Chem 107 - ughbanks Exam 3, April 21, 2015 Name (Print) Solutions UIN # Section 502 Exam 3, Version A On the last page of this exam, you ve been given a periodic table and some physical constants. You ll probably want to tear that page off the to use during the exam you don t need to turn it in with the rest of the exam. he exam contains 9 problems, with 6 numbered pages. You have the full 75 minutes to complete the exam. Please show ALL your work as clearly as possible this will help us award you partial credit if appropriate. Even correct answers without supporting work may not receive credit. You may use an approved calculator for the exam, one without extensive programmable capabilities or the ability to store alphanumeric information. Print your name above, provide your UIN number, and sign the honor code statement below. On my honor as an Aggie, I will neither give nor receive unauthorized assistance on this exam. SIGNAURE:
(1) (20 pts) rue or alse? (Please enter or in the blank in front of each statement. 2 points each.) BE SURE A YOU READ EAC SAEMEN VERY CAREULLY. (a) Insulators generally have large band gaps. (b) In silicon, the conduction band orbitals above the band gap have antibonding character. (c) If germanium (Ge) is doped with arsenic (As), there are as about many valence band holes as there are As atoms at room temperature. (d) Doping a semiconductor always produces a material with a larger band gap. (e) he standard molar entropy for any element in its standard state is zero. (f) S for a spontaneous endothermic reaction must be less than zero. (or grading) Scores 1 /20 2 /6 3 /6 4 /6 5 /8 6 /12 7 /21 8 /8 9 /13 ot. /100 (g) G for a spontaneous endothermic reaction must be less than zero. (h) for the reaction, 4 (g) + C(g) C 4 (g), is negative and less than the enthalpy of formation, f, of C 4 (g). (i) he conversion of a gas to a liquid will have a positive value for. (j) he conversion of a gas to a liquid will have a negative value for S. (2) (6 pts) or a given reaction or a process such as a phase change, which of the following is likely to be most sensitive to a change in temperature? Circle the one correct answer. (A) (B) (C) G (D) E (E) S (3) (6 pts) Doping a small amount of arsenic (As) into a crystal of pure germanium (Ge) produces a (I) n-type semiconductor (II) p-type semiconductor (III) metal (IV) semiconductor with a small number of electrons in the antibonding conduction band (V) semiconductor with a small number of holes in the bonding valence band (A) only I is correct (B) only II is correct (C) I and IV are correct (D) I and V are correct (E) II, III, and IV are correct Ans. 3 C 1
(4) (6 pts) or which one of the following processes/reactions is S largest? (Put the letter of the correct answer in the blank provided.) (A) CaCO 3 (s) CaO(s) + CO 2 (g) (B) NC(g) CN(g) (C) 2 O(s) 2 O(l) (D) C(s, diamond) C(s, graphite) (E) CaO(s) + SiO 2 (s) CaSiO 3 (s) Ans. 4 A (5) (8 points) Acetone (C 3 6 O) has a density of 0.791 g/cm 3 and a vapor pressure of 230 torr at 25 C. In a gas-tight, unventilated room that is 2.75 m high, 3.0 m wide, and 4.0 m long, will 10.0 L of acetone evaporate completely? (No credit unless you show your work.) 10.0 L of acetone has a mass of (0.791 g)(10000 cm 3 ) = 7910 g hat is (7910 g)/(58 g/mol) = 136.4 mol of acetone in the liquid Volume of the room = 2.75 3.0 4.0 m 3 = 33 m 3 = 33,000 L If the room is saturated with acetone vapor, the number of moles that the room can hold is n = (230/760 atm)(33,000 L)/(0.082059 L-atm/mol K)(298 K) = 408.4 mol possible in the vapor All the acetone will evaporate before the room is saturated with vapor. 2
(6) (12 points) Consider the following molecule. he circled numbers are meant to label specific atoms for some of the questions below. Put answers in the blanks provided. (a) (2 pts) ow many sigma (σ) bonds are present in this molecule? (Count all of them!) 20 C 3 1 (b) (2 pts) ow many pi (π) bonds are present in this molecule? 4 3 N 2 N 2 (c) (2 pts) What is the angle, CNC, (in degrees) at the nitrogen atom labeled as #3 in the figure? 