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Chapter 6Chapter 6

Principles of Reactivity: Energy Principles of Reactivity: Energy and Chemical Reactionsand Chemical Reactions

Read/Study:Read/Study: Chapter 6 in e-Textbook!Chapter 6 in e-Textbook!

Learn Key Definitions:Learn Key Definitions: Class Lecture NotesClass Lecture Notes

OWL Assignments: OWL Assignments: Chapter 6Chapter 6

OWL Quiz for Chapter 6: OWL Quiz for Chapter 6: NONE!NONE!

Read/Study:Read/Study: Chapter 6 in e-Textbook!Chapter 6 in e-Textbook!

Learn Key Definitions:Learn Key Definitions: Class Lecture NotesClass Lecture Notes

OWL Assignments: OWL Assignments: Chapter 6Chapter 6

OWL Quiz for Chapter 6: OWL Quiz for Chapter 6: NONE!NONE!

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1. INTRODUCTION1. INTRODUCTION

CHEMISTRY -CHEMISTRY - The study of the properties, composition, and structure of matter, the physical and chemical changes it undergoes, and the ENERGYENERGY liberated or absorbed during those changes.

THERMODYNAMICS -THERMODYNAMICS - Derived from the Greek words for heat and power; it is the study of all forms of energy and the interconversions among the different forms.

THERMOCHEMISTRY -THERMOCHEMISTRY - The study of the energy liberated (released) or absorbed by chemical or physical changes of matter.

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2SYSTEM AND SURROUNDINGS -SYSTEM AND SURROUNDINGS -

1) System -System - The part of the universe a scientist is interested in.

2) Surroundings -Surroundings - Everything else in the universe that is outside of the system.

3) Boundary -Boundary - A real or imaginary barrier betweenthe system and its surroundings through whichTHERMAL ENERGYTHERMAL ENERGY may flow, work mayappear or disappear, and matter may or maynot be exchanged.

4) Closed System -Closed System - A system that does notexchange matter with its surroundings.

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5) Open System -Open System - A system that may exchange THERMAL ENERGYTHERMAL ENERGY and MATTER MATTER with its surroundings.

6) Isolated System -Isolated System - A system that does not exchange MATTERMATTER or THERMAL THERMAL ENERGYENERGY with its surroundings.

2. CHANGE - WHY DOES IT HAPPEN?2. CHANGE - WHY DOES IT HAPPEN?

A. A. SPONTANEOUS CHANGE -SPONTANEOUS CHANGE - A change that takes place by itself.

B. B. NON-SPONTANEOUS CHANGE -NON-SPONTANEOUS CHANGE - Theopposite of a spontaneous change.

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C. Factors Affecting Change -C. Factors Affecting Change -

1) Energy -Energy - The ability to do work.

a) Thermala) Thermalb) Electricalb) Electricalc) Radiantc) Radiantd) Chemicald) Chemicale) Mechanicale) Mechanicalf) Nuclearf) Nuclearg) Kineticg) Kinetich) Potentialh) Potential

2) Entropy -Entropy - A measure of disdisorder.

3) The Rate of Change -The Rate of Change - (Kinetics)

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D. D. Predicting Change -Predicting Change - Requires a knowledge ofhow energy and entropy of a system will changeas a result of the change and whether the changewill take place at a practical speed.

3. Temperature, Thermal Energy, Heat3. Temperature, Thermal Energy, Heat

A. A. Temperature -Temperature - An indirect measure of theaverage kinetic energy of the molecules, atoms,or ions in the material; A measure of the “hot-ness” or “coldness” of a material; an INTENSIVEINTENSIVEproperty of matter.

B. B. Thermal Energy -Thermal Energy - The energy that is transferredfrom hotter objects to colder objects due to the kinetic energy of the molecules, atoms, or ions;an EXTENSIVEEXTENSIVE property of matter.

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C. C. Heat -Heat - The transfer of thermal energy thatresults from a difference in temperature; theprocess of transferring thermal energy fromhotter objects to colder objects.

