Thoughts and Links for Selected ACS Text Consider This, Investigate This Activities

Many of the activities have answers provided immediately following in the text.  So, if you donÕt find an answer here, keep reading in your text.

 

1.21(a):  How does molecular geometry affect molecular dipoles?  (Carbon Dioxide)   Below is a Lewis structure for carbon dioxide. 

 

Oxygen is more electronegative than carbon, so the C-O bond is polar.  This is shown by indicating partial charges

 

or with bond dipoles drawn with the points to oxygen

     

 

The C-O pairs have identical arrangements of bonding and non-bonding electron pairs.  So, even though each C-O bond is polar, the individual bond dipole vectors cancel and the molecule as a whole is nonpolar (has no net dipole moment).

 

 

 

1.23 Do boiling points correlate with the number of molecular electrons?

 

(a)          As more CH2 groups are added, more electrons are being added.  Induced-dipole attractions (London forces, dispersion forces) between one molecule and another increase as the number of molecular electrons increases.  The stronger these attractions are, the more energy it will take to break them.  To boil requires converting from liquid to the gas phase – molecules in the liquid phase must be pulled away from one another (must gain enough energy to overcome the attractive forces between them) to enter the gas phase.  Bigger molecules mean more attractive forces between molecules which means that more energy will be required to separate them.  So, at least for similar types of molecules, boiling point should go up with number of molecular electrons.

(b)          The reasoning described in (a) also applies to the pattern of boiling points described in Figure 1.21. The induced-dipole attractions increase going from CH4 to SnH4 (because the number of molecular electrons increases), and the boiling points increase in the same pattern.  Molecules with strong intermolecular attractions require more energy to overcome their intermolecular attractions than molecules with weak attractions for one another.

 

1.25:  How does polarity affect boiling point?  

 

(a)    The question asks us to compare polarities of a set of hydrides within one group (e.g., the group VI hydrides H2O, H2S, H2Se, H2Te).  Group IV hydrides (CH4, SiH4, GeH4, SnH4) are all nonpolar because of their great symmetry.  All the other hydrides are polar because they have at least one lone pair of electrons, and the bonding and lone pairs are not distributed symmetrically within the molecule.  Relative bond polarities are a function of electronegativity differences in the bonding atoms.  EN increases going across the periodic table left to right and from bottom to top.  The relative molecular polarities are determined by how those bonds and any lone pairs are oriented in the molecule.  (So, build models to compare.)  For the group VI hydrides, the polarities decrease in the order H2O > H2S  >  H2Se > H2Te.  (These are all bent molecules with two lone pairs at one side of the central atom.  H-O bonds are very polar, H2Te the least polar.  Another difference is that the size of the molecule increases going from H2O to H2Te, so the partial charges are distributed over an ever increasing volume.  The lower charge density helps to make H2Te is less polar than H2O.)  You would expect boiling points to increase as polarity increases (dipole-dipole interactions become stronger), so youÕd expect the boiling points to increase in the order  H2Te < H2Se  < H2S  < H2O.  ThatÕs not exactly what Figure 1.24 shows.

(b)    Dispersion forces increase going left to right in Figure 1.24 because the number of molecular electrons increases in this direction (e.g., H2Te has more molecular electrons than H2S, so H2Te has stronger dispersion forces).  Boiling points should increase in a series of hydrides going left to right in Figure 1.24 if dispersion forces dominate.

(c)     From Figure 1.24, itÕs apparent that most of the hydrides show increasing boiling point as the number of molecular electrons increases (when dispersion forces dominate).  The most interesting feature of Figure 1.24 is that the boiling points of certain row 2 hydrides (NH3, H2O, HF) are very out of line with the boiling points of their respective groups.  These are very polar molecules (HF > H2O > NH3), so apparently that facet of their behavior influences their boiling points.   The boiling points show the trend H2O >HF > NH3, which is not exactly the same as their relative polarities.  So, you might conclude that both polarity and number of molecular electrons are important (dipole-dipole and dispersion forces are both important).  It turns out that the explanation lies in whether hydrogen bonds (a special case of dipole-dipole interactions) are present.  Keep reading!

 

1.26:  Does hydrogen bonding explain the boiling points of NH3 and HF?  

 

(b)  Every hydrogen bond requires 1 lone pair on one molecule and a H bonded to N, O, or F on another.  If we have N ammonia molecules, we have 3N H atoms and 1N lone pairs.  So, even though an individual ammonia molecule can make four hydrogen bonds to other ammonia molecules, in a collection of ammonia N molecules at most N hydrogen bonds can form.  For N HF molecules, there are 3N lone pairs and 1N H atoms, so at most N hydrogen bonds can form (even though an individual HF molecule can hydrogen bond to four other HF molecules).

