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001 5.0 points Which of the following is not a definition of enthalpy or change in enthalpy?
- ∆H = Hf inal − Hinitial
- The heat transfer at constant volume cor- rect
- A state function that influences the spon- taneity of a reaction
- ∆H = ∆E + P ∆V
Explanation: The heat transfer at constant volume is equal to the change in internal energy, not the change in enthalpy. The other choices are all mathematical identities relating to enthalpy or change in enthalpy or prose restatements of these mathematical identities.
002 5.0 points What is the residual entropy of 0.5 moles of CH 3 Cl?
- 0 J · mol−^1 · K−^1
- 11 .52 J · mol−^1 · K−^1
- 9 .13 J · mol−^1 · K−^1
- 5 .76 J · mol−^1 · K−^1 correct
Explanation:
S = k · ln W = k · ln
40.^5 NA
= NA k · ln(2) = R ln(2) = 5.76 J · mol−^1 · K−^1
003 5.0 points The enthalpy of vaporization of acetone is 31. kJ/mol, and its boiling point is 56.3◦C. What
is the entropy change for the vaporization (boiling) of acetone?
- 556 J/mol K
- 556 kJ/mol K
- Not enough information is given.
- 10.5 kJ/mol K
- 95.0 J/mol K correct
Explanation:
004 5.0 points Which of the reactions below will likely have the largest increase in entropy (∆Srxn)?
- C 5 H 12 (ℓ) + 8 O 2 (g) → 6 H 2 O(g) + 5 CO 2 (g) correct
- N 2 H 4 (g) + H 2 (g) → 2 NH 3 (g)
- Na+(g) + Cl−(g) → NaCl(s)
- S 3 (g) + 9 F 2 (g) → 3 SF 6 (g)
- 2 CH 4 (g) + 2O 3 (g) → 4 H 2 O(g) + 2 CO(g)
Explanation: The reaction with the greatest positive value for ∆ngas will have the greatest value of ∆Srxn.
005 5.0 points Calculate ∆Suniverse for the condensation of methanol (CH 3 OH) at 10 ◦C if the S m◦ for CH 3 OH(g) and CH 3 OH(ℓ) are 240 J · mol−^1 · K−^1 and 127 J · mol−^1 · K−^1 , respectively. Assume the standard boiling point of methanol is 65◦C.
- 135 J · mol−^1 · K−^1
- −135 J · mol−^1 · K−^1
- 113 J · mol−^1 · K−^1
- 22 J · mol−^1 · K−^1 correct
- −113 J · mol−^1 · K−^1
- −22 J · mol−^1 · K−^1
Explanation: Since the system in this case is condensing methanol,
CH 3 OH(g) ←→ CH 3 OH(ℓ)
∆Ssystem =
Sm,products −
Sm,reactants
= 127 J · mol−^1 · K−^1 − 240 J · mol−^1 · K−^1 = −113 J · mol−^1 · K−^1
∆Hsystem = T ∆Ssystem = (338 K)(−113 J · mol−^1 · K−^1 ) = −38194 J · mol−^1
Since the only process during a phase change is heat exchange,
∆Ssurroundings = −
∆Hsystem Tsurroundings
=
38194 J · mol−^1 283 K = 134.96 J · mol−^1 · K−^1
∆Suniverse = ∆Ssystem + ∆Ssurroundings = −113 J · mol−^1 · K−^1
- 135 J · mol−^1 · K−^1 = 22 J · mol−^1 · K−^1
006 5.0 points Phosphine (the common name for PH 3 , a highly toxic gas used for fumigation), has a ∆H vap◦ = 14.6 kJ · mol−^1 and a ∆S vap◦ =
78 .83 J · mol−^1 · K−^1. What is the normal boiling point of phosphine?
1. − 0. 2 ◦C
- − 87. 8 ◦C correct
- 2 ◦C
- 273 ◦C Explanation: Because boiling is an equilibrium process,
∆G◦ vap = −T ∆S vap◦ = 0 T ∆S◦ vap = ∆H vap◦
T =
∆H vap◦ ∆S vap◦
14 , 600J · mol−^1 78 .83 J · mol−^1 · K−^1 = 185.2 K = − 87. 8 ◦C
007 5.0 points For which reactions I) O 2 (g) + H 2 (g) → H 2 O 2 (ℓ) II) C(s, diamond) + O 2 (g) → CO 2 (g) III) N 2 (ℓ) + 3 F 2 (g) → 2 NF 3 (ℓ) IV) C(s, graphite) +
O 2 (g) + H 2 (g) → CO 2 (g) + H 2 O(g) V) 2 Fe(s) +
O 2 (g) → Fe 2 O 3 (s) would ∆H r◦ = ∆H f◦?
