Sectional Solutions Key

Sectional Solutions Key 1. For the equilibrium: 2SO 2 (g) + O 2 (g) 2SO 3 (g) + 188 kj, the number of moles of sulfur trioxide will increase if: a. the temperature of the system is increased (at constant V) b. the volume of the reaction flask is decreased (at constant T) c. O 2 (g) is removed from the flask (leaving T and V constant) d. the pressure of the system is increased by adding an inert gas (leaving V and T constant) e. all of the above Solution: The total number of reactant moles is greater than the total number of product moles (3 compared to 2). According to Le Chatelier s principle, decreasing the volume (or increasing the pressure) of a system at equilibrium shifts the equilibrium position in the direction of fewer moles product side in this case. 2. Consider the reaction: N 2 (g) + 3H 2 (g) 2NH 3 (g). A mixture is prepared by combining 0.81 mol of N 2 and 0.53 mol of H 2 in a 2.0 L reaction vessel. At equilibrium, 0.70 mol of N 2 remained. What is the equilibrium concentration of H 2? a. 0.15 M b. 0.20 M c. 0.30 M d. 0.10 M e. 0.42 M Solution: n nitrogen = initial mol N 2 equilbrium mol N 2 = 0.81 mol 0.70 mol = 0.11 mol. From stoichiometry, the corresponding n hydrogen is 3 x n nitrogen or 0.33 mol. The moles of hydrogen at equilibrium = initial mol H 2 n hydrogen = 0.53 mol 0.33 mol = 0.20 mol. The equilibrium concentration of hydrogen is 0.20 mol H 2 /2.0 L = 0.10 M 3. If the value of the equilibrium constant K C for A (g) + 2B (l) 4C (g) is 0.125, the value of the equilibrium constant for 4C (g) A (g) + 2B (l) is: a. 0.125 b. -0.125 c. 0.975 d. 8.00 e. 12.5 Solution: The two equations are the same except for direction. The equilibrium constants for such equations are related as follows: K reverse = 1/K forward. K for the forward reaction is 0.125; hence, K for the reverse reaction is 1/0.125 = 8.00 4. In a particular reaction in nuclear chemistry, it was found that the total mass of the product nuclei is less than the total mass of the reactant nuclei. This indicates that: a. an alpha particle must have initiated the reaction b. energy is released in the reaction c. a nuclear chain reaction occurred d. the difference in mass is equal to the constant c e. nuclear activation occurred

Solution: A nuclear reaction involves the interconversion of mass and energy. Since the sum of the product masses is less than the sum of the reactant masses, energy is released. 5. When aqueous solutions of calcium nitrate and sodium carbonate are mixed: a. calcium carbonate precipitates b. calcium sodiumate precipitates c. sodium nitrate precipitates d. carbon nitrate precipitates e. no precipitate forms Solution: When these two solutions are combined two products are possible calcium carbonate and sodium nitrate. According to the solubility rules, calcium carbonate is insoluble, hence it will precipitate. The following information is to be used for the next two questions. Given two solutions of a nondissociating and nonvolatile solute (M = 40 g/mol). The solution makeup is as follows: Solution A: 20 g of solute + 1200 g of H 2 O Solution B: 30 g of solute + 900 g of H 2 O For the solvent water, K f = 1.86 C/m and K b = 0.51 C/m. 6. Which of the following is correct? a. the vapor pressure above Solution A is greater than that above solution B (both at equilibrium). b. the boiling point of Solution A is higher than that of Solution B. c. the freezing point of Solution A is lower than that of Solution B. d. the mole fraction of solute in Solution B is 0.025 e. the molality of solute in Solution B is 0.050. Solution: From the given information the molality of solution A is 0.42 m and the molality of solution B is 0.83 m. The solute concentration in solution B is greater than the solute concentration in solution A. Hence, the vapor pressure above solution B is smaller than the vapor pressure above solution A. 7. If half of solution A is mixed with all of solution B, the freezing point of the new solution would be: a. +1.34 C b. 0.00 C c. -0.78 C d. -1.24 C e. -1.34 C Solution: The mass of solute in the resulting solution is 40 g (= 1 mole) and the mass of solvent is 1500 g (1.5 kg). The molality is 0.667 m. Substituting into the equation K f x m = T f and solving for T yields 1.24. By definition, T = T o T soln where T o is the freezing point of pure water, 0 o C. Hence, T soln = -1.24 o C.

