Chemistry 201 FINAL EXAM Illinois Wesleyan University December 12, 2005 Fall 2005

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1 Chemistry 201 FINAL EXAM Name KEY Illinois Wesleyan University December 12, 2005 Fall 2005 There are 200 points on this exam, and you have exactly two hours to complete it. 1. (12 points) a. Give the name or formula of the following: Formula Na 2 Cr 2 7 Name sodium dichromate Ba(Cl3)2 Barium chlorate P 4 10 tetraphosphorous decaoxide b. Give the name or structure of the following: Structure Name H 2-hexanol H pentanoic acid H 3 C CH 3 methyl ethanoate /12

2 Chemistry 201 Fall 2005 FINAL EXAM page 2 of (12 points) For each of the following molecules, draw a valid Lewis structure. If there is more than one valid Lewis structure, draw two resonance structures. Draw structure(s) with the least separation of formal charge. If there is only one valid Lewis structure, you will use only one of the two boxes for each compound. List the formal charge on the central atom. (nly neat final answers in the boxes will be graded.) HS 4 (Hint: the hydrogen is attached to an atom of the more electronegative element) H S H S Formal charge on central atom: 0 Formal charge on central atom: 0 Br 2 Br Br Formal charge on central atom: _0 Formal charge on central atom: 0 3. (8 points) Draw a Lewis structure for three oxygen compounds: the peroxide anion ( 2 2 ), oxygen gas ( 2 ) and ozone ( 3 ) ther resonance form rder these species from longest to shortest oxygen-oxygen bond length. (longest bond) 2 2- > 3 > 2 (shortest bond) /20

3 Chemistry 201 Fall 2005 FINAL EXAM page 3 of (27 points) Multiple choice: CLEARLY circle the letter corresponding to the best answer. In which of the following compounds does the carbon atom have the highest oxidation number? a. methane b. carbon dioxide c. carbon monoxide d. formaldehyde (structure given) H e. formic acid (structure given) H H H formaldehyde formic acid Which of the following compounds are soluble in water: I. CaCl 2 II. K 2 S III. AgBr a. I and III b. I and II c. III only d. I, II, and III are all soluble in water e. none of them are soluble in water Which of the following compounds would you expect to be a weak electrolyte when dissolved in water? I. MgCl 2 II. HF III. AgBr a. II and III b. I and II c. II only d. I only e. none of them are weak electrolytes Which of the following are ordered correctly, in terms of ionization energy? a. > N > C b. Rb > K > Na c. Na + > Ne > Na d. S > Cl > Ar e. none of the above are ordered correctly How many electrons are present in a rhodium (II) ion? a. 42 b. 43 c. 44 d. 45 e. 47 /15

4 Chemistry 201 Fall 2005 FINAL EXAM page 4 of MULTIPLE CHICE CNTINUED Which of the following is false? a. l is the symbol for the angular momentum (azimuthal) quantum number b. n is the symbol of the principal quantum number c. l can have values from +m l to m l d. the l value indicates the shape of the orbital e. the magnetic quantum number indicates the orientation of the orbital. The hybridization of the carbon atom in carbon dioxide is a. sp b. sp 2 c. sp 3 d. sp 3 d e. sp 3 d 2 The F-N-F bond angle in the NF 3 molecules is a. slightly larger than b. exactly 120 c. slightly less than d. exactly e. exactly 90 The F-Xe-F bond angle in the XeF 4 molecule is about. a. 90 b c. 120 d. 180 e (10 points) The insecticide DDT has the following composition by mass: 47.43% carbon, 2.56% hydrogen, and 50.01% chlorine. Determine the empirical formula of DDT. Assume 100g: g C (1 molc/12.011gc) = mol C x5= g H (1molH/1.0079gH) = mol H g Cl (1molCl/35.453gCl) = mol Cl So this is a mole ratio of C:H:Cl of : : Divide all by to get the ratio of 2.8 : 1.8: 1.0, which is 14:9:5, so empirical formula is C 14 H 9 Cl 5 /22

