CHAPTER 10Liquids and Solids

LIQUIDS and IMFs(Intermolecular Forces)

In Chapters 8 and 9 we studied how atoms form stable units called molecules by sharing electrons. This is call intramolecular forces.

Now we will look to see how intermolecular forces form attractions between condensed states of matter (liquids and solids).

Key components to IMFs in liquids are: Dipole-dipole attractions Hydrogen bonding (which isn’t really a chemical

bond) London dispersion forces

van der Waals forces

Forces between nonpolar and noble gasesLondon dispersion forces are the weakest.

Dipole-dipole interactionsoccur between polar molecules

Hydrogen bonding occurs between hydrogensbound to highly electronegative atoms likenitrogen, oxygen and fluorine- (FON). These bonds are very strong.

Properties of liquids (11:03 min)

Surface tension: The resistance of a liquid to an increase in its surface area.

Adhesive forces are forces between liquid molecules and their containers.

Cohesive forces are the intermolecular forces among the molecules of the liquid.

Capillary action is the spontaneous rising of a liquid in a narrow tube.

Viscosity is the resistance to flow. (molasses is more viscous than water)

Structure and Types of Solids (9:05)

Crystalline solids have highly regular arrangements of their components represented by a lattice. The smallest repeating unit of a lattice is called a unit cell.

Amorphous solids have considerable disorder to their structures.

Network Structures of Carbon (8:20)

Network Structures of Silicon (9:26)

X-Ray Analysis of Solids An X-ray which reflects from the

surface of a substance has travelled less distance than an X-ray which reflects from a plane of atoms inside the crystal. The penetrating X-ray travels down to the internal layer, reflects, and travels back over the same distance before being back at the surface. The distance travelled depends on the separation of the layers and the angle at which the X-ray entered the material. For this wave to be in phase with the wave which reflected from the surface it needs to have travelled a whole number of wavelengths while inside the material. Bragg expressed this in an equation now known as Bragg's Law.

n = an integer (1,2,3 etc.)

θ is the angle of incident rays and surface of the crystal

d is the spacing between the layers of atoms.

λ is the wavelength

Let’s try it shall we?

X-rays from a copper X-ray tube (λ = 154 pm) were diffracted at an angle of 14.22 degrees by a crystal of silicon. Assuming first-order diffraction (n=1) what is the interplanar spacing in the silicon?

nλ = 2d sin θ d= nλ ÷ 2 sin θ d = 313 pm = 3.13 x 10-

10m

Again…

X-rays of wavelength 1.54 Å were used to analyze an aluminum crystal. A reflection was produced at θ = 19.3 degrees. Assuming n = 1, calculate the distance between the planes of atoms producing this reflection.

233 pm

Specific classifications of solids according to what component

occupies the lattice points. Atomic solids: atoms at the lattice points-form from

London dispersion forces between noble gases but only at very low temperatures—ie: solid Argon.

Molecular solid: Have lower melting and boiling points than other solids. Ex: ice

Ionic solid: attraction of ions to one another to form the crystal lattice. Ex: sodium chloride

Metallic solid: Held together by valence electrons. Are malleable, ductile, shiny and good conductors of heat and electricity. Ex: silver

Network solids: huge crystals held together by valence electrons. Ex: ruby, diamond, amethyst

Amorphous solids: do not have a crystalline structure. Ex: glass, plastic

Structure and Bonding in Metals

A metallic crystal can be seen as containing spherical atoms packed together and bonded to each other equally in all directions. This is called closest packing. There are two types, both having spheres with 12 neighbors. These arrangements are why metals have high thermal/electrical conductivity, are malleable and ductile.

Unit Cells Knowing the net number of spheres (atoms) in a

unit cell is important to applications involving solids. The number is defined by the centers of the spheres on each cube’s corners. 8 cubes share a given sphere so 1/8 of this sphere lies inside each unit cell. 8 x 1/8 = 1. The cubes at the center of each face are shared by 2 unit cells so ½ of each lies inside a particular unit cell. There are 6 sides so 6 x ½ or enough to make 3 whole spheres. The net number of spheres in a face-centered cube is (8 x 1/8) + (6 x ½ ) = 4

