Mean 5.64 out of 9 pts AP CHEMISTRY Question 1 CO(g) + 1 2 O 2 (g) CO 2 (g) 1. The combustion of carbon monoxide is represented by the equation above. (a) Determine the value of the standard enthalpy change, using the following information. H, for the combustion of CO(g) at 298 K rxn C(s) + 1 2 O 2 (g) CO(g) C(s) + O 2 (g) CO 2 (g) H 298 = 110.5 kj/mol rxn H 298 = 393.5 kj/mol rxn Reverse the first equation and add it to the second equation to obtain the third equation. CO(g) 1 2 O 2 (g) + C(s) + C(s) + O 2 (g) CO 2 (g) CO(g) + 1 2 O 2 (g) CO 2 (g) = H f of CO 2 (g) H f of CO(g) H rxn H 298 = + 110.5 kj/ mol rnx H 298 = 393.5 kj/mol rxn H rxn = 110.5 + ( 393.5) = 283.0 kj/mol rxn = 393.5 kj/mol rxn ( 110.5 kj/mol 1 ) = 283.0 kj/mol rxn One point is earned for reversing the first equation. One point is earned for the correct answer (with sign). Two points are earned for determining H rxn from the enthalpies of formation. (If sign is incorrect, only one point is earned.) (b) Determine the value of the standard entropy change, the information in the following table. S rxn, for the combustion of CO(g) at 298 K using Substance S 298 (J/K mol rxn ) CO(g) 197.7 CO 2 (g) 213.7 O 2 (g) 205.1 2006 The College Board. All rights reserved. 4
Question 1 (continued) S rx n = 213.7 J/K mol rxn (197.7 J/K mol rxn + 1 2 = 86.5 J /K mol rxn (205.1 J/K mol rxn)) One point is earned for taking one-half of S 298 for O 2 (g). One point is earned for the answer (with sign). (c) Determine the standard free energy change, your answer. G, for the reaction at 298 K. Include units with rxn G rxn = Hrxn T S rxn = 283.0 kj/mol rxn (298 K)( 0.0865 kj/k. mol rxn ) K G o rxn = 257.2 kj/mol rxn One point is earned for substituting the values from parts (a) and (b) into the equation. One point is earned for the answer (with sign and units). (d) Is the reaction spontaneous under standard conditions at 298 K? Justify your answer. Yes, the reaction is spontaneous because the value of G rxn for the reaction is negative ( 257.2 kj mol rxn 1 ). One point is earned for an answer with justification (consistent with the answer in part (c)). (e) Calculate the value of the equilibrium constant, K eq, for the reaction at 298 K. G rxn = RTln K eq -257,200 Jmol -1-1 -1 -(8.31 J mol K )(298 K) RT G rxn = ln K eq = ln K eq K eq = 1.28 10 45 One point is earned for correct substitution into the equation. One point is earned for the answer. 2006 The College Board. All rights reserved. 5
Mean 3.18 out of 10 pts AP CHEMISTRY Question 2 Total Score 10 Points H + (aq) + OH (aq) H 2 O(l) 2. A student is asked to determine the molar enthalpy of neutralization, H neut, for the reaction represented above. The student combines equal volumes of 1.0 M HCl and 1.0 M NaOH in an open polystyrene cup calorimeter. The heat released by the reaction is determined by using the equation q = mc T. Assume the following. Both solutions are at the same temperature before they are combined. The densities of all the solutions are the same as that of water. Any heat lost to the calorimeter or to the air is negligible. The specific heat capacity of the combined solutions is the same as that of water. (a) Give appropriate units for each of the terms in the equation q = mc T. q has units of joules (or kilojoules or calories or kilocalories) m has units of grams or kilograms c has units of J g 1 C 1 or J g 1 K 1 (calories or kilograms acceptable alternatives) T has units of C or K 1 point earned for any two units 2 points earned for all four units (b) List the measurements that must be made in order to obtain the value of q. volume or mass of the HCl or NaOH solutions initial temperature of HCl or NaOH before mixing final (highest) temperature of solution after mixing 1 point earned for any volume (mass of reactant) 1 point earned for initial and final (highest) temperature ( T is not a measurement) Copyright by College Entrance Examination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Examination Board.
(c) Explain how to calculate each of the following. Question 2 (cont d.) (i) The number of moles of water formed during the experiment Since there is mixing of equal volumes of the same concentration and the reaction has 1:1 stoichiometry, moles of H 2 O = moles of HCl = moles NaOH. To determine the number of moles of HCl: mol HCl 1 mol H2O (volume HCl) = mol H 2 O 1 L 1 mol HCl 1.0 mol NaOH (volume NaOH) 1 mol H 2O 1 L = mol H 2 O 1 mol NaOH n H 2 O = n HCl = n NaOH = V HCl 1 M = V NaOH 1 M 1 point earned for the number of moles of H 2 O using the stoichiometric relationship between HCl (or NaOH) and H 2 O (ii) The value of the molar enthalpy of neutralization, H neut, for the reaction between HCl(aq) and NaOH(aq) Determine the quantity of the heat produced, q, from q = mc T, where m = total mass of solution; divide q by mol H 2 O determined in part (c) (i) to determine H neut : H neut = q q mol H 2O mol H 2O (mol reactant can substitute for mol H 2 O) 1 point earned for q 1 point earned for H neut Copyright by College Entrance Examination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Examination Board.
