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CHEM 1001 2011 Final Exam

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Lawrence Murphy

Final Examination December 2011 CHEM 1001 A and T R. Burk Page 1 of 13 Part A.Answer all twelve questions with a few sentences or equations (5 marks each). 1. State Dalton’s law of partial pressures. Dalton’s law of partial pressures states that the total pressure of a mixture of gases is the sum of their partial pressures. 2. How does NO act (g)a catalyst in the production of ozone at ground level? Show the relevant reactions. NO + ½ O → NO (g) 2(g) 2(g) NO 2(g)+ hν → NO + O(g) (g) O (g) 2(g) O 3(g) Since NO is destroyed in the first step but regenerated in the second step, it is acting as a (g) catalyst. 3. Why does a bomb calorimeter actually measure E, and not H of a reaction? In a bomb calorimeter, the volume is constant, so no work is done. H is measured only under constant pressure conditions, like in a coffee cup calorimeter. 4. State Hund’s rule. Hund’s rule states that electrons will occupy degenerate orbitals so as to be as unpaired as possible. 5. Why is a substance with a large bandgap likely to be transparent? If the bandgap is large, then lower energy photons, such as those in the visible portion of the spectrum, do not have enough energy to excite electrons from the HOMO to the LUMO. These visible photons are therefore not absorbed, and are transmitted by the substance, causing it to be transparent (clear). 6. Why does CO have2a much lower boiling point than NO ? 2 CO i2 a linear molecule, and is not polar, and therefore experiences only dispersion forces with other CO mo2ecules. NO is ben2 and polar, and therefore interacts with other NO molecules 2 via a relatively strong dipole-dipole force. This means that it will require more thermal energy to break free from these intermolecular forces, i.e. the boiling point will be higher. 7. Why is hexane (C H )6no14soluble in water? Hexane has only dispersion forces. Water has H-bonds. The two types of forces are very different, and simply put, “like dissolves like.” These two are not alike, so the solubility is very low. Final Examination December 2011 CHEM 1001 A and T R. Burk Page 2 of 13 8. Why is the freezing point depression of a concentrated solution of NaCl (aq)ot as large as what is predicted by T = bKb? f Freezing point depression is a colligative property and therefore depends on the amount of solute in solution. In a concentrated salt solution, there is a lot of ion pairing, resulting in effectively fewer particles in solution, hence less depression of the freezing point, 9. The solubility of Li S2 4(s)ecreases as the temperature increases. Why? + Apparently the process of dissolving Li SO i2 exo4hermic. This is because hydrating the Li ion and the SO 42-ion releases more energy than separating these ions from one another in the solid crystal. In other words, the hydration energy is greater than the lattice energy. 10. What are enantiomers? These are molecules that are non–superimposable mirror images of one another. 11. What is wrong with the name 1,3,4-trifluorobenzene for the molecule shown below? What is the correct name? Numbering beginning at the bottom, we can use lower numbers: 1,2,4-trifluorobenzene. 12. How do we know that CHBrCl is achir2l? To be chiral, the central atom (carbon in this case) must be bonded to four different groups. In this molecule it is bound to only three different atoms (H, Br, and Cl), so it is achiral. Final Examination December 2011 CHEM 1001 A and T R. Burk Page 3 of 13 Part B. Answer all three questions B1, B2 and B3 (20 marks each). B1. (a) [10 marks] Place the following chemical species in increasing order of melting point and explain your reasoning: CH CH OH,3H O,2He, O ,2NaBr 2 He < O < 2H CH OH3< H 2 < NaBr 2 – He experiences only weak dispersion forces, so it has the lowest boiling point. – O experiences only dispersion forces, but these are larger than those in Ne because O is a 2 2 larger molecule. – CH C3 OH 2xperiences H–bond forces, which are larger than dispersion forces – H 2 also experiences H–bond forces. These are larger than those in CH CH OH because i3 2 H 2 there are two O–H bonds, each of which contributes to the dipole moment of the molecule. – NaBr is an ionic solid. Ionic forces are the largest, so boiling this