Benzene is a liquid under standard conditions. The standard enthalpy and entropy changes for the evaporation of liquid benzene to form benzene vapor are +33.9 kJ/mol and 96.3 J/mol · K, respectively. If liquid benzene is placed in a closed container at 78.85°C, which of the following statements is true? The liquid cannot attain equilibrium with its vapor in a closed container. a. b. The liquid will evaporate completely. Liquid benzene will evaporate until the vapor pressure is 1 bar. d. Liquid benzene will not evaporate. e. Liquid and vapor-phase benzene cannot coexist at temperatures greater than or equal to 78.85°C.

Benzene is a liquid under standard conditions. The standard enthalpy and entropy changes for the evaporation of liquid benzene to form benzene vapor are +33.9 kJ/mol and 96.3 J/mol · K, respectively. If liquid benzene is placed in a closed container at 78.85°C, which of the following statements is true? The liquid cannot attain equilibrium with its vapor in a closed container. a. b. The liquid will evaporate completely. Liquid benzene will evaporate until the vapor pressure is 1 bar. d. Liquid benzene will not evaporate. e. Liquid and vapor-phase benzene cannot coexist at temperatures greater than or equal to 78.85°C.

Chemistry: An Atoms First Approach

2nd Edition

ISBN:9781305079243

Author:Steven S. Zumdahl, Susan A. Zumdahl

Publisher:Steven S. Zumdahl, Susan A. Zumdahl

Chapter16: Spontaneity, Entropy, And Free Energy

Section: Chapter Questions

Problem 84AE: Calculate the entropy change for the vaporization of liquid methane and liquid hexane using the...

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For each process, tell whether the entropy change of the system is positive or negative, (a) A glassblower heats glass (the system) to its softening temperature, (b) A teaspoon of sugar dissolves in a cup of coffee. (The system consists of both sugar and coffee.) (c) Calcium carbonate precipitates out of water in a cave to form stalactites and stalagmites. (Consider only the calcium carbonate to be the system.)
Tetrachloromethane (carbon tetrachloride), CCl4, has a normal boiling point of 76.7C and an enthalpy of vaporization, Hvap, of 29.82 kJ/mol. Estimate the entropy of vaporization, Svap. Estimate the free energy of vaporization, Gvap, at 25C.
What is entropy? Why is entropy important?
Explain how the entropy of the universe increases when an aluminum metal can is made from aluminum ore. Thefirst step is to extract the ore, which is primarily a formof A12O3, from the ground. After it is purified by freeingit from oxides of silicon and iron, aluminum oxide ischanged to the metal by an input of electrical energy. 2Al2O3(s)electricalenergy4Al(s)+3O2(g)
A key component in many chemical engineering designs is the separation of mixtures of chemicals. (a) What happens to the entropy of the system when a chemical mixture is separated? (b) Are designs for chemical separation more likely to rely on spontaneous or nonspontaneous processes?
Which contains greater entropy, a quantity of frozen benzene or the same quantity of liquid benzene at the same temperature? Explain in terms of the dispersal of energy in the substance.
For mercury, the enthalpy of vaporization is 58.51 kJ/mol and the entropy of vaporization is 92.92 J/K mol. What is the normal boiling point of mercury?
For each process, tell whether the entropy change of the system is positive or negative. Water vapor (the system) deposits as ice crystals on a cold windowpane. A can of carbonated beverage loses its fizz. (Consider the beverage but not the can as the system. What happens to the entropy of the dissolved gas?) A glassblower heats glass (the system) to its softening temperature.
The standard molar entropy of iodine vapor, I2(g), is 260.7 J Kl mol-1 and the standard molar enthalpy of formation is 62.4 kJ/mol. a) Calculate the entropy change for vaporization of 1 mol of solid iodine (use data from Table 16.1 or Appendix J). b) Calculate the enthalpy change for sublimation of iodine. c) Assuming that rSdoes not change with temperature, estimate the temperature at which iodine would sublime (change directly from solid to gas).
Calculate the entropy change for the vaporization of liquid methane and liquid hexane using the following data. Boilling point (1atm) Hvap Methane 112K 8.20 KJ/mol Hexane 342K 28.9 KJ/mol Compare the molar volume of gaseous methane at 112 K with that of gaseous hexane at 342 K. How do the differences in molar volume affect the values of Svap for these liquids?
Consider the reaction of 2 mol H2(g) at 25C and 1 atm with 1 mol O2(g) at the same temperature and pressure to produce liquid water at these conditions. If this reaction is run in a controlled way to generate work, what is the maximum useful work that can be obtained? How much entropy is produced in this case?

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