Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat...
-
Upload
eleanore-nash -
Category
Documents
-
view
213 -
download
0
Transcript of Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat...
![Page 1: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/1.jpg)
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
![Page 2: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/2.jpg)
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.
![Page 3: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/3.jpg)
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
![Page 4: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/4.jpg)
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
![Page 5: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/5.jpg)
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.
![Page 6: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/6.jpg)
Conservation of EnergyWater flowing through turbine: potential energy converted to kinetic energy which is then converted into electrical energy
![Page 7: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/7.jpg)
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
![Page 8: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/8.jpg)
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)
![Page 9: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/9.jpg)
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
![Page 10: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/10.jpg)
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
![Page 11: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/11.jpg)
Conduction, Convection, & Radiation
![Page 12: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/12.jpg)
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
![Page 13: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/13.jpg)
Thermal ConductionProcess by which energy transferred as heat through material between two points at different temperatures (without material moving)
Metal rod conducting heat
![Page 14: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/14.jpg)
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)
![Page 15: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/15.jpg)
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
![Page 16: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/16.jpg)
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)
![Page 17: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/17.jpg)
Best – metals
Middle – nonmetals
Lowest - Gases
Dependence of Thermal Conductivity on Material
![Page 18: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/18.jpg)
Range of Thermal Conductivity of Various Materials at Room Temperature
![Page 19: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/19.jpg)
Styrofoam – Good Thermal
Insulator
![Page 20: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/20.jpg)
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
![Page 21: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/21.jpg)
Convection
Energy literally carried by fluid• Forced hot air heating• Antifreeze cooling of engine block
![Page 22: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/22.jpg)
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
![Page 23: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/23.jpg)
Convection Currents in Boiling Water
![Page 24: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/24.jpg)
Convection – Room Heating
![Page 25: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/25.jpg)
Natural vs Forced Convection
Sometimes natural convection inadequate or inappropriate
Use forced convection with pump or fan
![Page 26: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/26.jpg)
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
![Page 27: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/27.jpg)
Cooling Coffee
What modes of heat transfer involved?Conduction: (through cup walls)Convection: in coffee and in airRadiation:from all exterior surfaces (IR)
![Page 28: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/28.jpg)
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
![Page 29: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/29.jpg)
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
![Page 30: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/30.jpg)
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
![Page 31: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/31.jpg)
Practice
Conversion of energy units
Problems 1- 3 page 519
Problems 62 - 66 page 552
Problems 1 - 2 page 986
![Page 32: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/32.jpg)
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
![Page 33: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/33.jpg)
Specific Heat
Quantity 4.184 J/(gC) is the specific heat of water – measure of ability to absorb heat
Alternate term: specific heat capacity
![Page 34: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/34.jpg)
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
![Page 35: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/35.jpg)
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
![Page 36: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/36.jpg)
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
![Page 37: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/37.jpg)
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
![Page 38: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/38.jpg)
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
![Page 39: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/39.jpg)
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)
![Page 40: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/40.jpg)
Practice
Specific heat
Problems 4 - 6 page 521
Problem 10 page 522
Problem 67 page 552
Problems 3 - 5 page 986
![Page 41: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/41.jpg)
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
![Page 42: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/42.jpg)
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.
![Page 43: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/43.jpg)
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.
![Page 44: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/44.jpg)
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
![Page 45: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/45.jpg)
Measuring HeatCalorimeterInsulated device for measuring heat absorbed/released during a chemical or physical process
Coffee cup calorimeter with
stirrer and thermometer
![Page 46: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/46.jpg)
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
![Page 47: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/47.jpg)
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)
![Page 48: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/48.jpg)
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
![Page 49: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/49.jpg)
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)?
![Page 50: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/50.jpg)
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.
![Page 51: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/51.jpg)
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))
![Page 52: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/52.jpg)
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
![Page 53: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/53.jpg)
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
![Page 54: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/54.jpg)
Practice
Using specific heat / calorimetry
Problems 12 - 15 page 525
Problems 74 - 78, page 552
Problem 6 page 986
![Page 55: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/55.jpg)
Thermochemistry
Thermochemistry is the study of heat changes that accompany chemical reactions and phase changes
![Page 56: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/56.jpg)
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
![Page 57: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/57.jpg)
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
![Page 58: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/58.jpg)
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
![Page 59: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/59.jpg)
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
![Page 60: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/60.jpg)
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
![Page 61: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/61.jpg)
Enthalpy & Enthalpy ChangesReactants
Reactants
Products
Products
Endothermic Exothermic
Ent
halp
y
Ent
halp
y
![Page 62: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/62.jpg)
Enthalpy Change - Heat PackE
ntha
lpy
Exothermic Reaction H < 0
4Fe(s) + 3O2(g)
2Fe2O3(s)
H = - 1625 kJ
![Page 63: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/63.jpg)
Enthalpy Change - Cold Pack
NH4+(aq) + NO3
-(aq)
NH4NO3(s)
Endothermic Reaction H > 0
H = +27 kJ
Ent
halp
y
![Page 64: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/64.jpg)
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
![Page 65: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/65.jpg)
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
![Page 66: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/66.jpg)
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.