compound requires the highest temperature. (b) [10 marks] Titanium (Ti) metal crystallizes in a body-centred cubic (BCC) structure with an edge length of 331 pm. Calculate the radius of a titanium atom (in pm). You may use geometry to solve this problem, or the fact that the atoms in a BCC structure occupy 68% of the volume of the unit cell. MethodA: Note that the diagonal through the centre of the unit cell passes through three atoms, and has a length of 4r: l l l d 4r 2 2 2 2 2 2 d = l + l = (331 pm) + (331 pm) = 219122 pm but, (4r) = l + d = (331 pm) + 219122 pm 2 4r = 573.3 pm r = 143.3 pm Method B: 3 7 3 V cell (331 pm) = 3.62 x 10 pm But since the atoms occupy only 68% of the unit cell, V atoms= 0.68 x 3.62 x 10 pm = 2.47 x 10 7 3 pm In a body centred cubic unit cell, there are the equivalent of two atoms. Thus, the volume of each atom is ½ (2.47 x 10 pm ) = 1.23 x 10 pm 7 3 Each atom is a sphere of radius r and volume = (4/3)r 3 Final Examination December 2011 CHEM 1001 A and T R. Burk Page 4 of 13 7 3 3V 3(1.23×10 pm ) 3 = 3 =143.3pm Thus, r = 4π 4(3.14) B2. (a) [6 marks] A0.15 g sample of a protein is dissolved in water to make 2.0 mL of solution. The osmotic pressure of this solution is found to be 18.6 Torr at 25°C. Calculate the molecular weight of the protein. Π = MRT Π = 18.6 Torr = (18.6 Torr/750.1 Torr/bar) = 0.0248 bar Thus, M = Π/(RT) = 0.0248 bar / (0.08314 L bar K mol x 298 K) = 0.00100 mol L -1 In this 2.0 mL solution there would be 0.00100 mol L x 0.002 L = 2.00 x 10 mol of protein The molecular weight is therefore 0.15 g / (2.00 x 10 mol) = 7.45 x 10 g mol 4 -1 (b) [6 marks] Asolution is made by adding 3.0 mol Na SO 2 4(s)o 2 kg H 2 .(l)at is the vapour pressure of this solution at 30°C? The vapour pressure of pure water at 30°C is 31.82 mm Hg. 3.0 mol Na SO2 640 mol Na + 3.0 mol SO , or 9.0 4oles of solute all together. 2 kg H O2= 2,000 g = 2000 g / 18 g mol = 111.1 mol H O 2 Thus, X H2O = 111.1/(111.1 + 9.0) = 0.925 o p H2O = p H2OX H2O = 31.82 mm Hg (0.925) = 29.44 mm Hg (c) [8 marks] Asolution is made by adding 0.12 mole of iron chloride to 500 g water. The freezing point of this solution is found to be -1.79 C. Is the iron chloride FeCl or FeC2 ? (K for3 f water is 1.86 C kg mol ) -1 ΔT = iK b f f ΔT 1.79 C Thus, i = f = = 4 K f o −1 0.12mol  1.86 Ckgmol  ÷  0.500kg  The salt must therefore be FeCl . 3 Final Examination December 2011 CHEM 1001 A and T R. Burk Page 5 of 13 B3. (a) Name the following compounds [2 marks each]: 2–amino–3–fluoropentane 1,4-dichloro-2-ethyl-6-nitrobenzene N,N-dimethyl-3-chlorobutanamide 3–bromo-2-butanone (b) Draw the structure or give the correct name for each product in the following reactions [2 marks each]: or, propylbutanoate Air, 300 C or, pentanal Heat, HS2 4 or, diethyl ether (c) [3 marks] Assign either an E or Z stereochemical descriptor to the molecule shown below. Explain how you arrived at your answer. Final Examination December 2011 CHEM 1001 A and T R. Burk Page 6 of 13 On carbon number 3, the groups are C H and C2. 5l has the higher priority. On carbon number 4, the groups are CH and3C H . C2H 5as 2he5higher priority. The two higher priority groups are Cl and C H , 2 5 which are on the same side of the double bond (i.e. cis to one another). This is therefore the Z stereoisomer. (d) [3 marks] Is the molecule shown below the R or S enantiomer? Explain how you arrived at your answer. Rotating the molecule to put the lowest priority group (H) at the back, we have The other three groups from highest to lowest priority are Br, Cl and O, which are read counterclockwise, hence this molecule is the S enantiomer. Final Examination December 2011 CHEM 1001 A and T R. Burk Page 7 of 13 Part C.Answer any three of the five questions C1 – C5. If you answer more than three, the best three will be used to calculate your mark (20 marks each). C1. (a) [2 marks] Use the ideal gas law to calculate the pressure (in bar) exerted by 2.00 mol of He(g) in a 1.00 L container at 25 C. −1 −1 p = nRT = 2 mol(0.08314LbarK mol )(298.15 K) = 48.6 bar V 1L (b) [4 marks] Use the van der Waals equation to calculate the pressure exerted by 2.00 mol of o 2 -2 -1 He (g)a 1.00 L container at 25 C. For He, a = 0.0351 bar L mol and b = 0.0238 L mol . nRT n 2 p = −a  ÷ V −nb V  −1 −1 2 2 mol(0.08314LbarK mol )(298.15
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