Vapor Pressure and Phase Changes:

When a substance melts, part of the intermolecular forces are overcome.  Since liquids are usually just slightly less dense than solids, the majority of intermolecular forces are still "in place."  When the liquid evaporates, nearly all the intermolecular forces are overcome.  The energy needed to do this is usually much greater than the energy require for melting.  Hence, in nearly all cases, DH_vap > DH_fusion.

The initial and final states in both cases are solid and vapor, respectively.  Since H is a state function, the first law states that their difference must be independent of path.  Hence,

We make a couple of extra comments here.

• If you are curious, you might wonder why we did not write




The reason for this is that we have to be at 25°C to have the "°" superscript.
Also, you will note that the problem states "at the same temperature." DH 's actually change a little with temperature.  Since melting and boiling usually occur at different temperatures, these numbers would have to be adjusted if the separate processes were to occur and the normal melting and boiling points respectively.  However, they can be defined at the same temperature and, in this case, are rigorously additive.

Problem 10.48:  How much energy (in kJ) is needed to heat 5.00 g of ice from -10.0°C  to 30.0°C?  The heat of fusion of water is 6.01 kJ/mol and the molar heat capacity of ice is 36.6 J/mol·K and that for liquid water is 75.3 J/mol·K.

This is the sum of three processes:

• 1:  The heating of ice from -10°C to 0°C.
• 2:  Melting the ice at 0°C.
• 3:  Heating the resulting water from 0°C to 30°C.

The easiest way to do this is to add the three process together for a mole of water and then multiply by the number of moles.  We do this now:

This much like the previous problem except that, here, we have cooling instead of heating.  Note that, in the calculation, q will be negative since we are referencing to the system.  The actual amount of heat evolved will be positive, however, since, in that case, we are referring to things as seen by the surroundings.  We do the problem by first adding the appropriate molar quantities and then multiplying by the number of moles.  We dispense with units to make this easier to see on your screens.

(Well, it did fit just a little better.  It is suggested that you go back and put in the units and make sure that you understand the signs.  Note that temperture differences in Celcius and Kelvin are equivalent.)

We can then state that 36.5₇ kJ of heat are evolved.  Now, we are looking at things in terms of the surroundings!

Before we begin, please notice that the enthalpy of fusion does not enter into the calculation!  Sometimes, on exams, professors have been known to put in extra numbers to see if the students are truly on their toes!

We have three processes to consider here:

• Cooling the vapor to the boiling point.
• Condensing the liquid at the bp.
• Cooling the resulting liquid to the final temperature.

The MW of ethanol is 46.0684₄ g/mol.  Since this problem is just like 10.49, we shall "condense" our calculation even further.  It is left as an exercise for you to put in the missing steps.

From the standpoint of the surroundings, we say that 26.8₃ kJ of heat are evolved.

First, let us look at Figure 10.10 to see what we have to do.

Before you panic, I don't really think that we need to find a picture for sodium.

To draw a diagram we need to plot T vs. the heat added.  The easiest way to do this is per mole of sodium.  We calculate the heat needed for each step and keep a running tally of the total heat added and the temperatures at particular points.  When the temperatures are changing, we have straight lines of various slopes.  When undergoing phase changes, we have horizontal lines at the transition temperatures.  Since we are dealing with exactly one mole, the relevant equations are as follows:

We have "fancied up" these a little.  Each represents two parts of the curve:  the first equation is for the melting and vaporization processes and the second is for the two heating processes where the temperature changes.

If you look at the water diagram above, you see that 6 points are calculated.  For Na we need only 5 since the heating of the solid to the melting point is left out.  We now give our calculations at each point using one of the two formulas given above for each step.  The starting point is point 1 which is at 97.8°C.  We start there and explain each step in turn.  The heats reported at the total heats up to a given point.
 