How to calculate the density of a unit cell. You know you want to know how…

Density = mass ÷ volume For a face-centered cube, calculate the density of a unit cell of

silver. Since the atoms touch along the diagonal we’ll use that length. r will be the radius of the spheres. r + 2r + r = 4r Use the Pythagorean theorem: d2 + d2 = (4r)2 where d is the depth of the sides. 2d2 = 16r2 , d2 = 8r2

d = 8r2 = r8 since r = 144 pm for a Ag atom, d = (144 pm)(8 ) = 407 pm The volume of the unit cell is d3 = (407pm)3 = 6.74 x 107 pm3 6.74 x 107 pm3 x (1.00 x 10-10 cm/pm)3 = 6.74 x 10-23 cm3

D = (4 atoms)(107.9 g/mol)(1 mol/6.022 x 1023 atoms) 6.74 x 10-23 cm3 = 10.6 g/cm3

Okay, try it yourself

Calcium has a cubic closest packed structure as a solid. Assuming that calcium has an atomic radius of 197 pm, calculate the density of solid calcium in cm3.

r = 1.97 x 10-8cm d = (1.97 x 10-8 cm)(√8) = 5.57 x10-8cm V=d3=(5.57 x 10-8)3 = 1.73 x 10-22 cm3

D = (4 atoms)(40.08 g/mol)(1mol/6.022 x 1023atoms

1.73 x 10-22 cm3

1.54 g/cm3

Vapor Pressure and Changes of State

Vaporization is an endothermic process.

The heat of vaporization or enthalpy of vaporization is the amount of heat it takes to vaporize 1 mole of a liquid at 1 atm. ΔHvap

Consider the graph to the right. Which of the following would be true?

This liquid has weaker IMF’s than water.

The liquid’s normal boiling point is around 85oC.

The liquid boils at room temperature when the pressure is dropped to about 200 mm Hg.

Vaporization/Condensation

Describe what is happening to the liquid in the sealed Erhlenmeyer flask over time.

Did you know that a significant portion of the sun’s energy that reaches earth is spent evaporating water from the oceans, lakes, and rivers rather than warming the earth?

Equilibrium vapor pressure(vapor pressure)

Patm = Pvapor + PHg

Liquids with a high vapor pressure are volatile, meaning they vaporize readily.

The vapor pressure of a liquid depends on its IMFs. Those with high IMFs have low vapor pressures.

Of course with increased temperatures comes increased vapor pressures.

Notice the nonlinear increase in v.p. for all the liquids as temperature increases. How can we get a straight line?

You didn’t really think we were done with math stuff…

ln(Pvap) = - ΔHvap ÷ R (1/T) + C

T = Kelvin temperature

ΔHvap = enthalpy of vaporization

R= universal gas constant

R = 8.3145 J/K· mol ln(Pvap) = natural log

of vapor pressure

y = ln(Pvap) x = 1/T m = slope = - ΔHvap ÷

R b = intercept = C

Using the plots in the figure to the right, determine whether water, diethyl ether or ethanol has the larger enthalpy of vaporization.

Plot ln(Pvap) on y and 1/T on the x.

If you plot this, then ether ends up with a slope of -3230, ethanol -4450 and water -4504.

All three have a negative slope with ether having the smaller slope. Thus ether has a smaller ΔHvap.

Does that make sense? Consider the hydrogen bonding in water as you justify your answer.

ln(Pvap) = - ΔHvap ÷ R (1/T) + C If you know the values of ΔHvap and Pvap

at one temperature you can use the above equation to calculate Pvap at another temperature. Since C is a constant and is not dependent on temperature, you can write the equality:

ln(Pvap, T1) + ΔHvap1 ÷ RT1 = C = ln(Pvap, T2) + ΔHvap2 ÷ RT2

Rearrange:

ln(Pvap, T1) - ln(Pvap, T2) = ΔHvap ÷ R [ 1/T2 – 1/T1]

OR

ln [Pvap, T1 ÷ Pvap, T2] = ΔHvap ÷ R [ 1/T2 – 1/T1]

PRACTICE

The vapor pressure of water at 25oC is 23.8 torr and the heat of vaporization of water at 25o C is 43.9 kJ/mol. Calculate the vapor pressure of water at 50oC.

ln[23.8torr/PvapT2] = 43,900J/8.3145 J/mol K ( 1/298K – 1/323K)

ln (23.8 torr/ PvapT2) = -1.37

Take antilog of both sides:

ln (23.8 torr/ PvapT2) = 0.254

PvapT2 = 23.8/0.254 = 93.7 torr

More practice

In Breckenridge, Colorado the typical atmospheric pressure is 520. torr. What is the boiling point of water in Breckenridge? ΔHvap = 40.7 kJ/mol.

