Name: _____KEY______________ TA: Section:

Score: /20

Quiz 1: Tuesday, 2-1-11

Show all work to receive full credit. Circle your final answer. If you don’t know how to do a problem, try your best to explain how you might go about doing it – you might get partial credit!

Potentially useful equations:  ln !!!!

=  !!˚!"#!

!!!− !

!!

ln! =−ΔH˚vap!" +

ΔS˚vap!                                           ! +  

!!!

!! ! − !" = !"#

If you do not remember an equation that is not listed above, I will give it to you for the price of

one point.

1. Octane, CH3(CH2)6CH3, is a major component in gasoline. It has a vapor pressure of 13.95 Torr at 25.0˚C and 144.78 Torr at 75.0˚C, and a standard enthalpy of vaporization (ΔH˚vap) of 40.3 kJ/mol. Determine the following: a. What intermolecular forces would you expect to occur in a solution of octane? (1 pt) b. Standard entropy of vaporization (ΔS˚vap) (2 pts) c. Standard Gibbs free energy of vaporization (ΔG˚vap) (2 pts) d. Normal boiling point of octane (remember: gasoline is a liquid at the pump!) (1 pt)

a) London Dispersion Forces. No dipoles (only C-H bonds). No Hydrogen bonds. Remember Hydrogen bonding intermolecular forces only count if a hydrogen is covalently bonded to O, N or F. It does not mean that some atom is simply “bonded” to hydrogen

b) Let us use the equation that relates enthalpy and entropy to pressure and temperature.

ln ! +  ∆!!"#!

!" =∆!!"#!

!        =>      ∆!!"#! = !"# ! +  ∆!!"#!

!

∆!!"#! = 8.314!

!"# ∗ ! ∗ ln1.83×10!!!"#1.00  !"# +

40.3!3 !!"#

298.15  !

∆!!"#! = 135.2− 33.3  !

!"# ∗ !

∆!!"#! = 101.9  !

!"# ∗ !

c) Now that we have standard entropy and standard enthalpy, we can determine the standard free energy.

∆!!"#! = ∆!!"#! − !∆!!"#!

∆!!"#! = 40.3×10!   !!"#

− 298.15! 101.9 !!"#∗!

∆!!"#! = 40.3− 30.4 !"!"#

d) The “normal” boiling point is a substance’s boiling point at 1 atm of pressure. We can reuse the equation for free energy in terms of enthalpy and entropy, noting that there is no free energy when a system is at dynamic equilibrium, such as at the boiling point of a liquid. Remember, this is true for any phase boundary (the lines on the phase diagram).

∆!!"#! = ∆!!"#! − !∆!!"#!

∆!!"#! − !∆!!"#! = 0

!"#$%&'  !"#  !…

! =  ∆!!"#!

∆!!"#! =  40.3  ×  10! !

!"#

101.9   !!"#∗!

! = 395!

2. Calculate the pressure exerted by 1.00 mol C2H6(g) behaving as (a) an ideal gas; (b) a Van der Waals gas when it is confined to 250 mL at 100˚C. (4 pts) Van der Waals parameters for C2H6: a = 5.507 L2*atm*mol-2; b = 6.51 x 10-2 L*mol-1

(a) Solving for pressure in the ideal gas equation…

! =  !"#!

! =  1.00  !"# 0.08206  !∗!"#!"#∗! 373.15!

0.250  !

! =  1.22  ×10!!"#

(b) Solving for pressure in the ideal gas equation…

! =  !"#! − !" − !

!!

!!

! =  1.00  !"# (0.08206!∗!"#!"#∗!)(373.15!)0.250! − (1.00!"#)(6.51×10!! !

!"#)− (5.507!

!∗!"#!"!!

)(1.00!"#)!

(0.250  !)!

! =  1.656×10!!"# − 8.811×10!!"#

! =  77.5  !"#

3. Match each compound (A-E) with its boiling point by writing the compound’s letter designation under its boiling point. (5 pts) Analysis The only molecules with H-bonds are B and E. B has 2 H-bonding sites, and so has the strongest IM forces of all the molecules. C is the only molecule that has a net dopole with no H-bonds. A and D both participate ONLY in Van der Waals (VDW) interactions, and so are the weakest. But how do we compare between A and D? They have exactly the same formula weight, thus the same number of electrons and polarizability. Here we must recall the effect of shape on VDW interactions. Since D has more branching, its shape is more round, as opposed to A, which is more rod-like. Thus, A has a higher surface area and more sites for VDW contacts (remember: spheres are round, and they have the smallest surface area per volume of any 3D object…this is why bubbles form)

So we have ranked the molecules in terms of IM forces as D < A < C < E < B. The boiling points must follow this same trend

58°C 69°C 89°C 101°C 198°C

D A C E B

CC

CC

H

H

H

H

HH

H

H

H

H

OH

H

OH

H

OH

H

OH

H

O

H

HO

H

H

4. We say that the hydrocarbon butane (CH3CH2CH2CH3) is “insoluble” in water at 298K and 1 atm.

a. What does this imply about the sign of the standard Gibbs free energy of mixing

(ΔGmix)? (2 pts)

ΔGmix must be positive. “Insoluble” implies they do not mix. A more technical way of saying this is that their mixing is non-spontaneous. (ΔGmix > 0).

b. Using the equation that relates ΔG to enthalpy and entropy, explain why butane does not spontaneously dissolve in water, despite the fact that the mixing of butane in water is slightly exothermic at standard conditions. A simple figure may help! (Hint: don’t worry about absolute numbers. The equation and figure should guide your reasoning descriptively) (3 pts) In class, we learned that water molecules can “orient” themselves around the butanes in a cage-like structure. This ordering comes at a cost of energy, specifically by decreasing the entropy. Thus, ∆!!"#! < 0 and (−!∆!!"#! ) > 0.

Relating this idea to one of our favorite equations,

∆!!"#! = ∆!!"#! − !∆!!"#!

Since (−!∆!!"#! ) > 0 it overpowers the negative sign of enthalpy and makes ∆!!"#! > 0 (the entire process non-spontaneous), and thus we call the mixture insoluble.

Water molecules are more organized than in pure water.