[PDF] Intermolecular Forces

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Intermolecular Forces

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 2

Intermolecular Forces

Intermolecular forces are attractive forces between molecules. Intramolecular forces hold atoms together in a molecule. Intermolecular vs Intramolecular

• 41 kJ to vaporize 1 mole of water (inter) • 930 kJ to break all O-H bonds in 1 mole of water (intra)

Generally, intermolecular forces are much weaker than intramolecular forces. "Measure" of intermolecular force

boiling point melting point ΔH vap ΔH fus ΔH sub 3

Intermolecular Forces

Dipole-Dipole Forces Attractive forces between polar molecules Orientation of Polar Molecules in a Solid

4

Intermolecular Forces

Ion-Dipole Forces Attractive forces between an ion and a polar molecule Ion-Dipole Interaction 5 in solution

Interaction Between Water and Cations

6

Intermolecular Forces

Dispersion Forces Attractive forces that arise as a result of temporary dipoles induced in atoms or molecules ion-induced dipole interaction dipole-induced dipole interaction

7

Induced Dipoles Interacting With Each Other

8

Intermolecular Forces

Dispersion Forces Continued Polarizability is the ease with which the electron distribution in the atom or molecule can be distorted. Polarizability increases with:

• greater number of electrons • more diffuse electron cloud Dispersion forces usually increase with molar mass. 9 S What type(s) of intermolecular forces exist between each of the following molecules? HBr

HBr is a polar molecule: dipole-dipole forces. There are also dispersion forces between HBr molecules.

CH 4 CH 4 is nonpolar: dispersion forces. SO 2 SO 2 is a polar molecule: dipole-dipole forces. There are also dispersion forces between SO 2 molecules. 10

Intermolecular Forces

Hydrogen Bond The hydrogen bond is a special dipole-dipole interaction between they hydrogen atom in a polar N-H, O-H, or F-H bond and an electronegative O, N, or F atom. A H ... B A H ... A or A & B are N, O, or F

11

Hydrogen Bond HCOOH and water

12 Why is the hydrogen bond considered a "special" dipole-dipole interaction?

Decreasing molar mass Decreasing boiling point

13

Gases

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 14

Elements that exist as gases at 25

0

C and 1 atmosphere

15 16 • Gases assume the volume and shape of their containers.

• Gases are the most compressible state of matter. • Gases will mix evenly and completely when confined to

the same container. • Gases have much lower densities than liquids and solids.

Physical Characteristics of Gases

NO 2 gas 17

Units of Pressure 1 pascal (Pa) = 1 N/m

2

1 atm = 760 mmHg = 760 torr 1 atm = 101,325 Pa

Pressure = Force Area

(force = mass x acceleration)

Gases

Very weak in termolecular fo rces: Van der Wa als and London. Gases occupy all the volume of the container. Pressure: gas p ressure is due to the particle collisi ons onto recip ient surface. Pressure increases with T, because in this way increases the E

kin.

of the particles, or decreasing the recipient volume (or by adding gas inside the recipient). Four parameters a re necessary to characterize a gas: mass, vo lume, pressure and temperature.

Ideal gases

To characterize a gas we need a function, which depends on four parameters f(m, V, T, P) Model: Ideal gas The function describing the state of an ideal gas is called: ideal gas equation.

20

Ideal Gases

1. A gas is composed of molecules that are separated from each other by distances far greater than their own dimensions. The molecules can be considered to be points; that is, they possess mass but have negligible volume.

2. Gas molecules are in constant motion in random directions,

and they frequently collide with one another. Collisions among molecules are perfectly elastic.

3. Gas molecules exert neither attractive nor repulsive forces

on one another.

4. The average kinetic energy of the molecules is proportional

to the temperature of the gas in kelvins. Any two gases at the same temperature will have the same average kinetic energy KE = ½ mu

2 21

Ideal Gases

• The definition of ideal gas removes the chemical differences among gases, so allowing the definition of general laws.