ln [520 torr/760 torr] = 40700J/8.3145 J/mol K(1/373K – 1/T2)

T2 = 362 K or 89.3oC

Carbon tetrachloride has a vapor pressure of 213 torr at 40.oC and 836 torr at 80oC. What is the normal boiling point of CCl4

Solve for ΔHvap first. ΔHvap = 3.1 x 104 J/mol Using ΔHvap , 1.00 atm and 760. torr,

solve for T. Answer: 350.K or 77oC

Phase changes for water

A-B specific heat of ice: 2.09 J/g oC

B-C heat of fusion: 334 J/g C-D specific heat of water:

4.184 J/g oC D-E heat of vaporization:

2260 J/g E-F specific heat of water

vapor: 1.01 J/g oC Calculate the amount of

energy it would take to change the temperature of 50.0 grams of water from

-20oC to 120oC

Super “Cool” Water

The melting and boiling points of a substance are determined by the vapor pressures of the solid and liquid phases.

Notice how when temperature is increased, the vapor pressure of the ice increases more rapidly than the liquid.

When both the vapor pressure of the solid and liquid match, that is the melting point.

Super “Hot” Water

When a liquid is raised to a temperature above its boiling point and doesn’t actually boil, it is super heated. A superheated liquid has a vapor pressure in the liquid greater than the atmospheric pressure. When a bubble does form, since the internal pressure is greater than the external pressure, it can burst before rising to the surface, blowing the surrounding REALLY HOT liquid out of the container.

Case 1-A temperature at which the vapor pressure of the solid is greater than that of

the liquid

At this temperature, the solid requires a higher pressure than the liquid does to be in equilibrium with the vapor. As the vapor is released from the solid the liquid will absorb it in an attempt to reduce the vapor pressure to its equilibrium value.

The solid will decrease, the volume of liquid will increase and finally there will only be the liquid which will come to equilibrium with the vapor.

The temperature must be above the melting point of ice, since only the liquid state can exist.

Case-2: A temperature at which the vapor pressure of the solid is less than

that of the liquid. This is the opposite of the last scenario.

To reach equilibrium, the liquid is going to have to obtain a higher pressure than the solid in order to reach equilibrium with the vapor. The liquid will gradually disappear and the amount of ice will increase. Finally, only the solid will remain, which will achieve equilibrium with the vapor. The temperature must be below the melting point of ice since only the solid state can exist.

Case-3: A temperature at which the vapor pressures of the solid

and liquid are identical.

In this case, the solid and liquid states have the same vapor pressure, so they can coexist in the apparatus at equilibrium simultaneously with the vapor. The temperature would be at the freezing point where both the solid and liquid states can exist.

Definitions

The melting point is the temperature at which the solid and liquid states have the same vapor pressure under conditions where the total pressure is 1 atm.

The normal boiling point of a liquid is the temperature at which the vapor pressure of the liquid is exactly 1 atm.

Note: At temperatures where the vapor pressure of the liquid is less than 1 atm, no bubbles of vapor can form because the pressure on the surface of the liquid is greater than the pressure in any spaces in the liquid where the bubbles are trying to form.

Phase Diagrams In a closed system a phase diagram is a convenient

way of representing the phases of a substance with respect to pressure and temperature.

Line AB is the liquid-vapor line, showing the vapor pressure of the liquid. It represents the equilibrium between the liquid and gas phase.  

Line AC is the solid-vapor line, representing the variation in the vapor pressure of the solid as it sublimes at different temperatures.  

Line AD is the solid-liquid line, representing the change in melting point of the solid with increasing pressure. This line usually slopes slightly to the right as pressure increases, because the solid phase of a substance is usually more dense than the liquid phase.  

The critical temperature is the point at which any temperature above that is where the vapor cannot be liquefied no matter how much pressure is applied.

The critical pressure is the pressure required to liquefy the vapor at the critical temperature.

a)  Approximately what is the normal boiling point and what is the normal melting point of the substance?

300K What is the physical state

of the substance under the conditions in the table:

Solid Vapor Gas as a supercritical fluid

because you’re above the critical pt. which is 260 atm. and 450 K.

Temperature Pressure

150 K 0.5 atm

325 K 0.9 atm

450 K 165 atm