![Page 67: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/67.jpg)
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
![Page 68: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/68.jpg)
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)
![Page 69: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/69.jpg)
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
![Page 70: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/70.jpg)
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
![Page 71: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/71.jpg)
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.
![Page 72: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/72.jpg)
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.
![Page 73: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/73.jpg)
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)
![Page 74: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/74.jpg)
Practice
Energy released in chemical reaction
Problem 25 page 532
Problem 29 page 533
Problems 85 - 86, page 553
![Page 75: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/75.jpg)
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)
![Page 76: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/76.jpg)
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
![Page 77: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/77.jpg)
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
![Page 78: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/78.jpg)
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)
![Page 79: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/79.jpg)
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
![Page 80: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/80.jpg)
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
![Page 81: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/81.jpg)
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
![Page 82: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/82.jpg)
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
![Page 83: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/83.jpg)
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
![Page 84: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/84.jpg)
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
![Page 85: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/85.jpg)
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
![Page 86: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/86.jpg)
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
![Page 87: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/87.jpg)
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
![Page 88: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/88.jpg)
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
![Page 89: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/89.jpg)
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.
![Page 90: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/90.jpg)
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
![Page 91: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/91.jpg)
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
![Page 92: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/92.jpg)
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
![Page 93: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/93.jpg)
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
![Page 94: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/94.jpg)
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
![Page 95: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/95.jpg)
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
![Page 96: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/96.jpg)
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)
![Page 97: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/97.jpg)
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
![Page 98: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/98.jpg)
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
![Page 99: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/99.jpg)
Practice
Hess’s Law
Problems 32 - 33, page 537
Problem 42 page 541
Problems 93 - 94, page 553
Problems 9 - 10, page 987
![Page 100: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/100.jpg)
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)
![Page 101: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/101.jpg)
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
![Page 102: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/102.jpg)
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)
![Page 103: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/103.jpg)
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
![Page 104: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/104.jpg)
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
![Page 105: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/105.jpg)
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
![Page 106: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/106.jpg)
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
![Page 107: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/107.jpg)
Practice
Enthalpy Change from Standard Enthalpies of Formation
Problems 34 – 37, 41, page 541
Problem 92, page 553
Problems 11-12, page 987
![Page 108: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/108.jpg)
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
![Page 109: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/109.jpg)
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.
![Page 110: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/110.jpg)
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.
![Page 111: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/111.jpg)
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
![Page 112: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/112.jpg)
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
![Page 113: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/113.jpg)
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.
![Page 114: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/114.jpg)
Spontaneous Process (SP)
Wood burns spontaneously in an exothermic reaction to form CO2 and H2O
Wood not formed when CO2 and H2O mixed together.
![Page 115: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/115.jpg)
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
![Page 116: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/116.jpg)
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
![Page 117: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/117.jpg)
Expansion of a Gas into an Evacuated Bulb
Gas Vacuum
![Page 118: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/118.jpg)
#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
![Page 119: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/119.jpg)
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
![Page 120: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/120.jpg)
Entropy
Low Higher
![Page 121: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/121.jpg)
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
![Page 122: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/122.jpg)
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.
![Page 123: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/123.jpg)
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)
![Page 124: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/124.jpg)
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
![Page 125: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/125.jpg)
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
![Page 126: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/126.jpg)
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
![Page 127: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/127.jpg)
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
![Page 128: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/128.jpg)
S H2O(s) < S H2O(l) << S H2O(g)
![Page 129: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/129.jpg)
Increase in entropy from solid to liquid to gas
Large changes
in S occurat phase
transitions
![Page 130: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/130.jpg)
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
![Page 131: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/131.jpg)
Gas dissolving in liquid: Ssystem < 0
O2 gas
O2 dissolved
O2 molecule confined
by solvent “cage”
![Page 132: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/132.jpg)
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)
![Page 133: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/133.jpg)
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)
![Page 134: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/134.jpg)
NaCl(s) + H2O(l) Na+(aq) + Cl-(aq)
Ssystem>0
![Page 135: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/135.jpg)
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
![Page 136: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/136.jpg)
Predicting S
Ssystem = Sproducts – Sreactants
5. Temperature increase
T increase KE increase
Ssystem > 0 (increased disorder)
![Page 137: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/137.jpg)
Practice
Predicting sign of DS
Problems 44 (a-d), 45, page 545
Problems 96 - 97, page 554
Problems 13-14, page 987
![Page 138: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/138.jpg)
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
![Page 139: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/139.jpg)
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
![Page 140: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/140.jpg)
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
![Page 141: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/141.jpg)
Free Energy
G used in equilibrium (chapter 17) concepts in higher level courses
![Page 142: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/142.jpg)
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
![Page 143: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/143.jpg)
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
![Page 144: Chapt. 15 – Energy & Chemical Change 15.1Energy (& Modes of Heat Transfer - NIB) 15.2 Heat 15.3Thermochemical Equations 15.4Calculating Enthalpy Change.](https://reader037.fdocuments.us/reader037/viewer/2022110208/56649de95503460f94ae3add/html5/thumbnails/144.jpg)
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