• Definition of experimental laws: they were derived from experimental observations • Boyle, Charles and Volta Gay-Lussac

Ideal Gases

Real gases behave as ideal gases at low pressure (low concentration) and high temperature (high E kin ) Boyle Law - constant Temperature PV=cost P o V o = P 1 V 1 23

P x V = constant P

1 x V 1 = P 2 x V 2

Boyle's Law

Constant temperature Constant amount of gas P ∝ 1/V 24

A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL?

P 1 x V 1 = P 2 x V 2 P 1 = 726 mmHg V 1 = 946 mL P 2 = ? V 2 = 154 mL P 2 = P 1 x V 1 V 2 726 mmHg x 946 mL 154 mL = = 4460 mmHg P x V = constant

Ideal gases

Charles Law - constant Pressure V/T = cost V

o /T o = V 1 /T 1 Increasing T, increases V: V t =V 0 (1+αt) with α=1/273,15

Ideal gases: Charles law

V t =V 0 (1+αt)

Ideal gases: Charles law

Repeating the experiment with different gases or different starting V, we can observe different linear curves, but the intercept with X axis is always at -273,15 °C, corresponding to V = 0: this is the minimum value of T.

T(K)= t(°C) + 273,15

28
A sample of carbon monoxide gas occupies 3.20 L at 125 ° C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant? V 1 = 3.20 L T 1 = 398.15 K V 2 = 1.54 L T 2 = ? T 2 = V 2 x T 1 V 1 1.54 L x 398.15 K 3.20 L = = 192 K V 1 /T 1 = V 2 /T 2 T 1 = 125 ( °

C) + 273.15 (K) = 398.15 K

It is a con sequence of th e Charles law, althoug h it was independently defined. The volume of a gas is always the same with e qual number of moles (with the same conditions of T and P) In particur at 0 °C e 1 atm (normal conditions), the volume occupied by a mole of a gas is V

0 = 22,414 liters. V t /T=V 0 /T 0

Ideal gases: Avogadro law

30

Avogadro's Law

V ∝ number of moles (n) V = constant x n V

1 / n 1 = V 2 / n 2

Constant temperature Constant pressure

Ideal gases

Volta Gay-Lussac Law - constant Volume P/T = cost P o /T o = P 1 /T 1

Increasing T, increases P: P

t =P 0 (1+βt) with β=1/273,15 P t /T=P 0 /T 0 32

Ammonia burns in oxygen to form nitric oxide (NO) and water vapor. How many volumes of NO are obtained from one volume of ammonia at the same temperature and pressure?

4NH 3 + 5O 2 4NO + 6H 2

O 1 mole NH

3 1 mole NO At constant T and P 1 volume NH 3 1 volume NO 33

Summary of Gas Laws Boyle's Law

34

Charles Law

35

Avogadro's Law

36

Ideal Gas Equation

Charles' law: V ∝ T (at constant n and P) Avogadro's law: V ∝ n (at constant P and T) Boyle's law: P ∝ (at constant n and T) 1 V V ∝ nT P V = constant x = R nT P nT P R is the gas constant PV = nRT

37

The conditions 0

° C and 1 atm are called standard temperature and pressure (STP). PV = nRT R = PV nT = (1 atm)(22.414L) (1 mol)(273.15 K) R = 0.082057 L • atm / (mol • K) Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L. 38
What is the volume (in liters) occupied by 49.8 g of HCl at STP?

