Chapter 11

Question 1

What is a dispersion force?

Answer 1

An intermolecular force exhibited by all atoms and molecules that results from fluctuations in the electron distribution

Question 2

How do London dispersion forces work?

Answer 2

An electron will displace another in a different atom creating a temporary slight charge (denoted with the symbol for delta) called an instantaneous dipole / temporary dipole creating attraction. A greater # of electrons generally means a greater attraction.

Question 3

What is dipole-dipole attraction?

Answer 3

When two polar molecules are attracted to each other’s charged areas. Why polar water can’t interact with non-polar oils. More permanent and stronger than dispersion forces.

Question 4

What is ion-induced dipole force?

Answer 4

An ion inducing a dipole in a nonpolar atom.

Question 5

Miscibility:

Answer 5

The ability to mix two liquids together. Polar liquids can mix with other polar liquids, and non-polar can mix with non-polar. Extremely difficult/impossible to mix polar and non-polar.

Question 6

What is hydrogen bonding?

Answer 6

A stronger form of dipole-dipole attraction due to the difference in electronegativity.

Question 7

What are the species involved in hydrogen bonding?

Answer 7

Hydrogen: oxygen, nitrogen, or fluorine.

Question 8

What is dipole-induced dipole force?

Answer 8

Intermolecular force between a permanent dipole and a nonpolar molecule. The dipole distorts the nonpolar electrons to create partial charges and induce small dipole

Question 9

What is ion-dipole force?

Answer 9

Intermolecular force between a permanent dipole and an ion. The charges of the ions attract the partially charged polar atoms. Responsible for the ability of ionic substances to form solutions in water.
These are the strongest type of intermolecular force described in CH111.

Question 10

Why is water fucking with chemists:

Answer 10

Low electron count but high boiling point, much higher than some molecules with far more electrons. It also expands when freezing as opposed to contracting which is why most creatures get ganked when below freezing.

Question 11

Surface Tension

Answer 11

The energy required to increase the surface area of a liquid by a unit amount; responsible for the tendency of liquids to minimize their surface area, giving rise to membrane-like surface. Decreases as intermolecular forces decrease

Question 12

Viscosity

Answer 12

The resistance of a liquid to flow. Greater in substances with high intermolecular forces.
Also depends on the shape, increasing in larger molecules, and temperature. The unit for viscosity is centipoise, with room temp water at 1 centipoise.

Question 13

Capillary Action

Answer 13

The ability of a liquid to flow against gravity up a narrow tube due to adhesive and cohesive forces. It comes about due to the attraction between molecules of liquid called cohesive force, and the attraction between the molecules and the surface of the tube called adhesive forces. if af > cf the force draws the liquid up.

Question 14

Heat of Vaporization

Answer 14

Amount of heat needed to vaporize one mole of liquid denoted using Delta(vap)H. For water it is 40.7 kJ/mol. The heat of vaporization is always positive because the process is endothermic—energy must be absorbed to vaporize a substance. The reverse is true for condensation with the same heat just in reverse.

Question 15

Dynamic Equilibrium

Answer 15

The point where the reverse reaction or process equals the rate of the forward reaction or process. Happens in a sealed container.
When a system in dynamic equilibrium is disturbed, the system responds so as to minimize the disturbance and return to a state of equilibrium.

Question 16

Vapour Pressure

Answer 16

The partial pressure of a vapour in dynamic equilibrium with its liquid. Depends on the intermolecular forces present in the liquid and the temperature.

Question 17

Boiling point

Answer 17

The temperature at which its vapour pressure equals the external pressure. At lower pressure, there’s a lower boiling point

Question 18

Normal boiling point

Answer 18

temperature at which its vapour pressure equals 1.01325 bar = 1 atm = 760 Torr.

Question 19

Standard boiling point

Answer 19

The temperature at which the vapour pressure of a liquid equals 1 bar. For water the standard boiling point is 99.6 C slightly lower than the normal boiling point since 1 bar is slightly lower than 1 atm.

Question 20

The Clausius-Clapeyron Equation

Answer 20

The relationship between vapour pressure and temperature is not linear. With pressure raising faster than the temperature exponentially. This can be expressed using this formula:
ln(Pvap)=-DeltavapHR(1T)+lnbeta. Pvap is the vapour pressure. Beta is a constant that depends on gas. Deltavap H is the heat of vapourization. R is the gas constant. and T is the temperature in kelvin.

Question 21

Two-point form Clausius-Clapeyron Equation

Answer 21

ln(P2/P1)= -delavapH/R (1/t2 - 1/t1).
Generally inferior to plotting multiple points because fewer data points results in greater possible error.

Question 22

The Critical Point

Answer 22

The temperature and pressure above which a supercritical fluid exists.
Critical Temperature: The temperature above which a liquid cannot exist, regardless of pressure.
Critical Pressure: (Pc) The pressure required to bring about a transition to a liquid at the critical temperature.

Question 23

Sublimation

Answer 23

The phase transition from solid to gas.

