Notes on Thermodynamics - Principles of Chemistry 1 - Exam 4 | CH 301, Exams of Chemistry

Download Notes on Thermodynamics - Principles of Chemistry 1 - Exam 4 | CH 301 and more Exams Chemistry in PDF only on Docsity! 165 version last name first name signature Biberdorf CH301 Exam 4 Fall 2016 49965 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 165 – Exam 4 - F16 – biberdorf – (49965) 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 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) and He(g) correct 2. Ar(ℓ) and He(g) 3. N2(g), CO2(g), and He(g) 4. O2(ℓ) and Ar(ℓ) 5. N2(g), O2(ℓ), Ar(ℓ) , and He(g) 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. 002 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. −680 kJ/mol 2. 1300 kJ/mol 3. 0 kJ/mol 4. 251 kJ/mol 5. −1300 kJ/mol correct 6. −251 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 003 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. 298 K 2. 335 K correct 3. 59.3 K 4. 665 K 5. 0.34 K Explanation: ∆Hvap−T∆Svap = 31.4×1000−T×93.6 = 0 T = 335 K 004 4.0 points Which of the following statements is true? 1.The standard molar entropy of an element in its standard state is zero. 2. The Gibb′s free energy change of the sys- tem must be positive for a spontaneous pro- cess. 3. The magnitude of the entropy change of the system will always equal the magnitude Version 165 – Exam 4 - F16 – biberdorf – (49965) 5 Explanation: A reaction that is endothermic and has a negative change in entropy is nonspontaneous at all temperatures. This means that ∆Suniv is negative and ∆G is positive. Because it is never spontaneous, there will never be a point in which ∆G = 0. 013 4.0 points You have two liquids of identical mass, and both with initial temperatures of 15◦C. One is ethanol, C2H5OH, with a specific heat of 2.46 J/g◦C and the other is benzene, C6H6, with a specific heat of 1.74 J/g◦C. If both liquids absorb the same amount of heat, which one will have the highest final temperature? Assume that neither liquid reaches its boiling point. 1. Both liquids will have the same final tem- perature. 2. Cannot tell without more information given. 3. benzene correct 4. ethanol Explanation: Temperature rise (∆T ) is inversely propor- tional to the heat capacity. ∆T = q mCs Therefore, because benzene has a smaller heat capacity, Cs, it will have the larger tem- perature rise. 014 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. Both reactions have ∆H < ∆U . 2. Neither reaction has ∆H < ∆U . 3. 2NO(g) + O2(g) → 2NO2(g) correct 4. N2(g) + O2(g) → 2NO(g) Explanation: . 015 4.0 points An important reaction that takes place in the atmosphere is NO2(g) −→ NO(g) + O(g) which is brought about by sunlight. Calculate the standard enthalpy of the reaction from the following information reaction ∆H◦ (kJ) O2(g) → 2O(g) +498.4 NO(g) + O3(g) −→ NO2(g) + O2(g) −200.0 3 2O2(g) −→ O3(g) +142.7 1. 306.5 kJ correct 2. 449.2 kJ 3. 106.5 kJ 4. 320.2 kJ 5. 963.8 kJ 6. 820.5 kJ 7. 555.7 kJ Explanation: Using Hess’ Law we add the reverse (flip) of reaction 2; the reverse (flip) of reaction 3; and one half of reaction 1: NO2(g) + O2(g) −→ NO(g) + O3(g)+200 O3(g) −→ 3 2 O2(g) −142.7 1 2 O2(g) −→ O(g) +249.2 NO2(g) −→ NO(g) + O(g)+306.5 016 4.0 points Version 165 – Exam 4 - F16 – biberdorf – (49965) 6 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. +124 J/K 3. +104 J/K 4. −104 J/K correct 5. −124 J/K 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 017 4.0 points Consider a system where 2.50 L of ideal gas expands to 6.25 L against a constant external pressure of 330 torr. Calculate the work (w) for this system. 1. +165 J 2. +1238 J 3. −1238 J 4. −165 J correct 5. −1.63 J 6. +1.63 J Explanation: Convert torr to atm, and then convert an- swer in L·atm to joules. The answer will be negative due to expansion of the gas. w = −P∆V = −(330/760)(3.75L) w = −1.628 L atm ×101.325 J/(L atm) = −165 J 018 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. −137 kJ correct 2. −370 kJ 3. −104 kJ 4. −304 kJ 5. +136 kJ 6. +70 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 019 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? Version 165 – Exam 4 - F16 – biberdorf – (49965) 7 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. 020 4.0 points Calculate ∆G◦ for the following reaction at 298 K. NH4NO3(s) → N2O(g) + 2H2O(g) 1. +97.2 kJ 2. +130 kJ 3. −1.33× 105 kJ 4. −113 kJ 5. −169 kJ correct 6. −130 kJ 7. +169 kJ Explanation: Must use ∆H◦f and S ◦ values because the ∆G◦f ones are not available. Then to get free energy change use: ∆G = ∆H − T∆S ∆S = [220 + 2(189)]− 151 = 447 J/K ∆H = [82 + 2(−242)]− (−366) = −36 kJ ∆G = −36000− 298(447) = −169206 J ∆G = −169 kJ 021 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. 755 J; released 2. 1478 J; released 3. 977 J; absorbed 4. 755 J; absorbed 5. 977 J; released correct 6. 1478 J; absorbed Explanation: for 1 gram (cooling + freezing): 23.5(4.184) + 334 = 432.324 J/g scale up to 2.26 g : 432.324(2.26) = 977.052 J = 977 J released 022 4.0 points Consider a thermodynamic system that is si- multaneously releasing heat and doing work. The internal energy of this system will: 1. Stay exactly the same. 2. Increase 3. Increase, decrease, or stay the same de- pending on the magnitudes of heat and work 4. Decrease correct Explanation: The change in internal energy is equal to the sum of the heat absorbed by the system and work done on the system based on the equation: ∆U = q + w. In this case, q and w are both negative. Therefore the internal energy will be decreasing regardless of the magnitudes of heat and work. 023 4.0 points What is responsible for the solubility of sub- stances that dissolve endothermically?