8. A particular substance has the following properties: it is a colorless gas at ordinary temperature and pressure with a distinctive odor; it is very soluble in water and its aqueous solutions are basic. Which of the following substances has these characteristics? a. NaOH b. carbon dioxide c. ethanol d. CH 4 e. ammonia Solution: ammonia is a colorless gas with a distinctive odor and its aqueous solutions are basic. 9. The equilibrium partial pressure of water vapor over a 0.10 mole sample of liquid water is 25 mm Hg at 25 C. If the amount of liquid water is increased to 0.20 mole, the equilibrium partial pressure would be: a. 25 mm Hg b. 12.5 mm Hg c. 50 mm Hg d. 75 mm Hg e. cannot be determined without knowing the volume of the container Solution: The vapor pressure of a liquid does not depend on the quantity of liquid except that some liquid must be present in order to establish equilibrium. 10. All of the following are involved in linking water molecules together in the liquid and solid states EXCEPT: a. the polarity of the H O bonds within each water molecule b. dipole-dipole forces c. hydrogen bonding d. van der Waals forces e. attraction between hydrogen molecules and oxygen molecules Solution: A water molecule is composed of hydrogen and oxygen atoms linked together in a specific way. Hydrogen and oxygen molecules are not present in a water molecule or in a sample of water. 11. Which of the following pairs is not a conjugate acid-base pair? a. NH 4 + and NH 3 b. H 3 O + and OH - c. H 2 CO 3 and HCO 3 - d. H 2 O and OH - e. HF and F - Solution: By definition, an acid and its conjugate base differ by H +. Hence, H 3 O + and OH - are not a conjugate acid-base pair. 12. The label on a stock bottle of hydrochloric acid indicates that the solution is 36.7% HCl by mass with a solution density of 1.18 g/ml. What is the molarity of HCl (M = 36.45 g/mol) in this solution? a. 8.5 M

b. 11.9 M c. 18.6 M d. 43.3 M e. 311 M Solution: If 100 g of solution is assumed, the mass of HCl in this solution is 36.7 g (= 1.006 mol HCl). The corresponding solution volume is 100 g/1.18 g/ml = 84.77 ml (= 0.08477 L). Molarity is found by substituting these quantities into the molarity expression (M = n solute /L soln): 1.006 mol HCl/0.08477 L = 11.9 M 13. The three carbon isotopes 12 C, 13 C, and 14 C are the same in what way? a. they are found in the same abundance b. they have the same mass c. they have the same number of neutrons d. they have the same number of protons e. all of the above Solution: By definition, isotopes have the same number of protons but different numbers of neutrons and therefore different masses. 14. The atomic mass of aluminum on the carbon-12 scale is 27.0 amu. If an atomic mass scale was set up with carbon-12 assigned a mass of 4.00 amu, the atomic mass of aluminum on this new scale would be: a. 3.00 amu b. 9.00 amu c. 27.0 amu d. 81.0 amu e. 108 amu Solution: Aluminum is 2.25 times more massive than carbon. Hence, on the new scale, Al would have a mass of 9.00 amu (4.00 amu x 2.25) 15. The allowed subshells in the n = 3 shell are: a. s b. s and p c. s, p, and d d. s, p, d, and f e. s, p, d, f, and g Solution: In any shell the allowed subshells are determined by the azimuthal quantum number, l, and it can have integral values of 0, 1, n-1. When n = 3, l can be 0, 1, or, 2. This corresponds to s, p, and d subshells. 16. For a cobalt atom (Co, Z = 27) in its ground state, the total number of orbitals populated by one or more electrons is: a. 18 b. 15 c. 12 d. 9 e. 6

