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last name first name (Middle) Print Name: Instructor: Circle one: Dr. Balasubramanian Dr. Butherus Dr. Krishtal Section #: Student #: DIRECTIONS (PUT AWAY YOUR CELL PHONE. IF YOU ARE CAUGHT WITH A CELL PHONE OR PDA OUTSIDE DURING THE EXAM, YOU WILL BE EJECTED FROM THE EXAM IMMEDIATELY AND GIVEN A 0) 1. First fill in the information at the top of this paper. Print your last name as it appears on the class roster, then print your first name and middle initial. Then read and sign the honor code statement below. Finally fill in the requested information on the SCANTRON ANSWER SHEET according to the following scheme: NAME: Print LAST name, then first (print clearly) SUBJECT: Course and section DATE: 04/17/17 PERIOD: Your Student Number 2. The last page of this exam is a periodic table. You may carefully remove this last page, if you choose, but do not open the staple. 3. This exam consists of two parts, A and B. PART A has 19 multiple choice questions worth four points each for a total of 76 points. Each question has only one correct answer, and no credit is lost if you choose a wrong answer. Therefore, leave no question unanswered, even if you simply must guess. 4. PART B consists of open ended problems worth 24 points. Show all work, including how units cancel to give the final answer, along with appropriate sig. figs. Write your final answers in the spaces provided. INTEGRITY CODE STATEMENT: On my honor I pledge that I have not violated the provisions of the NJIT student integrity code as specified by http://www.njit.edu/academics/pdf/academic-integrity-code.pdf Signature: DO NOT WRITE BELOW THIS LINE PART A: (76) PART B: #1 (12) TOTAL SCORE: #2 (12) 1
Part A: 19 multiple choice questions worth 4 points each for a total of 76 points 1. Place the following in order of increasing entropy at 298 K. CO 2 He CO C 6 H 12 A. He < CO 2 < CO < C 6 H 12 B. C 6 H 12 < CO 2 < CO < He C. CO 2 < He < C 6 H 12 < CO D. CO 2 < CO < C 6 H 12 < He E. He < CO < CO 2 < C 6 H 12 2. For a particular exothermic reaction q = 175kJ and w = 21.0 kj. Choose the correct statement A. The system gains heat and does work on the surrounding B. The system loses heat and does work on the surroundings C. The system gains heat and the surroundings do work on the system D. The system loses heat and the surroundings do work on the system E. Change in internal energy = 196kJ 3. 1 mole of an ideal gas expands from a volume of 1.00L to 11.6 L against a constant external pressure of 1.00 atm. Calculate work (w) (1 L*atm = 101.3J) A. +1.28 x 10 3 J B. -1.28 x 10 3 J C. +1.07 x 10 3 J D. -1.07 x 10 3 J E. 10.6 J 4. When the reaction below is properly balanced how many electrons are transferred? A. 1 B. 2 C. 3 D. 4 E. 6 I 3 - (aq) + S 2 O 3 2- (aq) I - (aq) + S 4 O 6 2- (aq) 5. In the following equation that which element is getting reduced? Cu(s) + 4HNO 3 (aq) A. N B. O C. H D. Cu E. This is not a redox equation Cu(NO 3 ) 2 (aq) + 2NO 2 (g) +2H 2 O(l) 2
6. Which of the following statements is always TRUE for a spontaneous, irreversible reaction? I. entropy change, ΔSuniverse > 0 II. free energy change, ΔGrxn < 0 III. enthalpy change, ΔHrxn > 0 IV. free energy change, ΔGrxn > 0 A. I only B. I and II C. I and III D. II only E. II and III 7. What is the entropy change of the surroundings, ΔSsurr when 25.0g of liquid ammonia is completely converted to ammonia vapor at its normal boiling point of 33.4 C and 1.00 atm? ΔH vap = 23.35 kj/mol A. -17.5 J/K B. -143 J/K C. 0 J/K D. +97.4 J/K E. +1.03 kj/k 8. Will the reaction 3A(g) à B(g) + C(s), ΔH = +40 kj be spontaneous or not spontaneous? A. ΔG < 0, reaction spontaneous. B. ΔG < 0, reaction not spontaneous. C. ΔG > 0, reaction spontaneous. D. ΔG > 0, reaction not spontaneous. E. Incomplete information cannot determine 9. Which of the following statements is TRUE for the process that involves melting of ice at 25 C? A. entropy change, ΔSsys < 0 B. free energy change, ΔGsys = 0 C. enthalpy change, ΔHsys < 0 D. ΔGsys < 0 E. ΔSsurr > 0 10. The enthalpy of vaporization (ΔH vap º) of a liquid Z is 77 kj/mol and ΔS vap º is 175 J/K/mol. What is the boiling point of liquid Z under standard conditions? A. 250 K B. 440 K C. 400 K D. 2.3 K E. 0.44 K 3
