Short Answer Questions


  1. a. F.  Chemical potentials must be equal, but not necessarily concentrations.
    b. T  (this is one of the conditions of equilibrium)
    c. F  The Clapeyron equation is general
    d. T
    e. F  This equation involves at least two approximations
    f. T
    g. F  One of the phases must be a vapor
    h. F  The Clapeyron equation is general

  2. ra  < rb implies Vm,a  > Vm,b, so DabV < 0. Since DabH < 0 (exothermic), DabS < 0, and dP/dT = DabS / DabV  > 0. Thus, b is the lower-temperature phase and the phase boundary has a positive slope. In this case, lower T and higher P will favor b.

  3. The m curve for the liquid phase shifts downward upon addition of solute, so that the intersection point with msolid (melting point) shifts to lower T and that with mvapor (boiling point) shifts to higher T.

  4. Since k is the derivative of V with respect to P and V is the derivative of G with respect to P, k is a second derivative of G. Thus it will exhibit a singularity at the transition temperature for 1st order transitions, and a discontinuity (step) for 2nd order transitions. The value of k need not be the same before and after the transition, and it may change in a positive or negative direction at the transition, and also exhibit P-dependence in either phase.

 


  1. (a) Raoult’s Law: 

    (b) Henry’s Law

 

Problems

1.      n2 = 1000 g/(18.01 g mole-1) = 55.525 mole
V / cm3 = 1001.21 + 34.69(m-0.070)2


2.                 
 At the equilibrium vapor pressure P, DvapG=0. At the standard pressure Po=1bar, DvapGo = DfG(Br2,vap)-DfG(Br2,liq) = DfG(Br2,vap)-0 These set the bounds of integration for the above equation, which we can carry out after making the approximation that DVm = Vm,vapor = RT/P.:

This is an incredibly low vapor pressure. Extra credit was given for recognizing and mentioning this fact. In fact, the test problem contained an error: 82.4 kJ / mol is the DfG of Br(g), not Br2(g). The actual DfG of Br2(g) is 3.110 kJ /mol, which leads to a more believable vapor pressure of 0.285 bar at 25°C using the above method.

3.                  Using the Clausius-Clapeyron equation, we use each pair of P, T points to find the slope:
 
Thus we may write the equations (using the lower temperature as reference point)

The triple point is found by solving these two equations for P, and T. Setting the pressures equal,

DfusH = DsublH-DvapH = 11.04 kJ mol-1

 

4.                  (a) DvapH = Tboil DvapS » Tboil (88.8 J K-1 mol-1) for a spherical nonpolar molecule.
Thus, DvapH = (273.2+68.6)(88.8) J mol-1 = 30.35 kJ mol-1.
(b) The y-intercept of the Clausius-Clapeyron plot for a spherical nonpolar molecule (i.e., Trouton’s focus) is 88.8 J K-1 mol-1 / R = 10.68. The slope of the plot is -DvapH / R. Thus,