Given the thermodynamic data of ethanol below, calculate the chanee in entropy of 55.5 g of solid ethanol at -150.0 Cheated to the normal melting point, melted, heated to the normal boiling point, and vaporized at.constant pressure. T=-114.1 °C Cyohd = 0.97 J/g"c AH = 4.9 kl/mol T= 78.37°C Cuqud= 2.46 J/g°c AHap- 38.7 k/mol

Given the thermodynamic data of ethanol below, calculate the chanee in entropy of 55.5 g of solid ethanol at -150.0 Cheated to the normal melting point, melted, heated to the normal boiling point, and vaporized at.constant pressure. T=-114.1 °C Cyohd = 0.97 J/g"c AH = 4.9 kl/mol T= 78.37°C Cuqud= 2.46 J/g°c AHap- 38.7 k/mol

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 39E: For ammonia (NH3), the enthalpy of fusion is 5.65 kJ/mol and the entropy of fusion is 28.9 J/K mol....

See similar textbooks

Similar questions

10.69 If a sample of air were separated into nitrogen and oxygen molecules (ignoring other gases present), what would be the sign of for this process? Explain your answer. The next four questions relate to the following paragraph [Frank L. Lambert, Journal of Chemical Education, 76(10), 1999, 1385]. "The movement of macro objects from one location to another by an external agent involves no change in the objects' physical (thermodynamic) entropy. The agent of movement undergoes a thermodynamic entropy increase in the process."
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?
How does the second law of thermodynamics explain a spontaneous change in a system that becomes more ordered when that process is exothermic?
The reaction CO2(g)+H2(g)CO(g)+H2O(g) is not spontaneous at room temperature but becomes spontaneous at a much higher temperature. What can you conclude from this about the signs of H and S , assuming that the enthalpy and entropy changes are not greatly affected by the temperature change? Explain your reasoning.
What is the second law of thermodynamics? For any process, there are four possible sign combinations for Ssys and Ssurr. Which sign combination(s) always give a spontaneous process? Which sign combination(s) always give a non-spontaneous process? Which sign combination(s) may or may not give a spontaneous process?
For the process A(l) A(g), which direction is favored by changes in energy probability? Positional probability? Explain your answers. If you wanted to favor the process as written, would you raise or lower the temperature of the system? Explain.
Explain why the statement No process is 100 efficient is not the best statement of the second law of thermodynamics.
Calculate the entropy change that occurs when 1.00 mol of steam is converted to liquid water at 100 C in a reversible process. (qvap = 40. 7 kJ/mol)
For ammonia (NH3), the enthalpy of fusion is 5.65 kJ/mol and the entropy of fusion is 28.9 J/K mol. a. Will NH3(s) spontaneously melt at 200. K? b. What is the approximate melting point of ammonia?
Limestone is predominantly CaCO3, which can undergo the reaction CaCO3(s)CaO(s)+CO2(g). We know from experience that this reaction is not spontaneous, yet S for the reaction is positive. How can the second law of thermodynamics explain that this reaction is not spontaneous?
9.54 The phase change between graphite and diamond is difficult to observe directly. Both substances can be hurned, however. From these equations, calculate H for the conversion of diamond into graphite. C(s,graphite)+O2(g)CO2(g) H= -393.51 kJ C(s,diamond)+O2(g)CO2(g) H= -395.94 kJ