120 (d) (2 pts) What is the hybridization of the nitrogen atom labeled as #1 in the figure? sp 3 (e) (2 pts) What is the hybridization of the carbon atom labeled as #2 in the figure? sp 2 (f) (2 pts) What is the hybridization of the nitrogen atom labeled as #3 in the figure? sp 2 (7) (21 points) Data and answers for this question are on the following two pages. Be sure to complete the entire question. icl 4 is a dense, colorless, distillable liquid (at room temperature) that is an important precursor used in the industrial production of i metal. icl 4 is usually produced by the chloride process, which typically involves reaction of ilmenite (eio 3 ) with carbon under flowing chlorine at 900 C (the carbon is in excess and under such conditions, CO is formed instead of CO 2 ): 2 eio 3 (s) + 7 Cl 2 (g) + 6 C(s) 2 icl 4 (g) + 2 ecl 3 (g) + 6 CO(g) owever, a two-step process has been developed because the ecl 3 that is formed in the reaction above causes difficulty in separation of the product. In the first step, a rather impure form of io 2 (called rutile slag ) is made by reacting carbon with eio 3 without the chlorine: eio 3 (s) + C(s) io 2 (s, rutile) + e(s) + CO(g). (1) In the second step of the process, icl 4 is made by reaction of the crude rutile io 2 formed in the first step, after removing the iron, with more carbon plus chlorine gas: io 2 (s, rutile) + 2 C(s) + 2 Cl 2 (g) icl 4 (g) + 2 CO(g). (2) Use the data in the table on the top of the next page to answer the questions that follow. 3
Substance & state f, kj mol 1 G f, kj mol 1 S, J mol 1 K 1 C graphite (s) 5.74 CO(g) 110.52 137.15 197.56 O 2 (g) 205.03 Cl 2 (g) 223.08 e(s) 27.28 i(s) 30.63 icl 4 (g) 763 727 355 io 2 (s) 944.8 889.5 50.3 eio 3 (s) 1237 1159 106 (a) (7 pts) Compute and G for reaction (1) at 298 K. eio 3 (s) + C(s) io 2 (s, rutile) + e(s) + CO(g). (1) = 944.8 110.52 + 1237 = +182 kj G = 889.5 137.15 + 1159 = +132 J/K (b) (7 pts) Showing all your work, find the range of temperatures for which reaction (1) is thermodynamically spontaneous. Because > 0, at low temperature G > 0 and the reaction is not spontaneous. It will become spontaneous if G changes sign which it will since S > 0. Set G = S = 0 to find that temperature. S = 50.3 + 27.28 + 197.56 106 5.74 = +163 J/K S = 0 when = / S = (+182000 J)/(163 J/K) = 1117 K = 844 C In summary, reaction (1) is not spontaneous below 844 C (1117 K) and is spontaneous above that temperature. Problem 7 is continued on the next page 4
Problem 7, continued (c) (7 pts) Showing all your work, find the range of temperatures for which reaction (2) is thermodynamically spontaneous. io 2 (s, rutile) + 2 C(s) + 2 Cl 2 (g) icl 4 (g) + 2 CO(g). (2) = 763 + (2)( 110.52) ( 944.8) = 39 kj S = 355 + (2)(197.56) 50.3 (2)(5.74) (2)(223.08) = +242 J/K BO and S favor spontaneity, reaction (2) is spontaneous at all temperatures. (8) (8 points) he reaction shown below is involved in the chemistry of the stratosphere, and has been studied extensively due to concern over depletion of the ozone layer. S(s) + O 2 (g) SO 2 (g) = 296.8 kj 2 S(s) + 3 O 2 (g) 2 SO 3 (g) = 791.2 kj Use the data above to determine for the following reaction: 2 SO 3 (g) 2 SO 2 (g) + O 2 (g) he heats of formation for SO 2 and SO 3 can be obtained from the two reactions given: f (SO 2 ) = 296.8 kj/mol f (SO 3 ) = ( 791.2 kj)/2 mol = 395.6 kj/mol hese can be used to calculate for the reaction: = (2 mol)( 296.8 kj/mol) (2 mol)( 395.6 kj/mol) = 197.6 kj Alternatively, you can flip the second reaction and then 2 times the first reaction to it. 5
(9) (13 points) Molecular orbitals can be used to describe the bonding in any molecule, not just the homonuclear diatomics. or nitric oxide (NO), the MO treatment is much better than the Lewis model for handling the odd number of electrons. (a) (8 pts) Draw a molecular orbital energy level diagram for the NO molecule. Include all the orbitals formed from the valence atomic orbitals (i.e., 2s and 2p). Label the diagram completely, identifying all of the molecular orbitals, and use arrows to indicate the electron configuration. (IN: he molecular orbitals formed are the same ones as we've seen for O 2. So you can draw the orbital diagram for O 2, and then just use the correct number of electrons for NO.). σ* 2p π* 2p π 2s σ 2s σ* 2s σ 2s (b) (2 pts) Compare neutral NO to the NO + ion. Which will have the longer bond length and how do you know? (IN: use your diagram from part (a) to decide.) In forming the NO + ion, NO loses its π* electron. hus, the bond shortens in the NO + ion. (c) (3 pts) Rank the following molecules in order of increasing bond strength: N 2, O 2, NO. Give a BRIE explanation of your choice. (weakest bond) O 2 < NO < N 2 (strongest bond) bond orders: 2 2.5 3 he molecules have, in order, 2 π* electrons, 1 π* electron, and 0 π* electrons hence the bond orders. 6