Entropy Quiz
The question sheet
Reveal any answer as you study-
In the exercise, all four particles ending up in only one of two boxes relates to determining what quantity?
- Free energy
- Bond energy
- Enthalpy change
- Entropy change, ΔS
Reveal answer
Answer: Entropy change, ΔS
Source evidence
PDF page 683: two boxes. Determine the entropy change, ΔS, if the particles are initially evenly distributed between the two boxes, but upon redistribution all end up in Box (b).
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For the phase changes I₂(s) → I₂(g) and I₂(s) → I₂(l), the exercise asks whether ΔS is which of the following?
- Positive or negative
- Undefined
- Always constant
- Always zero
Reveal answer
Answer: Positive or negative
Source evidence
PDF page 683: temperature but somewhat higher pressure). I (s) ⟶ I (g) 2 2 I (s) ⟶ I (l) 2 2 Is ΔS positive or negative in these processes? In which of the processes will the magnitude of the entropy change be greater?
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Exercise 20 asks about the difference between which set of quantities for a chemical change?
- Kp, Kc, and Ksp
- ΔH, ΔH°, and ΔH°
- ΔG, ΔG°, and ΔG°
- ΔS, ΔS°, and ΔS°
Reveal answer
Answer: ΔS, ΔS°, and ΔS°
Source evidence
PDF page 684: 20. What is the difference between ΔS, ΔS°, and ΔS° for a chemical change?
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For the thermite reaction, the exercise states that during the reaction the surroundings absorb how much heat?
- 851.8 kJ/mol
- 100 kJ/mol
- 457.18 kJ/mol
- 27.95 kJ/mol
Reveal answer
Answer: 851.8 kJ/mol
Source evidence
PDF page 684: such thermite reaction is Fe O (s) + 2Al(s) ⟶ Al O (s) + 2Fe(s). Is the reaction spontaneous at room 2 3 2 3 temperature under standard conditions? During the reaction, the surroundings absorb 851.8 kJ/mol of heat.
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Exercise 27 asks whether the melting of 1 mole of NaCl(s) is spontaneous by calculating what at each temperature?
- Kp
- ΔG° of formation
- ΔH°
- ΔSuniv
Reveal answer
Answer: ΔSuniv
Source evidence
PDF page 684: 27. By calculating ΔSuniv at each temperature, determine if the melting of 1 mole of NaCl(s) is spontaneous at 500
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In Exercise 27, what is the given ΔH°fusion for NaCl?
- 72.11 kJ/mol
- 95.06 kJ/mol
- 27.95 kJ/mol
- 851.8 kJ/mol
Reveal answer
Answer: 27.95 kJ/mol
Source evidence
PDF page 684: °C and at 700 °C. J J S° = 72.11 S° = 95.06 ΔH° = 27.95 kJ/mol NaCl(s) NaCl(l) fusion mol·K mol·K What assumptions are made about the thermodynamic information (entropy and enthalpy values) used to solve this problem?
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For the reaction in Exercise 31 with ΔH° = 100 kJ/mol and ΔS° = 250 J/mol·K, the question asks whether it is spontaneous at what?
- High pressure only
- Low pressure only
- Absolute zero
- Room temperature
Reveal answer
Answer: Room temperature
Source evidence
PDF page 685: 31. A reactions has ΔH° = 100 kJ/mol and ΔS° = 250 J/mol·K. Is the reaction spontaneous at room
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Exercise 32 asks to explain what happens as a reaction starts with ΔG < 0 and reaches what point?
- ΔG > 100 kJ
- ΔH = 0
- ΔG = 0
- ΔS = 0
Reveal answer
Answer: ΔG = 0
Source evidence
PDF page 685: 32. Explain what happens as a reaction starts with ΔG < 0 (negative) and reaches the point where ΔG = 0.
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For the evaporation of water, the source gives ΔG° for H₂O(l) ⇌ H₂O(g) as which value?
- −17 kJ
- 1.7 kJ
- −30 kJ
- 8.58 kJ
Reveal answer
Answer: 8.58 kJ
Source evidence
PDF page 687: H O(l) ⇌ H O(g) ΔG° = 8.58 kJ 2 2 298
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The source notes that diamond spontaneously changing into graphite is described as which of the following?
- Observed rapidly
- Not observed
- Always endothermic
- Impossible thermodynamically
Reveal answer
Answer: Not observed
Source evidence
PDF page 687: diamond to graphite. Discuss the spontaneity of the conversion with respect to the enthalpy and entropy changes. Explain why diamond spontaneously changing into graphite is not observed.
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When ammonium chloride dissolves in water, the resulting solution is described as feeling how?
- Cold
- Hot
- Warm
- Neutral in temperature
Reveal answer
Answer: Cold
Source evidence
PDF page 688: 51. When ammonium chloride is added to water and stirred, it dissolves spontaneously and the resulting solution
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For the ATP-driven process, the source gives Glu + ATP → G6P + ADP with ΔG° equal to which value?
- −30 kJ
- −17 kJ
- +1.7 kJ
- +8.58 kJ
Reveal answer
Answer: −17 kJ
Source evidence
PDF page 687: described by the following equation: Glu + ATP ⟶ G6P + ADP ΔG° = −17 kJ 298 In this process, ATP becomes ADP summarized by the following equation: ATP ⟶ ADP ΔG° = −30 kJ 298 Determine the standard free energy change for the following reaction, and explain why ATP is necessary to drive this process: Glu ⟶ G6P ΔG° = ? 298
Chemistry: Atoms First
Chemistry: Atoms First by OpenStax, used under CC BY 4.0. Changes made by Stratacademy.
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