1.      Explain the structure of a hydrogen bond.

The electrons of an electronegative atom are attracted to a partially positively charged H which is bonded to another electronegative atom.



2.      Write the Lewis structures and give the shapes of each of the following molecules; predict which substance of each pair has the higher boiling point.  Explain your reasoning.


a)      Ethanol, CH3CH2OH or dimethyl ether, CH3-O-CH3?


Ethanol can hydrogen bond therefore the intermolecular forces are stronger and it has a higher boiling point.


b)      Butane, C4H10 or octanol, C8H18?


Octanol has a higher molecular weight and therefore is has stronger dispersion forces (London forces) leading to a higher boiling point.


c)      PF3 or PCl3?

Because F is more electronegative than Cl, PF3 is more polar and has stronger intermolecular forces therefore it has a higher boiling point.




d)      SO2 or CO2?


e)      See question below.  Polar SO2 has stronger intermolecular forces than nonpolar CO2.


f)        Extra question:  Explain why sulfur dioxide, SO2, has a dipole moment of 1.63D, but that of CO2 is zero.

SO2 is bent therefore it has a net dipole moment towards the oxygens.  CO2 is linear and the two bond dipoles cancel.




3.      The phase diagram of an unknown substance is shown below: 



















a)      Label the solid liquid and gas regions of the phase diagram.

b)      Label the triple point and the critical point.

c)      Label the normal freezing and boiling points.

There is not a normal boiling point or freezing point as the picture is drawn.  The normal sublimation point is the point where there is a transition between the solid and the gas phase at 1 atm pressure.


d)      Which phase is more dense the liquid or the solid?  Explain your reasoning.

The liquid is more dense.



4.      Explain why CH3OH is completely soluble in water but not completely soluble in toluene, C6H5CH3.


CH3OH is polar so it dissolves easily in polar solvent water but not so easily in nonpolar toluene.



5.      A room in which the humidity has been lowered feels cooler.  Yet the dehumidifier has not “cooled” the room.  Why is that?


As the humidity is lowered, more water can evaporate before the air is again saturated with water.  The evaporation of water is an endothermic process which means that the heat is drawn out of the room or the room is cooled as the water evaporates.



6.      How might one go about establishing whether a solution of sodium sulfate in water is saturated, unsaturated, or supersaturated?


Add a little more sodium sulfate.  If the solution is unsaturated, the salt will dissolve, if it is saturated the salt will settle on the bottom of the solution, and if the solution is supersaturated the addition of additional salt will cause the excess salt to precipitate out of solution.



7.      A 10.7 m solution of NaOH has a density of 1.33 g/mL at 20oC.  Calculate

a)      The mole fraction of NaOH



b)      The mass percentage of NaOH



c)      The molarity of the solution




8.      A solution of glucose (C6H12O6) is prepared by dissolving 100.0 g of glucose in 1000. g of water.  The density of the resultant solution is 1.050 g/mL.  Kb for water is 0.52 oC/m and kf for water is –1.86 oC/m.

a)      What is the vapor pressure of the solution at 100.0oC?


Vapor pressure = (Xwater)(Pwater) = (0.9901)(760 torr) = 752 torr


b)      What is the boiling point of the solution?


rTb = kb•m = (0.52 oC/m)(0.5549 m) = 0.29oC

Boiling point = 100.29oC

c)      What is the osmotic pressure of the solution at 25oC?




9.      A compound of carbon, hydrogen, and oxygen was burned in oxygen, and 1.000 g of the compound produced 1.434 g CO2 and 0.783 g H2O.  In another experiment, 0.1107 g of the compound was dissolved in 25.0 g of water.  This solution had a freezing point of -0.0894oC.  What is the molecular formula of the compound?


10.    The osmotic pressure of blood at 37oC is 7.7 atm.  A solution that is given intravenously must have the same osmotic pressure as the blood.  What should be the molarity of a glucose solution to give an osmotic pressure of 7.7 atm at 37oC?





11.    For the equilibrium         C(s)  +  2 H2(g)  <==>  CH4(g)  +  heat


a)      Write the equilibrium constant expression as Kc.               

b)      What are the units of the equilibrium constant Kp?

(pressure) -1

c)      How is Kc related to Kp (specify quantitatively)

Kc = Kp(RT)             

d)      For each of the following changes to the system at equilibrium, predict the direction of the shift and explain why it occurs:



Shift (left or right)


The volume of the reaction vessel is doubled.


volume change causes P to drop.  Shift left makes more mols of gas and thereforea higher pressure

The temperature is increased.


shifts left to use up heat

The pressure of H2(g) is increased.


shifts to the right to decrease the moles of H2.  Reaction rate to right increases, to the left stays the same so there is a net shift tothe right.

C(s) is added to the system.

no change

The concentration of C stays the same.

Adding a catalyst

no change

a catalyst does not affect the positin of the equilibrium.  (So then, just what does it do?)


