solids, liquids, energy & heat. intermolecular forces “inter” => between “molecular”...
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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