Solids, Liquids, Energy & Heat

Intermolecular Forces

“Inter” => between “molecular” => molecules

Intermolecular forces apply only to covalent bonds

Intermolecular Forces vs. Intramolecular Forces

Intermolecular Forces

Intramolecular Forces

• Intermolecular forces are between neighboring molecules

• Weaker than intramolecular forces

• Types:• London dispersion

forces• dipole-dipole forces• hydrogen bonding

• Intramolecular forces are within one molecule

• a.k.a. Bonding• Stronger than

intermolecular forces• Types:

• Ionic Bonding• Covalent Bonding

Intermolecular Forces• Intermolecular forces vary in strength and

properties:

Hydrogen bonds

Dipole-dipole

London dispersion

Boiling point, melting point, surface tension, and viscosity

and

London Dispersion Forces(van der Waals Forces)

• Very weak forces• Occurs when an electron cloud temporarily

gets distorted and forms a temporary positive and negative charge.

• The ONLY forces of attraction that exist between nonpolar molecules (e.g. Ar, C8H18 )

London Dispersion Forces

Dipole-Dipole Attraction

• Polar molecules that orient themselves so that + and – ends of the dipoles are close to each other.

Hydrogen Bonding• Strongest dipole-dipole attraction when

hydrogen is bound with either: N, O, F.• Molecule would contain –NH, -OH, or -FH

Hydrogen Bonding in DNA

Steps to Determine Intermolecular Forces

1) Draw & determine the VSEPR Shape for each molecule: I2 H2Se NH3

I I•••• • •

• • ••••Se H

H

• ••• NH H

H

• •

linear bent pyramid

2) Determine if each molecule is a polar or non-polar structure: Options: (a) There are no polar bonds non-polar structure(b) Every polar bond has an opposite (no lone pairs on center) non-polar structure (c) Not every polar bond has an opposite (lone pairs on center) polar structure

I2 H2Se NH3

non-polar polar polar

•••• • •

• • ••••I I

Se H

H

• ••• NH H

H

• •

• 3) Determine the type of intermolecular force for each molecule:

Options: (a) non-polar structure London dispersion forces(b) polar structure with H-O, H-N, or H-F bond(s) hydrogen bonding(c) polar structure without H-O, H-N, or H-F bond(s) dipole-dipole forces I2 H2Se NH3

London dispersion forces

dipole-dipole forces

hydrogen bonding

Energy

* + & - indicate whether the heat is flowing into or out of the system*

Energy is the capacity to do work. Energy allows us to “do things”, such as drive, cook eggs, and read the words on this page. Often energy is used to change the temperature of a substance. When you change the temperature of a substance, you are reflecting the random motion of the molecules. There are two common units of energy – the calorie and the joule.

A calorie (cal) is defined as the amount of energy required to raise the temperature of one gram of water by 1C. The joule (J) is the unit of energy in the SI System.

1 calorie = 4.184 J

Example – Express 22.4 cal of energy in joules.

1 kcal = 1000 cal 1 kJ = 1000 J

22.4 cal X _________________

cal

J

1

4.184= 93.7 J

The amount of energy required to raise the temperature of a substance depends on three things:

1. amount of substance (mass)2. the amount of the temperature change3. type or identity of the substance

Different substances respond to heat differently –

4.184 J raises 1 g of water - 1C BUT 4.184 J raises gold 32C

Specific Heat Capacity – the amount of energy required to change the temperature of one gram of a substance by 1C

Formula to calculate energy requirements-

Q = (s)(m)(T)

Q = energy required, Js = specific heat capacity, J/gC

m = mass of substance, gT = change in temperature, C

T = Tfinal - Tinitial

Include: Given, formula, setup, answer, SF, units

Specific Heat, Heat of Fusion, and Heat of Vaporization of Some Selected Substances

Substance Specific Heat(J/g°C)

Heat of FusionΔHfus ( J/g)

Heat of VaporizationΔHvap ( J/g)

alcohol - ethanol 2.51 109 879aluminum 0.900 395 10800benzene 1.42 126 548chloroform 0.971 79.5 264copper 0.385 205 5310ice 2.09 334 2260iron 0.460 267 7450lead 0.130 24.7 946magnesium 1.02 372 6070mercury 0.138 11.3 285steam 2.02 334 2260water 4.184 334 2260gold 0.130 63.0 1580silver 0.240 111 2360

Ex. – A 5.63 g sample of solid copper is heated from 21C to 32C. How much energy in joules is required?

m = 5.63 g

T = Tf – Ti

T = 32C - 21C = 11C

Q = ?

s = 0.385 J/gC

Q = s m T

Q = (0.385 J/gC) (5.63 g) (11 C )

Q = 24 J

Example – A 2.8 g sample of pure metal requires 10.1 J of energy to change its temperature from 21C to 36C. What is the metal?

m = 2.8 g

Q = 10.1 JT = Tf – Ti

T = 36C - 21C = 15Cs = ?

s = __Q__ m T

s = ___10.1 J___

(2.8 g)(15C)

s = 0.24 J/gC

Look on chart

Silver

Heating Curve for Water

When a substance melts or boils, there is no change in temperature (no ∆T). Heat of Fusion (∆Hfus) is the amount of energy required to melt a substanceHeat of Vaporization (∆Hfus) is the amount of energy required to boil a substance The formulas to use are:

For melting or freezing, Q=m∆Hfus

For boiling or condensing, Q=m∆Hvap

Molar Heat of Fusion and Molar Heat of Vaporization have units of J/molHeat of Fusion, Heat of Vaporization, and Specific Heat capacity are characteristic properties and can be used to identify a substance. Ex. How much energy is required to melt 8.93 moles of gold with no change in temperature? Ex. How much energy is absorbed when 5.34 g of chloroform condenses from a gas?

m = 8.93 mol Au

Hfus Au =

q=m∆Hfus

q = (1760 g)(63.0 J/g)

q = 111000 J

x _________g Au

1196.97

mol Au63.0 J/g

= 1760 g Au

m = 5.34 g

Hvap chloroform =

q=m∆Hvap

q = (5.34 g)(264 J/g)

q = 1410 J264 J/g

Enthalpy (H) – Heat of reaction

H = exothermic reaction - heat flows out of the system

+ H = endothermic reaction - heat flows into the system

feels hot or gives off light

feels cold

Entropy (S) – Disorder Second law of thermodynamics - the entropy of

the universe is always increasing

Solid liquid gas

Ordered disordered

Also: # of moles and number of bonds

Ex: Which of the following has more entropy?

a. solid or gaseous phosphorus

b. KBr (s) or KBr (aq)

c. CH4 (g) or C3H8 (g)

Ex: Does entropy increase or decrease for each of the following reactions?a.(NH4)2Cr2O7 (s) Cr2O3 (s) + 4 H2O (l) + N2 (g)

b. Mg(OH)2 (s) MgO (s) + H2O (g)

c. PCl3 (g) + Cl2 (g) PCl5 (g)

INCREASE

INCREASE

DECREASE

Phase Diagram

• Triple point is the temperature and pressure at which all three phases can exist in equilibrium.Above the critical point, molecules are unable to liquefy.

Phase Changes:Melting  Freezing  Boiling   Condensing  Sublimation  Deposition

Normal Boiling Point:    Normal Freezing Point 

the temperature at which a liquid boils at 1 atmosphere of pressure.

the temperature at which a liquid freezes at 1 atmosphere of pressure.

solid liquid

liquid solid

liquid gas

gas liquid

solid gas

gas solid