D. Potential Energy - D. Potential Energy - Storedenergy that is a related to anobjects relative position.

E. E. Kinetic Energy -Kinetic Energy - The energy ofmotion.

K.E. = ½ mvK.E. = ½ mv22

K.E. = ½(2 kg)(1 m/s)K.E. = ½(2 kg)(1 m/s)22 = 1kg-m = 1kg-m22/s/s22

= 1 Joule= 1 Joule

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4. The 4. The First LawFirst Law of Thermodynamics of Thermodynamics

Energy is neither created nor destroyed.Energy is neither created nor destroyed.

Energy given up by the systemthe system must be absorbedby the surroundingsthe surroundings. . This type of change isEXOTHERMICEXOTHERMIC..

Energy absorbed by the systemthe system must come fromthe surroundingsthe surroundings. . This type of change is ENDO-ENDO-THERMICTHERMIC..

A. A. Internal Energy, E -Internal Energy, E - The total energy of asystem.

E = E = q + wq + w = E = Efinalfinal - E - Einitialinitial

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Constant Volume CalorimetryConstant Volume Calorimetry E = q + wE = q + w(w = 0 at(w = 0 atconstant V)constant V)

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B. B. Enthalpy, H -Enthalpy, H - The Thermal Energy gainedThe Thermal Energy gainedor lost by a system when the system under-or lost by a system when the system under-goes a change under constant goes a change under constant pressure.pressure.

finalfinal - H - Hinitialinitial

We can only measure We can only measure The ChangeThe Change in in enthalpy, not the absolute enthalpy. enthalpy, not the absolute enthalpy. Enthalpy is a state function.Enthalpy is a state function.

Exothermic Change -Exothermic Change - Changes during Changes duringwhich the system gives off thermal energywhich the system gives off thermal energy

and and negative).negative).

Endothermic Change -Endothermic Change - Changes during Changes duringwhich the system absorbs thermal energywhich the system absorbs thermal energy

and and positive).positive).

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Constant Pressure CalorimetryConstant Pressure Calorimetry HH

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Examples ofExamples of ExothermicExothermic Changes -Changes -

HH22O (l) O (l) HH22O (s)O (s) H < 0H < 0

HH22O (l)O (l)

HH22O (s)O (s)

H = -H = -This is an This is an exothermicexothermic process processbecause the because the finalfinal energy state is energy state is lower than the lower than the initialinitial energy state. energy state.

22 HH22 (g) + O (g) + O22 (g) (g) 2 H2 H22O (l)O (l) H < 0H < 0

2 H2 H22O (l)O (l)

2 H2 H2 2 (g)(g) + O + O22 (g) (g)

H = -H = - Water is lower in enthalpy thanWater is lower in enthalpy thanhydrogen and oxygen are. hydrogen and oxygen are. ExothermicExothermic

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Examples ofExamples of EndothermicEndothermic Changes -Changes -

HH22O (l) O (l) HH22O (g)O (g) H > 0H > 0

HH22O (g)O (g)

HH22O (l)O (l)

H = +H = +

This is This is EndothermicEndothermic because the because the final state is higher in enthalpy thanfinal state is higher in enthalpy thanthe initial state.the initial state.

CaCOCaCO33 (s) (s) CaO (s) + CO CaO (s) + CO22 (g) (g) H > 0H > 0

CaO (s) and COCaO (s) and CO22 (g) are higher in (g) are higher in

enthalpy than CaCOenthalpy than CaCO33 (s). (s).

CaO (s) + COCaO (s) + CO22 (g) (g)

CaCOCaCO33 (s) (s)H = +H = +

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Class Exercises - Class Exercises - Endothermic or Exothermic ?Endothermic or Exothermic ?

