2009 SCORING GUIDELINES Question 2 (10 points) A student was assigned the task of determining the molar mass of an unknown gas. The student measured the mass of a sealed 843 ml rigid flask that contained dry air. The student then flushed the flask with the unknown gas, resealed it, and measured the mass again. Both the air and the unknown gas were at 23.0 C and 750. torr. The data for the experiment are shown in the table below. Volume of sealed flask Mass of sealed flask and dry air Mass of sealed flask and unknown gas 843 ml 157.70 g 158.08 g (a) Calculate the mass, in grams, of the dry air that was in the sealed flask. (The density of dry air is 1.18 g L 1 at 23.0 C and 750. torr.) m D V = (1.18 g L 1 )(0.843 L) = 0.995 g setup and calculation of mass. (b) Calculate the mass, in grams, of the sealed flask itself (i.e., if it had no air in it). 157.70 g 0.995 g = 156.71 g subtracting the answer in part (a) from 157.70 g. (c) Calculate the mass, in grams, of the unknown gas that was added to the sealed flask. 158.08 g 156.71 g = 1.37 g subtracting the answer in part (b) from 158.08 g. (d) Using the information above, calculate the value of the molar mass of the unknown gas. 750. atm (0.843 L) PV 760 n 0.0342 mol RT 1 1 (0.0821 L atm mol K )(296 K) 1.37 g molar mass = = 0.0342 mol OR molar mass = DRT P 40.1 g mol 1 1.37 g 1 1 (0.0821 L atm mol K )(296 K) 0.843 L 750. atm 760 = 40.0 g mol 1 the conversion of pressure (if necessary) and temperature and the use of the appropriate R. setup and calculation of moles of gas. setup and calculation of molar mass. OR If calculation is done in a single step, 1 point is earned for the correct P and T, 1 point is earned for the correct density, and 1 point is earned for the correct answer. 2009 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com.
2009 SCORING GUIDELINES Question 2 (continued) After the experiment was completed, the instructor informed the student that the unknown gas was carbon dioxide (44.0 g mol 1 ). (e) Calculate the percent error in the value of the molar mass calculated in part (d). percent error = 1 1 44.0 g mol 40.1g mol 44.0 g mol 1 100 = 8.9% the correct setup and answer. (f) For each of the following two possible occurrences, indicate whether it by itself could have been responsible for the error in the student s experimental result. You need not include any calculations with your answer. For each of the possible occurrences, justify your answer. Occurrence 1: The flask was incompletely flushed with CO 2 (g), resulting in some dry air remaining in the flask. This occurrence could have been responsible. The dry air left in the flask is less dense (or has a lower molar mass) than CO 2 gas at the given T and P. This would result in a lower mass of gas in the flask and a lower result for the molar mass of the unknown gas. reasoning and conclusion. Occurrence 2: The temperature of the air was 23.0 C, but the temperature of the CO 2 (g) was lower than the reported 23.0 C. This occurrence could not have been responsible. The density of CO 2 is greater at the lower temperature. A larger mass of CO 2 would be in the flask than if the CO 2 had been at 23.0 C, resulting in a higher calculated molar mass for the unknown gas. reasoning and conclusion. (g) Describe the steps of a laboratory method that the student could use to verify that the volume of the rigid flask is 843 ml at 23.0 C. You need not include any calculations with your answer. Valid methods include the following: 1. Find the mass of the empty flask. Fill the flask with a liquid of known density (e.g., water at 23 C), and measure the mass of the liquid-filled flask. Subtract to find the mass of the liquid. Using the known density and mass, calculate the volume. 2. Measure 843 ml of a liquid (e.g., water) in a 1,000 ml graduated cylinder and transfer the liquid quantitatively into the flask to see if the water fills the flask completely. for a valid method. 2009 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com.