Solution: The ground state electron configuration for cobalt is: 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 7. There is 1 orbital per s subshell, 3 orbitals per p subshell, and 5 orbitals per d subshell. All 5 d orbitals contain an electron. (4 x 1) + (2 x 3) + (1 x 5) = 15 17. Common table salt, NaCl, is best described as a(n): a. molecular substance b. polar molecule c. ionic compound d. metal e. semiconductor Solution: NaCl is a combination of a metal and non-metal. It is ionic. 18. The number of unshared electron pairs around the central sulfur atom in the H 2 S molecule is: a. 0 b. 1 c. 2 d. 3 e. 4 Solution: A total of 8 valence electrons are required for the Lewis structure of H 2 S. The skeletal structure uses four of these: H-S-H. The remaining 4 are distributed as two sets of unshared electron pairs on sulfur to satisfy the octet rule. 19. In which of the following molecules or ions is the octet rule violated: a. CO 2 b. NO c. NH 4 + d. F 2 e. CN - Solution: NO contains an odd number of electrons. 20. A possible set of quantum numbers for the outer most electron of a gallium atom (Z = 31) in its ground state is: n l m l m s a. 5 1-1 -½ b. 4 1-1 -½ c. 4 0 0 -½ d. 3 2 +2 +½ e. 3 1 +1 +½ Solution: The outermost electron is a 4p electron (n= 4, l =1) The following table of half-reactions is needed for the next three questions. Half-Reaction E o, V MnO - 4 (aq) + 8H + (aq) + 5e - Mn 2+ (aq) + 4H 2 O (l) +1.51

Ag + (aq) + e - Ag (s) +0.799 Cu 2+ (aq) + 2e - Cu (s) +0.337 2H + (aq) + 2e - H 2 (g) +0.000 Cd 2+ (aq) + 2e - Cd (s) -0.403 Fe 2+ (aq) + 2e - Fe (s) -0.440 Zn 2+ (aq) + 2e - Zn (s) -0.763 21. Which listed species will convert Cu 2+ to Cu but will not convert Fe 2+ to Fe. a. Ag + b. H + c. Cd d. Zn e. Ag Solution: Cu 2+ can be reduced to Cu by any product species that is part of a half-reaction having a more negative reduction potential than that for the Cu 2+ /Cu half-reaction. H 2, Cd, Fe, and Zn in the table. Of these, Cd will reduce Cu 2+ but not Fe 2+ since the Cd 2+ /Cd reduction potential is more positive than the Fe 2+ /Fe reduction potential. 22. What is the value of G for the Daniell cell Zn (s) + Cu 2+ (aq) Zn 2+ (aq) + Cu (s)? a. -106 kj/mol b. 3.18 x 10 4 J/mol c. 212,000 kj/mol d. -53.1 kj/mol e. -212 kj/mol Solution: the cell potential, E o is 1.10 V (= 0.337 V- (-0.763 V)), and 2 electrons are transferred. The free energy change is found by substituting these quantities into the following equation and solving: G o = nfe o = 2(9.65x10 4 C/mol)(1.1 J/C) = 212,000 J/mol = 212 kj/mol 23. What is the value of E at 298 K for the half cell: MnO 4 - (0.010 M) + 8H + (0.20 M) + 5e - Mn 2+ (0.020 M) + 4H 2 O? a. 1.44 V b. 1.50 V c. 1.52 V d. 1.58 V e. 1.86 V Solution: 4 2+ [Mn ] 0.02 Q = = = 7.81x10 + 8 8 [MnO ][H ] (0.010)(0.20) 5

