Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat...

Chapt. 15 – Energy & Chemical Change

15.1 Energy (& Modes of Heat Transfer - NIB)

15.2 Heat

15.3 Thermochemical Equations

15.4 Calculating Enthalpy Change

15.5 Reaction Spontaneity

Section 15.1 Energy

• Define energy.

• Distinguish between potential and kinetic energy.

• Relate chemical potential energy to the heat lost or gained in chemical reactions.

• Calculate the amount of heat absorbed or released by a substance as its temperature changes.

• Describe the three ways that heat can be transferred.

Energy can change form and flow, but it is always conserved.

Key Concepts• Energy is the capacity to do work or produce heat.

• Chemical potential energy is energy stored in the chemical bonds of a substance by virtue of the arrangement of the atoms and molecules.

• Chemical potential energy is released or absorbed as heat during chemical processes or reactions.

• The amount of heat required to change the temperature of a given mass of a substance can be computed using

q = c × m × ∆T

• Heat can be transferred through convection, conduction or radiation.

Section 15.1 Energy

Nature of EnergyEnergy: Ability to do work or produce heat

Two forms

Potential• Due to composition or

position of material or object• Water stored behind dam

Kinetic• Energy of motion: ½ m v2

• Water falling over dam

Conservation of Energy

Water flowing over dam• Potential energy converted to kinetic energy

Wood burning• Chemical potential energy converted to heat

Law of conservation of energy

In any chemical reaction or physical process, energy can be converted from one form to another, but is neither created or destroyed.

Conservation of EnergyWater flowing through turbine: potential energy converted to kinetic energy which is then converted into electrical energy

Potential Energy (PE)Energy stored by virtue of positionGravitational• Higher up object is, higher the PE• Plays very little role in chem. processes

Electrical• Two + charges close together - high PE• Ions involved in storing electrical PE

Magnetic• Two N poles close together - high PE

PE in Chemical Systems

Chemical Potential energy – Energy stored in a substance because of its composition

• Types of atoms• Number, type (strength) of bonds• Atomic/molecular arrangement in space

Chemical potential energy converted to heat in many reactions (e.g., combustion)

Heat and TemperatureName of transfer process is heat

• Energy gets transferred• Heat NOT a substance

Temperature: property which • is directly proportional to KE of substance

under examination• determines direction heat will flow when two

objects brought into contact

Temperature and heat are not the same

Heat & Heat TransferHeat (q) – energy in the process of flowing from a warmer object to a cooler one (textbook definition)

• Raises T of cooler and lowers T of hotter

Transfer of KE from one medium or object to another, or from an energy source to a medium or object (alternate definition)

Heat transfer can occur in 3 ways: radiation, conduction, and convection

Conduction, Convection, & Radiation

Heat Transfer by RadiationHeat transferred by electromagnetic radiation – no contact

Any form: visible light, infrared, microwaves, radio waves, etc.

Examples – heat lamp for fast foods, sun warming Earth

Thermal ConductionProcess by which energy transferred as heat through material between two points at different temperatures (without material moving)

Metal rod conducting heat

Thermal Conduction

Atoms nearest higher temperature jostle less energetic neighbors and transfer energy in process

Rate of thermal conduction depends on material – thermal conductivity (k)

Heat Conduction – Electric StoveAtoms in pan vibrate about their equilibrium positions

Those near stove coil vibrate with larger amplitudes

These collide with adjacent atoms and transfer some energy

Eventually, energy travels entirely through pan & its handle

Thermal Conduction

Materials giving high rates of heat transfer are thermal conductors• Metals

Those giving low rates are thermal insulators• Ceramics, anything that’s mostly air

(cork, fiberglass, etc)

Best – metals

Middle – nonmetals

Lowest - Gases

Dependence of Thermal Conductivity on Material

Range of Thermal Conductivity of Various Materials at Room Temperature

Styrofoam – Good Thermal

Insulator

Conduction

When metal block and wooden block (both at same temperature) picked up, metal block feels cooler due to faster conduction of heat away from hand

Convection

Energy literally carried by fluid• Forced hot air heating• Antifreeze cooling of engine block

Convection

Heat transport due to movement of fluid (gas or liquid)Natural convection occurs due to circulation driven by differing densities - caused by uneven heatingWarm air expands and rises

