Name CHM 1051 Spring 2018 February 4 EXAMINATION ONE TENTATIVE SOLUTIONS I II III IV V

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1 Name CHM 1051 Spring 2018 February 4 EXAMINATION ONE TENTATIVE SOLUTIONS I II III IV V Total Glance over the entire exam, and then attempt the problems in the order of your choice. Rough point values are given for each problem. The total will be scaled to 100 points after the exams are marked. For questions with multiple parts, you do not necessarily need the answer to part A in order to work part B, etc. For calculations, give your answer to the correct number of significant figures, and be sure to include the correct units for your answer. You must show your work to receive any credit for a calculated answer. Additional information is provided in a separate information packet; you can use the back for scratch work. Good luck! I. (20 points) The direct measurement of hydrogen bond in a single molecule using an atomic force microscopy was recently reported ( The hydrogen bond in this study occurred between carbon monoxide on the tip of the atomic force microscope and a propellane hydrocarbon attached to the surface. The synthesis of one of the propellanes used in this study (TNP) is shown below. A. Write the formula of the first compound in this synthesis? C 12 H 6 O 2 B. Which of the three compounds shown above do you expect to be most soluble in water? the cmpd labeled 1 Explain your reasoning. The OH groups can form multiple H-bonding interactions with water, although an argument can be made for the first compound because it also has H-bond accepting sites (the lone pairs on oxygen) and a smaller nonpolar region. Which of the three compounds shown above do you expect to be least soluble in water? TNP Explain your reasoning. TNP is nonpolar (no polar or H-bonding sites to interact with water). C. Would you expect TNP to be able to participate in hydrogen bonding? No Explain why or why not. No N H, O H, or F H bonds. Only C H bonds. D. Sketch a H-bond that might form between water and compound #1, redrawn on the right. Two possible H-bonds are shown.

2 CHM 1051, Exam One, Spring 2018 page 2 II. (30 points) The vapor pressure of THC, primary psychoactive compound in marijuana, has been studied with the goal of developing a reliable breathalyzer for this drug (sciencedaily.com/releases/2017/07/ htm). The structure of THC is shown below. A. The challenge of developing a breathalyze for marijuana is that the vapor pressure of THC is much lower than that of ethanol, CH 3 CH 2 OH. Why is the vapor pressure of THC so much lower than the vapor pressure of ethanol? A complete answer will include a discussion of the important intermolecular forces for both compounds. Ethanol, CH 3 CH 2 OH, has an OH group that can form H-bonds with another CH 3 CH 2 OH. THC also has an OH group, as well as an O group that is polar and can form H-bonds, plus a much larger hydrocarbon section than ethanol, giving it significantly more London forces (random-dipole induced-dipole). Overall, THC has greater intermolecular forces, so it will have a lower vapor pressure B. One solution might be to increase the temperature of the sample to increase the vapor pressure of THC. Why does increasing temperature increase vapor pressure? Increasing the temperature => increasing the kinetic energy which can overcome the energy of attraction due to intermolecular forces. C. The vapor pressure of THC at 25 C is mm Hg. Calculate the vapor pressure of THC at 140 C. The enthalpy of vaporization for THC is 93.0 kj/mol. First, set up the equation you will solve including all of the values and any necessary unit conversions. temperature vapor pressure P H 1 25 C = K 4.63 x 10-8 vap mmhg ln = - P1 R T1 T C = K P kj 93.0 P2 mol 1 1 ln = x 10 mmhg J 1 kj K K K mol 1000 J Now solve for the vapor pressure of THC at 140 C. P2 ln = P -8 2 = mmhg 4.63 x 10 mmhg D. The graph on the right is from the published work described above. P sat is the vapor pressure (VP) of THC. Divide the number on the y-axis by 10 4 to get the VP in Pa. What is the VP of TCC at 140 C in mmhg? 10,000/10 4 = 1 Pa 1 Pa x (760 mmhg/101,325 Pa) = mmhg How well does it agree with your answer to Part C? Off by a factor of five. That s pretty darn good.

