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Page 1 of 11 Chemistry 1A, Spring 2009 KEY Midterm 3 April 13, 2009 (90 min, closed book) Name:___________________ SID:_____________________ TA Name:________________ ? There are 20 Multiple choice questions worth 3 points each. ? There are 3, multi-part short answer questions. ? For the multiple choice section, fill in the Scantron form AND circle your answer on the exam. ? Put your written answers in the boxes provided. Full credit cannot be gained for answers outside the boxes provided. ? The lecture, homework, chemquizzes, discussion or experiment that each question is based upon is listed after the question e.g. [L3, HW 1.13, CQ 7.3] Question Points Score Multiple Choice Section 60 Question 21, K and ?G 8 Question 22, Multiple K 13 Question 23abc, Haber 9 Question 23defg, Haber 10 Total 100 Page 2 of 11 Thermodynamics: ?G° = ?H° ? T?S° ?H° = ? ?H° f (products) ? ? ?H° f (reactants) ?S° = ? S° (products) ? ? S° (reactants) ?G° = ? ?G° f (products) ? ? ?G° f (reactants) S = k B lnW ?S = q rev /T ?E = q + w w = - P ext ?V for aA + bB cC + dD ba dc BA DC Q ][ ][ = At equilibrium, Q = K ?G =?G° + RTln Q G = G° + RTln(a); a = activity = ?P/P° or ?[A]/[A]° ?G° = ? RTln K ?G° = ? nF??º ?? = ??º - (RT/nF) lnQ R S TR H K °? + °? ?= 1 ln ?T = ik b,f m ? = iMRT P total = P A + P B = X A P A ° + X B P B ° Acid Base: pH = - log[H 3 O + ] pX = ? log X ][ ][ log HA A pKpH a ? += Kinetics: [A] t = [A] 0 e -kt ln[A] t = ln[A] 0 ? kt Quantum: E = h? ?? = c ? deBroglie = h / p = h / mv E kin (e-) = h? - ? = h? - h? 0 ? ?= R n Z E n 2 2 ?x ?p ~ h p = mv Particle in a box (1-D Quantum): E n = h 2 n 2 /8mL 2 ; n = 1, 2, 3... Vibrational: E v = (v + ½) hA/2?; A =(k/m) ½ Rotational: E n = n(n + 1) hB; B = h/8? 2 I; I = 2mr 2 m = m A m B /(m A + m B ) Ideal Gas: PV = nRT Constants: N 0 = 6.02214 x 10 23 mol -1 R ? = 2.179874 x 10 -18 J R ? = 3.28984 x 10 15 Hz k = 1.38066 x 10 -23 J K -1 h = 6.62608 x 10 -34 J s m e = 101939 x 10 -31 kg c = 2.99792 x 10 8 m s -1 T (K) = T (C) + 273.15 F = 96,485 C / mol 1 V = 1 J / C Gas Constant: R = 8.31451 J K -1 mol -1 R = 8.20578 x 10 -2 L atm K -1 mol -1 1 nm = 10 -9 m Page 3 of 11 MULTIPLE CHOICE One mole each of gaseous He, CO 2 and NH 3 are placed in a sealed container. The volume of the container is initially 20 L and the temperature is 298K. 1) Which gas has the lowest kinetic energy? [CQ 20.1] A) He B) CO 2 C) NH 3 D) they are all the same 2) Which gas has the most frequent number of collisions with the container walls? [HW 72] A) He B) CO 2 C) NH 3 D) they are all the same 3) If the container volume was reduced to 50 mL, which of the gases is most likely to still behave ideally? [L 21] A) He B) CO 2 C) NH 3 D) they are all the same 4) The valve on the container is opened and the gases expand adiabatically. The temperature of the gas _______ because ______? [CQ 23.1, HW 6.6] A) Increases, the gases are traveling faster B) Increases, the internal energy of the gas is increasing C) Stays the same, adiabatic means that temperature is constant D) Decreases, the system loses heat to the surroundings E) Decreases, the system does work 5) A 125.0 g piece of iron is heated to 400 ?C and placed in 300.0 g of water at 25 ?C. If the specific heat capacities of iron and water are 0.449 J/K?g and 4.184 J/K?g respectively, what will be the final temperature of the water? [Lab 6, HW 6.58] A) 32 ?C B) 41?C C) 101?C D) 383?C E) None of the above Page 4 of 11 6) A CD player and its battery together do 500 kJ of work, and the battery also releases 250 kJ of energy as heat and the CD player releases 50 kJ as heat due to friction from spinning. What is the change in internal energy of the system, with the system regarded as the battery and CD player together? [HW 6.52] A) +200 kJ B) ?700 kJ C) ?800 kJ D) ?200 kJ E) ?750 kJ The decomposition of NO and NO 2 to the elements proceeds via the following two reactions. 2 NO (g) ? N 2 (g) + O 2 (g) ?H?rxn= -180 kJ 2 NO 2 (g) ? N 2 (g) + 2 O 2 (g) ?H?rxn= -67 kJ 7) Calculate the enthalpy change for the reaction: [HW 6.68] 2 NO (g) + O 2 (g) ? 