Ethanol and water mixed together to form a solution at 25 degrees celsius. 100 g
ID: 825346 • Letter: E
Question
Ethanol and water mixed together to form a solution at 25 degrees celsius. 100 grams of ethanol (EtOH) and 150 grams of water are mixed.
a) Calculate the mole fractions: X (EtOH) and X water
b) The vapor pressure of pure water at 25 degrees celsius is 23.78 Torr and the vapor pressure of pure EtOH is 59.20 Torr. In the solution the water vapor pressure is 20.25 Torr and the Ethanol vapor pressure is 27.02 Torr. Calculate the activity of water and the activity of EtOH.
c) Calculate the activity coefficients of water and EtOH.
d) Is the delta H mix of these two liquids postive or negative? why, from a molecular perspective?
e) Calculate the delta G mix, the free energy ehange of mixing the liquids, in Joules, not J/mol.
f) under the assumption that this is a regular solution( delta S mix regular = delta S mix ideal, calculate the beta parameter.
Explanation / Answer
n(EtOH) = 100/46 =2.173
n(water) = 150/18 =8.333
a)
X(EtOH) = 2.173/(2.173+8.333) =0.206
X(water) =1- 0.206 =.794
b)
activity of water =p/p0 =20.25/23.78=0.851
activity of EtOH = 27.02/59.20 =0.456
c)
activity coefficient of water = activity/X(water) = 0.851/0.794 =1.071
activity coefficient of EtOH = 0.456/ 0.206 = 2.213
d) positive, because due to intermolecular interaction ..... the strong interaction of water and ethanol caused the volume to drop slightly .water and ethanol in solutions alone have their own intereactions once combined together .... the interaction is slightly greater than either individual liquids .
e)
g mixture=RT{X(water)*ln(activity of water) + X(EtOH) *ln(activity of EtOH)}
g mix =8.314 * 298 { 0.794*-0.161 + 0.206*-0.785}
g mix = -717.366 J
f)
beta coefficient = 1/(k*T)
k= boltzmann constant =1.38*10 to power 0f -23
beta coefficient = 2.43* 10 topower of 20.
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