Chemistry 112 Exam 3C Harwood/Fall 04 KEY 1. Which order is correct for increasing polarity (least polar first)? C—C C—Cl C—F C—I A B C D (a) A<D<B<C (d) A < D < C < B (b) D<C<B<A (e) A < B < C < D (c) C<B<A<D – 2. How many valence electrons are shown in the Lewis structure of OH ? (a) 8 (b) 7 (c) 6 (d) 9 (e) 10 3. Which of the following molecules is polar? (a) CO2 (d) CO (b) O2 (e) CH4 (c) C2H2 (H–C–C–H is skeleton structure) 4. Which pair is mismatched? (a) polarizability — dispersion forces (b) N–H or O–H bonds — hydrogen bonding (c) polarity — dipole moment (d) dipole moment — dipole-dipole forces (e) polar — low boiling point 5. Drawing (1) shows the equilibrium vapor pressure of a pure liquid. Which drawing 1 Chemistry 112 Exam 3C Harwood/Fall 04 KEY 6. The following pictures represent the equilibrium state for four different reactions of the type A2 + X2 ↔ 2 AX (X = B, C, D, E). A atoms are open circles. X atoms are closed circles. Which reaction has the smallest equilibrium constant? (a) A2 + B2 ↔ 2 AB (b) A2 + C2 ↔ 2 AC (c) A2 + D2 ↔ 2 AD (d) A2 + E2 ↔ 2 AE 7. A2 + B2 ↔ 2 AB A2 + C2 ↔ 2 AC A2 + D2 ↔ 2 AD A2 + E2 ↔ 2 AE A 6.0 m H2SO4 solution has d=1.34 g/mL. What is the mass percent H2SO4 in the solution? " 98.0794 g % mass H2SO 4 = ( 6.0 mol H2SO 4 )$ ' = 588 g # 1 mol & total mass of solution = 1000 g + 588 g = 1588 g " 588 g % $ ' = 0.370 # 1588 g & 0.370 ( 100% = 37.0% (a) 37.0 (b) 43.9 (c) 63.0 (d) 56.1 (e) 58.8 8. Which of the following solutions has the lowest freezing point? Molality of particles ! (a) 1.0 m glucose in water 1.0 lowest f.p. is sol’n with the largest molality of (b) 1.0 m CaCl2 in water 3.0 particles (c) 1.0 m NaBr in water 2.0 (d) 1.0 m NaCl in water 2.0 (e) water 0.0 9. KP for the reaction H2(g) + I2(g) ↔ 2 HI(g) is 54.3 at 698°C. What is the value of KC at this temperature? (a) 85.6 –2 (b) 1.80 × 10 –4 (c) 2.30 × 10 KP = KC (RT) Δn = 2-2 = 0 Δn so KP = KC 3 (d) 4.33 × 10 (e) 54.3 2 Chemistry 112 Exam 3C Harwood/Fall 04 KEY 10. Which direction will the following reaction (in a 5.0 L flask) proceed if the pressure of CO2(g) is 1.0 atm? CaCO3(s) ↔ CaO(s) + CO2(g) KP = 1.9 × 10 (a) to the left because Q < KP –23 Q = 1.0 (b) to the right because Q < KP Q>K (c) to the left because Q > KP (d) to the right because Q > KP (e) The reaction is at equilibrium. 11. For the system CaO(s) + CO2(g) ↔ CaCO3(s), the equilibrium constant expression is (a) [CO2] (d) [CaCO3] / [CaO] [CO2] (b) 1 / [CO2] (e) 1 / [CaO] [CO2] (c) [CaO] [CO2] / [CaCO3] 12. Which change will increase the quantity of product in the following reaction? 2 SO2(g) + O2(g) ↔ 2 SO3(g) ΔH° = -198 kJ (a) an increase in temperature (b) an increase in the volume of the container (c) an increase in pressure (d) a decrease in amount of SO2 Increasing T will cause shift to left (exothermic). Increasing V (decreasing P) will cause shift toward most number of moles of gas (left). Decreasing amount of reactant will cause shift to left. Increasing P will cause shift toward least number of moles of gas (right). 