Convection Currents in Boiling Water

Convection – Room Heating

Natural vs Forced Convection

Sometimes natural convection inadequate or inappropriate

Use forced convection with pump or fan

Thermos BottleMinimizes convection, conduction, and radiation heat transfer

Double-walled glass vessel with evacuated space between walls minimizes energy losses due to conduction and convection

Silvered surfaces reflect most radiant energy that would otherwise enter or leave liquid in thermos

Cooling Coffee

What modes of heat transfer involved?Conduction: (through cup walls)Convection: in coffee and in airRadiation:from all exterior surfaces (IR)

Heat PipeA Heat absorbed in evaporator section

B Liquid boils to produce vapor

C Heat released to environment as vapor condenses

D Liquid returns by wicking or gravity

Heat Pipe Characteristics

Able to transport large amounts of heat across very small temperature differences•

Thermal “super conductors”• 1000 x or more effective than solid Cu

Energy, Heat, and WorkKE and PE (including chemical PE) can do work and be converted to heat

Energy, heat, work: have same units• SI unit – joule (J)• Energy expended when force of one newton

is applied over displacement of one meter

Metric unit (not SI) of heat – calorie (cal)• 1 cal = 4.184 J (exactly, by definition)• 1 nutritional Calorie = 1 kcal = 4184 J• Must be able to convert between units

Practice

Conversion of energy units

Problems 1- 3 page 519

Problems 62 - 66 page 552


The calorie (cal) & Specific Heat

One calorie = 4.184 J

(exactly, by definition)

calorie – approximately equal to amount of heat needed to raise T of 1 g pure H2O 1 C

Quantity 4.184 J/(gC) is the specific heat of water – measure of ability to absorb heat

Specific Heat

Quantity 4.184 J/(gC) is the specific heat of water – measure of ability to absorb heat

Alternate term: specific heat capacity

Specific Heat

Each substance has its own specific heat

Symbol c used for specific heat (specific heat capacity)

c depends on T, but generally very weakly• Water at 0 C – 4.218 J/(gC) • Water at 40 C – 4.179 J/(gC)

c depends strongly on phase of substance• c for water very different than for ice – see

following slide

Specific Heat and Liquid Water

Water has highest specific heat capacity of common liquids

Water’s ability to store and release large quantities of heat plays important role in many natural phenomena and engineering applications

• Large bodies of water moderate temperatures (absorbs sun during day)

• Water used for cooling systems Thermal conductivity also plays a role

Heat Release/Absorption Calculations

Heat (q) depends not only on specific heat (c) but also upon mass (m) of substance and the size of the temperature change (T)

q = c m Tm in grams, T = Tfinal – Tinitial in C

Heat Release/Absorption Calculation

Problem: Solar pond made of 14,500 kg of granite and contains 22,500 kg of water. T rises 22 C during daylight and decreases by same amount at night. q stored during day?

q = c m Tqwater = 4.184 J/(gC) 2.25x107 g 22 C

= 2.1x109 J

qgranite = 0.803 J/(gC) 1.45x107 g 22 C

= 2.6x108 J

Heat Release/Absorption Calculation

qtotal = qwater+ qgranite = 2.1x109 J + 0.26x109 J

= 2.4x109 J = 2.4x106 kJ = 2.4x103 MJ

= 2.4 GJ

2.4 GJ absorbed during day and released at night

Specific Heat Calculation

Example problem 15-2Temperature of sample of iron with mass of 10.0 g changed from 50.4 C to 25.0 C with release of 114 J of heat. Calculate the specific heat of iron.

ciron = q m T = 114 J 10.0 g 25.4 C = 0.449 J/(gC)

Practice

Specific heat


Problem 10 page 522

Problem 67 page 552




15.2 Heat




Section 15.2 Heat

• Describe how a calorimeter is used to measure energy that is absorbed or released

• Calculate the various quantities that are involved in a calorimetry experiment, especially the specific heat of an unknown substance

The enthalpy change for a reaction is the enthalpy of the products minus the enthalpy of the reactants.

Section 15.2 Heat

• Explain the meaning of enthalpy and enthalpy change in chemical reactions and processes and identify a reaction as endo- or exothermic.

• Calculate the heat for a process given an associated enthalpy change for that process and the amount of substance.

Key Concepts

• In thermochemistry, the universe is defined as the system plus the surroundings.

• The heat lost or gained by a system during a reaction or process carried out at constant pressure is called the change in enthalpy (∆H).