11. CH 4 (g) + 4Cl 2 (g)à CCl 4 (g) + 4HCl(g), ΔH = 434 kj Based on the above reaction, what energy change occurs when 1.2 moles of methane (CH 4 ) reacts? A. 5.2 x 10 5 J are released. B. 5.2 x 10 5 J are absorbed. C. 3.6 x 10 5 J are absorbed. D. 3.6 x 10 5 J are released. E. 4.4 x 10 5 J are released. 12. When the equation Cl 2 à Cl + ClO 3 (in basic solution) is balanced using the smallest wholenumber coefficients, the coefficient of OH is: A. 1 B. 12 C. 3 D. 6 E. 4 13. Which of the following is TRUE for a galvanic (voltaic) cell? I. The anode is the negative electrode II. Electrons flow from the cathode to the anode III. Reduction happens at the cathode IV. The cathode is the negative electrode A. I only B. I and III C. II and IV D. III and IV E. IV only 14. Consider the two half-reactions below. When used in a galvanic cell to generate a voltage, which electrode represents the anode and what is the cell potential? Ni 2+ + 2e Ni (s) E = 0.23 V Fe 3+ + 3e Fe (s) E = -0.036V A. Ni(s) /Ni 2+( aq), E cell = -0.27V B. Ni(s) /Ni 2+( aq) E cell = +0.27V C. Fe(s) /Fe 3+( aq) E cell = +0.19V D. Fe(s) /Fe 3+( aq) E cell = +0.27V E. Ni(s) /Ni 2+( aq) E cell = +0.19V 15. What is the oxidation state of Hg in Hg 2 Cl 2? A. +2 B. 1 C. 2 D. +1 E. 0 4
16. Consider the redox reaction: 2Al(s) + 3Fe 2+ (aq) 3Fe + 2Al 3+ (aq) Given: Al 3+ (aq) + 3e - Al(s) E = -1.66 V Fe 2+ (aq) + 2 e - Fe(s) E = -0.45 V Calculate ΔG for the reaction (pick closest) A. -260 kj B. -700 kj C. +1221kJ D. -961kJ E. -1221kJ 17. For the following redox reaction in a voltaic cell, calculate E when [Zn 2+ ] = 0.05M and [Ag + ] = 11.26M. (Hint: Balance equation for electrons first) Zn + Ag + (aq) à Zn 2+ (aq) + Ag (s) E o = 1.56V A. +1.66 V B. +1.56 V C. +1.46 V D. +1.76 V E. +1.63 V Use this information to answer questions 18-19 Zn 2+ + 2e Zn Ni 2+( aq) + 2 e Ni (s) I 2 (s) + 2e 2I (aq) Ag+(aq) + e Ag(s) Br2 (l) + 2 e 2 Br (aq) E = 0.76 V E = -0.23 V E = +0.54 V E = +0.80 V E = +1.07 V 18. Which of the following is the strongest oxidizing agent? A. I 2 (g) B. Br 2 (l) C. Zn(s) D. Cu(s) E. Ag(s) 19. Which of the following reactions is spontaneous? A. I 2 (s) + Ni 2+ (aq) 2I - (aq) + Ni(s) B. Zn (s) + Ni 2+ (aq) Zn 2+ (aq) + Ni(s) C. 2Ag(s) + Ni 2+ (aq) Ni (s) + 2Ag + (aq) D. Br 2 (l) + Ni 2+ (aq) 2Br - (aq) + Ni(s) E. 2AgBr(s) + Ni 2+ (aq) NiBr 2 (aq) + 2Ag + (aq) 5
Part B: Two questions worth 24 points 1. Clearly write the equation and show the steps for all questions Given the reaction: 4 HNO3(g) + 5 N2H4(l) 7 N2(g) + 12 H2O(l) ΔH f (kj/mol) -133.9 50.6-285.8 S (J/mol K) 266.9 121.2 191.6 70.0 A. Calculate the entropy change for the reaction ΔSº rxn. ( 3 points) S =7mol (191.6 J/mol*K)+ (12mol (70.0 J/mol*K ) 5mol (121.2 J/mol*K)+ 4 mol(266.9 J/mol*K] S = (2181.2 1673.6) J/K S = 508 J/K B. Calculate the enthalpy change for the same reaction ΔHº rxn.( 3 points) H =7mol (0)+ (12mol (-285 kj/mol ) [5mol (50.6 kj/k)+ 4 mol( -133.9 kj/k] H = -3420 - (-282.6 )kj H = -3137 kj C. Calculate the entropy change for the surroundings at 298 K. (2 points) o o ΔH rxn ( 3137000) J 4 Δ Ssurr = = = 1.05 10 J / K T 298K J/K D. Does this reaction satisfy the entropy condition (ΔS univ > 0) for spontaneity? Justify with your calculations.(2 points) ΔS univ= ΔS sys + ΔS surr = 505J/K + 1.05 x10 4 J/K >>0, spontaneous E. Calculate the value of ΔGº rxn value and verify the criteria for spontaneity. ( 2 points) G = H - T S = -3137kJ (298K x 505J/K) = -3137 kj +(- 150.5 kj) = -3287kJ G < 0, reaction spontaneous 6
II. Consider the following balanced redox reaction at 25 C and answer the questions below: 2 Al(s) + 3 Mn 2+( aq) 2 Al 3+ (aq) + 3 Mn(s) Mn 2+ + 2e Mn E = -1.18V Al 3+( aq) + 3 e Al(s), E = -1.66 V (anode) A. Use the half-cell potentials to calculate the E cell for the balanced redox reaction. (3 points) E o cell = E o cathode- E o anode E o cell = -1.18V (-1.66V)= 0.48V B. How many electrons are transferred in the overall balanced reaction: (2 points) 6 electrons C. Calculate the value of the equilibrium constant (K), for the reaction (3 points) G = -nf E o cell = 6mol (96500 C/mo))(0.48V ) = -277920J 277920J= - (8.314J/mol.K ) x 298 K lnk lnk = -112.1 K = 5.2 x 10 48 D. Calculate E cell when the [Al 3+ ] is 7.50M and the [Mn 2 + ] = 0.50 M (4 points) 2 [ Al] Q = 3 [ Mn ] [0.75] = 3 [0.5] 2 = 450 E cell = 0.48V (0.0591/n) log 450 E cell = 0.45 V 7
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