 

(c)  HF and NH3 make fewer hydrogen bonds than H2O does, so H2O is expected to have a higher boiling point.

 

1.28:  What kinds of hydrogen-bonded networks can you construct?   This problem is part of the Expt 1 problem set.  Be sure to do it in lab and check your answer.

 

1.30:  What are the temperatures in an ice-water mixture?   See PPT1.30 to view the experimental set up and results.  Use Safari or Firefox if you can.  There are often glitches in how things are displayed in Explorer.

 

1. 31:  How do you interpret the temperatures in an ice-water mixture?  (At what temperature do you think liquid water is most dense?)  As long as ice is present, the top where the ice is stays at ~0 degrees C.   As heat (energy) enters the ice-water container (graduated cylinder), some of the ice melts to form water.  Because liquid water is more dense than ice, the water formed from ice melting sinks.  As it sinks and gets further away from the ice, it warms (you see that the temp probe near the bottom registers an increase). As the water in the bottom of the graduated cylinder absorbs more heat and warms up, it becomes less dense and rises.  Then it hits the cooler water sinking from the top and the heat is exchanged, and the water sinks again (i.e., convection currents are set up in the water).  Eventually, the temp of the water in the bottom ~levels off at ~5.5 degree C.  This must be the temp at which water has its maximum density. (In our expt, the temp probe registered ~5.5 degrees when the temp leveled off, but the actual temp when water reaches a maximum density is ~4 degrees.  Perhaps there was a calibration error with the probe.)  

 

1.32:  What are the connections between water density and temperature?  (See 1.31)  Ice is less dense than water, so it floats atop of liquid water.  Liquid water itself sinks or rises depending on its density which varies with temperature.  The most dense liquid water will be at the bottom of a container (graduated cylinder, lake, whatever) and its temperature will be at ~4 degrees C (we measured 5.5 degrees in our expt). 

 

1.33  How do the DNA bases fit together? This problem is part of the Expt 1 problem set.  Be sure to do it in lab and check your answer. 

 

1.34  How does hydrogen bonding affect the thermal stability of DNA?  Hydrogen bonds hold the DNA helix together.  It takes energy to disrupt the H-bonds – added heat will provide enough energy to overcome the H-bond attraction between the DNA bases.  The more H-bonds, the greater the thermal stability will be (the more heat required to disrupt the H-bonds).  From Investigate This 1.33, we know that 3 H-bonds form per C-G pair and 2 H-bonds form per A-T pair.  So if the C-G to A-T ratio is higher, thermal stability should be higher.

 

1.35-1.39  Sweating and energy diagrams.  Your finger feels cooler when wet than when dry.  This is because the water on your skin absorbs energy from your skin as it evaporates (goes from l to g phase).  The energy change for the water as it is vaporized is + (endothermic, heat is absorbed).  The energy change for the skin is negative (exothermic, heat leaves).  For 1.39(a), your energy level diagram should show an arrow starting at a high energy and ending at a lower energy (pointing down).  This energy change is negative because the skin loses energy (cools down).  The length of the energy change arrow should be the same length as the one in 1.35.  The energy required to vaporize a certain amount of water is equal in magnitude but opposite in sign to the amount of energy that must be lost by the skin to accomplish the vaporization.

 

1.40-1.41 Do all liquids evaporate at the same rate?  This will be discussed in lecture in Week 3.

 

1.50 Which compounds in Table 1.2 form hydrogen bonds?  Water, methanol, and ethanol can form hydrogen bonds.  An individual water molecule can form 4 H-bonds (see Figure 1.27), though not that many will form in a collection of water molecules (see Consider This 1.26).  An individual methanol or ethanol molecule can form 3 H-bonds  to other like molecules (one at the H bonded to O, two through the lone pairs on O).  Dimethyl ether cannot H-bond to other dimethyl ether molecules.  Even though it contains both H and O, all the HÕs are involved in ~non-polar bonds to C.  So the HÕs arenÕt +enough to be attractive to OÕs on another molecule.  (For H-bonding to occur, the H must be bound directly to N, O, or F).  Hexane contains only C and H, so no H-bonding is possible. 

 

1.53-1.54  What is observed when liquids are heated at the same rate??  This will be discussed in lecture in Week 3.

 

1.60  How fast do different liquids flow from a pipet?  Equal volumes of water and hexane are placed into pipets and the time required to completely drain is recorded.  The hexane always drains out a few seconds faster (its viscosity – resistance to flow – is lower).  This makes sense because viscosity should decrease as the strength of IMFs decrease and hexane has relatively weak IMFs (only London forces).