Thermodynamics data and Single Energy Equations - Examination 4 | CH 301, Exams of Chemistry

Download Thermodynamics data and Single Energy Equations - Examination 4 | CH 301 and more Exams Chemistry in PDF only on Docsity! 202 version last name first name signature LaBrake CH301 Exam 4 Fall 2016 49945 / 49950 Remember that the bubble sheet has the periodic table on the back. Thermodynamic Data at 25◦C ∆H◦f S ◦ Substance kJ/mol J/mol K Br2 (ℓ) — 152 Br2 (g) 31 245 C3H8 (g) -104 270 C5H12 (ℓ) -174 263 Cl2 (g) — 223 HNO3 (aq) -207 146 H2O (ℓ) -286 70 H2O (g) -242 189 NH4NO3 (s) -366 151 NO2 (g) 33 240 NO (g) 90 211 N2H4 (ℓ) 51 12 N2O (g) 82 220 O2 (g) — 205 Single Bond Energies (kJ/mol) H C N O S Br H 436 C 413 346 N 391 305 163 O 463 358 201 146 S 347 272 — — 226 Br 366 285 — 201 217 193 Multiple Bond Energies (kJ/mol) C=C 602 C=N 615 C=O 799 C≡C 835 C=S 577 C≡O 1072 N=N 418 O=O 498 N≡N 945 Some Physical Properties property H2O CH3OH density g/mL 1.000 0.792 Cs,solid J/g K 2.09 — Cs,liquid J/g K 4.184 2.533 Cs,gas J/g K 2.03 — ∆Hfus J/g 334 102 ∆Hvap J/g 2022 1226 Tmp ◦ C 0 −98 Tbp ◦ C 100 65 NOTE: Please keep your Exam copy intact (all pages still sta- pled). You must turn in your exam copy, bubble sheet, and scratch paper. Version 202 – Exam 4 - F16 – labrake – (49945) 2 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 4.0 points Which one of the processes listed below (if any) have a positive value for ∆S ? 1. The condensation of water droplets on an ice cold drink. 2. The formation of ice crystals from water in a freezer compartment. 3. None of the choices here have a positive ∆S. 4. Rubbing alcohol (isopropanol) evaporat- ing from your skin. correct Explanation: Evaporation is liquid to gas which has a +∆S value. Freezing and condensation have negative values for ∆S. 002 4.0 points Calculate the standard reaction enthalpy (∆H◦rxn) for the final stage in the production of nitric acid, when nitrogen dioxide dissolves in and reacts with water: 3NO2(g) + H2O(ℓ) → 2HNO3(aq) + NO(g) 1. −304 kJ 2. −370 kJ 3. −137 kJ correct 4. +136 kJ 5. +70 kJ 6. −104 kJ Explanation: Values for ∆H◦f from external table are in order (from reaction) +33, -286, -207, and +90 ∆H◦rxn = ( ∑ n∆H◦j ) products − ( ∑ n∆H◦j ) reactants = [ 2∆H◦f, HNO3(aq) +∆H ◦ f, O(g) ] − [ 3∆H◦f, NO2(g) +∆H ◦ f, H2O(ℓ) ] = [ 2 (−207) + 90 ] − [ 3 (33) + (−286) ] = −137 kJ 003 4.0 points What is the change in entropy (∆S) for the heating of 20.0 grams of methanol (CH3OH, liquid) from 34 ◦C to 62◦C? 1. 0 J/K 2. 30.42 J/K 3. 4.42 J/K correct 4. 168.81 J/K 5. 0.22 J/K 6. 1418 J/K 7. -30.42 J/K Explanation: The specific heat capacity of methanol is equal to 2.533J/g◦C via table of data. Use the equation: ∆S = mCs ln ( T2 T1 ) ∆S = 20(2.533) ln(335/307) = 4.42 004 4.0 points Consider a chemical reaction that is endother- mic and has a negative change in entropy. Which of the following is/are true? I) ∆Suniv is negative at all temperatures. II) This reaction will reach equilibrium when T = ∆H/∆S . Version 202 – Exam 4 - F16 – labrake – (49945) 5 3. Thermodynamics does not apply to living systems. 4. This reaction is characterized by an en- ergy change so close to zero that it is essen- tially reversible. 5. Thermodynamics does not apply to pho- tochemical reactions. Explanation: . 011 4.0 points When water condenses, what are the signs for q, w, and ∆Ssys, respectively? 1. +, +, − 2. +, −, + 3. −, +, − correct 4. +, −, − 5. +, +, + 6. −, +, + Explanation: 012 4.0 points The two reactions shown below are both en- dothermic. For which reaction is ∆H < ∆U? N2(g) + O2(g) → 2NO(g) 2NO(g) + O2(g) → 2NO2(g) 1. 2NO(g) + O2(g) → 2NO2(g) correct 2. Both reactions have ∆H < ∆U . 3. N2(g) + O2(g) → 2NO(g) 4. Neither reaction has ∆H < ∆U . Explanation: . 013 4.0 points A system that can exchange energy but not matter with the surroundings is termed: 1. Open 2. Isolated 3. Closed correct Explanation: A closed system can exchange energy but not matter with the surroundings. 014 4.0 points What is responsible for the solubility of sub- stances that dissolve endothermically? 1. The negative value of qsys for the dissolu- tion process. 