13.E: Intermolecular forces (Exercises) (2023)

Q13.1a

List all intermolecular interactions that occur in each of the following kings of molecules: \(CCl_3F\), \(CCl_2F_2\), \(CClF_3\), and \(CF_4\).

P13.1b

Determine what kind of intermolecular forces exist in the following molecules: LiF, MgF₂, h₂O e HF.

S13.1b

• H₂O: London force, dipole-dipole interaction, hydrogen bonds.
• HF: Dipole-dipole intermolecular forces, hydrogen bonds.
• MgF₂and LiF: strong ionic attraction.

Q13.2a

Arrange the following species in order of decreasing melting point: CsBr, KI, KCL, MgF2.

P13.2b

Which has the highest boiling point?_2,brother_2,y Cl₂. Explain why?

S13.2b

The atomic weight of iodine = 127, bromine = 80 and chlorine = 35.5. The weight is proportional to the London dispersion strength, and the greater the molecular weight, the greater the strength. So, I₂has a higher boiling point.

Q13.3

1-Propanol C₃H₇OH e methoxyetano CH₃OC₂H₅they have the same molecular weight. Which has the highest boiling point?

S13.3

1-propanol can form London force, dipole-dipole and H bonds due to the H bonded to the O atom of the OH group, while methoxyethane cannot form H bonds. Therefore, 1-propanol has a higher intermolecular force of attraction and therefore a higher boiling point.

Q13.3

Why do lighter compounds like NH3, H2O and HF have the highest boiling points?

Q13.4a

What kind of attractive interactions exist between atoms and between nonpolar molecules?

P13.4b

Why natural gas CH₄Is it a good option for winter storage tank?

S13.4b

It's about the boiling point. Methane has a boiling point of -161°C, making it a good winter choice.

Q13.4

Explain why methane (CH__4\)It is used as the main heating gas in Alaska during the winter, instead of the more commonly used butane or propane gases in the lower 48 states.

S13.4

Methane (\(CH_4\)) is still a gas because its boiling point is around -160 °C. Other gases, such as propane or butane, liquefy under freezing conditions. Therefore, methane is more likely to be used during the Alaskan winter.

Q13.5

Define the types of intermolecular forces and give an example of each.

S13.5

There are 3 types of intermolecular force: London Dispersion, Dipole-Dipole (Example: Two \(NaCl\)) and Ion-Dipole (Example: \(Mg^+\) and \(HCl\))

• Dipole: The dipole occurs between polar molecules.
• The ion-dipole occurs between an ion and polar molecules
• London scattering occurs between nonpolar molecules.

Q13.6

How does the intermolecular determine the boiling point?

S13.6

The weaker intermolecular, the lower boiling point.

P13.7a

What is the dipole-dipole interaction of two HCl molecules that are head-to-tail collinear?

Given: The dipole moment of HF is 1.86 D. The dipole moment of HCl is 1.05 D. The distance between the two is 1.78

H⁺-F^-- - - - H⁺-Cl^-

S13.7a

\[V=-\dfrac{2\mu _{A}\mu _{B}}{4\pi \varepsilon _{0}r^{3}}\]

\[V=-\dfrac{2(1.05)(1.86)}{4\pi(8.854187817\cdot 10^{-12})(1.78)^{3}}\]

P13.7b

Calculate an ion-ion interaction energy between \(K^+\) and \(Cl^-\) at a distance of 600 pm.

S13.7b

The ion-ion interaction energy is given by Coulomb's law.

\[V = \dfrac{q_1q_2}{4 \pi \epsilon_or}\]

\[V = \dfrac{- (1,602 \times 10^{-19}\;\cancel{C})(1,602 \times 10^{-19} \cancel{C})}{4 \pi (8,853 \ veces 10^{-12} \cancel{C^2} \cdot N^{−1} \cdot m \cancel{^{−2}})(6 \times 10^{-10}\; \cancel{ m})} = -3,84 \times 10^{-19} \; J\]

Q13.7

Calculate an ion-dipole interaction energy between \(K^+\) and \(HCl\) at a distance of 600 pm. \(HCl\) has a dipole moment of \(1.08\;D\).

