CH 15HW: CH 15: 1,5,13-17SUGGESTED: 4, 9, 11

Liquids, Solids, and Van der waals (Intermolecular) Forces

States of Matter Differ By Intermolecular Distance

• The state of a substance at a given temperature and pressure is determined by two factors:• Thermal energy of the molecules• Intermolecular forces (called Van der walls forces) between molecules

States of Matter• Gases:

– thermal energy is greater than the energy of attraction between the gas molecules, so molecules have enough energy to separate

– have completely free motion (translational, rotational, and vibrational)

• Liquids:– the thermal energy is somewhat less than the intermolecular

attractive forces, so the molecules are slightly separated– the thermal energy available allows “tumbling” of molecules, which is

why liquids can be poured– restricted translational, rotational, and vibrational movement

• Solids:– the thermal energy is much less than the energy of attraction. – the molecules are completely fixed in space– vibrational motion only

• Since thermal energy is required to overcome intermolecular forces, we can observe how the phase and temperature of a substance changes as heat is added (constant pressure).

• A heating curve for water is shown below, going from -10o C to 125o C

Heat rate = 100Heat of fusion

Heat of vaporization

Energy is Required to Change Phase

• The fusion (melting) of water can be represented by:

• Therefore, the energy (heat) required to melt n moles of water would be:

• The vaporization of water can be represented by:

• The energy (heat) required to vaporize n moles of water would be:

𝐻2𝑂 (𝑠 )→𝐻2𝑂 (𝐿 )∆𝐻 𝑓𝑢𝑠=6.01𝑘𝐽𝑚𝑜𝑙

𝑞 𝑓𝑢𝑠=𝑛∆𝐻 𝑓𝑢𝑠

𝐻2𝑂 (𝐿 )→𝐻2𝑂 (𝑔)∆𝐻𝑣𝑎𝑝=40.65𝑘𝐽𝑚𝑜𝑙

𝑞𝑣𝑎𝑝=𝑛∆𝐻𝑣𝑎𝑝

Step 1: Raise to melting temp.

Example

• Calculate the heat required to heat 28 g of H2O(s) at -10oC to H2O(L) at 50oC, given that the heat capacities of ice and liquid water are 37.7 and 75.3 J/mol K, respectively?

𝒒𝒑=𝒏𝑪𝒑 , 𝒊𝒄𝒆∆𝑻

-10oC 0oC

Step 2: Fusion

0oC

𝒒𝒑=𝒏∆𝑯 𝒇

Step 3:

Raise to 50o C

50oC

𝒒 𝒑=𝒏

∆𝑪 𝒑 , 𝒍𝒊𝒒𝒖𝒊𝒅∆𝑻

total heat=𝑞1+𝑞2+𝑞3

Example

• Calculate the heat required to heat 28 g of H2O(s) at -10oC to H2O(L) at 50oC, given that the heat capacities of ice and liquid water are 37.7 and 75.3, respectively?

𝑞𝑡𝑜𝑡=(1.556mol )(( 37.7 Jmol K ) (10 K )+(6010 Jmol )+(75.3 J

mol K )  (50  K )  )

total heat=𝑞1+𝑞2+𝑞3=𝑛𝐶𝑝 ,𝑖𝑐𝑒 ∆𝑇+𝑛∆𝐻 𝑓+𝑛𝐶𝑝 , 𝑙𝑖𝑞𝑢𝑖𝑑∆𝑇

heating ice melting ice heating water

𝑞𝑡𝑜𝑡=15796 𝐽 𝑜𝑟 15.8𝑘𝐽

Sublimation

• Certain substances, like “dry ice” (CO2), convert straight from solid to gas without passing through a liquid phase. This is called sublimation.

𝐶𝑂2 (𝑠 )→𝐶𝑂2 (𝑔)∆𝐻 𝑠𝑢𝑏=25.2𝑘𝐽𝑚𝑜𝑙

Intermolecular Forces: Coulombic Attractions

• As you recall, ionic compounds are solids at room temperature. There are ion-ion attractions in ionic compounds.

• The coulombic force that holds ions together is very strong. Coulombic attractions are the strongest of all intermolecular forces.

• Therefore, all ionic compounds have very high melting/boiling points.

Na+ Cl-𝐸𝑐

Intermolecular Forces: Dipole-Dipole Forces

• The values of ΔHvap and ΔHsub reflect how strongly the molecules attract one another in the liquid and solid phases. The more strongly the molecules attract, the greater the values of ΔH.

• Recall polarity from chapter 8. Any molecule with a net dipole is polar.

H Clδ-δ+

Partial positive character

Partial negative character

Dipole-Dipole Forces

• Polar molecules attract one another. This type of intermolecular force is called dipole-dipole attraction.

δ+ δ- δ+ δ-

Covalent bond: Very Strong

Dipole-dipole interaction: Weaker than intra-molecular forces

• Polar molecules will orient themselves in a way to maximize these attractions. The strength of these attractions increases with increasing polarity. Polar molecules have higher melting points than non polar ones.

London Dispersion Forces

• With nonpolar molecules, there are no dipoles, so we would not expect to see dipole-dipole interactions. Despite this, intermolecular interactions have still been observed.

• For example, nonpolar gases like Helium can be liquified, but how can this happen? What force brings the He atoms together?

• Fritz London, a physicist, proposed that the motion of electrons in a nonpolar molecule can create instantaneous dipoles

• Lets take a Helium atom. At some moment in time, the electrons are spread out within the atom

• However, because electrons are constantly moving, electrons can end up on the same side of the atom, creating a charge gradient (instantaneous dipole). This temporary dipole can induce a temporary dipole on another atom, yielding a weak dipole-dipole interaction called a London dispersion force.

e-

e-2+

e-

e-2+

δ+δ-e-

e-2+

e-

e-2+

e-

e-2+

δ+δ- δ+δ-

London Dispersions

• Because London dispersion forces depend on electron motion, the strength of these forces increases with the number of electrons.

• The ease of the electron distortion is called polarizability. The more polarizable an atom/molecule, the more likely it is to induce instantaneous dipoles.

• Hence, London dispersion forces increase with increasing molar mass because heavier atoms/molecules are more polarizable. All substances have dispersion forces.

• In general, for covalently bonded molecules, boiling/melting point increases with molar mass.

C15H32 C18H38C5H12

Boiling Points Increase With Increasing Strength of London Dispersion Forces

Hydrogen Bonding

• A special, and very strong type of dipole-dipole interaction is hydrogen bonding.

• Because hydrogen atoms are so small, the partial positive charge on H is highly concentrated. Therefore, it strongly attracts very electronegative elements.

• Hydrogen bonds exist only between the H atom in an H—F, H—O, or H—N bond and an adjacent lone electron pair on another F, O, or N atom in another molecule

Structure and Density of Ice

• Water is one of the few compounds that is less dense in its solid phase than its liquid phase.

• This is due to hydrogen bonding.

• In liquid water, 80% of the atoms are H-bonded. In ice, 100% are H-bonded.

• Complete H-bonding creates gaps in the crystal structure. This causes the water to expand.

• Therefore, we have the same mass of water, with a larger volume. Since ρ=(mass/volume), ρ decreases.

Hydrogen Bonding Causes Abnormalities in Boiling Point Trend