2009 SCORING GUIDELINES Question 3 (8 points) CH 4 (g) + 2 Cl 2 (g) CH 2 Cl 2 (g) + 2 HCl(g) Methane gas reacts with chlorine gas to form dichloromethane and hydrogen chloride, as represented by the equation above. (a) A 25.0 g sample of methane gas is placed in a reaction vessel containing 2.58 mol of Cl 2 (g). (i) Identify the limiting reactant when the methane and chlorine gases are combined. Justify your answer with a calculation. Cl 2 is the limiting reactant because, in order to react with the given amount of CH 4, more moles of Cl 2 are required than the 2.58 moles of Cl 2 that are present. 1 mol CH4 25.0 g CH 4 16.04 g CH 2 mol Cl 4 1 mol CH 24 = 3.12 mol Cl 2 answer with supporting calculation. (Alternative methods are acceptable.) (ii) Calculate the total number of moles of CH 2 Cl 2 (g) in the container after the limiting reactant has been totally consumed. 1 mol CH2Cl2 2.58 mol Cl 2 2 mol Cl 2 = 1.29 mol CH 2 Cl 2 answer. Initiating most reactions involving chlorine gas involves breaking the Cl Cl bond, which has a bond energy of 242 kj mol 1. (b) Calculate the amount of energy, in joules, needed to break a single Cl Cl bond. 242 kj mol 1,000 J 1 kj 1 mol 6.02 10 23 = 4.02 10 19 J answer with appropriate setup. (c) Calculate the longest wavelength of light, in meters, that can supply the energy per photon necessary to break the Cl Cl bond. For electromagnetic radiation, c = and E = h. = = 19 E h = 4.02 10 J 34 6.63 10 J s = 6.06 10 14 s 1 8 1 c n = 3.0 10 ms 14 1 6.06 10 s = 4.9 10 7 m a correct setup that is consistent with part (b). (Both appropriate equations or the combined equation E = hc/ are required.) answer. 2009 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com.
2009 SCORING GUIDELINES Question 3 (continued) The following mechanism has been proposed for the reaction of methane gas with chlorine gas. All species are in the gas phase. Step 1 Cl 2 2 Cl fast equilibrium Step 2 CH 4 + Cl CH 3 + HCl slow Step 3 CH 3 + Cl 2 CH 3 Cl + Cl fast Step 4 CH 3 Cl + Cl CH 2 Cl 2 + H fast Step 5 H + Cl HCl fast (d) In the mechanism, is CH 3 Cl a catalyst, or is it an intermediate? Justify your answer. CH 3 Cl is an intermediate because it is produced in step 3 and consumed in step 4 of the reaction mechanism. identification of CH 3 Cl with appropriate justification. (e) Identify the order of the reaction with respect to each of the following according to the mechanism. In each case, justify your answer. (i) CH 4 (g) The order of the reaction with respect to CH 4 is 1. The rate law for the slowest step in the reaction, step 2, is rate = k [CH 4 ] [Cl]. Because the exponent of CH 4 in the rate law is 1, the order of the reaction with respect to CH 4 is 1. answer with appropriate justification. (ii) Cl 2 (g) The order of the reaction with respect to Cl 2 is 1 2. For step 1, 2 2 [Cl] K [Cl] = K 1/2 [Cl [Cl ] 2 ] 1/2 Substituting into the rate law for step 2 (the slowest step in the mechanism): rate = k [CH 4 ] [Cl] = k [CH 4 ](K 1/2 [Cl 2 ] 1/2 ) = (k)(k 1/2 )[CH 4 ] [Cl 2 ] 1/2 answer with appropriate justification. Because the exponent of Cl 2 in the rate law is 1/2, the order of the reaction with respect to Cl 2 is 1/2. 2009 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com.
2007 SCORING GUIDELINES Question 3 An external direct-current power supply is connected to two platinum electrodes immersed in a beaker containing 1.0 M CuSO 4 (aq) at 25 C, as shown in the diagram above. As the cell operates, copper metal is deposited onto one electrode and O 2 (g) is produced at the other electrode. The two reduction half-reactions for the overall reaction that occurs in the cell are shown in the table below. Half-Reaction E (V) O 2 (g) + 4 H + (aq) + 4 e 2 H 2 O(l) +1.23 Cu 2+ (aq) + 2 e Cu(s) +0.34 (a) On the diagram, indicate the direction of electron flow in the wire. The electron flow in the wire is from the right toward the left (counterclockwise). for the correct direction. (b) Write a balanced net ionic equation for the electrolysis reaction that occurs in the cell. 2 H 2 O(l) + 2 Cu 2+ (aq) 4 H + (aq) + 2 Cu(s) + O 2 (g) for all three products. for balancing the equation. (c) Predict the algebraic sign of G for the reaction. Justify your prediction. The sign of G would be positive because the reaction is NOT spontaneous. indicating that G is greater than zero and supplying a correct explanation. 2007 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents).
