Chem 107 - Hughbanks Final Exam, May 11, 2011 Name (Print) UIN # Section 503 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. The exam contains 13 problems, with 9 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. SIGNATURE:
(1) (10 pts) Consider the elements in the second, third, and fourth rows of the periodic table (lithium through krypton, elements #3 through #36). Answer the following questions by putting the symbol(s) of the element in the blanks provided. (2 pts. each) Ne Zn K Mg, Ca (a) Which one element has the greatest ionization energy? (b) Which one element has a 3d 10 4s 2 valence electronic configuration? (c) Which one element has the lowest electronegativity? (d) List two of these elements (called X ) that might form an ionic chloride compound with the formula XCl 2? C, Si (e) List two of these elements (called X ) that might form a covalent compound with hydrogen with the formula XH 4? (Assume that the Lewis diagram for XH 4 obeys the octet rule.) (2) (10 pts) Write T or F in the blanks to indicate whether the statements are true F F or false. (2 pts each.) (a) Stronger bonds tend to be longer bonds. (b) Exothermic reactions are always spontaneous. (For grading) Scores 1 /10 2 /10 3 /8 4 /6 5 /8 6 /8 7 /7 8 /8 9 /8 10 /15 11 /20 12 /26 13 /16 Tot. /150 T (c) In the reaction MgCO 3 (s) MgO(s) + CO 2 (g), ΔS > 0. F T (d) According to the Second Law of Thermodynamics, any reaction for which ΔS < 0 can not occur. (e) In a reaction where no gases are consumed or formed, ΔE and ΔH are nearly equal. (3) (8 pts) By the use of Lewis diagrams and considering hybridization at the central atom, decide which of the following molecules or ions are bent. (I) NO 2 + (the nitrogen atom is in the middle) (II) NO 2 (the nitrogen atom is in the middle) (III) HCN (IV) SO 2 (the sulfur atom is in the middle) (V) HCO + (the carbon atom is in the middle) (A) I and V (B) II and V (C) III and IV (D) II and IV (E) I, II, and IV Ans. 3 D 1
(4) (6 pts) Arrange the following sets of anions in order of increasing ionic radii: (a) S 2, P 3, Cl : Cl < S 2 < P 3. (b) S 2, O 2, Se 2 : O 2 < S 2 < Se 2. (c) Br, I, Cl : Cl < Br < I. (5) (8 pts) Classify the following compounds as acids or bases, weak or strong, in accord with their behavior in aqueous solution (circle the correct answer in each case). (A) HClO 4 : weak acid weak base strong acid strong base (B) CsOH : weak acid weak base strong acid strong base (C) H 2 CO 3 : weak acid weak base strong acid strong base (D) CH 3 CH 2 NH 2 : weak acid weak base strong acid strong base (6) (8 pts) Given the following data for the reaction A + B C, which of the expressions below is a fully correct statement of the rate law? (A) rate = [C] t (C) rate = [B] t (E) rate = [C] t Expt. Initial [A] Initial [B] Initial Rate of Formation of C 1 0.20 M 0.10 M 5.0 10 6 M/s 2 0.30 M 0.10 M 1.13 10 5 M/s 3 0.30 M 0.20 M 1.13 10 5 M/s = k[a] 2 (B) rate = [A] t = k[a][b] 2 (D) rate = [C] t = k[b] 2 = k[a] 2 [B] = k[b] Ans. 6 A 2
(7) (7 pts) An atom in its ground state absorbs a photon with a 200 nm wavelength. The atom then absorbs a second photon with a wavelength of 250 nm. The atom then returns to its ground state with emission of a photon. What is the wavelength of the emitted photon? (A) 50 nm (B) 300 nm (C) 111 nm (D) 450 nm (E) 222 nm Ans. 7 C (8) (8 pts) Put the following molecules in order of increasing boiling point and clearly explain the reasons that justify your answer: H 2 O, H 2 S, H 2 Se, H 2 Te. lowest boiling point highest boiling point H 2 S < H 2 Se < H 2 Te < H 2 O. Explanation: Hydrogen bonding in H 2 O causes water to have the strongest intermolecular forces, hence its relatively high boiling point. The other three increase as the strength of the dispersion forces for the chalcogenide increases, from sulfur to tellurium. (9) (8 pts) Two resonance structures are shown for the cation below. How many hydrogen atoms for this cation must lie in the plane of the paper? (a) 1 (b) 3 (c) 4 (d) 5 (e) 7 Ans. 9 C 3
Questions 10-12 show all work (61 pts) (10) (15 pts) Nitrous oxide (N 2 O) decomposes according to the reaction shown below. The rate of this reaction is exceedingly slow in pure N 2 O, but chlorine gas accelerates it significantly. 2 N 2 O(g) 2 N 2 (g) + O 2 (g) The rate law depends on the pressures of both N 2 O and Cl 2 and can be written as rate = k [P(N 2 O)] m [P(Cl 2 )] n where the partial pressures P(N 2 O) and P(Cl 2 ) are used instead of concentrations. Initial rate date were collected on this reaction at T = 800 K: Expt. P(N 2 O) (torr) P(Cl 2 ) (torr) initial rate (torr min 1 ) 1 300 40 3.0 2 150 40 1.5 3 300 10 1.5 (a) (3 pts) What is the specific term used to describe the role of chlorine, Cl 2, in this reaction? Cl 2 is a catalyst (b) (7 pts) Determine the order of this reaction with respect to N 2 O and Cl 2. m = 1 n = ½ (c) (5 pts) Give the value of the rate constant, k, with the correct units. k = (rate)/{[p(n 2 O)] [P(Cl 2 )] 1/2 } = (3.0 torr min 1 ) /{[300 torr] [40 torr] 1/2 } = 0.00158 torr 1/2 min 1 4
