Every problem in the chapter beside the book’s answer, a chapter at a time as you scroll.
Chapter 12: Thermodynamics
36 problems · 4 figuresExample 12.1 §12.1 source ↗
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Example 12.2 §12.2 source ↗
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Example 12.3 §12.2 source ↗
(a)
(b) the freezing of liquid water
(c)
(d)
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Example 12.4 §12.3 source ↗
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Example 12.5 §12.3 source ↗
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Example 12.6 §12.3 source ↗
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Example 12.7 §12.4 source ↗
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Example 12.8 §12.4 source ↗
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(b) 141.5 kJ/mol, nonspontaneous
Example 12.9 §12.4 source ↗
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12-1 §12.1 source ↗
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12-3 §12.1 source ↗
Indicate whether the following processes are spontaneous or nonspontaneous.
(a) Liquid water freezing at a temperature below its freezing point
(b) Liquid water freezing at a temperature above its freezing point
(c) The combustion of gasoline
(d) A ball thrown into the air
(e) A raindrop falling to the ground
(f) Iron rusting in a moist atmosphere
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12-5 §12.1 source ↗
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12-7 §12.2 source ↗
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12-9 §12.2 source ↗
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12-11 §12.2 source ↗
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12-13 §12.2 source ↗
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12-15 §12.2 source ↗
Indicate which substance in the given pairs has the higher entropy value. Explain your choices.
(a) C2H5OH(l) or C3H7OH(l)
(b) C2H5OH(l) or C2H5OH(g)
(c) 2H(g) or H(g)
Answer
12-17 §12.2 source ↗
Predict the sign of the entropy change for the following processes. Give a reason for your prediction.
(a)
(b)
(c)
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12-19 §12.2 source ↗
Answer
There are 7.5 moles of gas initially, and 3 + 6 = 9 moles of gas in the end. Therefore, it is likely that the entropy increases as a result of this reaction, and ΔS is positive.
12-21 §12.3 source ↗
Calculate for the following changes.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
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12-23 §12.3 source ↗
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12-25 §12.3 source ↗
Using the relevant values listed in Appendix G, calculate for the following changes:
(a)
(b)
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12-27 §12.3 source ↗
By calculating ΔSuniv at each temperature, determine if the melting of 1 mole of NaCl(s) is spontaneous at 500 °C and at 700 °C.
What assumptions are made about the thermodynamic information (entropy and enthalpy values) used to solve this problem?
Answer
12-29 §12.3 source ↗
Use the standard entropy data in Appendix G to determine the change in entropy for each of the following reactions. All the processes occur at the standard conditions and 25 °C.
(a)
(b)
(c)
(d)
(e)
(f)
Answer
12-31 §12.4 source ↗
Answer
Above 400 K, ΔG will become negative, and the reaction will become spontaneous.
12-33 §12.4 source ↗
Use the standard free energy of formation data in Appendix G to determine the free energy change for each of the following reactions, which are run under standard state conditions and 25 °C. Identify each as either spontaneous or nonspontaneous at these conditions.
(a)
(b)
(c)
(d)
(e)
(f)
Answer
12-35 §12.4 source ↗
Given:
(a) Determine the standard free energy of formation, for phosphoric acid.
(b) How does your calculated result compare to the value in Appendix G? Explain.
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12-37 §12.4 source ↗
Consider the decomposition of red mercury(II) oxide under standard state conditions.
(a) Is the decomposition spontaneous under standard state conditions?
(b) Above what temperature does the reaction become spontaneous?
Answer
12-39 §12.4 source ↗
Calculate ΔG° for each of the following reactions from the equilibrium constant at the temperature given.
(a)
(b)
(c)
(d)
(e)
(f)
Answer
12-41 §12.4 source ↗
Calculate the equilibrium constant at 25 °C for each of the following reactions from the value of ΔG° given.
Answer
12-42 §12.4 source ↗
Determine ΔGº for the following reactions.
(a) Antimony pentachloride decomposes at 448 °C. The reaction is:
An equilibrium mixture in a 5.00 L flask at 448 °C contains 3.85 g of SbCl5, 9.14 g of SbCl3, and 2.84 g of Cl2.
(b) Chlorine molecules dissociate according to this reaction:
1.00% of Cl2 molecules dissociate at 975 K and a pressure of 1.00 atm.
Answer
(b) 98.9 kJ/mol
12-44 §12.4 source ↗
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12-46 §12.4 source ↗
The evaporation of one mole of water at 298 K has a standard free energy change of 8.58 kJ.
(a) Is the evaporation of water under standard thermodynamic conditions spontaneous?
(b) Determine the equilibrium constant, KP, for this physical process.
(c) By calculating ∆G, determine if the evaporation of water at 298 K is spontaneous when the partial pressure of water, is 0.011 atm.
(d) If the evaporation of water were always nonspontaneous at room temperature, wet laundry would never dry when placed outside. In order for laundry to dry, what must be the value of in the air?
Answer
12-48 §12.4 source ↗
One of the important reactions in the biochemical pathway glycolysis is the reaction of glucose-6-phosphate (G6P) to form fructose-6-phosphate (F6P):
(a) Is the reaction spontaneous or nonspontaneous under standard thermodynamic conditions?
(b) Standard thermodynamic conditions imply the concentrations of G6P and F6P to be 1 M, however, in a typical cell, they are not even close to these values. Calculate ΔG when the concentrations of G6P and F6P are 120 μM and 28 μM respectively, and discuss the spontaneity of the forward reaction under these conditions. Assume the temperature is 37 °C.
Answer
12-50 §12.4 source ↗
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12-52 §12.4 source ↗
What happens to (becomes more negative or more positive) for the following chemical reactions when the partial pressure of oxygen is increased?
(a)
(b)
(c)
Answer
(b) Increasing will shift the equilibrium toward the products, which increases the value of K. ΔG° therefore becomes more negative.
(c) Increasing will shift the equilibrium the reactants, which decreases the value of K. ΔG° therefore becomes more positive.
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Problems and answers are adapted from Chemistry: Atoms First 2e by Paul Flowers, Edward J. Neth, William R. Robinson, Klaus Theopold and Richard Langley, OpenStax, and are used under the CC BY-NC-SA 4.0 licence. Access for free at https://openstax.org/books/chemistry-atoms-first-2e/pages/1-introduction. Illustrations: Copyright Rice University, OpenStax, under CC BY-NC-SA 4.0 license. This adaptation is released under the same licence, is non-commercial, and is neither affiliated with nor endorsed by OpenStax or Rice University.