S13.7

\[\mu = 1,08 \cancel{D} \times \dfrac{3,3356 \times 10^{30} \; C \cdot m}{1\;\cancel{D}} = 3,6 \times 10^{-30}\; C \cdot m\]

\[V = \dfrac{-q\;\mu}{4 \pi \epsilon_o r^2} = \dfrac{- (1,602 \times 10^{-19}\;\cancel{C})(3, 6 \ veces 10^{-30} \cancel{C} \cdot \cancel{m})}{4 \pi (8,853 \times 10^{-12} \cancel{C^2} \cdot N^{− 1} \cdot m \cancel{^{−2}})(6 \times 10^{-10}\; \cancel{m})^2} = -1,44 \times 10^{-20} \ ; J\]

As expected, this is considerably lower in energy than covalent bonds (eg \(HCl\) has a binding enthalpy of \(7.0 \times 10^{-19}\;J\)).

Q13.8a

Rank the interactions from weakest to strongest:

1. H₂O--OH₂
2. li⁺- - F^-
3. li⁺- - OH₂

S13.8a

1. ion-ion interaction: Li⁺- - F^-
2. ion-dipole interaction: Li⁺- - OH₂
3. dipole-dipole interaction: H₂O--OH₂

P13.8b

A low concentration electrolyte solution behaves non-ideally, while a high concentration of the same solution behaves ideally. Explain this phenomenon in terms of forces, noting that the Coulomb forces depend on 1/r²while van der Waals forces depend on 1/r⁷. Low concentration electrolyte solutions are likely to follow which of these forces? High concentration?

S13.8b

The interatomic distances in a low-concentration electrolyte solution are greater than in a high-concentration solution. They follow van der Waals forces and therefore behave less ideally. High concentration electrolyte solutions follow Coulomb forces.

Q13.9a

Calculate the ion-dipole interaction between H₂medical condition⁺. They give the dipole moment of H₂O is 1.82 D. The distance between these two is 2 Å.

S13.9a

\[V=-\frac{q\mu }{4\pi \varepsilon _{0}r^{2}}\]

\[=\frac{1,82D\cdot(\frac{3,3356\cdot 10^{-30}Cm)}{1D}}{4\pi (8,85\cdot 10^{-12})(2\cdot 10 ^{-10}m)^{^{2}}} =1,36\; kJ/mol\]

P13.9b

Calculate the potential energy of interaction between a Cl^-ion located 120 pm from a \(H_2O\) molecule with a dipole moment of 1.85 D.

S13.9b

\[\mu = 1,85 \cancel{D} \times \dfrac{3,3356 \times 10^{30} \; C \cdot m}{1\;\cancel{D}} = 6,18 \times 10^{-30}\; C \cdot m\]

\[r = 1,2 \times 10^{-10}\; metro\]

\[V = \dfrac{q\mu}{4\pi \varepsilon _{o}r^{2}} = \dfrac{(-1.602\times10^{-19}\;C)(6.18\ times^{-30}\;C\cdot m)}{4\pi(8,851\times 10^{-12}\; C^{-2}N^{-1}m^{-2})( 1 \ times 10^{-10}\;m)^2}\]

Q13.10

Do you expect a larger dipole-dipole interaction between two molecules that are antiparallel or between two molecules that are collinear head-tail?

S13.10

Expect a stronger interaction when the two are collinear from head to tail. This can be seen by looking at the formula or in the photos of the two.

Q13.10

Express the equilibrium distance r_miin the ð term and show V = - €

Q13.12

1. Determine the Vander Waals radius of argon.
2. Use this radius to find the volume fraction for 2 moles of argon at room temperature at 1 atm.