12.    The equilibrium constant for the reaction N2O4(g)  <==>  2 NO2(g)  is 0.212 mol/L at 100o C.  What is the value of Kc at 100oC for:


a)      2 NO2(g)  <==>  N2O4(g)


reaction is reversed so Krxn = 1/(Kc) = 1/.212 = 4.72


b)      NO2(g)  <==>  1/2 N2O4(g)


reaction is halved and reversed so Krxn = = 2.17

13.    For the reaction NO(g)  +  NO2(g)  +  H2O(g)  <==>  2 HNO2(g), occurring at 28oC, [NO]i = [NO2]i = 44.1 torr and [H2O]i = 17.5 torr.  If the total pressure at equilibrium is 95.6 torr.


a)      What are the equilibrium pressures of all species?


NO(g)  +  NO2(g)  +  H2O(g)  <==>  2 HNO2(g)               total pressure


Initial   44.1 torr        44.1 torr           17.5 torr                      0 torr                         105.7 torr


D          -x                         -x              -x                        +2x                                   -10.1 torr


Equil      44.1-x               44.1-x        17.5-x                  2x                                      95.6 torr



final pressure = P(NO)  +  P(NO2)  +  P(H2O)  +  P(HNO2)

95.6 torr   =  (44.1-x)torr +  (44.1-x)torr + (17.5-x)torr  +  (2x)torr

x = 10.1 torr


Alternate method

pressure change = -10.1 torr = -x + -x + -x + +2x = -x

x = 10.1 torr


P(NO) = 44.1 torr - 10.1 torr = 34.0 torr


P(NO2) =  44.1 torr - 10.1 torr = 34.0 torr


P(H2O) = 17.5 torr - 10.1 torr = 7.4 torr


P(HNO2) = 2(10.1 torr) = 20.2 torr



b)      Calculate Kp for the reaction?




14.    At some temperature the system  2 SO2(g)  +  O2(g)  <==>  2 SO3(g)  is at equilibrium when   [SO2] = 0.0100 M,           [O2] = 0.200 M and [SO3] = 0.100 M.  What is the value of Kc at this temperature?






If at the same temperature, 3.00 mol of SO3 is added to an empty 1.00 L vessel, what will be the value of [O2] at equilibrium?  (hint: check your answer - iterate to within 5%)

2 SO2(g)  +  O2(g)  <==>  2 SO3(g)


Initial              0                    0                         3.00 M


D                +2x                  +x                       -2x


Equilibrium  2x                    x                         3.00-2x



1st iteration

x = 0.165 M


check answer: 

the assumption that 2x << 3.00 is incorrect.


2nd iteration

x = 0.153 M






3rd iteration

x = 0.154 M



4th iteration

x = 0.154 M


x’s have converged we are finished



[SO2] = 2(0.15 M) = 0.30 M

[O2] =  0.15 M

[SO3] =  {3.00 - 2(0.15)}M = 2.70 M


15.    The reaction  2 NO(g)  +  Br2(g)  <==>  2 NOBr(g)  has a Kp = 1.17 atm-1 at 25oC.  If 1.10 atm of NOBr, 0.100 atm of NO, and 0.0100 atm of Br2 are mixed at 25oC, what reaction will occur?  Explain.



a)      When 5.00 atm of NOBr is allowed to equilibrate at 50oC, the equilibrium pressure of NOBr is measured to be 4.30 atm.  What is the value of Kp at 50oC?  Compare with the value of Kp at 25oC and explain.


2 NO(g)  +  Br2(g)  <==>  2 NOBr(g)

Initial      0 atm        0 atm                 5.00 atm

D            +2x                 +x                 -2x                     

Equil           2x                 x                4.30 atm               x = 0.35 atm

= 0.70 atm     = 0.35 atm



The Kp increases as the temperature is increased.  This means the reaction shifts to the right when heat is added so it is an endothermic reaction.


16.    Assume that you place a fresh water plant into a salt solution and examine it under   a microscope.  What happens to the plant cells?  What if you placed a salt-water plant in pure water?  Explain.  Draw pictures to illustrate your explanation. Assume that you place a fresh water plant into a salt solution and examine it under   a microscope.  What happens to the plant cells?  What if you placed a salt-water plant in pure water?  Explain.  Draw pictures to illustrate your explanation.



A fresh water plant in salt water will lose water from the plant cells to dilute the salt solution and all of the cells will shrivel.


A salt water plant in pure water will draw water into the cells to dilute the salt in them. The cells will burst.



17.    Pure iodine (100.0 g) is dissolved in 300.0 g of CCl4 at 65oC.  Given that the vapor pressure of CCl4 at this temperature is 504 mm Hg, what is the vapor pressure of the CCl4/I2 solution at 65oC?  (Assume that I2 does not contribute to the vapor pressure.)