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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 Br 2 (ℓ) — 152 Br 2 (g) 31 245 C 3 H 8 (g) -104 270 C 5 H 12 (ℓ) -174 263 Cl 2 (g) — 223 HNO 3 (aq) -207 146 H 2 O (ℓ) -286 70 H 2 O (g) -242 189 NH 4 NO 3 (s) -366 151 NO 2 (g) 33 240 NO (g) 90 211 N 2 H 4 (ℓ) 51 12 N 2 O (g) 82 220 O 2 (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 H 2 O CH 3 OH density g/mL 1.000 0. Cs,solid J/g K 2.09 — Cs,liquid J/g K 4.184 2. 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.
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?
N 2 (g) O 2 (ℓ) Ar(ℓ) CO 2 (g) He(g)
- N 2 (g) and He(g) correct
- Ar(ℓ) and He(g)
- N 2 (g), CO 2 (g), and He(g)
- O 2 (ℓ) and Ar(ℓ)
- N 2 (g), O 2 (ℓ), Ar(ℓ) , and He(g)
Explanation: Standard state for all of these should be gas state. CO 2 is not an element. Only elements in their standard states will have ∆G◦ f equal to zero. Only N 2 (g) and He(g) match this criteria.
002 4.0 points For the combustion reaction of ethylene (C 2 H 4 )
C 2 H 4 + 3 O 2 → 2 CO 2 + 2 H 2 O
assume all reactants and products are gases, and calculate the ∆Hrxn^0 using bond energies.
- −680 kJ/mol
- 1300 kJ/mol
- 0 kJ/mol
- 251 kJ/mol
- −1300 kJ/mol correct
- −251 kJ/mol
- 680 kJ/mol
Explanation:
∆Hrxn^0 =
BE (^) reactants −
BE (^) products
[
(C C) + 4 (C H)
+3 (O O)
]
[
4 (C O) + 4 (H O)
]
[(
kJ mol
kJ mol
(^498) molkJ
)]
[
kJ mol
kJ mol
)]
kJ mol
003 4.0 points Calculate the approximate boiling point of chloroform, CHCl 3 , given the following data: ∆Hvap = 31.4 kJ mol−^1 ∆Svap = 93.6 J mol−^1 K−^1
- 298 K
- 335 K correct
- 59.3 K
- 665 K
- 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?
- The standard molar entropy of an element in its standard state is zero.
- The Gibb′s free energy change of the sys- tem must be positive for a spontaneous pro- cess.
- The magnitude of the entropy change of the system will always equal the magnitude
For which of the following chemical equations would ∆H rxn◦ = ∆H f◦?
- N 2 (ℓ) + 3 F 2 (g) → 2 NF 3 (ℓ)
- CO(g) + 12 O 2 (g) → CO 2 (g)
- O 2 (g) + H 2 (g) → H 2 O 2 (ℓ) correct
- C(s, graphite) + 32 O 2 (g) + H 2 (g) → CO 2 (g) + H 2 O(g)
Explanation: For O 2 (g) + H 2 (g) → H 2 O 2 (ℓ), ∆H f◦ of O 2 (g) and H 2 (g) are 0. Therefore, ∆H rxn◦ = ∆H◦ f (H 2 O 2 (ℓ))
009 4.0 points A system that can exchange energy but not matter with the surroundings is termed:
- Closed correct
- Open
- Isolated
Explanation: A closed system can exchange energy but not matter with the surroundings.
010 4.0 points When water condenses, what are the signs for q, w, and ∆Ssys, respectively?
- −, +, +
- +, +, −
- +, +, +
- +, −, −
- +, −, +
- −, +, − correct
Explanation:
011 4.0 points The oxidation of sugar to carbon dioxide
and water is a spontaneous chemical reac- tion. Since we know that reactions that occur spontaneously in one direction cannot occur spontaneously in the reverse direction, how can we understand photosynthesis?
- Thermodynamics does not apply to living systems.
- Thermodynamics deals only with closed systems; photosynthesis is an open system.