3 H3 H2 2 (g) + N(g) + N22 (g) (g) 2 NH2 NH33 (g) (g)

H = - 46.1 kJ/molH = - 46.1 kJ/mol

NN22 (g) + 2 O (g) + 2 O22 (g) (g) 2 NO2 NO22 (g) (g)

H = + 33.2 kJ/molH = + 33.2 kJ/mol

CC33HH88 (g) + 5 O (g) + 5 O22 (g) (g) 3 CO3 CO22 (g) + (g) + 4 H4 H22O (g)O (g)

Sign of Sign of H ??H ??NegativeNegative

NHNH44Cl dissolves in water with a Cl dissolves in water with a decreasedecrease in in

temperature.temperature. Sign of Sign of H ??H ??PositivePositive

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Thermochemical Equations - Thermochemical Equations - A chemical equa-A chemical equa-tion that includes an enthalpy change explicitly.tion that includes an enthalpy change explicitly.

2 H2 H22 (g, 1 atm) + O (g, 1 atm) + O22 (g, 1 atm) (g, 1 atm) 2 H2 H22O (l)O (l)

HH298 K298 K = - 571.7 kJ = - 571.7 kJ

A. A. HH subscript indicates temperature of rxn. subscript indicates temperature of rxn.

B. B. H H represents thermal energy evolved whenrepresents thermal energy evolved when2 moles2 moles of H of H22O (l) are formed!O (l) are formed!

HH22 (g, 1 atm) + 1/2 O (g, 1 atm) + 1/2 O22 (g, 1 atm) (g, 1 atm) HH22O (l)O (l)

HH298 K298 K = - 285.8 kJ = - 285.8 kJ

C. The sign of C. The sign of H H is reversed when the equationis reversed when the equationis reversed!is reversed!

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HH22O (l) O (l) H H22 (g, 1 atm) + 1/2 O (g, 1 atm) + 1/2 O22 (g, 1 atm) (g, 1 atm)

HH298 K298 K = + 285.8 kJ = + 285.8 kJ

D. The enthalpy change during a reaction can be D. The enthalpy change during a reaction can be

considered as a reactant (considered as a reactant (endothermicendothermic) or as ) or as

a product (a product (exothermicexothermic).).

285.8 kJ + H285.8 kJ + H22O (l) O (l) H H22 (g, 1 atm) + 1/2 O (g, 1 atm) + 1/2 O22 (g, 1 atm) (g, 1 atm)

HH22 (g, 1 atm) + 1/2 O (g, 1 atm) + 1/2 O22 (g, 1 atm) (g, 1 atm) H H22O (l) + 285.8 kJO (l) + 285.8 kJ

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Practice Problem:Practice Problem: When 2.0 moles of isooctane are When 2.0 moles of isooctane are burned, 10 930.9 kJ of thermal energy are liberated under burned, 10 930.9 kJ of thermal energy are liberated under constant temperature conditions.constant temperature conditions. How many kJ will be How many kJ will be liberated when 369 g of isooctane are burned?liberated when 369 g of isooctane are burned?

Problem:Problem: 369 g369 gEquation: Equation: 2 C2 C88HH1818 (l) + 25 O (l) + 25 O22 (g) (g) 16 CO 16 CO22 (g) + 18 H (g) + 18 H22O (l)O (l)

H = -10 930.9 kJH = -10 930.9 kJMolar Masses: Molar Masses: 114.231 114.231 ? kJ? kJ

““When in doubt, calculate MOLES!”When in doubt, calculate MOLES!”

(369 g C(369 g C88HH1818)(1 mol C)(1 mol C88HH1818//114.231 g C114.231 g C88HH1818) =) =

3.233.2300 mol C mol C88HH1818

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(3.23(3.2300 mol C mol C88HH1818)(-10 930.9 kJ/2 mol C)(-10 930.9 kJ/2 mol C88HH1818) )

== - 1.77 x 10- 1.77 x 1044 kJ kJ

Practice Problem:Practice Problem: The The HHrxnrxn for the burning of H for the burning of H22 (g) to (g) to

form Hform H22O (l) is –285.83 kJ/mol HO (l) is –285.83 kJ/mol H22O. How many grams of O. How many grams of

HH22 (g) are needed to produce 539.63 kJ of thermal (g) are needed to produce 539.63 kJ of thermal

energy?energy?

Problem:Problem: ? g ? g - 539.63 kJ - 539.63 kJ

Equation: Equation: HH22 (g) + ½ O (g) + ½ O22 (g) (g) H H22O (l) O (l) H = -285.83 kJH = -285.83 kJ

Molar Masses: 2.015 88 uMolar Masses: 2.015 88 u

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(- 539.63 kJ)(1 mol H(- 539.63 kJ)(1 mol H22/-285.83 kJ) = 1.887 9/-285.83 kJ) = 1.887 944 mol H mol H22

(1.887 9(1.887 944 mol H mol H22)(2. )(2. 015 88 g H015 88 g H22/mol H/mol H22) = 3.8059 g H) = 3.8059 g H22

MOLESMOLESMASSMASS PARTICLESPARTICLES

VOLUMEVOLUME P, V, TP, V, T

MolarMolar

MassMass

Avogadro’sAvogadro’s

NumberNumber

MolarityMolarity PV = nRTPV = nRT

Thermal EnergyThermal Energy

Molar HeatMolar Heat

of Reactionof Reaction

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Standard Enthalpy Changes - Standard Enthalpy Changes - A standard A standard enthalpy change, enthalpy change, HH00, is the enthalpy change , is the enthalpy change for a reaction in which each reactant and each for a reaction in which each reactant and each product is in its product is in its “Standard State”.“Standard State”.

- Solids - Pure solid at 1 atmSolids - Pure solid at 1 atm

- Liquids – Pure liquid at 1 atmLiquids – Pure liquid at 1 atm

- Gas – Ideal gas at 1 atm partial pressureGas – Ideal gas at 1 atm partial pressure

- Solute – Ideal solution at 1 M conc.Solute – Ideal solution at 1 M conc.

HH22 (g, 0.5 atm) + Cl (g, 0.5 atm) + Cl22 (g, 0.5 atm) (g, 0.5 atm) 2 HCl (g, 1 atm) 2 HCl (g, 1 atm)

HHrxnrxn, NOT , NOT HHoorxnrxn

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Standard Enthalpy of Formation - Standard Enthalpy of Formation - ff

The enthalpy change accompanying the formation The enthalpy change accompanying the formation of one mole of a substance in its standard state of one mole of a substance in its standard state from its elements, each in their standard states and from its elements, each in their standard states and most stable form.most stable form.Hg (l) + ClHg (l) + Cl22 (g) (g) HgCl HgCl22 (s) (s) ff

oo = -221.3 kJ = -221.3 kJ

S (s) + OS (s) + O22 (g) (g) SO SO22 (g) (g) ffoo = -296.8 kJ = -296.8 kJ

HH22 (g) (g) H H22 (g) (g) HHffoo = 0 kJ = 0 kJ

Elements in their standard states have zero Elements in their standard states have zero enthalpies of formation!enthalpies of formation!

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Standard Enthalpy of Combustion - Standard Enthalpy of Combustion - combcombThe enthalpy change accompanying the The enthalpy change accompanying the

combustion of one mole of a substance in Ocombustion of one mole of a substance in O22 in its in its

standard state from its elements, each in their standard state from its elements, each in their standard states and most stable form.standard states and most stable form.

HH22 (g, 1 atm) + ½ O (g, 1 atm) + ½ O22 (g, 1 atm) (g, 1 atm) H H22O (l, 1 atm) O (l, 1 atm)

HHcombcomboo = -285.8 kJ = -285.8 kJ

CC88HH1818 (l) + (l) + 25/225/2 O O22 (g) (g) 8 CO 8 CO22 (g) + 9 H (g) + 9 H22O (l)O (l)

HHcombcomboo = -5455.6 kJ = -5455.6 kJ

CC22HH55OH (l) + 7/2 OOH (l) + 7/2 O22 (g) (g) 2 CO 2 CO22 (g) = 3 H (g) = 3 H22O (l)O (l)

HHcombcomboo = -1366.8 kJ = -1366.8 kJ