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
Problems 1 - 2 page 986
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
Problems 4 - 6 page 521
Problem 10 page 522
Problem 67 page 552
Problems 3 - 5 page 986
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 15.2 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
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 @ 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)?
Temperature and 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.
Temperature and Heat
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))
Temperature and Heat
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
Temperature and Heat
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 12 - 15 page 525
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
Chemical Energy and the Universe
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
Chemical Energy and the Universe
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
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 15.3 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
Problems 85 - 86, page 553
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
Energy Requirements for Heating Substance with Phase Changes
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
Energy Requirements for Heating Substance with Phase Changes
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
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 15.4 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
Problems 32 - 33, page 537
Problem 42 page 541
Problems 93 - 94, page 553
Problems 9 - 10, page 987
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)
Standard Enthalpy of Formation
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
Standard Enthalpy of Formation
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
Using Std. Enthalpies of Formation
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
Using Std. Enthalpies of Formation
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
Using Std. Enthalpies of Formation
Practice
Enthalpy Change from Standard Enthalpies of Formation
Problems 34 – 37, 41, page 541
Problem 92, page 553
Problems 11-12, page 987
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 15.5 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.
Section 15.5 Reaction Spontaneity
• 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
Section 15.5 Reaction Spontaneity
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
Second Law of Thermodynamics
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
Second Law of Thermodynamics
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
Ssystem = Sproducts – Sreactants
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
Ssystem = Sproducts – Sreactants
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
Ssystem = Sproducts – Sreactants
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
Ssystem = Sproducts – Sreactants
5. Temperature increase
T increase KE increase
Ssystem > 0 (increased disorder)
Practice
Predicting sign of DS
Problems 44 (a-d), 45, page 545
Problems 96 - 97, page 554
Problems 13-14, page 987
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
Problems 15-16, page 987
Top Related