3 CHM 1051, Exam One, Spring 2018 page 3 III. (22 points) A refreshing carbonated beverage is refreshing for a complex mixture of reasons. The some of the dissolved carbon dioxide reacts with water to form carbonic acid, H 2 CO 3. A study of this process in real beverages was recently reported ( A. The paper was published in the Journal of Physical Chemistry, and sometimes physical chemists, for good reasons) use concentration units that we would consider unconventional. For example, the control in this study used the following concentration of CO 2 in water: 10 molecules of CO 2 were introduced in a cubic box consisting of 10 4 water molecules Convert this concentration to molality (mol of solute per kg of solvent). 10 CO 2 x (1 mol/6.022 x ) = 1.6 x mol 10 4 H 2 O x (18 g H 2 O/1 mol H 2 O) x (1 mol/6.022 x ) x (1 kg/1000 g) = 3.0 x kg 1.6 x mol/ 3.0 x kg = mol/kg This concentration will be very close to the molarity (mol/l). Why? For a dilute aqueous solution, 1 L solution 1 kg solvent, so (mol solute)/(l solution) (mol solute)/(kg solvent). B. Carbon dioxide is very soluble in water. The Henry s law constant for CO 2 (g) at 20 C is mol/l atm. Calculate the solubility (mol/l) of carbon dioxide in water at 20 C, assuming an atmosphere containing 4.5% CO 2. [gas] = kp gas = 3.91 x 10 2 mol/l atm x 1 atm x (4.5% N 2 /100% air) = 1.8 x 10 3 mol/l C. Do you expect carbon dioxide to be more soluble or less soluble in water at a higher temperature? Less soluble. IV. (12 points) For each of the following compounds, identify the strongest intermolecular force LiCl ion-ion CH 3 CH 3 London dispersion CH 3 OH H-bonding CH 3 F dipole-dipole and then arrange them from lowest boiling point to highest boiling point CH 3 CH 3 < CH 3 F < CH 3 OH < LiCl

4 CHM 1051, Exam One, Spring 2018 page 4 V. (16 points) I could not find the concentration of p Glu in jellyfish in time to explore this claim, so let s see when kind of osmotic pressure we get for a low concentration of solute. Assuming a p Glu concentration of 0.5 M, calculate the osmotic pressure. = MRT = 0.5 mol L ( L atm )( K) = = 12 atm (1 sf) Kmol 12 atm!! What other assumptions did you make in your calculation? assumed p Glu is a nonelectrolyte (so the van t Hoff coefficient, i, is 1) assumed a temperature of 20 C

5 CHM 1051, Exam One, Spring 2018 page 5 exo-thc, H vap = kj/mol J/mol = Vapor Pressure 4.63X10-8 mm Hg at 25 deg C (est) mmhg = Pa US EPA; Estimation Program Interface (EPI) Suite. Ver Jan, Available from, as of Oct 19, 2011:

6 CHM 1051, Exam One, Spring 2018 page 6 B. Currently the D. How many E. Circle the critical point.

7 CHM 1051, Exam One, Spring 2018 page 7 III. (18 points) We used Henry s law A. Turns out nitrogen 10 molecules of CO 2 were introduced in a cubic box consisting of 10 4 water molecules B. Our brains C. Estimate the D. Given that the E. Convert the

PLEASE DO NOT MARK ON THE EXAM. ALL ANSWERS SHOULD BE INDICATED ON THE ANSWER SHEET. c) SeF 4

Chem 130 EXAM 4 Fall 99 PLEASE DO NOT MARK ON THE EXAM. ALL ANSWERS SHOULD BE INDICATED ON THE ANSWER SHEET QUESTIONS 1-5 MAY HAVE MORE THAN ONE POSSIBLE ANSWER CIRCLE ALL CORRECT RESPONSES TO EACH QUESTION