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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.
Typology: Quizzes
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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
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?
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
mol·K
) ( (^) kJ mol
Using this data, which of the following an- swers includes the compounds that are ther- modynamically unstable?
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?
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.
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?
Explanation:
∆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
∆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)