Enthalpy of vaporization from Henry's law measurements - Journal of

Principles of Solution Thermodynamics: Demonstration of Nonideal Behavior of Henry's Law. An Undergraduate Laboratory Experiment. Luke Chandler Short ...
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JOHN J. ALEXANDER University of Cincinnati Cincinnati.OH 45221

Enthalpy of Vaporization from Henry's Law Measurements

method or a linear regression routine on, e.g., Texas Instrument SR-51 calculator may he employed. The result is: 2840

In K M (atm) = - + 11.6 T

Wiktor M. Raldow The Royal Institute of Technology Stockholm 70, Sweden Wayne E. Wentworth University of Houston Houston, Texas 77004 The solubility of NH3 in n-heptanol was investigated at different temoeratures. The results of the measurements can be summarized as follows: T (K)

P (atm)

311.9 317.1 322.6 328.5 333.0 337.4 342.6 346.6 352.1

1.83 2.11 2.88 3.66 4.01 4.50 5.41 6.05 7.87

From this equation the vaporization enthalpy can he calculated to he AHo = R.2840 = 24.5 kJ/mol This can he compared with the value obtained for pure NH3 from the equation given in the question: LW, = 23.0 kJ1mal The entropy terms (intercept = R.ASo) are equal. The entropy of vaporization is AS,, = 23.05 cal1K-mol = 96.4 JIK-mol which is in reasonable agreement with Trouton's Rule. In conclusion: The vaporization enthalpy for NH3 from the solution is somewhat higher indicating a weak interaction with the solvent (presumably with the s o l v e n t O H group). The entropy changes are practically equal and describe the vaporization of 1 mole of (ammonia) gas to a pressure of one atmosphere. The deviations from the ideal solution are thus small.

where T = temperature (K), P = equilibrium vapor pressure, X = mole ratio of NH3 in the liquid phase. Calculate the enthalpy of vaporization of NH3 in the temperature range of the experiments. You may assume that the gas phase consists solely of NH3 and that Henry's law is obeyed. From the literature i t was found that the vapor pressure above liquid NH3 can he described as: 2770

In P (atm) = - --ir + 11.6

Use the information derivable from this equation to comment on your own results. Acceptable Solution The analysis of the temperature dependence of Henry's law constant can be approached in a straightforward manner by assuming K H is an equilibrium constantwith the temperature dependence given AS" I n K H = - - +AHo R. T ..

..R

In this equation it is assumed that AH" and AS" are temperature independent (ACp = 0). Obviously ln KH should he linearly related to reciprocal absolute temperature. I t should be realized that the AH" and ASo for this vaoorization denend on the standard states of NH? in the heotnnol and in the gas phase. I>or Ntl:~in heptunoi the standard owoerries of the infinitels dilute state is X