2007 SCORING GUIDELINES Question 3 (continued) (d) Calculate the value of G for the reaction. E = 1.23 V + 0.34 V = 0.89 V = 0.89 J C 1 G = n F E = 4 (96,500 C mol 1 )( 0.89 J C 1 ) = +340,000 J mol 1 = +340 kj mol 1 calculating E. calculating G (consistent with the calculated E ). An electric current of 1.50 amps passes through the cell for 40.0 minutes. (e) Calculate the mass, in grams, of the Cu(s) that is deposited on the electrode. q = (1.50 C s 1 )(40.0 min) mass Cu = (3,600 C) 1mol e 96,500 C = 1.19 g Cu 60 s 1 minute = 3,600 C 1molCu 2mole 63.55 g Cu 1mol Cu calculating q. calculating the mass of copper deposited. OR Two points are earned for calculating the mass of copper in one step. (f) Calculate the dry volume, in liters measured at 25 C and 1.16 atm, of the O 2 (g) that is produced. n O2 = (1.19 g Cu) V = 1mol Cu 63.55 g Cu 1molO2 2molCu = 0.00936 mol O 2 1 1 nrt P = (0.00936 mol)(0.0821 L atm mol K )(298 K) 1.16 atm = 0.197 L calculating the number of moles of O 2. calculating V (consistent with previous calculations). 2007 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents).
Answer the following problems about gases. AP CHEMISTRY 2007 SCORING GUIDELINES (Form B) Question 2 (a) The average atomic mass of naturally occurring neon is 20.18 amu. There are two common isotopes of naturally occurring neon as indicated in the table below. Isotope Mass (amu) Ne-20 19.99 Ne-22 21.99 (i) Using the information above, calculate the percent abundance of each isotope. Let x represent the natural abundance of Ne-20. 19.99 x + 21.99(1 x) = 20.18 19.99 x + 21.99 21.99 x = 20.18 19.99 x 21.99 x = 20.18 21.99 2 x = 1.81 x = 0.905 answer. percent abundances are: Ne-20 = 90.5% Ne-22 = 9.5% (ii) Calculate the number of Ne-22 atoms in a 12.55 g sample of naturally occurring neon. 12.55 g Ne 1molNe 20.18 g Ne 0.095 mol Ne-22 1mol Ne = 3.6 10 22 Ne-22 atoms 23 6.02210 Ne-22 atoms 1 mol Ne-22 the correct molar mass. the correct fraction of Ne-22 in Ne. the number of atoms. 2007 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents).
2007 SCORING GUIDELINES (Form B) Question 2 (continued) (b) A major line in the emission spectrum of neon corresponds to a frequency of 4.34 10 14 s 1. Calculate the wavelength, in nanometers, of light that corresponds to this line. c = λν λ = c ν λ = 8 1 30. 10 ms 1nm 14 1 9 434. 10 s 10 m = 690 nm setup. the answer. (c) In the upper atmosphere, ozone molecules decompose as they absorb ultraviolet (UV) radiation, as shown by the equation below. Ozone serves to block harmful ultraviolet radiation that comes from the Sun. O 3 (g) UV O 2 (g) + O(g) A molecule of O 3 (g) absorbs a photon with a frequency of 1.00 10 15 s 1. (i) How much energy, in joules, does the O 3 (g) molecule absorb per photon? E = hν = 6.63 10 34 J s 1.00 10 15 s 1 = 19 6.63 10 J per photon for the correct answer. (ii) The minimum energy needed to break an oxygen-oxygen bond in ozone is 387 kj mol 1. Does a photon with a frequency of 1.00 10 15 s 1 have enough energy to break this bond? Support your answer with a calculation. 19 23 6.63 10 J 6.022 10 photons 1 kj = 399 kj mol 1photon 1mol 3 10 J 399 kj mol 1 > 387 kj mol 1, therefore the bond can be broken. 1 calculating the energy. the comparison of bond energies. 2007 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents).