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Week 12 CHEM 1310 - Sections L and M 1

CHEM 1310: Update

Exam 1 Exam 2 Exam 3

Final Exam: Mon, Dec 10 (2:50 - 5:40 pm)

Chapters 1 - 5

Chapters 12, 13, 16, 17

Chapters 6-9

Chapters 10, 15

Chemical Equilibrium Acids and Bases Applications of Aqueous Equilibria Energy, Enthalpy, and Thermochemistry

Exam 3Fri., Nov 16

Week 12 CHEM 1310 - Sections L and M 2

Energy, Enthalpy, & Thermochemistry

9.1 The Nature of Energy 9.2 Enthalpy 9.3 Thermodynamics of Ideal Gases 9.4 Calorimetry 9.5 Hess’s Law 9.6 Standard Enthalpies of Formation 9.7 Present Sources of Energy 9.8 New Energy Sources

Week 12 CHEM 1310 - Sections L and M 3

What is Energy? Electromagnetic Energy: inversely related to wavelength

∆E = (h x c)/ λ From Ch. 12 - Quantum Mechanics

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Week 12 CHEM 1310 - Sections L and M 4

What is Energy? Chemical (Mechanical) Energy: the capacity to do work

or to produce heat

ΔE = q + wwhere E is the internal energy of a system

• q = Heat Absorbed by a system If q > 0 , heat is absorbed If q < 0 , heat is given off

• w is the work done on the “body”Recall

w = F x d w = ΔEKinetic= Δ (½ mv2) w = ΔEpotential= mgΔh

In Ch. 9,focus on the transfer of E

via heat

Week 12 CHEM 1310 - Sections L and M 5

Units of Energy

Joule = 1 N m = kg x m2 x s-2

SI Unit

Calories Food energy 1 calorie = 4.186 J

British Thermal Units (BTU) Heating & Air, Power, etc.

James Prescott Joule(1818-1889)

Week 12 CHEM 1310 - Sections L and M 6

Nature of Energy

Law of Conservation of Energy Energy can be converted from one form to

another, but energy cannot be created ordestroyed.

Kinetic: energy of an object due to motion

Potential: “stored” energy; energy an objecthas the capability of using if it were inmotion

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Week 12 CHEM 1310 - Sections L and M 7

Kinetic vs. Potential Energy

Work = Force over a distance

Potential energy (e.g., ΔEpotential= mgΔh) Kinetic energy (e.g., ΔEKinetic= Δ (1/2mv2) Energy transfer Through heat (frictional heating) Through work

Week 12 CHEM 1310 - Sections L and M 8

Endothermic Reactions

Reactants + Heat Products

Week 12 CHEM 1310 - Sections L and M 9

Exothermic Reactions

Reactants Products + Heat2 Al (s) + Fe2O3 (s) → Al2O3 (s) + 2Fe (s)

Thermite Reaction

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Week 12 CHEM 1310 - Sections L and M 10

Internal Energy Flow

An Endothermic ProcessAn Exothermic Processq = – x q = + x

Week 12 CHEM 1310 - Sections L and M 11

Energy as Work

Work = F x dWork = F x ΔhWork = P x A x Δh

Since ΔV = A ΔhWork = P x ΔV

Week 12 CHEM 1310 - Sections L and M 12

Energy as Work

Force exerted by piston. Expansion vs. Compression

w = −Pext ΔV

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Week 12 CHEM 1310 - Sections L and M 13

Energy as Work

Expansion ΔV > 0 therefore w < 0

The system does work on the surroundings

Compression ΔV < 0 therefore w > 0 The surroundings have done work on the system

w = – Pext ΔV

Week 12 CHEM 1310 - Sections L and M 14

Enthalpy

Normally measure change in enthalpy, ΔH

H = U + PVEnthalpy Internal Energy

of System

qpressure = ΔH

If ΔH positive then q > 0 Heat is absorbed Endothermic

If ΔH negative then q < 0 Heat is given off Exothermic

Enthalpy:Heat contentof a system

Week 12 CHEM 1310 - Sections L and M 15

Enthalpy is a State Function

A state function is a property that depends on itspresent state, not on any path leading to that state.

Analogy

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Week 12 CHEM 1310 - Sections L and M 16

Enthalpy of a Reaction

ΔHrxn = Hproducts – Hreactants

CO (g) + ½ O2 (g) → CO2 (g) ΔH= -283 kJ Exothermic!

If molar ratios are doubled, then enthalpy doubles…

2 CO (g) + 1 O2 (g) → 2 CO2 (g) ΔH= -566 kJ

CO2 (g) → CO (g) + ½ O2 (g) ΔH= +283 kJ

If the rxn is reversed, then enthalpy sign changes…

Endothermic!

Week 12 CHEM 1310 - Sections L and M 17

Example Problem

H2 + Br2 → 2HBr (g)

ΔH = -72.8 kJ mol-1

2 HBr (g) → H2 + Br2 ΔH = +72.8 kJ mol-1

Note: decomposition = reverse reaction

How much heat energy is needed to decompose 9.74g ofHBr (MW = 80.9 g/mol) into its elements?

Week 12 CHEM 1310 - Sections L and M 18

Example Problem

�

q = DH =72.8 kJ

2 moles HBr

Ê Ë Á

ˆ ¯ ˜

1 mol HBr

80.91g HBr

Ê

Ë Á

ˆ

¯ ˜ 9.74 g HBr( )

= 4.38 kJ

Solution2HBr (g) → H2 + Br2 ΔH = +72.8 kJ mol-1

How much heat energy is needed to decompose9.74g of HBr (M =80.9 g/mol) into its elements?