(11) (20 pts) Consider the reaction between reactants A and B to give products D and E the energy diagram below applies. C is an intermediate product that is not present when the reaction is over. (Note: There are five parts to this question, part (e) is on the top of the next page.) A + B Step 1 C Step 2 D + E (a) (3 pts) Give the values of ΔE and the energy of activation (E a ) for the reactions indicated. E(A + B C) = +50 kj E(A + B D + E) = 10 kj E a (A + B C) = +90 kj (b) (3 pts) Which step is rate determining (circle the correct answer here and in part c)? Step 1 Step 2 Both Neither No way to tell (c) (3 pts) The overall reaction is Exothermic Endothermic Neither No way to tell (d) (5 pts) Catalyst P increases the rate of the first step (A + B C) but does not affect the second step (C D + E). Catalyst Q increases the rate of the second step, but does not affect the rate of the first step. Which catalyst will be most effective in increasing the rate of the overall reaction (A + B D + E)? (You must give a correct brief explanation for your answer to receive credit.) Since the first step is rate determining, catalyst P will be more effective because that is the step it affects. 5
(10) continued (e) (6 pts) The rate constant for the reaction (A + B D + E) is 0.0072 M -1 s -1 at a temperature of 30.0 C. Use the Arrhenius equation to calculate the rate constant at a temperature of 45.0 C? k = A exp{ E a /RT} or ln(k 2 /k 1 ) = (E a /R)(1/T 1 1/T 2 ) ln(k 2 /k 1 ) = (E a /R)(1/T 1 1/T 2 ) = (90000 J mol -1 )/(8.314 J mol -1 K -1 ) (1/(303 K) 1/(318 K)) ln(k 2 /k 1 ) = 1.685 k 2 /k 1 = (0.0072 M -1 s -1 )e 1.685 = 0.0388 M -1 s -1 (12) (26 pts) The following two reactions are involved in alternative methods of purification of iron from its ores. Some relevant thermodynamic data is shown below. Reaction # 1 Fe 2 O 3 (s) + 3 CO(g) 2 Fe(s) + 3 CO 2 (g) S = +15.2 J/K Reaction # 2 2 Fe 2 O 3 (s) + 3 C(s, graphite) 4 Fe(s) + 3 CO 2 (g) S = +557.98 J/K Substance H f (kj mol -1 ) G f (kj mol -1 ) S (J K -1 mol -1 ) Fe 2 O 3 (s) 824.2?? 87.40 CO(g) 110.5 137.2 197.6 C(s, graphite) 5.740 Fe(s)?? CO 2 (g) 393.5 394.4 213.6 NOTE: You do not need to fill in all the blanks in the table - just answer the questions below! (a) (5 pts) Calculate H for the reaction #1. H = 2(0) + 3( 393.5) 3( 110.5) ( 824.2) = 24.8 kj Continued on next page 6
(b) (5 pts) Calculate S for solid iron. S = +15.2 = 2(S Fe ) + 3(+213.6) 3(197.6) 87.40 = 2S Fe + 3(+213.6) 3(197.6) 87.40 = 2S Fe 39.4 J K -1 2S Fe = 39.4 + 15.2 = 54.6 J K -1 mol -1 (c) (8 pts) Calculate G f for Fe 2 O 3. For reaction #1, H = 24.8 kj and S = +15.2 J/K G = 24.8 kj (298 K)(+.0152 kj) = 20.27 kj 20.27 kj = 2(0) + 3( 394.4) 3( 137.2) G f (Fe 2 O 3 ) = 771.6 kj G f (Fe 2 O 3 ) G f (Fe 2 O 3 ) = 771.6 kj + 20.27 kj = 751.23 kj mol -1 (d) (8 pts) At what temperatures is reaction #2 spontaneous? H = 4(0) + 3( 393.5) 2(0) 2( 824.2) = +467.9 kj set G = 0 = H T S T = H/ S = (+467900 J)/(557.98 J/K) = 839 K Above 839 K, the process should be spontaneous. 7
(13) (16 pts) On the next page is a temperature-pressure phase diagram for uranium hexafluoride (UF 6 ). UF 6 is an important compound used by nuclear chemists and engineers in uranium isotope separation (of 235 UF 6 from 238 UF 6 ). Note that the pressure axis is logarithmic and expressed in SI units (1.0 kpa = 1.0 kilopascal = 1000 Pa = 1000 N m 2 = 0.01 bar = 0.009869 atm). Read the directions below carefully before proceeding with this problem. (a) (12 pts) There are twelve (12) missing labels on this diagram. From the following list of words, supply the correct labels. Put the LETTER before each word (A, B, C, Q) into the correct box in each case. The three boxes with black borders are labels for the entire region in which they sit. Six of the labels refer to the processes indicated by arrows next to the labels that cross phase boundaries. There are three more labels: one of them refers to the dashed horizontal line in the diagram, one refers to the point where the dashed line intersects one of the phase boundaries, and the third refers to the point where the three regions of the diagram meet. Note: Five labels do not belong on the diagram just skip those! (A) Condensation (to a liquid) (B) Gas (C) Liquefaction (melting) (D) Normal boiling point (E) Plasma (F) Vaporization (boiling) (G) Solid (H) Desublimation (opposite of sublimation) (I) Normal melting point (J) Liquid (K) Equivalence point (L) Triple Point (M) Sublimation (solid vaporization) (N) Normal sublimation point (O) Atmospheric pressure (P) Equilibrium point (Q) Solidification (freezing) (b) (4 pts) At 10 atm. pressure, give approximate values for the temperatures (in C) at which UF 6 melts and boils. Melts: 60 C Boils: 141 C 8
C Q F J A G N L M O H B 9