S13.12

a. r = σ/2 = 3.40 A⁰/2 = 1,70 A⁰

b. Volume of 2 mol of Ar

4/3 πr^3 ((6.022 x 10^23)/(2 mol))= 4/3 π (1.70 x 10^(-10) m)^3 ((6.022 x 10^23)/( 2 mol ))

= 6.19 x 10-6 L mol-1

V/n=RT/P= ((0,08206 L atm·K^(-1) mol^(-1) (298,2 K))/1 atm

= 24.5 L mol-1

The fraction of this volume occupied by 2 moles of Ar

(1.239 x 10^-2 L mol^-1)/ 24.47 L mol^-1= 2,5x10^-7

Q13.15

a) What is the original polarity in a molecule?

b) es CO₂polar? Explain.

Q13.16

Which of the following compounds is (are) soluble?

1. CH₄
2. NCl₃
3. C₆H6
4. CO(SMALL)₂)₂

Q13.17

What makes a compound soluble in water? Explain using examples.

Q13.18

Explain why water has a high specific heat.

Q13.20

The energy of one hydrogen bond for each pair of bases in DNA is 15 kJ/mol. Two complementary strands are 50 base pairs each. What is the relationship of the two different strands to the hydrogen double helix in a solution at a temperature of 300 K?

S13.20

First calculate the ratio of the two different wires to a single pair.

\[ e^{\Delta E/RT}=exp[(15 \times 10^3\; J/mol)/(8,314\; J/K*mol)(300\; K) = 2,4 \times 10^ {-3}\]

Since there are 50 base pairs, we must multiply by 50 to account for all base pairs.

exp[100X(15X10³J/mol)/(8.314 J/K*mol)(300K) = 0

P13.24

Consider two pure liquids. One has strong intermolecular interactions and the other has relatively weak intermolecular interactions. For the following properties, indicate which of the liquids you would expect to have a higher value (answer with "strong" or "weak").

1. goop
2. Steam pressure
3. freezing point
4. superficial tension

S13.24

1. Strong. Higher viscosity results from stronger interactions between liquid molecules.
2. Weak. The weaker bound liquid needs less energy to turn into vapor, so it will exert a higher vapor pressure.
3. Strong. The freezing point is the same as the melting point; more energy is required to melt a solid with stronger intermolecular interactions.
4. Strong. Surface tension is the result of intermolecular interactions. The stronger these interactions, the higher the surface tension.

P13.25

Fun fact: if the DNA of a single human cell were stretched out (but still in its familiar double helix conformation), it would be about 2 meters long. The distance along the helix between the nucleotides is 3.4 Å.

1. Estimate the number of base pairs in the haploid human genome, from the 2-meter fun fact.
2. The human body contains about 100 trillion cells. About a quarter of them are erythrocytes (red blood cells) and do not contain genomic DNA. Use the average molar mass for a base pair, 650 grams per mole, to estimate how much of the human mass is human genomic DNA.
3. At its closest point, Pluto is 4.28 billion kilometers from Earth. Do you have enough DNA to get to Pluto?

Hint: humans are diploid.

S13.25

(a)

$$ 2\ m/cell \times \dfrac{bp}{3.4\ Å} \times \dfrac{10^{10}\ Å}{m} \times \dfrac{cell}{2\ genomes\ haploids } = 3 \times 10^{9} \dfrac{bp}{genome\haploid} \]

Three billion base pairs.

(b) 75 billion human cells in your body have genomic DNA.

$$ 75 \times 10^{20}\ cells \times \dfrac{genomes\haploid}{cell} \times \dfrac{3 \times 10^9\ bp}{genoma\haploid} \times \dfrac{mol} {6.022 \times 10^{23}} \times 650 \dfrac{g}{mol\bp} = 200\ g \]

That's about half a kilo.

(C)

$$ \dfrac{2\ m}{cell} \times 75 \times 10^{12}\ cells \times \dfrac{km}{1000\ m} = 2 \times 10^{11}\ km \]

Yes, you have a lot more DNA than you need to stretch from Earth to Pluto.