑥

+∞

−∞ =

1𝛼𝛼√𝜋𝜋

� 𝑥𝑥2𝑒𝑒𝑥𝑥𝑝𝑝 �−𝑥𝑥2

𝛼𝛼2� 𝑑𝑑𝑥𝑥

+∞

−∞

= 𝛼𝛼2

√𝜋𝜋� 𝑦𝑦2𝑒𝑒𝑥𝑥𝑝𝑝(−𝑦𝑦2)𝑑𝑑𝑦𝑦+∞

−∞ =

2𝛼𝛼2

√𝜋𝜋� 𝑦𝑦2𝑒𝑒𝑥𝑥𝑝𝑝(−𝑦𝑦2)𝑑𝑑𝑦𝑦+∞

0 =

2𝛼𝛼2

√𝜋𝜋14√

𝜋𝜋 = 𝛼𝛼2

2

Consider this: The result implies that the average potential energy of the particle while in this state is:

⟨𝑉𝑉⟩ = ⟨𝑘𝑘2𝑥𝑥2⟩ =

𝑘𝑘2⟨𝑥𝑥2⟩ =

𝑘𝑘𝛼𝛼2

4

CHEMISTRY 444.10 Spring, 2018 QUIZ 6 April 6, 2018 NAME: SUFJAN STEVENS Score ______/20

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 393.366 nm and 396.847 nm. These arise from the 2P3/2 and 2P1/2 components of the excited state that emit to go to the 2S1/2 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 = 𝐸𝐸𝑆𝑆𝑆𝑆 � 𝑃𝑃32

2 � − 𝐸𝐸𝑆𝑆𝑆𝑆� 𝑃𝑃1/22 �

The formula for the spin-orbit energy is 𝐸𝐸𝑆𝑆𝑆𝑆� 𝐿𝐿𝐽𝐽𝑆𝑆 � = 𝐴𝐴2�𝐽𝐽(𝐽𝐽 + 1) − 𝐿𝐿(𝐿𝐿 + 1) − 𝑆𝑆(𝑆𝑆 + 1)�

The energy difference is

∆𝐸𝐸 = 𝐴𝐴2�

32�

32

+ 1� − 1(1 + 1) − 12�

12

+ 1� −12�

12

+ 1� + 1(1 + 1) +12�

12

+ 1� �

= 𝐴𝐴2�

32�

32

+ 1� −12�

12

+ 1� � = 𝐴𝐴2�

154

−34

� = 32𝐴𝐴

Then, the experimental value is

∆𝐸𝐸 = 109𝑛𝑛𝑛𝑛/𝑛𝑛

393.366 𝑛𝑛𝑛𝑛 −

109𝑛𝑛𝑛𝑛𝑛𝑛

396.847 𝑛𝑛𝑛𝑛 = 2.54216 × 106 𝑛𝑛−1 − 2.51986 × 106 𝑛𝑛−1

= 2.23 × 104 𝑛𝑛−1 = 2.23 × 102 𝑐𝑐𝑛𝑛−1 Then 𝐴𝐴 = 2

3∆𝐸𝐸 = 2

32.23 × 102 𝑐𝑐𝑛𝑛−1 = 148.7 𝑐𝑐𝑛𝑛−1

2. (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 1s1 2S

He 1s2 1S

N 1s22s22p3 4S

F 1s22s22p5 2P

Ne 1s22s22p6 1S

CHEMISTRY 444.10 Spring, 2018 QUIZ 7 April 20, 2018 NAME: ROBERT LOPEZ Score ______/20

1. (10 points) The equilibrium bond length of 7Li1H has been found to be 159.49 pm. Using data in the Handbook, find the value of Be for 7Li1H, 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 of light. So, the appropriate equation is: 𝐵𝐵𝑒𝑒 = ℎ

8𝜋𝜋2𝑐𝑐𝑐𝑐𝑅𝑅𝑒𝑒2. So, we need to calculate the reduced mass, using the

nuclear mases given in Table 12.2.

𝜇𝜇 = 𝑚𝑚𝐿𝐿𝐿𝐿𝑚𝑚𝐻𝐻

𝑚𝑚𝐿𝐿𝐿𝐿 + 𝑚𝑚𝐻𝐻 =

7.01600 𝑔𝑔𝑚𝑚𝑚𝑚𝑚𝑚 �1.007825 𝑔𝑔

𝑚𝑚𝑚𝑚𝑚𝑚�

7.01600 𝑔𝑔𝑚𝑚𝑚𝑚𝑚𝑚 + 1.007825 𝑔𝑔

𝑚𝑚𝑚𝑚𝑚𝑚 = 0.881238

𝑔𝑔𝑚𝑚𝑚𝑚𝑚𝑚

= 1.463337 × 10−27𝑘𝑘𝑔𝑔/𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚

Then, by substitution

𝐵𝐵𝑒𝑒 = 6.6260693 × 10−34𝐽𝐽𝐽𝐽

8𝜋𝜋2(2.99792458 × 108𝑚𝑚 𝐽𝐽−1)(1.463337 × 10−27𝑘𝑘𝑔𝑔 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚−1)(0.15949 × 10−9𝑚𝑚)2 = 752.028 𝑚𝑚−1

= 7.52028 𝑚𝑚𝑚𝑚−1 2. (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 2. P, Q, R, S, and T branches.

