Core 3: Molecular Interactions

Question 1

How are real gases modelled from a perfect gas?

Answer 1

A compressability factor, z, is required which is the V_m of gas/V_m of a perfect gas at the same temperature and pressure.

Question 2

What factors does the state of matter depend on?

Answer 2

The attraction between molecules and the kinetic energy of the system.

Question 3

How can dipole strength be calculated? What are the units? How is this related to electronegativity?

Answer 3

μ=qR where q is the charge on one atom and R is the distance between atoms/charges. Cm or D where D is a Debye and represents 3.335x10^-30Cm (databook)

The dipole in a molecule is roughly equal to the difference in electronegativity

μ≈ΔΧ

Question 4

How is percentage ion character calculated?

Answer 4

(actual dipole/+,- approximation of bond dipole) x100

Question 5

What must be considered when comparing dipoles?

Answer 5

They have a direction as well as a value.

Question 6

How can the overall dipole in more complex molecules be considered for strength?

Answer 6

The dipoles can be considered to be vectors. Opposite vectors of the same value cancel out, parallel vectors add.

Question 7

What is polarisability? How are induced dipoles represented?

Answer 7

The ability to have an electron cloud distorted by a weak electric field. Induced dipole, μ*=αE where α is the polarisability and E is the electric field.

Strong fields require require more complex equations.

Question 8

What are the units and alternate forms of polarisability?

Answer 8

Units of α is C²m²J^-1

Polarisability volume is a useful measure, α’=α/4πε₀ units=m³

Question 9

What is the Coulombic potential energy?

Answer 9

V=q₁q₂/4πε₀r

Question 10

What four key steps are there for deriving the energy between charges?

Answer 10

• multiply by r/r to remove a factor of r
• substitute l/nr for x where n is a number and l is the bond lenth
• expand 1/1±x assuming r>>l
• μ=ql

Question 11

Why are Van der Waals forces always attractive?

Answer 11

Molecules are nearly freely rotating, if they were freely rotating the interactions would cancel out. They will adopt lower energy configurations more favourably meaning all interactions overall are attractive.

Question 12

What is the name for dipole-dipole interactions and what are its important features?

Answer 12

Keesom interaction.

Negative sign, 1/r⁶ distance dependance, inverse temperature dependance.

Question 13

What is the name and key features for dipole-induced dipole interactions?

Answer 13

Debye interactions.

Negative sign, 1/r⁶ distance dependance, no temperature dependance(cannot be unalligned).

Question 14

What is the name and key features for induced dipole-induced dipole interactions?

Answer 14

London (dispersion) interaction.

Negative sign, 1/r⁶ distance dependance, no temperature dependance.

Question 15

What is the Van der Waals interaction?

Answer 15

The sum of the 3 types of intermolecular interactions. V=-C/r⁶ where C depends on the molecule.

Question 16

What additionally need to be considered for molecules without dipoles?

Answer 16

Electrostatic interactions which decay faster than dipole interactions but are significant in non-polar systems. They are called multipoles or n-poles.

V_n∝1/r^n+m-1

n and m depend on the molecule being considered, monopoles are only single atoms, dipoles are two atoms, quadrupoles are CO₂ equvalents or 4 atom squares, octupoles are cube shape or tetrahedral with a central heteroatom.

Question 17

How are three body interactions accounted for? How significant are these interactions?

Answer 17

Using the Axilrod-Teller formula. Not large but not insignificant, they account for 10% of the total interactions of argon.

Question 18

How does shape affect polarisability and boiling point? Describe using n-pentane and neopentane.

Answer 18

n-pentane has a higher boiling point as the chain can stack better than the neopentane molecules and its polarisability is more spread out meaing it can shift around easier.

Question 19

Give an approximate strength of attaction using the boiling point of a substance.

Answer 19

Strength of attraction, ε≈k_BT_BP

Question 20

How does strength of attraction relate to troutons rule?

Answer 20

ΔH_vap∝ε as the energy required to vapourise a molecule depends on its attractions to surrounding molecules.

Therefore ΔH_vap≈constant x T_BP which is troutons rule.

