Question

o.18 crm , at 95 %c.1. 3) Data for the vapor pressure of an unknown liquid are given T(C) 84 622, 1.433 694.,542 740 775 398 →5.4 40.4 01 43.4 27.0 30.1 32.8 36.5 484. .182 45.2 550,3 1O 469 The above data can be analyzed by the Clausius-Clapeyron equation According to that equation d (I/T)R where p is the vapor pressure, T is the absolute temperature (in K), and AH ap is the enthalpy of vaporization for the liquid. Therefore the slope in a plot of In(p) vs 1/T can be used to find the enthalpy of vaporization for the liquid a) Plot the above data by hand according to eq 3.I. You may use the data plot in the sample lab report as guide. You will normally be allowed to use EXCEL or other plotting programs for data plots in your lab report, but for this problem I am requiring that you do the plot by hand. I will provide you with graph paper for your plot. b) Find the best fiting line to the above data (slope and intercept, including correct units) c) Based on your answer in b, find AH vap (including correct units)

Graph should be ln(p) vs 1/T

Please show graph and full work for problem

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Answer 1

Clausius - Clapeyron equation: ln (P_1/P_2) = -Delta H degree_vop/R (1/T_1 - 1/T_2) rightarrow Graph should be lnp vs 1/T y = mx m = -Delta H degree_vop/R Inp rightarrow Torr (or) mm Hg YT rightarrow 1/k rightarrow k rightarrow because R = 8.314 J/mole . k \r\n Taking (1) & (2) points from best let to calculate slope. alpha_1 = 3.33 (m/k) y_1 = 5.986 (Torr) alpha_2 = 3.125 (m/k) y_2 = 6.65 (Torr) m = y_2 - y_1/x_2 - x_1 = -3.239 m = -3.239 times 10^3 (torr/k) or mm of Hg Therefore m = -3.239 times 10^3 (Torr/k) (c) + Delta H degree_vap/R = + 3.239 times 10^3 (Torr/k) + Delta H degree_vap = 3.239 times 8.314 (J/mol . k) times k Middot Torr times 10^3 Delta H degree_vap = 2.6 Delta H degree_vap = 26.92 times 10^3 J/mol Delta H degree_vap = 26.92 KJ/mol

Continues....