 

Point	Heat for Step (kJ/mol)	T (°C)	Total Heat (kJ/mol)
1	0	97.8	0
2	2.64	97.8	2.64
3	22.14	883	24.78
4	89.6	883	114.38
5	6.12	1100	120.50

These numbers come from simple additions for the the melting and vaporization steps and multiplying the heat capacity (molar) times the temperature difference.  Note that, in the latter case, we had to divide by 1000 since the heat capacities had Joules as the energy unit whereas were are plotting kJ/mol on the abscissa.

The graph is now fairly easy to draw.  Here, for instance, is the one from the instructor's manual:

It is probably wise to do this by hand also.  To do this, do the following steps:

• Select some good graph paper.  My personal preference is to use an engineering scratch pad.
• Look at your numbers and draw convenient and easy to read scales.
• Put down the points (1-5) in succession and number them.
• Draw the lines between the points.

The following effort is quite crude but is easy to do.  I won't do things this way on all problems, but this is as good a way as any to get accustomed to drawing a graph.  Anyway, here is the (beautiful?) result:

This was clipped and rescaled to fit the screen gracefully.  This could be done better, probably, but this is good enough to get the general idea!

This is a relatively easy plug-in problem.  Here are the equations and the solution.  Note that we first make use of the fact that DG = 0 at equilibrium.

This is much like the previous problem.  We skip the "preliminaries" and just tive the result.

Here is the table again, with the values filled in.  This is all just plug-in and should not be too mysterious.
 

We take a few liberties with the sig. figs. for the logarithms since we have not yet discussed how to handle them.  The main thing is that the final answers for plots and applications will have three sig. figs. because of the temperatures.  Note that I multiplied the latter by 10³ to make them easy to plot.  (For instance, in the first row the actual value is 3.80₂ x 10^-3.)  This is a common practice when compiling long lists of numbers.

Rather than subject you to my poor art work, here are the graphs as draw in the instructor's manual.

On the left side, the points are connected by straight lines.  This is a smooth enough variation to make the plot look almost like a simple curve instead of a bunch of straight line segments.  The plot to the right is a pure straight line.  If you wish, print these out and put in the points by hand.

Problem 10.61:  The following table gives the vapor pressure of mercury at various temperatures.  Fill in the rest of the table and use the data to plot curves of P_vap vs. T and ln P_vap vs. 1/T.
 

This is pretty much like the previous problem.  We shall fill in the blanks and present the curve and line without comment.
 

The numbers have been given and are good to 3 sig. figs. overall.  Here is the graph.

(Looks like after you've seen one, you've seen them all.)

Problem 10.62:  Use the plot you made in Problem 10.60 to find a value (in kJ/mol) for DH_vap for dichloromethane.

For any two points on the straight lines, you have the following equation:

You can use the two end-points to determined the heat of vaporization.  Probably more relevant here is to use the slope of the line as plotted.  In this case, you would use

It is left as an exercise for you to find the heat of vaporization by either method.  The slope of the line is -3628K and the result from this is DH_vap = 30.1₇ kJ/mol.  In this case, I am giving you the answer but making you do the work.

This goes pretty much as did the previous problem.  To save time and space--and to make you do the work--the slope is -7219K and DH_vap = 60.0₂ kJ/mol.  Have pHun doing this!

Problem 10.64:  Choose any two temperatures and corresponging vapor pressures in the table given in Problem 10.60 and use those values to calculate DH_vap for dichloromethane (in kJ/mol).  How does the value you calculated compare to the value you read from you plot in Problem 10.62?

Here, I shall show a little mercy and actually do the calculation for you.  The two-point formula given with Problem 10.62 can be rearranged to give

Some people prefer the second formula (at least I do) and we shall use it to give the final answer as follows:

Here, we use the first and last points since these are usually the best choices.

As before, we use the first and last points.  Given those, the answer is

On both of these problems, you might wish to try other combinations of points to see what happens.  The numbers won't necessarily be the same but should be close if the data are good.