• When ∆H is positive, the reaction is endothermic. When ∆H is negative, the reaction is exothermic.

Section 15.2 Heat

Measuring HeatCalorimeterInsulated device for measuring heat absorbed/released during a chemical or physical process

Coffee cup calorimeter with

stirrer and thermometer

Using Calorimeter to Determine c

Measure Ti of 50.0 g H2O

22.0 C

Heat 150.0 g Pb to 100 C

Lead

Shot

Measure Tf of Pb + H2O

28.8 C

Determining c from Calorimeter

150.0 g Pb, Ti = 100 C, Tf = 28.8 C

50.0 g H2O, Ti = 22.0 C, Tf = 28.8 C

qlead = - qwater (heat loss = heat gain)

qwater = mH2O cH2O TH2O =

50.0 g 4.18 J/(g °C) 6.8 °C = 1.14 x103 J

clead = qlead / (mlead Tlead) =

-1.14 x103 J/(50.0 g -71.2 °C) = 0.13 J/(g°C)

Temperature and Heat

c Li = 3.58 J/(gC) c Pb = 0.128 J/(gC)

Have 1.00 g each of Li (lithium) & Pb (lead)

Li is at 20.0 C Pb at 80.0 CWhich metal is hotter (has higher temperature)?

Pb


c Li = 3.58 J/(gC) c Pb = 0.128 J/(gC)

Have 1.00 g each of Li (lithium) & Pb (lead)

Li @ 20.0 C Pb @ 80.0 CEach metal put into separate beakers containing 1.00 g of water at 10.0 C and allowed to equilibrate.

Which one raises the temperature of the water more (provides more heat)?


Note: the following 3 slides show a general algebraic solution to the question asked. This is done to illustrate this type of approach. However, the problem could have been solved by doing the calculation for each metal separately.


m = metal w = water i = initial f = final

Heat for metal: qm = mmcmDT DT = Tf – Tim

Heat for H2O: qw = mwcwDT DT = Tf – Tiw

Energy conserved: – qm = qw

– mmcm(Tf – Tim) = mwcw(Tf - Tiw)

mm = mw (both 1.00 g)

Tf –Tim = – (cw/cm)(Tf - Tiw)

Tf = (Tim + (cw/cm)Tiw) / (1 + (cw/cm))


m = metal w = water i = initial f = final

c Li = 3.58 J/(gC) c Pb = 0.128 J/(gC)

Li @ 20.0 C=Tim1 Pb @ 80.0 C=Tim2

cw = 4.184 J/(gC)

Tf = (Tim + (cw/cm)Tiw) / (1 + (cw/cm))

Li: cw/cm = 4.184/3.58 = 1.17

Pb: cw/cm = 4.184/0.128 = 32.7


c Li = 3.58 J/(gC) c Pb = 0.128 J/(gC)

Li @ 20.0 C=Tim1 Pb @ 80.0 C=Tim2

Tf = (Tim + (cw/cm)Tiw) / (1 + (cw/cm))Tf(Li) =(20.0 C + 1.1710.0 C) / 2.17 = 14.6 C

Tf(Pb) =(80.0 C + 32.710.0 C) / 33.7 = 12.1 C Li raises water temperature 4.6 C

Pb raises water temperature 2.1 C

Colder metal (Li) provides much more heat

Temperature & heat are different quantities

Practice

Using specific heat / calorimetry


Problems 74 - 78, page 552

Problem 6 page 986

Thermochemistry

Thermochemistry is the study of heat changes that accompany chemical reactions and phase changes

Chemical Energy and the Universe

System: Specific part of universe containing the reaction or process you wish to study

Surroundings: Everything else in the universe except the system

Universe = System + Surroundings


Heat pack – iron reacts with O2 in air

4Fe(s) + 3O2(g) 2Fe2O3(s) + 1625 kJ

System - Heat pack

Surroundings?

Hands, air, etc

Exothermic reaction• Heat flows from system to

the surroundings


Cold pack – dissolution of ammonium nitrate

27 kJ + NH4NO3(s) NH4+(aq) + NO3

-(aq)

System - Cold pack

Surroundings?