Question 24

Deposition

Answer 24

The phase transition from gas to solid

Question 25

Melting point

Answer 25

The temperature at which the molecules of a solid have enough thermal energy to overcome intermolecular forces and become a liquid.

Question 26

Melting or fusion

Answer 26

The phase transition from solid to liquid the opposite of freezing. Once the melting point of a solid is reached, additional heating only causes more rapid melting; it does not raise the temperature of the solid above its melting point.

Question 27

Heat of Fusion

Answer 27

(DeltafusH) The amount of heat required to melt 1 mole of solid.

Question 28

Heating Curve of Water (segment 1)

Answer 28

Solid ice is warmed from -25 C to 0 C. Since no transition between states occurs here, the amount of heat required to heat the solid ice is given my q=mCsdeltaT Where Cs is the specific heat capacity of ice. 0.9941kJ added.

Question 29

Heating Curve of Water (segment 2)

Answer 29

In segment 2, the added heat does not change the temperature of the ice and water mixture because the heat is absorbed by the transition from solid to liquid. The amount of heat required to convert the ice to liquid water is given by q=ndeltafusH. 6.01 kJ added.

Question 30

Heating Curve of Water (segment 3)

Answer 30

In segment 3, the liquid water is warmed from 0 C to 100 C. Since no transition between states occurs here, the amount of heat required to heat the liquid water is given my q=mCdeltaT, as in segment 1. However now using the specific heat capacity of liquid water. 7.52 kJ added.

Question 31

Heating Curve of Water (segment 4)

Answer 31

In segment 4, the water undergoes a second transition between states, this time from liquid to gas. The amount of heat required to convert the liquid to gas is given by q=nDeltavapH. 40.7 kJ added.

Question 32

Heating Curve of Water (segment 5)

Answer 32

In segment 5, the steam is warmed from 100 C to 125C. Since not transition between states occurs here, the amount of heat required to heat the steam is given by q=mCsDeltaT, except we must use the specific heat capacity of steam. 0.904 kJ is added.

Question 33

X-ray Diffraction

Answer 33

A powerful lab technique that allows for the determination of the arrangement of atoms in a crystal and the measuring of the distance between them.

Question 34

Diffraction patterns

Answer 34

Light of similar wavelengths creates interference patterns. The exact pattern reveals the spacing between planes of atoms. Criterion for constructive interference: nLambda=2a. Using trig we can see the angle of reflection (theta) related to the distance (a) and the seperation between layers (d) by the following relation: sin(theta)=a/d, a=dsin(theta)

Question 35

Bragg's Law

Answer 35

nLambda = 2d sin (theta). For a given wavelength of light incident on atoms arranged in layers, we can measure the angle that produces constructive interference (which appears as a bright spot on the X-ray diffraction pattern) and then compute (d), the distance between atomic layers.

Question 36

Crystalline lattice.

Answer 36

Nature's way of aggregating the particles to minimize their energy. Represented using a small collection of atoms, ions, or molecules called the unit cell. A unit cell is the smallest divisible unit of a crystal that, when repeated in three dimensions, reproduced the entire crystal lattice.

Question 37

Simple Cubic Cell:

Answer 37

1 atom per unit cell, cubic structure, coordination number: 6, Edge length (in terms of r) 2r, Packing efficiency: 52%

Question 38

Body centered cubic cell

Answer 38

2 atoms per unit cell, cube with sphere in middle and further apart corners, coordination number of 8. Edge of length: 4r/sqrt(3), packing efficiency: 68%

Question 39

Face centered cubic

Answer 39

4 atoms per unit cell, cube with strange sphere in centre corners even further apart, coordination number of 12, Edge Length: (2sqrt(2))r, packing efficiency: 74%

Question 40

Coordination Number

Answer 40

The number of atoms with which each atom in a crystal lattice is in direct contact

Question 41

Packing Efficiency

Answer 41

The percentage of volume of a unit cell occupied by the atoms, assumed to be spherical.

Question 42

Hexagonal Closest Packing

Answer 42

A closest-packed arrangement in which the atoms of the third layer align exactly over those in the first layer.

Question 43

Cubic closest Packing

Answer 43

A closest-packed arrangement in which the third layer of atoms is offset from the first; the same structure as the face centred cubic.

Question 44

Molecular Solids:

Answer 44

Solids whose composite units are molecules. The lattice sites in a crystalline molecular solid are therefore occupied by molecules. Held together by dispersion forces, dipole-dipole, and hydrogen bonding. Low to moderately low melting points.

Question 45

Ionic Solids

Answer 45

Solids whose composite units are ions. (NaCl)(CaF2). Held together by coulombic interactions that occur between anions and cations occupying the lattice sites in the crystal.

Question 46

Atomic Solids

Answer 46

Solids whose composite units are individual atoms are atomic solids. Atomic solids themselves can be classified into three categories - nonbonding atomic solids(dispersion), metallic atomic solids(metallic bonds), and network covalent atomic solids(covalent bonds).