E is found by substituting the value for Q, the standard reduction potential (from table), and the number of electrons transferred (5) into the Nernst equation and solving: E = E o 0.0592 0.0592 5 log Q = + 1.51V - log(7.81x10 ) = 1.44 V n 5 24. Which of the following isoelectronic species will lose an electron most easily? a. S 2- b. Cl - c. Ar d. K + e. Ca 2+ Solution: the most negative species, S 2-, will lose an electron most easily since it has the highest ratio of electrons to protons. 25. An alkane: a. has the general formula C n H 2n. b. has at least one double bond. c. is a saturated hydrocarbon. d. is a binary compound of carbon and oxygen. e. all of the above. Solution: By definition an alkane is a saturated hydrocarbon. 26. The enthalpy change, H, for the combustion of one mole of methane (CH 4, M = 16.0 g/mol) is -891 kj. What is the enthalpy change when 1.5 kg of methane is combusted? a. -83.5 kj b. -1.34 x 10 3 kj c. -8.35 x 10 4 kj d. -1.34 x 10 6 kj e. -9.50 x 10 6 kj Solution: H = (1.5 x 10 3 g CH 4 )(1 mol CH 4 /16.0 g CH 4 )(-891 kj/mol CH 4 ) = -8.35 x 10 4 kj 27. For a spontaneous endothermic process: a. G > 0 b. G = 0 c. H < 0 d. S = 0 e. S > 0 Solution: G = H - T S. In a spontaneous process, G < 0. The described process is endothermic ( H > 0). Hence, S must be positive (> 0). 28. The organic starting materials for the preparation of an ester would be: a. a carboxylic acid and an alcohol b. a ketone and an alcohol

c. an alkane and a ketone d. only a carboxylic acid e. an amine and a carboxylic acid Solution: The classic preparation route for esters is the acid catalyzed reaction between a carboxylic acid and an alcohol Fischer esterification. 29. The number of collisions of N 2 molecules against a container wall per unit time is: a. increased by cooling the gas. b. increased by decreasing the pressure. c. increased by increasing the average kinetic energy of the gas molecules. d. increased by increasing the volume. e. increased by adding helium to the container. Solution: An increase in temperature or pressure or both will increase the number of collisions with the container wall. According to the kinetic molecular theory of gases, temperature is a measure of the average kinetic energy of a collection of gaseous molecules. 30. A 0.198 g sample of a noble gas exerts a pressure of 0.461 atm in a 0.250 L vessel at 15 o C. Which noble gas is it? a. argon b. chlorine c. helium d. hydrogen e. neon Solution: n gas = (0.461 atm)(0.250 L)/((0.0821 L-atm-mol -1 -K -1 )(288.2 K)) = 4.87 x 10-3 mol. The molar mass is (0.198 g)/(4.87 x 10-3 mol) = 40.6 g/mol. Argon. 31. A solid substance decomposes on heating. Two products are formed. One product is a white solid and the other is a colorless gas. From this data, it can be concluded that: a. at least one of the products is an element. b. both products are elements. c. the original solid is a compound. d. the original solid is an element and the gas is a compound. e. the original solid is a compound and the gas is an element. Solution: A compound can be decomposed into two or more simpler substances. Hence, the original solid is a compound. There is insufficient information to ascertain the nature of the products. 32. A mixture of 4.00 moles of H 2 and 3.00 moles of O 2 is ignited, forming water. What is the composition of the system by mass after the reaction is complete? a. 16.0 g of oxygen and 72.0 g of water b. 2.02 g of hydrogen and 54.0 g of water c. 1.01 g of hydrogen and 36.9 g of water d. 4.04 g of hydrogen, 64.0 g of oxygen and 36.0 g of water e. 32.0 g of oxygen and 72.0 g of water

