1. Suppose dn moles of A flow from phase b to phase a. Then the change in total G will be dG = -m_A^a dn + m_A^bdn. For a spontaneous process, dG < 0. So (m_A^b-m_A^a)dn < 0 , and since dn is positive, we have -m_A^b < m_A^a .

2. The major source of inaccuracy is nonideality in the solute (all of the colligative properties were derived assuming ideal behavior). Various types of nonideality are possible: attractive or repulsive interactions between solutes, association between molecules, ion pairing, excluded volume, etc. The best way to work around this type of nonideality is to measure limiting behavior as solvent concentration goes to zero.

3. Since catalysts lower the activation energy of a reaction, the Arrhenius law says that catalyzed reactions will become less sensitive to changes in temperature (neglecting any change in catalyst activity with temperature, e.g. enzyme denaturation).

4. In general, one determines the rate law and compares it to the stoichiometry of the reaction. Full credit was given for mentioning any 2 of the following possibilities. (1) Reaction involves 3 or more molecules in either forward or reverse directions, (2) Rate law does not correspond to molecularity or stoichiometry of the net reaction, (3) Temperature dependence of reaction does not obey Arrhenius law (or has a negative apparent E_a) (4) Rate law depends on species not in the net reaction, (5) Reaction involves changes to more than 1 part of the molecule that cannot occur in a concerted way (6) One can’t always distinguish complex from elementary reactions.

5. (a) The solid line cannot have negative slope since
(b) The solid line cannot have positive slope since

6. (a) Since A-A and B-B interactions are identical, the pure liquids will have the same vapor pressure. Then, if the mixture obeys Raoult’s Law, P = P_A^* + (P_B^*-P_A^* )x_B = P_A^* =P_B^* ; i.e., the pressure will be the same, independent of composition.
(b) The composition of the gas will be the same as that of the liquid. Since P_tot = P_A^* =P_B^*, we have y_A = P_A/P_tot = P_A/P_A^* = x_A, and similarly for B.

Problems

1. (a) Find molality from freezing pt . depression:
Molar mass is therefore
(b) Total volume of solution is mass/density:

(c) Phenol has a MW of 6´12.01+6*1.01 + 16.00 = 94.12. The overestimate of MW by colligative properties suggests that phenol molecules are associating in solution. However, since both osmotic pressure and freezing-point depression both depend on the number of solute particles instead of the solute molecular weight, the error will not affect the estimate of the osmotic pressure.

2. Since m is a partial molar quantity, we can write

From the definitions of chemical potential in terms of g,

Then, substituting into the Gibbs-Duhem equation for m,

But dx_A + dx_B=0, since x_A + x_B = 1, d(x_A+x_B) = d(1) Thus,

3. The second-order rate constants for the reaction of oxygen atoms with benzene are 1.44´10⁷ M^-1 sec^-1 at 300.3 K and 3.03´10⁷ M^-1 sec^-1 at 341.2 K.
(a) Estimate the pre-exponential factor and activation energy for this reaction. (10 pts)
(b) At 400 K, the DG^o for this reaction is -35.2 kJ mol^-1. Estimate the equilibrium constant and the rate constant for the reverse reaction at this temperature. (10 pts)

-R times the slope of this line gives E_a.

Plug in a pair of k, 1/T values to get pre-exponential

(b)

4.

Propagators are CH₃ and CH₃CO.
Rate expression is

Steady state approximations for propagators:

Adding steady state approximations gives

Substitution into rate expression gives

5. From Rule of Thumb 2b, the composition of the reactants in the rate-determining step is Ce⁴⁺ + Tl⁺ + Ag⁺ - Ce³⁺ = Tl + Ag (net charge 3+). Also, Ce³⁺ is a product of an equilibrium preceding the rds, but is not a reactant. So the rds must be
Tl⁺ + Ag²⁺ ® Tl²⁺ + Ag⁺ (slow)
The pre-equilibrium needs to produce Ce³⁺ and the intermediate Ag²⁺:
(1) Ce⁴⁺ + Ag⁺ = Ce³⁺ + Ag²⁺ (fast)
Finally, Tl²⁺ must be oxidized to Tl³⁺. The oxidant could be either Ce⁴⁺ or Ag²⁺ :
(3) Ce⁴⁺ + Tl²⁺ ® Ce³⁺ + Tl³⁺ OR Ag²⁺ + Tl²⁺ ® Ag⁺ + Tl³⁺ (fast)

In the second case, two equivalents of the first reaction must react for each Tl³⁺ produced.

6. (a)

(b)Steady-state approximations for [ES] and [ESI]:

(c) From the steady-state approximation for d[ESI]/dt, we have

Since we assume we can substitute into the steady-state approximation for d[ES]/dt:

Substituting the expressions for [E] and [ESI] into the conservation law gives

Then, substituting into the expression for d[P]/dt,