5 Chemistry 201 Fall 2005 FINAL EXAM page 5 of (9 points) "Glove warmers" are small packets of ground metallic iron that, by reacting with oxygen, form iron (III) oxide, Fe 2 3. If the standard molar enthalpy of formation of iron (III) oxide is kj/mol, calculate the heat produced when g of Fe reacts with excess oxygen g Fe 1molFe gFe 1molFe kJ = kj 1molFe 1molFe (16 points) Magnesium hydroxide, Mg(H) 2, is an insoluble salt. However, when solid Mg(H) 2 reacts with aqueous hydrochloric acid, the products of the reaction are soluble. a. Write the balanced chemical equation describing this reaction. Mg(H)2 (s) + 2 HCl (aq) MgCl2 (aq) + 2 H2 (l) b. Imagine a solid sample containing both Mg(H) 2 and some water-soluble non-reacting impurity. This sample is put in ml of water, and then ml of M HCl are added. The solid dissolves completely, indicating that the solid Mg(H) 2 was the limiting reactant in the reaction above. The excess HCl in the solution was then titrated with M NaH. It took ml of the NaH solution to completely neutralize the excess HCl. How many grams of Mg(H) 2 were present in the initial sample? How much HCl was added? L mol/l = mol HCl How much excess HCl was left to titrate? L NaH 0.150molNaH = mol NaH L - NaH 1molHCl mol NaH = mol HCl excess 1molNaH How much HCl reacted with Mg(H) 2 before the titration? mol HCl initial mol HCl excess = mol HCl reacted How much Mg(H) 2 does this amount of HCl react with? mol HCl 1molMg(H) 2 = mol Mg(H) 2 2molHCl mol Mg(H) gMg(H) 2 = g Mg(H) 2 1molMg(H) 2 /25

6 Chemistry 201 Fall 2005 FINAL EXAM page 6 of (8 points) Show the structure and name of the organic product of the reactions: H + Cr propanone or acetone H H NaBH 4 1-butanol 9. (15 points) Predict the products and write balanced net reactions, indicating the correct phases: Na 2 (s) + H 2 (l) Na 2 (s) + H 2 (l) 2 NaH (aq) Ba(H) 2 (aq) + H 3 P 4 (aq) 3 Ba(H) 2 (aq) + 2 H 3 P 4 (aq) Ba 3 (P 4 ) 2 (s) + 6 H 2 (l) Na 2 S 4 (aq) + BaCl 2 (aq) Na 2 S 4 (aq) + BaCl 2 (aq) 2 NaCl (aq) + BaS 4 (s) KF(aq) + HN 3 (aq) KF(aq) + HN 3 (aq) HF (aq) + KN 3 (aq) Pb(C 2 H 3 2 ) 2 (aq) + HI(aq) Pb(C 2 H 3 2 ) 2 (aq) + 2 HI(aq) PbI 2 (s) + 2 HC 2 H 3 2 (aq) /23

7 Chemistry 201 Fall 2005 FINAL EXAM page 7 of (10 points) Balance the following unbalanced redox reaction. You must show your work to receive partial credit. IN ACID: I3 - + SbH3 I2 + Sb antimony half reaction: SbH 3 Sb + 3 H e iodine half reaction: 2 I H e I H 2 Final Balanced Equation (put inside box) (SbH 3 Sb + 3 H e ) 10 (2 I H e I H 2 ) 3 10 SbH I H e 10 Sb + 30 H e + 3 I H 2 10 SbH I H + 10 Sb + 3 I H (9 points) The behavior of a substance in a magnetic field provides insight into how its electrons are arranged. A substance with one or more unpaired electrons is attracted into a magnetic field and is said to be paramagnetic. A substance with no unpaired electrons is repelled by the magnetic field and is said to be diamagnetic. Write the electron configurations of the atoms/ions listed below and, using what you know about the filling of the orbitals, determine if the atoms/ ions are paramagnetic or diamagnetic. You may use abbreviated forms for the electron configuration (i.e., [Ne] for 1s 2 2s 2 2p 6 ). Electron configuration Paramagnetic or diamagnetic? a. Zn 2+ [Ar] 3d 10 diamagnetic b. Bi [Xe] 6s 2 5d 10 4f 14 6p 3 paramagnetic c. Cuprous ion [Ar] 3d 10 diamagnetic /19