Question 21

How do the strengths of hydrogen bonds vary for NH₃, H₂O and HF?

Answer 21

HF and NH₃ both can form two hydrogen bonds per molecule whereas H₂O can form four per molecule, meaning it has stronger intermolecular forces.

HF has stronger hydrogen bonds than NH₃ as it has a greater dipole over the X-H bonds.

Question 22

How can hydrogen bond strengths be estimated?

Answer 22

By using the trend of ΔH_vap for the other hydrides in the group, the value for the first vaule can be predicted. The difference should be large and can be estimated to be the influence of the hydrogen bonds.

Question 23

For a plot of ΔH_vap vs T, what do liquids that lie above and below the line mean for their liquid and gas structures?

Answer 23

Above the line, such as alcohols and water, means there is additional structure in the liquid such as hydrogen bonds.

Below the line, such as methan/ethanoic acid, form additional structure in the gas e.g methanoic and ethanoic acid form dimers in the gas phase.

Question 24

What is special about sodium polyacrylate?

Answer 24

In dry conditions it packs tightly together as each O^- has a Na⁺ and the groups are happy to be close. In water the Na⁺ ions dissolve into solution and the chains straighten as the O^- ions repel each other. They also bind up water as the gel like structure holds water molecules.

Question 25

What is the strengths of typical bonds and interactions?

Answer 25

Covalent bond=ion ion interaction= 200-800kJmol^-1

Ion-dipole= 10-100kJmol^-1

H-bond= 10-40kJmol^-1

Dipole-dipole= 2-10kJmol^-1

London= 0.05-10kJmol^-1

Question 26

Explain the origins of surface tension.

Answer 26

At the surface of the liquid there is fewer neighbours so the molecules are at a higher potential energy (fewer intermolecular bonds). This means the liquid wants to minimise the molecules on the edge. When an object is introduced to the top of the liquid, it creates a greater area of higher potential molecules. They provide a restoring force to get to their ideal position. If the restoring force is large enough it will counteract the objects weight and allow it to float on the surface.

Question 27

What are the three ways to represent the total interaction of two atoms? When do each apply?

Answer 27

Hard spheres - no attraction, inifinite repulsion when atoms collide. This applies when the atoms are hot so attractions are negliable or for ideal gases.

Square well - set attraction between two values, infinite repulsions when atoms collide. Models simple attraction.

Mie potential - curve of increasing attraction until a minimum is reached and fast increase in repulsion as atoms collide to a practially infinite value.

Question 28

What is the Mie potential and how has it been adapted?

Answer 28

V=(C_n/rⁿ)-(C_m/r^m)

Lennard-Jones V=4ε{(σ/r)¹²-(σ/r)⁶} where ε is the well depth and σ is the seperation where V=0

The repulsion, (σ/r)¹², term can be improved with an exponential term but this not significantly different.

Question 29

How can interactions between two nobel gases be estimated?

Answer 29

ε_He-Ar≈(ε_He-Heε_Ar-Ar)^½

Question 30

What has to be considered when describing all interactions between large molecules? Why doesn’t this apply to small molecules?

Answer 30

Sum of bond vibrations, angle vibrations, dihedrals(bonds rotating) and interactions between molecules.

For small molecules, they rotate so fast that they can be estimated to be spherical.

Question 31

Why do non-polar molecules not mix with water despite the entropy increase?

Answer 31

They disrupt the hydrogen bonding and produce a cage structre around the non-polar molecules which decreases the entropy of the water molecules overall.

Question 32

Explain the benefits and drawbacks of forming micelles.

Answer 32

In micelles - Pros: Chains are free to move, H-bond structure retained

Cons: Chains are confined, head groups close together

In solution - Pros: Chains dispersed, headgroups dispersed

Cons: Chains immobile, H-bond structure broken

Question 33

How does the shape of the micelle formed realate to the size of the surfactant?

Answer 33

Ratio used is V/al where V is the volume of the tail, a is the width of the head group and l is the length of the tail.

If: V/al≈1 then a bilayer is formed

1/3<v></v>

<p>V/al&lt;1/3 then spheres are formed</p>

<p> </p>

</v>