- IV and V only
- I, III and V only
- II, III and IV only
- I and V only correct
- I, II, III, IV and V
- II and IV only
- I, III, IV and V only
- I and II only Explanation:
008 5.0 points
- The change in enthalpy after correction for P V work
- The heat transferred at constant pressure correct
- The difference between the final and ini- tial internal energy of a system
- ∆E = q + w
Explanation: Internal energy is not equal to heat at con- stant pressure. The other statements are all mathematical identities describing inter- nal energy or prose restatements of the same.
012 5.0 points Which of the following statements concerning internal energy is/are true? I) If the expansion work is small, ∆H and ∆E are close in value. II) ∆E for a system is equal to q at constant volume. III) Assuming no heat is exchanged, when pressure-volume work is done on the sys- tem, ∆E is positive.
- I only
- I, III
- I, II, III correct
- III only
- I, II
- II, III
- II only
Explanation: Statement I follows from the identity ∆H = ∆E + p∆V , because p∆V is the expansion work. Statement II follows from ∆E = q + w, because w = 0 for processes that occur at constant volume. Statement III also follows from ∆E = q + w, because w is the pressure-
volume work. (Note: for reversible processes, expansion work and pressure-volume work are identical.)
013 5.0 points Calculate ∆S◦ surr at 298 K for the reaction
6 C(s) + 3 H 2 (g) → C 6 H 6 (ℓ) ∆H r◦ = +49.0 kJ·mol−^1 and ∆S◦ r = − 253 J·K−^1 ·mol−^1.
- − 417 J·K−^1 ·mol−^1
- − 164 J·K−^1 ·mol−^1 correct
- +253 J·K−^1 ·mol−^1
- − 253 J·K−^1 ·mol−^1
- +164 J·K−^1 ·mol−^1 Explanation: ∆H r◦ = 49000 J · mol−^1 T = 298 K
∆S◦ surr =
qsurr T
−q T
−∆H r◦ 298 K
=
−(+49000 J · mol−^1 ) 298 K = − 164 .43 J · K−^1 · mol−^1.
014 5.0 points Rank the standard molar entropy of the fol- lowing from lowest to highest. I) H 2 O(ℓ) II) H 2 O(g) III) H 2 O 2 (ℓ) IV) H 2 O 2 (aq)
- I, III, II, IV
- IV, I, III, II
- II, IV, III, I
- I, III, IV, II correct
- III, I, IV, II Explanation:
The most ordered state is a solid, then a liquid, then an aqueous solution, then a gas.
015 5.0 points Calculate ∆G◦ r for the decomposition of mer- cury(II) oxide
2 HgO(s) → 2 Hg(ℓ) + O 2 (g) ∆H f◦ − 90. 83 − − (kJ · mol−^1 ) ∆S◦ m 70. 29 76. 02 205. 14 (J · K−^1 · mol−^1 )
at 298 K.
- +117.1 kJ · mol−^1 correct
- − 64 .5 kJ · mol−^1
- − 117 .1 kJ · mol−^1
- +246.2 kJ · mol−^1
- − 246 .2 kJ · mol−^1
Explanation: In order to find ∆G◦ r at 298 K, we must first calculate ∆H r◦ and ∆S r◦.
∆S◦ r = 2 · S Hg(◦ ℓ) + S O◦ 2 (g) − 2 · S HgO(s)◦
= 2
J
K · mol
J
K · mol
− 2
J
K · mol
J
K · mol
∆H r◦ = 2 · H Hg(◦ ℓ) + H O◦ 2 (g) − 2 · H HgO(s)◦ = 0 − 2 (− 90 .83 kJ/mol) = 181.64 kJ/mol
∆G◦ r = ∆H r◦ − T ∆S r◦ = 181.64 kJ/mol − (298 K)
×
J
K · mol
1 kJ 1000 J
= 117.093 kJ/mol
016 5.0 points If an MP3 player does 200 kJ of work and releases 100 kJ of heat, what is the change in internal energy for the MP3 player?
- 100 kJ
- 300 kJ
- −100 kJ
- −300 kJ correct
Explanation:
∆U = q + w = −100 kJ + (−200 kJ) = −300 kJ
017 5.0 points Which of the following would require the most energy to raise its temperature by 1 K?
- 1 gram of water
- 10 grams of iron metal
- 1 gram of iron metal
- 10 grams of water correct
Explanation: Heat capacity is an extensive property, that reflects the amount of energy required to raise the temperature of an object. Because water has a much higher heat capacity than metals (to some extent a consequence of stronger IMF), and because 10 grams is more than 1 gram, the 10 gram smaple of water would require the greatest amount of enegy to raise its temperature by 1 K.