Knee, air, etc

Endothermic reaction• Heat flows from

surrounding to the system

Enthalpy & Enthalpy Changes

Enthalpy (H) is the heat content of a system at constant pressure• “Coffee cup” style calorimeter is open to

atmosphere and is at constant P• Many reactions of interest take place at

constant (atmospheric) pressure

Really want to know change in enthalpy for a chemical process, H

H for Process (reaction /phase change)

Hprocess = enthalpy change of process

= Hfinal - Hinitial

= Hfinal state – Hinitial state

Exothermic processes• Hfinal state < Hinitial state

• Hprocess < 0

Endothermic processes• Hfinal state > Hinitial state

• Hprocess > 0

Enthalpy & Enthalpy ChangesReactants

Reactants

Products

Products

Endothermic Exothermic

Ent

halp

y

Ent

halp

y

Enthalpy Change - Heat PackE

ntha

lpy

Exothermic Reaction H < 0

4Fe(s) + 3O2(g)

2Fe2O3(s)

H = - 1625 kJ

Enthalpy Change - Cold Pack

NH4+(aq) + NO3

-(aq)

NH4NO3(s)

Endothermic Reaction H > 0

H = +27 kJ

Ent

halp

y

Enthalpy Change and qq – heat gained or lost in a general chemical reaction or process

qp – as above but for reaction or process occurring at constant pressure

For constant pressure reactions:

qp = Hrxn

If H units are per mole, then

qp = n Hrxn n = # moles



15.2 Heat




Section 15.3 Thermochemical Equations

• Write thermochemical equations for chemical reactions and other processes.

• Describe how energy is lost or gained during changes of state and calculate this energy.

• Calculate the heat absorbed or released in a chemical reaction.

• Determine the enthalpy for a given phase change by using the additivity principle for thermochemical equations.

Thermochemical equations express the amount of heat released or absorbed by chemical reactions.

Key Concepts

• A thermochemical equation includes the physical states of the reactants and products and specifies the change in enthalpy.

• The molar enthalpy (heat) of vaporization, ∆Hvap, is the amount of energy required to evaporate one mole of a liquid.

• The molar enthalpy (heat) of fusion, ∆Hfus, is the amount of energy needed to melt one mole of a solid.

Section 15.3 Thermochemical Equations

Writing Thermochemical Equations

Thermochemical equation - balanced chemical equation that includes physical state of all reactants and products and the energy change expressed as HNature of reaction or process written as subscript for H

(Hvap for vaporization)

Writing Thermochemical Equations

For reactions/processes carried out under standard conditions (1 atm and 298 K), superscript 0 is used - H0

Note: “standard conditions” are not STP

Enthalpy Change and Combustion

Combustion reaction of glucoseC6H12O6(s) + 6O2(g)

6CO2(g) + 6H2O(l) H0comb = -2808 kJ

Standard enthalpy of combustion (H0comb) is

enthalpy change for complete burning of one mole of a substance at 1 atmosphere pressure and 298 K – standard conditions

H0comb values in table 15.3, page 529

Bomb Calorimeter (constant V)

Sample of known mass ignited by spark and burned in excess of O2. Heat released transferred to H2O in outer chamber.

Bomb Calorimeter (constant V)

Because they operate at constant V and not constant P, heat change measured in a bomb calorimeter is not in general equal to H.

For some reactions (those with same number of moles of gas on both sides of the equation) the pressure is nearly constant and so the results can be used as a measure of H.

Example Problem 15.4Combustion reaction of glucose C6H12O6(s) + 6O2(g) (excess high P O2)

6CO2(g) + 6H2O(l) H0comb = -2808 kJ

How much heat evolved when 54.0 g glucose is burned?

Key issue: H0comb is per mole glucose

54.0 g glucose 0.300 mol glucose

0.300 mol C6H12O6 2808 kJ/mol C6H12O6

= 842 kJ (fudging constant P & standard conditions)

Practice

Energy released in chemical reaction

Problem 25 page 532

Problem 29 page 533


Enthalpy Change & State ChangesVaporization, sublimation, melting all require energy (endothermic, H > 0)• In thermochemistry, melting process is

called fusion

Hvap= molar enthalpy (heat) of vaporization• Heat required to vaporize one mole of liquid

Hfus = molar enthalpy (heat) of fusion• Heat required to melt one mole of a solid

qprocess = n Hprocess (n=# moles)

Thermochemical Equations for State Changes

H2O(l) H2O(g) Hvap = 40.7 kJ

H2O(s) H2O(l) Hfus = 6.01 kJ

The reverse processes (condensation, freezing) release the same amount of energy as were absorbed in the above

