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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 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 one of the processes listed below (if any) have a positive value for ∆S?
- The condensation of water droplets on an ice cold drink.
- The formation of ice crystals from water in a freezer compartment.
- None of the choices here have a positive ∆S.
- 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:
3NO 2 (g) + H 2 O(ℓ) → 2HNO 3 (aq) + NO(g)
- −304 kJ
- −370 kJ
- −137 kJ correct
- +136 kJ
- +70 kJ
- −104 kJ
Explanation: Values for ∆H f◦ from external table are in order (from reaction) +33, -286, -207, and
∆H rxn◦ =
n ∆H j◦
products −
n ∆H j◦
reactants
[
2 ∆H f◦, HNO 3 (aq) + ∆H f◦, O(g)
]
[
3 ∆H f◦, NO 2 (g) + ∆H f◦, H 2 O(ℓ)
]
[
]
[
]
= −137 kJ
003 4.0 points What is the change in entropy (∆S) for the heating of 20.0 grams of methanol (CH 3 OH, liquid) from 34◦C to 62◦C?
- 0 J/K
- 30.42 J/K
- 4.42 J/K correct
- 168.81 J/K
- 0.22 J/K
- 1418 J/K
- -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 = m Cs ln
T 2
T 1
∆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.
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.
- -2286 kJ/mol
- -2748 kJ/mol
- -3362 kJ/mol correct
- -1957 kJ/mol
- -3560 kJ/mol
- -4293 kJ/mol
- -3120 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
008 4.0 points Which of the following statements is true?
- The Gibb′s free energy change of the sys- tem must be positive for a spontaneous pro- cess.
- The standard molar entropy of an element in its standard state is zero.
- The magnitude of the entropy change of the system will always equal the magnitude of the entropy change for the surroundings for all phase changes.
- The change in the entropy of the system is always positive for a spontaneous process.
- The magnitude of the heat released by the system will equal the magnitude of the heat absorbed by the surroundings. correct
Explanation: .
009 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.
- −1238 J
- +1.63 J
- −165 J correct
- − 1 .63 J
- +1238 J
- +165 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. 75 L) w = − 1 .628 L atm × 101 .325 J/(L atm) = −165 J
010 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 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
- Thermodynamics does not apply to living systems.
- 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: .
011 4.0 points When water condenses, what are the signs for q, w, and ∆Ssys, respectively?
- +, +, −
- +, −, +
- −, +, − correct
- +, −, −
- +, +, +
- −, +, +
Explanation:
012 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)
- 2NO(g) + O 2 (g) → 2NO 2 (g) correct
- Both reactions have ∆H < ∆U.
- N 2 (g) + O 2 (g) → 2NO(g)
- Neither reaction has ∆H < ∆U.
Explanation: .
013 4.0 points A system that can exchange energy but not
matter with the surroundings is termed:
- Open
- Isolated
- 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?
- The negative value of qsys for the dissolu- tion process.
- The large amount of heat absorbed by the surroundings for the process.
- The increase in entropy of the system. correct
- The decrease in the entropy of the sys- tem. Explanation: .
015 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
- 335 K correct
- 298 K
- 665 K
- 59.3 K
- 0.34 K Explanation: ∆Hvap −T ∆Svap = 31. 4 × 1000 −T × 93 .6 = 0 T = 335 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
021 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.
- 0 kJ/mol
- 251 kJ/mol
- 1300 kJ/mol
- −1300 kJ/mol correct
- −251 kJ/mol
- 680 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
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.
- I and III only
- I and II only correct
- II only
- II and III only
- I only
- I, II, and III
- 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?
- 977 J; released correct
- 755 J; released
- 1478 J; released
- 1478 J; absorbed
- 977 J; absorbed
- 755 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
024 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.
- ethanol
- Both liquids will have the same final tem- perature.
- Cannot tell without more information given.
- benzene correct
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.
025 4.0 points
Consider a thermodynamic system that is si- multaneously releasing heat and doing work. The internal energy of this system will:
- Decrease correct
- Increase
- Stay exactly the same.
- Increase, decrease, or stay the same de- pending on the magnitudes of heat and work 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.