8 Chemistry 201 Fall 2005 FINAL EXAM page 8 of (16 points) Consider the combustion of ethane. a. Write the balanced chemical equation describing this reaction. All reactants and products will be present in the gas phase. 2 C 2 H 6 (g) (g) 4 C 2 (g) + 6 H 2 (g) b. Combustion reactions are oxidation-reduction reactions. For the combustion of ethane, which element is oxidized, and which element is reduced? carbon is oxidized, oxygen is reduced c. Calculate the standard enthalpy of the reaction, using standard enthalpy of formation data, given below. Your answer should be appropriate for the equation as you have balanced it in part a. DH = S (DH f ) products S (DH f ) reactants = 4 (-393.5) + 6 (-241.8) - [ 2 (-84.7) + 7 (0) ] = [ ] = [ 169.4] = kj (note that kj is also a correct answer, if you balanced the equation in part a as: C 2 H 6 (g) + 7/2 2 (g) 2 C 2 (g) + 3 H 2 (g) Enthalpy of formation data compound DH f (kj/mol) CH 4 (g) 74.8 C 2 H 6 (g) 84.7 C 2 H 4 (g) C 3 H 8 (g) C 3 H 6 (g) C 2 (g) C (g) H 2 (g) H 2 (l) /16

9 Chemistry 201 Fall 2005 FINAL EXAM page 9 of (16 points) You and a classmate are working together on a chemistry worksheet and one of the questions asks you to estimate the lattice energy of sodium iodide from published thermodynamic values. Your classmate does not yet have a very clear understanding of the thermodynamic processes involved, but wanted to help out by doing the library research. He left the following data on your desk and went to orchestra practice. Use his data to estimate the standard enthalpy of formation for solid sodium iodide. YU WILL USE NLY SME F THE PRVIDED VALUES. There is more than one way to solve the problem. Show your work. lattice energy for sodium iodide (expressed for both the forward and reverse process): Na + (g) + I (g) NaI (s) ; 702 kj/mol NaI (s) Na + (g) + I (g) ; +702 kj/mol first ionization energy of Na = 496 kj/mol first ionization energy of I = 1008 kj/mol electron affinity of Na = kj/mol electron affinity of I = kj/mol DH f 0 Na (g) = kj/mol DH f 0 Na + (g) = kj/mol DH f 0 I (g) = kj/mol DH f 0 I (g) = 197/mol Na + (g) + I (g) LE NaI (s) IE Na (g) DH f EA I (g) DH f DH f this is the goal Na (s) + 1/2 I 2 (s) From the diagram above, it is clear that: DH f (NaI) = DH f (Na) + DH f (I) + IE (Na+) + EA (I) + LE = = 287 kj (There are other acceptable ways to do this problem) /16

10 Chemistry 201 Fall 2005 FINAL EXAM page 10 of (16 points) Assuming that the reaction shown below goes to completion, calculate the final amount (in moles) of all species, if initially there are g of InN, g of HF, g of H 2 and g of N 2. You must prove any assumptions. HINT: build a table. InN(s) + 6 HF(g) + 3 H2(g) + N2(g) (NH4)3InF6(s) InN + 6 HF + 3 H2 + N2 (NH 4 ) 3 InF 6 initial grams MW (g/mol) initial moles change x 6x 3x x +x (mole ratio) final moles x x x x x value of x, if limiting HF has lowest value of x, therefore HF is limiting, and x = final moles x = x = x = x = x = So the final answer: mol InN left 0 mol HF left (HF was limiting reagent) mol H 2 left mol N 2 left mol (NH 4 ) 3 InF 6 formed 15. (4 points) List the following compounds in order of increasing absolute magnitude of lattice energy: LiI, LiF, Ca, RbI. Use the blanks provide. (lowest absolute magnitude) _RbI_ < _LiI_ < _LiF_ < _Ca_ (highest absolute magnitude) /20

11 Chemistry 201 Fall 2005 FINAL EXAM page 11 of (12 points) Given the following Lewis dot structures, determine the molecular geometry of the molecule and state whether it is polar or nonpolar. (Note that the way the Lewis structure is drawn is not meant to represent the geometry.) Molecule Lewis Structure Geometry Polar or non-polar? S 3 S trigonal planar non-polar (plus two more equivalent resonance structures) CH 3 Cl H Cl C H H tetrahedral or trigonal pyramidal polar PCl 3 Cl Cl P Cl trigonal pyramidal polar BrF 5 F F Br F F F square pyriamdial polar /12