Hvap = -Hcond

Hfus = -Hsolid

Temperature Change of Ice, Water, and Steam with Added Energy

Ice / WaterWater

Water/Steam Steam

Heat (x103 J)

Changes for heating 10.0 g ice

Relating q and DH

Example for simple phase change

Calculate energy released when 64.08 g of methanol (CH3OH) freezes

From table 15.4, page 530

DHfus = 3.22 kJ/mol

Freezing (solidification) is opposite process to melting (fusion)

DHsolid = - 3.22 kJ/mol (sign changed)

Relating q and DH

Calculate energy released (in J) when 64.08 g of methanol (CH3OH = MeOH) freezes

q = Hsolid n

n = # moles MeOH = 64.08 g MeOH 1 mol MeOH/32.04 g MeOH

= 2.000 mol MeOH

Relating q and DH

Calculate energy released (in J) when 64.08 g of methanol (CH3OH = MeOH) freezes

DHsolid = - 3.22 kJ/mol

n = 2.000 mol MeOH

q = DH x n = - 3.22 kJ/mol 2.000 mol 1x103 J/kJ = - 6.44x103 J

negative sign for energy release

Energy Requirements for Heating Substance with Phase Changes

Determine energy required to raise 225 g of water from 46.8 C to 173.0 C

225 g 1 mol H2O/18.02 g = 12.5 mol H2O

Have two phase changes and 3 states

cice, cwater, csteam – table 15.2, page 520

Enthalpies fusion, vaporization in table 15.4

Hfus = 6.01 kJ/mol

Hvap = 40.7 kJ/mol


m = 225 g n = 12.5 mol H2O

T1= 46.8 C T3= 100 C T5= 73.0 C

q1 = cice m T1

q2 = Hfus n (heat required to melt ice)

q3 = cwater m T3

q4 = Hvap n (heat required to boil water)

q5 = csteam m T5


Total q = q1 + q2 + q3 + q4 + q5

= 21.4 + 75.0 + 94.1+ 508 + 33.0 kJ

= 732 kJ

Note that majority of energy was used to boil the water

State Changes: Sublimation, Deposition

DH for sublimation & deposition not tabulated on page 530

Figure shows:

DHsub = DHfus + DHvap

D

D

DD

DHcond

DHsolid

40.7 kJ

6.01 kJ

DHdep = DHcond + DHsolid DHdep = –DHsub

DHsub

DHdep

Adding Thermochemical Equations

Have two phase changes for water

H2O(l) H2O(g) Hvap = 40.7 kJ

H2O(s) H2O(l) Hfus = 6.01 kJ

Can add them together to getH2O(s) H2O(g) Hsubl = 46.7 kJ

The resulting H is for the phase change from a solid to a gas = sublimation

State Changes: Sublimation, Deposition

DH for sublimation & deposition not tabulated on page 530Figure shows:

DHsub = DHfus + DHvap

= 6.01 kJ + 40.7 kJ

DHsub = 46.7 kJ

D

D

DD

DHcond

DHsolid

40.7 kJ

6.01 kJ

DHdep= DHcond + DHsolid

DHdep= – DHsub

DHdep= – 46.7 kJ

DHsub

DHdep

Practice

Energy in changes of state

Problems 23 – 24 page 532

Problems 27, 28, 30 page 533

Problems 83, 84, 87, 88 page 553

Problems 7 – 8, page 986



15.2 Heat




Section 15.4 Calculating Enthalpy Change

• Apply Hess’s law to calculate the enthalpy change for a reaction.

• Explain the basis for the table of standard enthalpies of formation.

• Calculate ∆Hrxn using thermochemical equations.

• Determine the enthalpy change for a reaction using standard enthalpies of formation data.

The enthalpy change for a reaction can be calculated using Hess’s law.

Key Concepts

• The enthalpy change for a reaction can be calculated by adding two or more thermochemical equations and their enthalpy changes.

• Standard enthalpies of formation of compounds are determined relative to the assigned enthalpy of formation of the elements in their standard states.

• The standard enthalpy change for a reaction can be computed from the standard enthalpies of formation using

Section 15.4 Calculating Enthalpy Change

Calculating Enthalpy Change

Calorimetry cannot be used to measure H for some reactions, including:• Slow reactions

C(s, graphite) C(s, diamond)

occurs on time scale of 106 years• Conditions difficult to reproduce in

laboratory• Ones that produce side products

Hess’s Law

If you add 2 or more thermochemical equations to produce a final equation for a reaction, then the sum of the enthalpy changes for the individual reactions is the enthalpy change for the final reaction

Hfinal = Hindividual

= summation operation

Hess’s LawCombustion of sulfur to form sulfur trioxide

2S(s) + 3O2(g) 2SO3(g) H=?

