Short Answer Questions

1.

2.

3.

C_P is a 2^nd derivative of m. For a 2^nd order transition.It should exhibit a discontinuity at the phase transition (which appears as a step in the calorimeter)

4. Which of the following pairs of compounds would you expect to exhibit markedly nonideal mixing, and which should exhibit more nearly ideal mixing?
(a) H₂O and n-propanol, CH₃(CH₂)₃CH₂OH nonideal (dissimilar sizes, polarity)
(b) toluene (methyl benzene) and ethyl benzene ideal
(c) solid D-alanine, CH₃CH(NH₂)COOH, and its stereoisomer, L-alanine nonideal
(d) acetone and carbon tetrachloride nonideal (not similarly enough shaped)
(e) ethanol and methanol ideal

5. Yes. D_mixS is greatest for a 50:50 mixture in an ideal solution.

Answers to Problems

Problem 1.

(a)

(b)

(c)

n₂ = 0.01 mol; V and V₁ were calculated above.
n₁ = 1000. g / 18.00 g mol^-1 = 55.56 mol

Problem 2.

(a) Note that P_A* = P(x_C=0)=347 torr, P_C* = P(x_C=1)=293 torr, and x_A = 1 - x_C = 0.4

(b) The Henry’s law constant for acetone in chloroform is the limiting behavior of P_A in the limit of low acetone (i.e., the linear terms) extrapolated to x_C = 0:
K_A = 0.048 + 89.96(1-0) = 90.01 torr

Similarly, the Henry’s law constant for chloroform in acetone is obtained by extrapolating the the lowest-order terms in the expression for P_C to x_C = 1:
K_C = 0.643 + 116.1(1) = 116.7 torr

You could also have extrapolated graphically, as shown below:

Problem 3.

since the chemical potentials for pure substances 1 and 2 are constants. Plugging this result into the G-D equation gives

But dx₁ = -dx₂, since x₁ + x₂ = 1. Thus these terms cancel and what remains is the G-D equation for chemical activity coefficients:

Problem 4

Use

(a) T_f^* = 276.97 K DH_m,fus = 6.23 kJ mol^-1. M₁ = 2(2.014)+16.00 = 0.0200 kg mol^-1

(b) M₂ = 6(1.008)+3(12.00)+16.00 = 58.05 g mol^-1. Then

.

(c) The expression used assumes (1) ideal behavior in the solution, and (2) behavior in the limit of zero molality of acetone. It will lose accuracy as the molality increases.

Bonus

From the Clapeyron Equation, . Observe that dP/dT changes sign for the b®a and the a®b transitions. This is most likely because DV changes sign as the pressure is increased (i.e., because the b phase is more compressible)