2. The large amount of heat absorbed by the surroundings for the process. 3. The increase in entropy of the system. correct 4. The decrease in the entropy of the sys- tem. Explanation: . 015 4.0 points Calculate the approximate boiling point of chloroform, CHCl3, given the following data: ∆Hvap = 31.4 kJ mol −1 ∆Svap = 93.6 J mol −1 K−1 1. 335 K correct 2. 298 K 3. 665 K 4. 59.3 K 5. 0.34 K Explanation: ∆Hvap−T∆Svap = 31.4×1000−T×93.6 = 0 T = 335 K Version 202 – Exam 4 - F16 – labrake – (49945) 6 016 4.0 points A student runs a reaction in a closed system. In the course of the reaction, 64.7 kJ of heat is released to the surroundings and 14.3 kJ of work is done on the system. What is the change in internal energy (∆U) of the reaction? 1. -79.0 kJ 2. -50.4 kJ correct 3. 79.0 kJ 4. 50.4 kJ 5. 90.4kJ Explanation: The change in internal energy is given by the formula: ∆U = q + w. In this reaction, q is -64.7 kJ and w is 14.3 kJ. The answer is -50.4 kJ. 017 4.0 points Which of the following compounds would you expect to have the highest S◦ ? 1. CH3F(g) correct 2. H2O2(ℓ) 3. C6H14(ℓ) 4. Ar(g) 5. CH4(g) Explanation: You should compare first based on the phase and second based on the complexity of the molecule. The most complex molecule in the gas phase at standard conditions is CH3F(g). 018 4.0 points For which of the following chemical equations would ∆H◦rxn = ∆H ◦ f ? 1. N2(ℓ) + 3 F2(g) → 2 NF3(ℓ) 2. O2(g) + H2(g) → H2O2(ℓ) correct 3. CO(g) + 12O2(g) → CO2(g) 4. C(s, graphite) + 32O2(g) + H2(g) → CO2(g) + H2O(g) Explanation: For O2(g) + H2(g) → H2O2(ℓ), ∆H ◦ f of O2(g) and H2(g) are 0. Therefore, ∆H ◦ rxn = ∆H◦f (H2O2(ℓ)) 019 4.0 points Which of the following have standard Gibbs free energy of formation values equal to zero? N2(g) O2(ℓ) Ar(ℓ) CO2(g) He(g) 1. N2(g), O2(ℓ), Ar(ℓ) , and He(g) 2. N2(g) and He(g) correct 3. N2(g), CO2(g), and He(g) 4. Ar(ℓ) and He(g) 5. O2(ℓ) and Ar(ℓ) Explanation: Standard state for all of these should be gas state. CO2 is not an element. Only elements in their standard states will have ∆G◦f equal to zero. Only N2(g) and He(g) match this criteria. 020 4.0 points Calculate the ∆Ssurr for the following reaction at 25◦C and 1 atm. Br2(ℓ) → Br2(g) ∆H ◦ rxn = +31 kJ 1. −93 J/K 2. −104 J/K correct 3. −124 J/K 4. +124 J/K 5. +104 J/K Version 202 – Exam 4 - F16 – labrake – (49945) 7 6. +93 J/K Explanation: In general for any process: ∆Ssurr = −∆Hsys Tsurr This is because the heat flow in the surround- ings is just the opposite of the heat flow for the system (qsurr = −qsys and at constant pressure the heat is equal to ∆H. therefore ∆Ssurr = −31000/298 = −104 J/K 021 4.0 points For the combustion reaction of ethylene (C2H4) C2H4 + 3O2 → 2CO2 + 2H2O assume all reactants and products are gases, and calculate the ∆H0rxn using bond energies. 1. 0 kJ/mol 2. 251 kJ/mol 3. 1300 kJ/mol 4. −1300 kJ/mol correct 5. −251 kJ/mol 6. 680 kJ/mol 7. −680 kJ/mol Explanation: ∆H0rxn = ∑ BE reactants − ∑ BE products = [ (C C) + 4 (C H) +3 (O O) ] − [ 4 (C O) + 4 (H O) ] = [( 602 kJ mol ) + 4 ( 413 kJ mol ) +3 ( 498 kJmol )] − [ 4 ( 799 kJ mol ) + 4 ( 463 kJ mol )] = −1300 kJ mol 022 4.0 points The absolute entropy of a system (S mea- sured in J/K) is related to the number of microstates in that system. Consider the three processes listed below. Which one(s) will result in an increase in the number of microstates in the system? I) The temperature of a gas is raised by 3◦C. II) A fixed amount of gas is allowed to ex- pand to a slightly larger volume. III) The total number of gas molecules in a system is reduced to a smaller number. 1. I and III only 2. I and II only correct 3. II only 4. II and III only 5. I only 6. I, II, and III 7. III only Explanation: Raising the temperature will always add to the number of available energy states in a system. More volume allows more states as well. Reducing the number of molecules however, will lower the number of microstates. 023 4.0 points 2.26 g of liquid water at 23.5 ◦C was com- pletely converted to ice at 0 ◦C. How much heat was (absorbed/released) by the system during this process? 1. 977 J; released correct