In lab, mostly sulfur dioxide produced

S(s) + O2(g) SO2(g) H = -297 kJ

Need to use Hess’s Law to compute H

Rules for Applying Hess’s Law

1. If you multiply/divide the coefficients of the chemical equation by some factor, the corresponding H must be multiplied/divided by the same factor

2S(s) + 3O2(g) 2SO3(g) H = -792 kJ

S(s) + 3/2 O2(g) SO3(g) H = -396 kJ

2. If you reverse the direction of a reaction, the sign of H must also be reversed

SO3(g) S(s) + 3/2 O2(g) H = +396 kJ

Hess’s Law – Sulfur Trioxide

2S(s) + 3O2(g) 2SO3(g) H = ?

Reactions with known H a. S(s) + O2(g) SO2(g) H = -297 kJ b. 2SO3(g) 2SO2(g) + O2(g) H = 198 kJ

Multiply reaction a by 2; reverse reaction b

c. 2S(s) + 2O2(g) 2SO2(g) H = -594 kJd. 2SO2(g) + O2(g) 2SO3(g) H = -198 kJ

Add c and d, canceling common terms2S(s) + 3O2(g) 2SO3(g) H = -792 kJ

Hess’s Law – Sulfur TrioxideE

NT

HA

LPY

H = -198 kJ

Overall enthalpy change

H = -792 kJ

H = -594 kJ

2S(s) + 2O2(g)

2SO2(g) 2SO2(g) + O2(g)

2SO3(g)

Hess’s Law – Sulfur Trioxide

2S(s) + 3O2(g) 2SO3(g) H = -792 kJ

Often written on per mole of product basis

S(s) + 3/2 O2(g) SO3(g) H = -396 kJ

Reactions with known H a. 2H2(g) + O2(g) 2H2O (l) H = -572 kJ b. H2(g) + O2(g) H2O2(l) H = -188 kJReverse reaction b and multiply by 2a. 2H2(g) + O2(g) 2H2O (l) H = -572 kJ c. 2H2O2(l) 2H2(g) + 2O2(g) H = 376 kJ

Hess’s Law – H2O2 Decomposition

2H2O2(l) 2H2O(l) + O2(g) H = ?

2H2O2(l) 2H2O(l) + O2(g) H = -196 kJ

Add a and c, canceling common terms

Practice

Hess’s Law


Problem 42 page 541



Standard Enthalpy of Formation

H0f = change in enthalpy that accompanies

the formation of one mole of the compound in its standard state from its constituent elements in their standard statesAKA Standard Heat of FormationStandard state means the normal physical state of the substance at 1 atm & 298 K

• Fe(s)• Hg(l)• O2(g)


Examples:

Formation of 1 mole SO3 from its elements

S(s) + 3/2 O2(g) SO3(g) H0f = - 396 kJ

Formation of 1 mole CO2 from its elements

C(s,graphite) + O2(g) CO2(g) H0f = -394 kJ

Note: the standard state of carbon has been assigned to be the graphite allotrope

Table 15.5, page 538 has examples


Arbitrary (but key) standard

Every free element in its standard state is assigned H0

f = 0.0 kJ (exactly)

• Fe(s), C(s,graphite), S(s), Al(s)• H2(g), O2(g), N2(g)

Using Std. Enthalpies of Formation

Key data to compute H0rxn using Hess’s Law

H2S(g) +4F2(g) 2HF(g) + SF6(g) H0rxn ?