Question 2 (cont d.) (d) The student repeats the experiment with the same equal volumes as before, but this time uses 2.0 M HCl and 2.0 M NaOH. (i) Indicate whether the value of q increases, decreases, or stays the same when compared to the first experiment. Justify your prediction. The T will be greater, so q increases. There are more moles of HCl and NaOH reacting so the final temperature of the mixture will be higher. 1 point earned for direction and explanation Note: Arguments about increased mass are not acceptable because the total mass increase is negligible (the solutions have virtually the same density) and is not the driving force for increases in q. (ii) Indicate whether the value of the molar enthalpy of neutralization, H neut, increases, decreases, or stays the same when compared to the first experiment. Justify your prediction. Both q and mol H 2 O increase proportionately. However, when the quotient is determined, there is no change in H neut Molar enthalpy is defined as per mole of reaction, therefore it will not change when the number of moles is doubled. 1 point earned for correct direction and explanation (e) Suppose that a significant amount of heat were lost to the air during the experiment. What effect would this have on the calculated value of the molar enthalpy of neutralization, H neut? Justify your answer. Heat lost to the air will produce a smaller T. In the equation q = mc T a smaller T will produce a smaller value for q (heat released) than it should. In the equation q H neut = mol H 2O 1 point earned for correct direction and explanation the smaller magnitude of q and the constant mol H 2 O means that H neut will be less negative (more positive). Notes: H decreases because q decreases earns 1 point T decreases because H decreases earns 1 point No points earned for T decreases therefore q decreases Copyright by College Entrance Examination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Examination Board.
Originally, mean of 3.71 out of 8 pts, Therefore expect a mean of 1.86 of 4 pts AP CHEMISTRY Question 3 Total Score 4 Points C(s) + CO 2 (g) 2 CO(g) 3. Carbon (graphite), carbon dioxide, and carbon monoxide form an equilibrium mixture, as represented by the equation above. (a) In the table below are data that show the percent of CO in the equilibrium mixture at two different temperatures. Predict the sign for the change in enthalpy, H, for the reaction. Justify your prediction. Temperature % CO 700 C 60 850 C 94 H = + More CO at the higher temperature indicates that the reaction shifts to the right with increasing temperature. For this to occur, the reaction must be endothermic. 1 point earned for indicating that H is positive 1 point earned for explanation (b) Appropriately complete the potential energy diagram for the reaction by finishing the curve on the graph below. Also, clearly indicate H for the reaction on the graph. 1 point earned for completing the graph according to the information in part (a) 1 point earned for appropriately labeling H rxn for the reaction as drawn
Originally, mean of 3.93 out of 10 pts, therefore expect a mean of 1.97 out of 5 points N 2 (g) + 3 F 2 (g) 2 NF 3 (g) DH 298 = 264 kj mol 1 ; DS 298 = 278 J K 1 mol 1 4. The following questions relate to the synthesis reaction represented by the chemical equation in the box above. (a) Calculate the value of the standard free energy change, DG 298, for the reaction. The enthalpy change in a chemical reaction is the difference between energy absorbed in breaking bonds in the reactants and energy released by bond formation in the products. (b) How many bonds are formed when two molecules of NF 3 are produced according to the equation in the box above? (c) Use both the information in the box above and the table of average bond enthalpies below to calculate the average enthalpy of the F F bond. Bond Average Bond Enthalpy (kj mol 1 ) NN 946 N F 272 F F? 2007 The College Board. All rights reserved. -7- GO ON TO THE NEXT PAGE.
Question 4 N 2 (g) + 3 F 2 (g) 2 NF 3 (g) ΔH 298 = 264 kj mol rxn 1 ; ΔS 298 = 278 J K 1 mol rxn 1 The following questions relate to the synthesis reaction represented by the chemical equation in the box above. (a) Calculate the value of the standard free energy change, G Δ 298, for the reaction. ΔG 298 = ΔH 298 T S Δ 298 One point is earned for the value of = 264 kj mol 1 (298 K)( 0.278 kj mol rxn 1 K 1 ) ΔG 298 (including kj or J per mol rxn). = 181 kj mol rxn 1 (b) How many bonds are formed when two molecules of NF 3 are produced according to the equation in the box above? There are six N F bonds formed. One point is earned for the correct answer. 2007 The College Board. All rights reserved.
Question 4 (continued) The enthalpy change in a chemical reaction is the difference between energy absorbed in breaking bonds in the reactants and energy released by bond formation in the products. (c) Use both the information in the box above and the table of average bond enthalpies below to calculate the average enthalpy of the F - F bond. Bond Average Bond Enthalpy (kj mol 1 ) N N 946 N F 272 F F? ΔH 298 = Σ E bonds broken Σ E bonds formed = 264 kj mol rxn 1 = [ BE N N + (3 BE F-F )] (6 BE N-F ) = [946 kj mol 1 + (3 BE F-F )] 6(272 kj mol 1 rxn ) = 264 kj mol 1 rxn 3 mol BE F-F = ( 264 946 + 1,632) kj mol 1 rxn BE F-F = 141 kj mol 1 rxn One point is earned for the correct number of bonds in all three compounds multiplied by the average bond enthalpies. One point is earned for the answer (including kj or J). Note: A total of one point is earned if an incorrect number of bonds is substituted in a correct equation and the answer is reasonable (i.e., positive). 2007 The College Board. All rights reserved.