13. Consider the reaction CO(g) + H2O(g) ↔ CO2(g) + H2(g) ΔH° = -41 kJ All of the following changes would shift the equilibrium to the right except one. Which one would not cause the equilibrium to shift to the right? (a) Add some CO. (b) Remove some CO2. (c) Decrease the container volume. (d) Decrease the temperature. (e) Increase the partial pressure of H2O. Decreasing T will cause shift to right (exothermic). Decreasing V (increasing P) will cause shift toward least number of moles of gas (no effect). Increasing amount of reactant will cause shift to right. Decreasing amount of product will cause shift to right. END OF SCANTRON PORTION OF EXAM 3 Chemistry 112 Exam 3C Harwood/Fall 04 KEY 14. (a) What is the shape (molecular geometry) of CHCl3? (Circle one and show work.) 4 pts tetrahedral linear bent Cl | Cl – C – Cl | H (b) Is PCl3 polar or nonpolar (C is the central atom)? (Show work) ! Polar •• Cl – P – Cl | 4 pts Cl (c) List all intermolecular forces present in a sample of PCl3? dipole-dipole dispersion 4 pts ! 15. Consider the following endothermic reaction: NH4HS(s) ↔ NH3(g) + H2S(g) (a) NH4HS(s) is placed in a vessel and decomposes at 22°C. After equilibrium is reached, 0.32 atm of both NH3 and H2S are present along with solid NH4HS. What is the equilibrium constant, KP, for this reaction? 4 pts KP = (0.32)(0.32) KP = 0.10 (b) Suppose that 0.128 atm of pure NH3(g) and 0.128 atm of pure H2S(g) are placed in a 5.00 L flask at 50°C (KP @ 50°C = 0.52). What are the partial pressures of NH3 and H2S at equilibrium? 5 pts NH4HS(s) NH3(g) ↔ + H2S(g) Initial ---- 0.128 0.128 Change --- –x –x Equilibrium --- 0.128 – x 0.128 – x 2 0.52 = (0.128 – x) = 0.016384 – 0.256x + x 2 2 x – 0.256x – 0.503616 = 0 x = 0.849, –0.593 PNH3 = PH2S = 0.128 – (–0.593) = 0.721 atm (c) Will a decrease in temperature favor the decomposition of NH4HS? 4 pts No (d) Will an increase in volume favor the decomposition of NH4HS? 4 pts Yes 4 Chemistry 112 Exam 3C Harwood/Fall 04 KEY BONUS QUESTION 16. Lysine, one of the amino acid building blocks found in proteins, contains 49.29% C, 9.65% H, 19.16% N, and 21.89% O by elemental analysis. A solution prepared by dissolving 30.0 mg of lysine in 1.200 g of the organic solvent biphenyl, gives a freezingpoint depression of 1.37°C. What is molecular formula of lysine? (Lysine does not dissociate.) (Kf for biphenyl is 8.00°C/m) 10 pts C: H: N: O: (49.29 g)(1 mol/12.011 g) = 4.104 mol ==> 3 (9.65 g)(1 mol/1.0079 g) = 9.5744 mol ==> 7 (19.16 g)(1 mol/14.0067 g) = 1.36792 mol ==> 1 (21.89 g)(1 mol/15.9994 g) = 1.36818 mol ==> 1 Empirical formula: C3H7NO Molar mass = 73 g/mol ΔTf = Kfm 1.37°C = (8.00°C/m) m m = 0.17125 mol / kg (1.200 × 10 –3 kg)(0.17125 mol/ kg) = 2.055 × 10 Molar mass = (30.0 × 10 –3 g) / (2.055 × 10 –4 –4 mol mol) = 146 g/mol (146 g/mol)/(73 g/mol) = 2 Molecular formula = C6H14N2O2 5