Use a H0f for every species in the equation

that is not an element in its standard state

a. ½ H2(g) + ½ F2(g) HF(g) H0f = -273 kJ

b. S(s) + 3F2(g) SF6(g) H0f = -1220 kJ

c. H2(g) + S(s) H2S(g) H0f = -21 kJ

Multiply eqn. a by 2 and reverse eqn. c


H2S(g) +4F2(g) 2HF(g) + SF6(g) H0rxn ?

d. H2(g) + F2(g) 2HF(g) H0f = -546 kJ

b. S(s) + 3F2(g) SF6(g) H0f = -1220 kJ

f. H2S(g) H2(g) + S(s) H0f = 21 kJ

Add and cancel terms common to both sides

H2S(g) + 4F2(g) 2HF(g) + SF6(g)

H0rxn= -1745 kJ


H0rxn can be obtained from H0

f in a simpler way than the method just illustrated by using H0

rxn = H0f(products)

- H0f(reactants)

H2S(g) +4F2(g) 2HF(g) + SF6(g)H0

rxn = [2 H0f(HF) + H0

f(SF6)] - [H0

f(H2S) + 4 H0f(F2)]

Must use proper coefficients in formula

CH4(g) +2O2(g) CO2(g) + 2H2O(g)

H0rxn = H0

f(products)

- H0f(reactants)

H0rxn = [H0

f(CO2) + 2 H0f(H2O)]

- [H0f(CH4) + 2 H0

f(O2)]Note that O2(g) is element in standard state, H0

f = 0; rest are from tableH0

rxn = [(-394) + 2(-286)]-[(-75) + 0] kJ = -891 kJ


Practice

Enthalpy Change from Standard Enthalpies of Formation

Problems 34 – 37, 41, page 541

Problem 92, page 553

Problems 11-12, page 987



15.2 Heat




Section 15.5 Reaction Spontaneity

• Differentiate between spontaneous and nonspontaneous processes.

• Explain the meaning of entropy and of the Gibb’s free energy.

• State the second law of thermodynamics.

Changes in enthalpy and entropy determine whether a process is spontaneous.


• Predict the sign of the entropy change for various processes.

• Explain how changes in entropy and free energy determine the spontaneity of chemical reactions and other processes and determine if a reaction is spontaneous by calculating the free energy at a given temperature.

• Predict the spontaneity of certain reactions at low and high temperature extremes from information about the associated enthalpy and entropy changes.

Key Concepts

• Entropy is a measure of the disorder or randomness of a system.

• Spontaneous processes always result in an increase in the entropy of the universe.

• Free energy is the energy available to do work. The sign of the free energy change indicates whether the reaction is spontaneous.

• The following equation relates the system free energy change to the changes in enthalpy and entropy.

∆Gsystem = ∆Hsystem – T∆Ssystem


Spontaneous Process (SP)SP = physical or chemical change that once begun, occurs with (no outside intervention)** except perhaps for some small amount of energy needed to get process started• Spark to light Bunsen burner flame

Formation of rust is spontaneous

4Fe(s) + 3O2(g) 2Fe2O3(s) H = -1625 kJ

Spontaneous Process (SP)

Ball rolls downhill. Never spontaneously rolls uphill.

A gas fills a container uniformly. Never spontaneously collects at one end.

Heat flow always occurs from hot object to a cooler one. Reverse never spontaneously occurs.

Spontaneous Process (SP)

Wood burns spontaneously in an exothermic reaction to form CO2 and H2O

Wood not formed when CO2 and H2O mixed together.

Spontaneous Processes and H

Many but not all endothermic reactions are nonspontaneous

H2O(s) H2O(l) H = 6.01 kJ

Spontaneous at T 0 CExothermic or endothermic nature of reaction not the sole determinant of reaction spontaneity

Entropy (S) must also be considered

Entropy

Measure of disorder or randomness of particles that make up a system

Measure of possible ways that energy of system can be distributed; related to freedom of system’s particles to move and number of ways they can be arranged

Expansion of a Gas into an Evacuated Bulb

Gas Vacuum

#1

#2

#3

Three Possible Arrangements (states) of Four Molecules in a Two-Bulbed Flask#1 abcd : 0 1 way#2 bcd : a 4 ways acd : b abd : c abc : d#3 ab : cd x2 6 ways ac : bd x2 ad : bc x2

Probability of All Molecules Being in Left Hand Bulb of a Two-Bulbed Flask

# Molecules Relative Probability

1 1/2

2 1/22 =1/4

3 1/23 = 1/8

100 1/2100 = 7.9x10-31

6.0x1023 0

Entropy

Low Higher

Second Law of Thermodynamics

2d Law = Spontaneous processes always proceed in such a way that the entropy of the universe (Suniv) increases

DSuniv > 0 for any spontaneous process

Universe = System + Surroundings


2d Law = Spontaneous processes always proceed in such a way that the entropy of the universe (Suniv) increases

Equivalent statement:

Nature spontaneously proceeds towards the states that have the highest probability of existing.