PV = nRT V = nRT P

T = 0

° C = 273.15 K P = 1 atm n = 49.8 g x 1 mol HCl HCl = 1.37 mol

V = 1 atm 1.37 mol x 0.0821 x 273.15 K

L•atm mol•K

V = 30.7 L

39

Density (d) Calculations d = m V = PM RT m is the mass of the gas in g M is the molar mass of the gas Molar Mass (M ) of a Gaseous Substance dRT P M = d is the density of the gas in g/L Related density

d A /d B = M A / M B 40
A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and 27.0 °

C. What is the molar mass of the gas?

dRT P M = d = m V 2.10 L = = 2.21 g L

M = 2.21

g L

1 atm x 0.0821 x 300.15 K

L•atm mol•K

M =

54.5 g/mol

41

Gas Stoichiometry

What is the volume of CO

2 produced at 37 0 C and 1.00 atm when 5.60 g of glucose are used up in the reaction: C 6 H 12 O 6 (s) + 6O 2 (g) 6CO 2 (g) + 6H 2 O(l) g C 6 H 12 O 6 mol C 6 H 12 O 6 mol CO 2 V CO 2 C 6 H 12 O 6 1 mol C 6 H 12 O 6 C 6 H 12 O 6 x 6 mol CO 2 1 mol C 6 H 12 O 6 x = 0.187 mol CO 2 V = nRT P 0.187 mol x 0.0821 x 310.15 K L•atm mol•K 1.00 atm = = 4.76 L

Gas mixtures

For ideal gases there is no differences in the application of gas laws for pure gases or mixtures of them, since there are no chemical differences. Each component of a mixture can access to the recipient volume and

contributes to the total pressure by a partial pressure.

The partial pressure of a gas is the pressure exerted by the gas present alone into the recipient: Dalton Law P = P

1 + P 2 + P 3 + ......P n = ∑ P i

Gas mixture

For the l'i-th component: P

i V = n i

RT (1) For the mixture: ∑P

i

V = ∑n

i

RT (2) con ∑n

i = n 1 + n 2 + n 3 + ......n n = n Dividing (1) with (2) P i

V/ PV = n

i

RT/ nRT P

i / P = n i / n = χ i = molar fraction P i = χ i P 44

Dalton's Law of Partial Pressures

V and T are constant P

1 P 2 P total = P 1 + P 2 45
Consider a case in which two gases, A and B, are in a container of volume V. P A = n A

RT V P

B = n B RT V n A is the number of moles of A n B is the number of moles of B P T = P A + P B X A = n A n A + n B X B = n B n A + n B P A = X A P T P B = X B P T P i = X i P T mole fraction (X i ) = n i n T 46

A sample of natural gas contains 8.24 moles of CH

4 , 0.421 moles of C 2 H 6 , and 0.116 moles of C 3 H 8 . If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C 3 H 8 )? P i = X i P T X propane = 0.116 8.24 + 0.421 + 0.116 P T = 1.37 atm = 0.0132 P propane = 0.0132 x 1.37 atm = 0.0181 atm 47

The distribution of speeds for nitrogen gas molecules at three different temperatures The distribution of speeds of three different gases at the same temperature

u rms = 3RT M √ 48

Gas diffusion is the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties.

NH 3

17 g/mol HCl

36 g/mol NH

4 Cl r 1 r 2 M 2 M 1 √ = molecular path 49

Gas effusion is the is the process by which gas under pressure escapes from one compartment of a container to another by passing through a small opening. = r

1 r 2 t 2 t 1 M 2 M 1 √ = Nickel forms a gaseous compound of the formula Ni(CO) x What is the value of x given that under the same conditions methane (CH 4 ) effuses 3.3 times faster than the compound? r 1 = 3.3 x r 2 M 1 = 16 g/mol M 2 = r 1 r 2 ( ) 2 x M 1 = (3.3) 2 x 16 = 174.2 58.7 + x • 28 = 174.2 x = 4.1 ~ 4 50

Deviations from Ideal Behavior 1 mole of ideal gas PV = nRT n = PV RT = 1.0 Repulsive Forces Attractive Forces

51
Effect of intermolecular forces on the pressure exerted by a gas. 52
Van der Waals equation nonideal gas P + (V - nb) = nRT an 2 V 2 ( ) } corrected pressure } corrected volume

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