2 NO 2 (g) A) 247 kJ B) -247 kJ C) -113 kJ D) -293 kJ E) None of the above 8) Calculate the amount of energy necessary to raise the temperature of 1 cup of water at 23 ?C to steam at 105 ?C. The specific heat capacity of water is 4.184 J·K -1 ·g -1 and the specific heat capacity of steam is 4.215 J·K -1 ·g -1 . The density of water can be assumed to be 1.00 g·ml -1 . For water, ?H vap =40.66 kJ·mol -1 . 1 cup = 0.2366 L [HW 6.38] A) 81.2 kJ B) 122 kJ C) 534 kJ D) 616 kJ E) None of the above 9) At room temperature, an element has a standard state molar entropy that is always______. [HW 7.24] A) negative B) zero C) positive D) not enough data to answer the question Page 5 of 11 10) Which of the following equations are true for a system at thermodynamic equilibrium? [HW 9.2] A) ?H=T?S B) ?G = 0 C) ?H=-T?S D) a and b E) b and c Consider a saturated solution of potassium chromate, K 2 CrO 4 . K 2 CrO 4 (s) null 2 K + (aq) + CrO 4 -2 (aq) K ~10 yellow yellow 2 H + (aq) + 2 CrO 4 -2 (aq) null Cr 2 O 7 -2 (aq) + H 2 O (l) K~ 2 ?H < 0 yellow orange 11) Which of the following will result in a solution of potassium chromate, K 2 CrO 4 , turning from yellow to orange? [Lab 7, CQ 28.4] A) Adding solid potassium chromate, K 2 CrO 4 B) Heating the solution C) Adding KCl D) Adding a few drops of concentrated NaOH E) Adding a few drops of concentrated HCl 12) For the following reaction predict the sign of ?H° and ?S°: H 2 (g) ? H . (g) + H . (g) Under what temperature conditions are the products favored? Explain your reasoning. [Disc 9] A) low temperature because the reaction is exothermic B) low temperature because entropy decreases with decreasing temperature C) high temperature because the reaction is exothermic D) high temperature because the entropy term becomes the dominant factor E) No temperature because enthalpy and entropy terms do not favor products Page 6 of 11 A CO 2 cylinder for carbonating beverages contains carbon dioxide liquid coexisting at equilibrium with carbon dioxide gas according to the following chemical equation: CO 2 (l) null CO 2 (g) The system is the CO 2 inside the tank. The phase diagram shown describes how the equilibrium depends on pressure and temperature (the straight lines show 1 atm and 298 K): 13) Using the phase diagram above, estimate the pressure inside the CO 2 cylinder (in atm) at standard temperature (298 K)? [L21] A) 1 atm B) 5 atm C) 50 atm D) 100 atm 14) At room temperature (298 K) and standard conditions, which is true for the system? [CQ 27.4] A) ?G° < 0 B) ?G° = 0 C) ?G° > 0 15) At room temperature (298 K) and standard conditions, which is true for the system? A) Q < K B) Q = K C) Q > K 16) If the tank were cooled to 250K, which is true for the pressure inside the tank? A) P 298 < P 250 B) P 298 = P 250 C) P 298 > P 250 Temperature (K) © 2007 Thomson Higher Education Page 7 of 11 CO 2 tank, continued 17) If the tank were cooled to 250K, which is true for the equilibrium constant? A) K 298 < K 250 B) K 298 = K 250 C) K 298 > K 250 The tank valve is opened and some CO 2 gas is released from the system. 18) Which is true for the system immediately after the valve is opened? [HW 9.80] A) Q < K B) Q = K C) Q > K 19) Which is true for the system immediately after the valve is opened for the instantaneous change in free energy (?G)? [CQ 30.3] A) ?G < 0 B) ?G = 0 C) ?G > 0 20) Which is true for the system as the gas is expanding? [HW 7.18]* A) ?S sys < 0 B) ?S sys = 0 C) ?S sys > 0 * Note: All answers were accepted because the question was determined to be too confusing. Page 8 of 11 Question 21) Free Energy and the Equilibrium Constant (8 points) Examine the data below for benzene and chlorobenzene. When placed in a sealed container, the liquid can reach equilibrium with its vapor phase. [CQ 30.2] Molecule Formula Structure Molar Mass (g/mol) T boiling (°C) Benzene C 6 H 6 78.11 80.1 Chlorobenzene C 6 H 5 Cl Cl 112.56 131.8 The graph below shows the relationship between temperature and the equilibrium constant for the vaporization of benzene in a sealed container. C 6 H 6 (l) null C 6 H 6 (g) a) On the graph above, draw a line corresponding to the vaporization for chlorobenzene. b) Explain your choice of slope based on the data and the molecular structures. The slope is negative because the reaction is endothermic and the slope = -?H/R. The reaction is endothermic because it takes energy to break attractions between molecules in the liquid to evaporate. Chlorobenzene is polar and benzene is nonpolar. The