2nd Law of Thermodynamics

No cyclic process that converts heat entirely into work is possible (engine)

No cyclic process can transfer energy as heat from a low-T body to a high-T body without work being done (fridge)

2nd Law of Thermodynamics

A cyclic process must transfer heat from a hot to cold reservoir if it is to convert heat into work

Work must be done to transfer heat from a cold to a hot reservoir

A useful perpetual motion machine does not exist


Conservation of energy tells us that total amount of energy in the universe is constant• Strictly speaking: mass + energy

Second law of thermodynamics says that the entropy of the universe is constantly increasing

Predicting Entropy Change, S

Entropy changes associated with changes in state, with creating certain types of solutions, with reactions of gases, and with increasing the temperature can be predicted

Predicting S

Ssystem = Sproducts – Sreactants

1.State Changes

Entropy increases as a substance changes from a solid to a liquid to a gas

• Increased mobility allows more states (positions of molecules) to be accessible

S H2O(s) < S H2O(l) << S H2O(g)

Increase in entropy from solid to liquid to gas

Large changes

in S occurat phase

transitions

Predicting S


2. Dissolving gas in a liquid

Entropy decreases – gas molecules more limited in movements

CO2(g) CO2(aq) Ssystem < 0

Gas dissolving in liquid: Ssystem < 0

O2 gas

O2 dissolved

O2 molecule confined

by solvent “cage”

Predicting S


3. Reactions with gaseous reactants and products

Entropy increases if # of product particles > # of reactant particles

2SO3(g) 2SO2(g) + O2(g) Ssystem>0

(2 mol of gas generates 3 mol of gas)

Predicting S


4. Solids/liquids dissolving to form solutions

Entropy generally increases

NaCl(s) Na+(aq) + Cl-(aq) Ssystem>0

(S of water decreases slightly due to ordering of H2O around hydrated ions)

NaCl(s) + H2O(l) Na+(aq) + Cl-(aq)

Ssystem>0

Small Increase in S When Ethanol Dissolves in Water

Freedom of movement remains ~ unchanged; S increase due solely to random mixing

ethanol water solution of ethanol and

water

Predicting S


5. Temperature increase

T increase KE increase

Ssystem > 0 (increased disorder)

Practice

Predicting sign of DS

Problems 44 (a-d), 45, page 545



DS, the Universe, Spontaneity

Suniverse > 0 for spontaneous process

Suniverse = Ssystem + Ssurroundings

Suniverse tends to be positive when

a. Hsystem<0 • Heat released by exothermic process

raises T of surroundings and makes Ssurroundings > 0

b. Ssystem > 0

Exothermic reactions with Ssystem >0 always spontaneous

DS, the Universe, and Free Energy

Gibbs Free Energy G• Combines enthalpy and entropy• Commonly called Free Energy• For processes that occur at constant

P and T, G is energy available to do work

Gsystem= Hsystem - T Ssystem T in K

For standard conditions (1 atm, 298 K)

G0system= H0

system - T S0system

Free Energy and SpontaneityGsystem = Hsystem - T Ssystem

Sign of Gsystem indicates if reaction is spontaneous at specified T and P

Allows information about system only to predict entropy change of universe

Reaction/process Gsystem Suniverse

Spontaneous < 0 > 0

Nonspontaneous > 0 < 0

Free Energy

G used in equilibrium (chapter 17) concepts in higher level courses

Calculating Free Energy Change

N2(g) + 3H2(g) 2NH3(g)

H0system= - 91.8 kJ S0

system= -197 J/KGaseous reaction – system S decreases because # moles gas decreases

G0system determines spontaneity

G0system= H0

system - T S0system

= -9.18x104 J – 298 K (-197 J/K)= -9.18x104 J + 5.87x104 J = -3.31x104 J

G0system< 0 so reaction is spontaneous

Reaction Spontaneity and the Signs of DHo, DSo and DGo (= DHo – T DSo)

DHo

DSo -TDSo DGo Reaction or process

- + - - Spont. all T

+ - + + Nonspont. all T

+ + - + or -

Spont. at higher TNonspont. at lower T

- - + + or -

Spont. at lower TNonspont. at higher T

All values for system See also table 15.6, page 547

Practice

Determine Reaction Spontaneity

Problems 46 (a-c), 47, 51, page 548

Problems 95, 97 (a-c), 98 - 103 page 554


Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat...

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