The balanced equation is: 2H 2 + O 2 2H 2 O. Hydrogen limits (3 moles of O 2 requires 6 moles of H 2 but only 4 moles are available). By stoichiometry, 2 moles of O 2 are consumed (1 mole = 32 g is in excess) and 4 moles (72 g) of water are produced. 33. What is the most likely formula of a binary compound of calcium and phosphorus? a. CaP b. Ca 3 P 2 c. CaK d. CaPO 4 e. Ca(PO 3 ) 3 Solution: The ions are Ca 2+ and P 3- which gives Ca 3 P 2 34. When a sample of water changes to ice its: a. mass decreases b. mass increases c. volume decreases d. volume increases e. mass and volume do not change Solution: Sample mass does not change in a phase change, but volume does change. In the case of water, its volume increases. Experience shows that water expands as it freezes. 35. The density of aluminum is 2.70 g/ml and the density of lead is 11.3 g/ml. What mass of lead occupies the same volume as 54.0 g of aluminum? a. 0.565 g b. 12.9 g c. 56.5 g d. 146 g e. 226 g Solution: The volume in question is 54.0 g/2.70 g/ml = 20.0 ml. The mass of lead in 20.0 ml of lead is 20.0 ml x 11.3 g/ml = 226 g. 36. Which of the following species has the largest radius? a. O b. S + c. S d. S 2- e. Cl Solution: S 2- has the largest radius since it has the most electrons and it possesses more electrons than protons. 37. Given the following data: 2SO 2 (g) + O 2 (g) 2SO 3 (g) H = -196.7 kj SO 3 (g) + H 2 O (l) H 2 SO 4 (l) H = -130.1 kj What is the enthalpy change for

2SO 2 (g) + O 2 (g) + 2H 2 O (l) 2H 2 SO 4 (l) a. 66.6 kj b. 326.8 kj c. -326.8 kj d. 456.9 kj e. -456.9 kj Solution: Apply Hess law in conjunction with the rules governing thermochemical equations to determine H for the target equation. First multiply the second given equation by 2 and add the result to the first given equation. The net is the target equation. In terms of H, when the second equation is multiplied by 2, H becomes 260.2 kj. When this equation is added to the first equation, H for the target equation is (-260.2 kj + (-196.7 kj) = -456.9 kj. 38. The temperature of a 2.0 x 10 2 g metal bar increases from 0.0 o C to 100.0 o C when it absorbs 5.0 kj of heat. What is the specific heat of the metal? a. 2.5 x 10-4 J/g- o C b. 0.25 J/g- o C c. 0.10 J/g- o C d. 10 J/g- o C e. 70 J/g- o C Solution: c = q/m T = (+5.0 x 10 3 J)/((2.0 x 10 2 g)(100.0 o C)) = 0.25 J/g- o C. 39. Potassium chlorate decomposes on heating to potassium chloride and oxygen gas. When 1.83 g of potassium chlorate is decomposed, 1.12 g of potassium chloride is obtained. What is the mass percent of oxygen in potassium chlorate? a. 38.8% b. 61.2% c. 63.4% d. 75.0% e. 163% mass of oxygen 1.83 g 1.12 g Solution : %O = x100 = x100 = 38.8% mass of potassium chlorate 1.83 g 40. Given at 25 C: G f (H 2 O(l)) = -237.13 kj/mol, G f (H 2 O(g)) = -228.57 kj/mol. Calculate the equilibrium vapor pressure of water at 25 C. H 2 O(l) H 2 O (g) a. 0.0413 atm b. 0.0395 atm c. 0.0317 atm d. 0.0264 atm e. 0.0118 atm

Solution: G o = -228.57 kj (-237.13 kj) = +85.6 kj. Next, K p for this reaction is found by substituting this quantity into the following equation and solving: G = -RTlnK. K p = exp (- G/RT) = exp (-85.6 kj/((298.2 K)(8.314 x 10-3 kj/mol-k))) = exp (-3.453) = 0.03165. The equilibrium constant expression for this reaction is K p = P water vapor. Hence, P = 0.03165 atm.