(a) Temperature dependence of the vapor pressure of ethylene log p (mm Hg) = - 834.13 / T + 1.75 log T - 8.375 x 10-3 T + 5.3234 , which gives the temperature dependence of the molar enthalpy of vaporization derive the equation. (b) Calculate the molar enthalpy of evaporation at -103.9 oC, the normal boiling point of ethylene. Note: Use the equation that gives the temperature dependence of ΔHvap= a + bT + cT2+ ….enthalpy.

(a) Temperature dependence of the vapor pressure of ethylene log p (mm Hg) = - 834.13 / T + 1.75 log T - 8.375 x 10-3 T + 5.3234 , which gives the temperature dependence of the molar enthalpy of vaporization derive the equation. (b) Calculate the molar enthalpy of evaporation at -103.9 oC, the normal boiling point of ethylene. Note: Use the equation that gives the temperature dependence of ΔHvap= a + bT + cT2+ ….enthalpy.

Chemistry: Principles and Reactions

8th Edition

ISBN:9781305079373

Author:William L. Masterton, Cecile N. Hurley

Publisher:William L. Masterton, Cecile N. Hurley

Chapter10: Solutions

Section: Chapter Questions

Problem 67QAP

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