slope for chlorobenzene is steeper than for benzene because the molecules have greater attractive forces. c) Explain your choice of y-intercept. The y-intercept corresponds to ?S/R. For both molecules this value should be positive because disorder is increasing as you go from a liquid to a gas. The change in entropy for a phase change is similar for similar compounds. lnK 1/T Page 9 of 11 Question 22) Multiple Equilibria [CQ 29.5, HW 9.24] (13 points) Observing colorful reactions are a great way to study chemical equilibrium. These reactions are similar to a demo done in class. Reaction #1) NiCl 4 2- (aq) + 6 H 2 O (l) null Ni(H 2 O) 6 2+ (aq) + 4 Cl - (aq) K 1 = 10 yellow green Reaction #2) Ni(H 2 O) 6 2+ (aq) + 6 NH 3 (aq) null Ni(NH 3 ) 6 2+ (aq) + 6 H 2 O (l) K 2 = 10 8 green blue Reaction #3) NiCl 4 2- (aq) + 6 NH 3 (aq) null Ni(NH 3 ) 6 2+ (aq) + 4 Cl - (aq) K 3 = ? yellow blue a) Write the expression for the equilibrium constant (K 3 ) for Reaction #3. K = [Ni(NH 3 ) 6 2+ ] [Cl - ] 4 [NiCl 4 2- ] [NH 3 ] 6 b) What is the value of the equilibrium constant (K 3 ) for Reaction #3? K 3 = K 1 × K 2 = 10 × 10 8 = 10 9 A solution is made by mixing 20 mL of 1 M solutions of NiCl 4 2- , NH 3 , Ni(NH 3 ) 6 2+ and Cl - . c) What is the value of Q, the reaction quotient for this solution? When the solutions are mixed, each will have a concentration of ¼ M because they dilute each other. 1M (20mL/80mL) = ¼ M Q = [Ni(NH 3 ) 6 2+ ] [Cl - ] 4 Q = [¼] [¼] 4 = 16 [NiCl 4 2- ] [NH 3 ] 6 [¼] [¼] 6 d) What color will this solution be when it reaches equilibrium? yellow green blue colorless e) Explain your answer. Since K is very large (10 9 ), at equilibrium the reaction will be very product favored so the solution will be blue. Page 10 of 11 Question 23) Haber Process [L25, CQ 25.4] (9 points) Ammonia, NH 3 , is an important fertilizer for plants. The Haber process combines atmospheric nitrogen with hydrogen to form ammonia by the following reaction: N 2 (g) + 3 H 2 (g) null 2 NH 3 (g) a) Draw the Lewis Structures for N 2 , H 2 , and NH 3 . b) Use the following table of bond dissociation energies to calculate the change in enthalpy (?H) for the overall balanced reaction. Bond Standard Bond Dissociation Energy, BDE (kJ/mol) N?N 932 N=N 409 N-N 163 H-H 424 N-H 388 ?H RXN = ? BDE reactants - ? BDE products ?H RXN = [1 mol(932 kJ/mol) + 3 mol(424 kJ/mol)] ?[6 mol(388 kJ/mol)] ?H RXN = -124 kJ ?H = -124 kJ c) Use the following table of molar entropies at 298 K to calculate the change in entropy (?S) for the overall balanced reaction. Gas Molar entropy, S m ° (J/K·mol) Nitrogen, N 2 (g) 191.61 Hydrogen, H 2 (g) 130.68 Ammonia, NH 3 (g) 192.45 ?S RXN = ? S products - ? S reactants ?S RXN = [2 mol(192.45 J/K·mol)] ? [1 mol(191.61 J/K·mol) + 3 mol(130.68 J/K·mol)] ?S RXN = -198.75 J ?S = -198.75 J Page 11 of 11 Question 23) Haber Process, continued [L29, HW 9.82] (10 points) N 2 (g) + 3 H 2 (g) null 2 NH 3 (g) d) Is the reaction reactant-favored or product-favored at room temperature and standard conditions? product favored reactant favored e) Explain your reasoning. ?G = ?H ? T?S Since ?H is negative and ?S is negative, the ?G will be negative at low temperatures. If you calculate it ?G is -64 kJ at room temperature. For systems with a negative ?G, they are considered product favored. Some gases can dissolve in water, the extent to which is called molar solubility. Use the following table of molar solubility data to answer the questions below. Gas Molar solubility (mol·L -1 ·atm -1 ) Nitrogen, N 2 (g) 7.0 x 10 -4 Hydrogen, H 2 (g) 8.5 x 10 -4 Ammonia, NH 3 (g) 1.8 x 10 -2 f) How would adding water to the bottom of the reaction vessel affect the reaction? becomes more product favored becomes more reactant favored g) Explain. Since the molar solubility of Ammonia is 2 orders of magnitude greater than nitrogen and hydrogen, many more moles of ammonia will dissolve into the water as it is produced in the reaction. By removing the product, ammonia gas, as it is formed, the reaction becomes more product driven by Le Chatelier?s principle. So, by adding water to the reaction vessel, more ammonia will be produced and drive the reaction more towards favoring products. Michelle Microsoft Word - MT3_KEY.doc
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