Quiz 6 in Msci 15: Thermodynamics - Prof. David A. Laude, Quizzes of Chemistry

The questions and answers for quiz 6 in the msci 15 course, focusing on thermodynamics concepts such as energy changes, thermodynamic data, and entropy. Students are required to identify the correct answers for multiple-choice questions related to topics like heat absorption, work done, thermodynamically unstable compounds, and entropy changes.

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Create assignment, 54705, Quiz 6, Nov 27 at 9:12 am 1
This print-out should have 8 questions.
Multiple-choice questions may continue on
the next column or page find all choices
before answering. The due time is Central
time.
Msci 15 0108
19:03, general, multiple choice, >1 min, .
001
If a system absorbs heat and also does work
on its surroundings, its energy
1. must increase.
2. must decrease.
3. must not change.
4. may either increase or decrease, depend-
ing on the relative amounts of heat absorbed
and work done. correct
Explanation:
E=q+w
q > 0 because heat is absorbed and w < 0 be-
cause the system does work on its surround-
ings. Therefore E= (+) + (). Ecan
be positive only if q > w, and negative only if
w > q.
ChemPrin3e T06 14
19:03, general, multiple choice, <1 min, .
002
A system had 150 kJ of work done on it and
its internal energy increased by 60 kJ. How
much energy did the system gain or lose as
heat?
1. The system lost 90 kJ of energy as heat.
correct
2. The system lost 210 kJ of energy as heat.
3. The system gained 60 kJ of energy as
heat.
4. The system gained 90 kJ of energy as
heat.
5. The system gained 210 kJ of energy as
heat.
Explanation:
DAL Thermo Instability
20:06, general, multiple choice, >1 min, .
003
Consider the following compounds and
their thermodynamic data:
Compound H
fSG
f
¡kJ
mol ¢ ¡ J
mol·K¢ ¡ kJ
mol ¢
CH475 186 50
CH2O108 218 102
C6H5NH287 166 319
C2H452 68 219
Using this data, which of the following an-
swers includes the compounds that are ther-
modynamically unstable?
1. CH4, CH2O, C2H4
2. CH2O, C6H5NH2
3. CH4, C2H4
4. C6H5NH2, C2H4correct
5. Cannot be determined from the data pro-
vided.
6. All of the compounds are thermodynami-
cally stable.
Explanation:
ChemPrin3e T07 15
20:04, general, multiple choice, <1 min, .
004
The enthalpy of fusion of H2O(s) at its normal
melting point is 6.01 kJ ·mol1. What is the
entropy change for freezing 1 mole of water at
this temperature?
1. +20.2 J ·K1·mol1
2. 0 J ·K1·mol1
3. 20.2 J ·K1·mol1
pf3

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Download Quiz 6 in Msci 15: Thermodynamics - Prof. David A. Laude and more Quizzes Chemistry in PDF only on Docsity!

This print-out should have 8 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time.

Msci 15 0108 19:03, general, multiple choice, > 1 min,. 001 If a system absorbs heat and also does work on its surroundings, its energy

  1. must increase.
  2. must decrease.
  3. must not change.
  4. may either increase or decrease, depend- ing on the relative amounts of heat absorbed and work done. correct

Explanation:

∆E = q + w

q > 0 because heat is absorbed and w < 0 be- cause the system does work on its surround- ings. Therefore ∆E = (+) + (−). ∆E can be positive only if q > w, and negative only if w > q.

ChemPrin3e T06 14 19:03, general, multiple choice, < 1 min,. 002 A system had 150 kJ of work done on it and its internal energy increased by 60 kJ. How much energy did the system gain or lose as heat?

  1. The system lost 90 kJ of energy as heat. correct
  2. The system lost 210 kJ of energy as heat.
  3. The system gained 60 kJ of energy as heat.
  4. The system gained 90 kJ of energy as heat.
  5. The system gained 210 kJ of energy as heat. Explanation:

DAL Thermo Instability 20:06, general, multiple choice, > 1 min,. 003 Consider the following compounds and their thermodynamic data: Compound (^) (∆ H f◦ S◦^ ∆G◦ f kJ mol

) ( J

mol·K

) ( (^) kJ mol

CH 4 − 75 186 − 50

CH 2 O − 108 218 − 102

C 6 H 5 NH 2 87 166 319

C 2 H 4 52 68 219

Using this data, which of the following an- swers includes the compounds that are ther- modynamically unstable?

  1. CH 4 , CH 2 O, C 2 H 4
  2. CH 2 O, C 6 H 5 NH 2
  3. CH 4 , C 2 H 4
  4. C 6 H 5 NH 2 , C 2 H 4 correct
  5. Cannot be determined from the data pro- vided.
  6. All of the compounds are thermodynami- cally stable. Explanation:

ChemPrin3e T07 15 20:04, general, multiple choice, < 1 min,. 004 The enthalpy of fusion of H 2 O(s) at its normal melting point is 6.01 kJ · mol−^1. What is the entropy change for freezing 1 mole of water at this temperature?

  1. +20.2 J · K−^1 · mol−^1
  2. 0 J · K−^1 · mol−^1
  3. − 20.2 J · K−^1 · mol−^1
  1. +22.0 J · K−^1 · mol−^1
  2. − 22.0 J · K−^1 · mol−^1 correct

Explanation:

ChemPrin3e T07 04a 20:04, general, multiple choice, < 1 min,. 005 The temperature of 2.00 mol Ne(g) is in- creased from 25◦C to 200◦C at constant pres- sure. Assume the heat capacity of Ne is 20. J/K-mol. Calculate the change in the entropy of neon. Assume ideal behavior.

  1. +7.68 J·K−^1
  2. +19.2 J·K−^1 correct
  3. − 7.68 J·K−^1
  4. − 19.2 J·K−^1
  5. +9.60 J·K−^1

Explanation:

Msci 15 1509 20:06, general, multiple choice, > 1 min,. 006 If you have an endothermic process in which the change in entropy is positive, how can you make it spontaneous?

  1. Increasing the pressure
  2. Decreasing the volume
  3. Increasing the temperature correct
  4. Decreasing the temperature
  5. Reducing the entropy change

Explanation:

∆G = ∆H − T ∆S

∆H > 0 for endothermic processes. ∆G < 0 for spontaneous processes.

T is always positive, so

∆G = ∆H − T ∆S = (+) − T ∆S

∆G is negative if T is very large, so in- creasing the temperature makes the process endothermic.

Msci 15 1412 20:05, general, multiple choice, > 1 min,. 007 Calculate ∆G at 298 K for the reaction

2 Ag 2 O(s) → 4 Ag(s) + O 2 (g).

Species ∆Hf^0 S^0 kJ/mol J/mol ·K Ag(s) 0. 0 42. 55 Ag 2 O(s) − 30. 57 121. 7 O 2 (g) 0. 0 205. 0

  1. 21.9 kJ/mol rxn correct
  2. 38.2 kJ/mol rxn
  3. 52.7 kJ/mol rxn
  4. −69.85 kJ/mol rxn
  5. 81.2 kJ/mol rxn Explanation:

∆Hrxn^0 =

n ∆Hf prod^0 −

n ∆Hf rct^0 = 0 kJ/mol − 2(− 30 .57 kJ/mol) = 61.14 kJ/mol

∆Srxn^0 =

n ∆Sf prod^0 −

n ∆Sf rct^0

=

[

4(42.55 J/mol · K)

  • (205.0 J/mol · K)

]