018 5.0 points Which law of thermodynamics governs the spontaneity of reactions?
reaction.) T is always positive. We thus have ∆G = (+) − T (−). For ∆G to be neg- ative, T would have to be negative, which is not physically possible, so the reaction is nonspontaneous at any temperature.
022 5.0 points In a bomb calorimetry measurement, 5 g of a tasty Cheetos snack are combusted at a temperature of 1200 K. A container of water surrounding the bomb calorimeter contains 2010 mL of water that rises in temperature by 25◦C. Determine the ∆H for the 5 g of Cheetos. Assume a water density of 1 g/mL and a specific heat for water of 1 cal/g◦C. Also, assume that the water absorbs all of the heat energy released in the combustion process.
- 50000 kcal
- 6 kcal
- 2400 kcal
- 50 .25 kcal correct
- 6000 kcal
Explanation: V = 2010 mL T = 25◦C density = 1 g/mL SH = 1 cal/g◦C
The heat given off by the reaction is ab- sorbed by the water. Heat absorbed by the water: ( 1
cal g◦C
(25◦C)
( (^) 1 g
mL
(2010 mL)
= (50000 cal)
( (^) 1 kcal
1000 cal
= 50.25 kcal Therefore, the heat released by the com- bustion of 5 g of Cheetos is 50.25 kcal.
023 5.0 points Assuming you had 1 mole of each of the species below, for which species would the actual residual entropy be closest to the pre- dicted residual entropy? (Hint: think in terms of S = k · ln W and ideality)
- FCl
- ClI
- FBr
- BrI correct
Explanation: All of the species have dipole-dipole inter- actions that make completely random orien- tations less likely as they form a crystal near absolute zero. The species with the weakest dipole moment will most closely approximate completely random orientations and will have an actual residual entropy closer to to the predicted residual entropy. The ∆EN of the species BrI, FCl, FBr and ClI are 0.3, 0.82, 1.02 and 0.5 respectively. The relative value of ∆EN can be inferred based on knowledge of the electronegativity trend.
024 5.0 points Which of the following statements concerning the second and third laws of thermodynamics is/are true? I) ∆Hsystem/T = ∆Ssurroundings II) In a perfect, pure crystal at absolute zero, entropy would be a small negative value. III) The entropy of the universe is always increasing.
- II, III
- II only
- I, II, III
- I only
- I, III
- I, II
- III only correct
Explanation: Statement I is false; ∆Gsystem/T = −∆Suniverse. Statement II is false; the third
law states that as the temperature of a sys- tem approaches absolute zero, the entropy of thee system approaches its minimum, which in the case of a perfect crystalline solid, is zero
- moreover, entropy can never have a negative value. Statement III is true; the second law of thermodynamics states that the entropy of the universe is always increasing.
025 5.0 points Which of the following state functions are intensive? I) Pressure (P) II) Entropy (S) III) Temperature (T)
- I, II
- III only
- I, II, III
- I only
- I, III correct
- II, III
- II only
Explanation: Examples of extensive state functions in- clude: mass, volume, enthalpy, entropy, in- ternal energy, free energy etc. Intensive state functions include pressure and temperature.
026 5.0 points For the reaction
2 H 2 (g) + O 2 (g) → 2 H 2 O(g)
find the value for the work done at 300 K.
- − 7 .5 kJ
- 7.5 kJ
- 2.5 kJ correct
- − 2 .5 kJ
Explanation: T = 300 K R =
8 .314 J
mol · K w = −P ∆ V = −∆ n R T
= −(2 − 3) mol
8 .314 J
mol · K
(300 K)
= 2494.2 J
027 5.0 points The molecules in the reaction
NH 3 (g) + 3 F 2 (g) → NF 3 (g) + 3 HF(g)
contain only single bonds. What is the heat evolved or absorbed per mole of NH 3 that reacts? These bond energy values might be useful: H-F 566 kJ/mol, F-F 158 kJ/mol, N-H 391 kJ/mol, N-F 272 kJ/mol.
- +151 kJ/mol
- -289 kJ/mol
- None of these
- +1105 kJ/mol
- −1105 kJ/mol
- +867 kJ/mol
- −867 kJ/mol correct
- −566 kJ/mol Explanation: 3(391)+3(158)-3(272)-3(566) = -
028 5.0 points What is the change of enthalpy associated with the following reaction?
CO(g) + 2 H 2 (g) → CH 3 OH(g)
species ∆H f◦ CO(g) − 110 .5 kJ · mol−^1 CH 3 OH(g) − 201 .0 kJ · mol−^1