__6__ Carbon dioxide 3. Theoretical method of specifying selection rules.

__3__ Dipole approximation 4. Strength of absorption of light with sample size.

_10__ Raman effect 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.

CHEMISTRY 444.10 Spring, 2018 QUIZ 8 April 27, 2018 NAME: DIANE WARREN Score ______/20

CHEMISTRY 444.10 Spring, 2018 QUIZ 9 May 4, 2018 NAME: BENJ APSEK Score ______/20

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 Symmetry Symbol

𝑁𝑁�Ψ1𝑠𝑠,𝐴𝐴 + Ψ1𝑠𝑠,𝐵𝐵� 𝜎𝜎𝑔𝑔

𝑁𝑁�Ψ1𝑠𝑠,𝐴𝐴 − Ψ1𝑠𝑠,𝐵𝐵� 𝜎𝜎𝑢𝑢

𝑁𝑁�Ψ2𝑝𝑝𝑧𝑧,𝐴𝐴 − Ψ2𝑝𝑝𝑧𝑧,𝐵𝐵� 𝜎𝜎𝑔𝑔

𝑁𝑁�Ψ2𝑝𝑝1,𝐴𝐴 + Ψ2𝑝𝑝1,𝐵𝐵� 𝜋𝜋𝑢𝑢

𝑁𝑁�Ψ2𝑝𝑝1,𝐴𝐴 − Ψ2𝑝𝑝1,𝐵𝐵� 𝜋𝜋𝑔𝑔

2. (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 (a) [bonding electrons – antibonding electrons]/2

__e__ Basis functions (b) A measure of the degree to which AOs have nonzero values in the same region.

__a__ Bond order (c) Atomic orbital.

__h__ Born-Oppenheimer approximation (d) For system subject to a Coulomb potential, the average kinetic and potential energies are related.

__f__ Delocalization (e) The set from which one forms MOs.

__g__ Molecular configuration (f) Extension of an MO over the whole molecule.

__b__ Overlap integral (g) Obtained by putting electrons into MOs, in order of energy and obeying Hund’s rule.

__i__ Secular determinant (h) The procedure that assumes the positions of nuclear centers are fixed.

__j__ Symmetry designation (i) Expression for finding the eigenvalues of the Hamiltonian matrix.

__d__ Virial theorem (j) Classification of a molecular orbital according to the physical layout of the function.

CHEMISTRY 444.10 Spring, 2018 QUIZ 10 May 11, 2018 NAME: LONNIE R. LYNN Score ______/20

1. (10 points) For each object below, give its point group. [Please use the Schoenfliess symbol, not the Hermann-Mauguin symbol.]

Object Point Group

𝑯𝑯𝟐𝟐 𝐷𝐷∞ℎ

𝑯𝑯𝑯𝑯𝑯𝑯 𝐶𝐶∞𝑣𝑣

𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺 𝒃𝒃𝒃𝒃𝑯𝑯𝑯𝑯 𝐼𝐼ℎ

𝑯𝑯𝑯𝑯𝟒𝟒 𝑇𝑇𝑑𝑑

𝑯𝑯𝟐𝟐𝑯𝑯𝟒𝟒 𝐷𝐷2ℎ

2. (6 points) In group theory, the action of two sequential operations (indicated as a product) is also an operation of the group. For C2v symmetry, determine the operation that is equivalent to the product in Column A and put that in column B.

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 C2v symmetry? [C and α are constants in each case.]

Possible Molecular Orbital Representation

𝑯𝑯(𝟏𝟏𝟏𝟏𝑯𝑯𝟏𝟏 + 𝟏𝟏𝟏𝟏𝑯𝑯𝟐𝟐 + 𝜶𝜶 𝟐𝟐𝟏𝟏𝑶𝑶) 𝑎𝑎1

𝑯𝑯(𝟏𝟏𝟏𝟏𝑯𝑯𝟏𝟏 − 𝟏𝟏𝟏𝟏𝑯𝑯𝟐𝟐) 𝑏𝑏1

Column A Column B

𝑯𝑯𝟐𝟐(𝒛𝒛)𝑯𝑯𝟐𝟐(𝒛𝒛) 𝐸𝐸

𝑯𝑯𝟐𝟐(𝒛𝒛)𝑬𝑬 𝐶𝐶2(𝑧𝑧)

𝑯𝑯𝟐𝟐(𝒛𝒛)𝝈𝝈(𝒙𝒙𝒛𝒛) 𝜎𝜎(𝑦𝑦𝑧𝑧)