- It is not a spontaneous chemical reaction; it is driven by an external source of energy – light. correct
- This reaction is characterized by an en- ergy change so close to zero that it is essen- tially reversible.
- Thermodynamics does not apply to pho- tochemical reactions. Explanation: .
012 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. III) The reaction is spontaneous only at rela- tively high temperatures. IV) ∆G is positive at all temperatures.
- II and III only
- I and IV only correct
- III only
- I, II, III, and IV
- I and II only
- I, III, and IV only
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, C 2 H 5 OH, with a specific heat of 2.46 J/g◦C and the other is benzene, C 6 H 6 , 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.
- Both liquids will have the same final tem- perature.
- Cannot tell without more information given.
- benzene correct
- 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? N 2 (g) + O 2 (g) → 2NO(g) 2NO(g) + O 2 (g) → 2NO 2 (g)
- Both reactions have ∆H < ∆U.
- Neither reaction has ∆H < ∆U.
- 2NO(g) + O 2 (g) → 2NO 2 (g) correct
- N 2 (g) + O 2 (g) → 2NO(g) Explanation: .
015 4.0 points An important reaction that takes place in the atmosphere is
NO 2 (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) O 2 (g) → 2 O(g) +498. 4
NO(g) + O 3 (g) −→ NO 2 (g) + O 2 (g) − 200. 0 3 2 O^2 (g)^ −→^ O^3 (g)^ +142.^7
- 306.5 kJ correct
- 449.2 kJ
- 106.5 kJ
- 320.2 kJ
- 963.8 kJ
- 820.5 kJ
- 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:
NO 2 (g) + O 2 (g) −→ NO(g) + O 3 (g)+
O 3 (g) −→
O 2 (g) − 142. 7 1 2
O 2 (g) −→ O(g) +249. 2
NO 2 (g) −→ NO(g) + O(g)+306. 5
016 4.0 points
- 79.0 kJ
- -50.4 kJ correct
- -79.0 kJ
- 50.4 kJ
- 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.
NH 4 NO 3 (s) → N 2 O(g) + 2H 2 O(g)
- +97.2 kJ
- +130 kJ
- − 1. 33 × 105 kJ
- −113 kJ
- −169 kJ correct
- −130 kJ
- +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?
- 755 J; released
- 1478 J; released
- 977 J; absorbed
- 755 J; absorbed
- 977 J; released correct
- 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:
- Stay exactly the same.
- Increase
- Increase, decrease, or stay the same de- pending on the magnitudes of heat and work
- 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?
- The increase in entropy of the system. correct
- The large amount of heat absorbed by the surroundings for the process.
- The decrease in the entropy of the sys- tem.
- The negative value of qsys for the dissolu- tion process.
Explanation: .
024 4.0 points Which of the following compounds would you expect to have the highest S◦^?
- CH 3 F(g) correct
- Ar(g)
- H 2 O 2 (ℓ)
- CH 4 (g)
- C 6 H 14 (ℓ)
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 CH 3 F(g).
025 4.0 points Methyl tert-butyl ether or MTBE is an oc- tane booster for gasoline. The combustion of 0.9211 grams of MTBE (C 5 H 12 O(ℓ), 88. g/mol) is carried out in a bomb calorimeter. The calorimeter’s hardware has a heat capac- ity of 1.540 kJ/◦C and is filled with exactly 2.022 L of water. The initial temperature was 26.336◦C. After the combustion, the temper- ature was 29.849◦C. Analyze this calorimeter data and determine the molar internal energy of combustion (∆U ) for this octane booster.
- -3560 kJ/mol
- -2748 kJ/mol
- -3120 kJ/mol
- -1957 kJ/mol
- -3362 kJ/mol correct
- -2286 kJ/mol
- -4293 kJ/mol Explanation: ∆T = 29. 849 − 26 .336 = 3. 513 ◦ Ccal = [2022(4.184) + 1540 ]/ = 10.00 kJ/◦C qcal = 10(3.513) = 35.130 kJ moles = 0.9211/88.15 = 0.01045 mol ∆U = -35.130 kJ / 0.01045 mol = -3362 kJ/mol
