Chapter 61 Principles of Reactivity: Energy and Chemical Reactions Chapter 6.

Chapter 6 1

Principles of Reactivity:Principles of Reactivity: Energy and Chemical Energy and Chemical

ReactionsReactions

Chapter 6Chapter 6

Chapter 6 2

Energy: Some BasicsEnergy: Some Basics

From Physics:

Force – a kind of push or pull on an object.

Energy – the capacity to do work.

Work – force applied over a distancew = F d

Heat – energy transferred from a warmer object to a cooler object.

Chapter 6 3

Kinetic and Potential Energy

Kinetic Energy (Thermal Energy) – energy due to motion.

221 mvEk


Chapter 6 4

Kinetic and Potential Energy

Potential Energy (Stored Energy) – the energy an object possesses due to its position.

- Potential energy can be converted into kinetic energy.

Example: a ball of clay dropping off a building.


Chapter 6 5

First Law of Thermodynamics

“The total amount of energy in the universe is fixed.”

Also referred to as the “Law of Conservation of Energy”


Chapter 6 6

Temperature and Heat

Temperature is a measure of heat energy• Heat is not the same as temperature.• The more thermal energy a substance has the greater

its molecular motion (kinetic energy).• The total thermal energy in an object is the sum of the

energies of all the “bodies” in the object.


Chapter 6 7

Systems and Surroundings

System – portion of the universe we wish to study.

Surroundings – everything else.

Universe = System + Surroundings


Chapter 6 8

Directionality of Heat

Heat energy always flows from the hot object to the cold object.


- this flow continues until the two objects are at the same temperature (thermal equilibrium).

Chapter 6 9

Directionality of Heat

Exothermic – Heat is transferred from the system to the surroundings (object will feel “hot”).

Endothermic – Heat is transferred to the system from the surroundings (object will fell “cold”).


Chapter 6 10

Energy UnitsSI Unit for energy is the joule, J:

A more traditional unit is the Calorie

Calorie (cal) – amount of energy required to raise 1.0 g of water 1oC.

1cal = 4.184J

22 s/m kg 1 J1


Chapter 6 11

Specific Heat CapacitySpecific Heat Capacity

The amount of heat transferred is dependant on three quantities:

– Quantity of material

– Size of temperature change

– Identity of the material

Chapter 6 12


Tmcq

q = energy c = specific heat capacityT = temperature change

initialfinal TTT

Chapter 6 13


Tmcq

exothermic -T -qendothermic +T +q

Chapter 6 14


Tmcq

• Specific heat capacity can be either per gram (J/g(oC) or per mole (J/mol(oC).

• The smaller a substances specific heat capacity, the better a thermal conductor it is.

Chapter 6 15

Energy and Changes of StateEnergy and Changes of State

Chapter 6 16


• In the previous slide there is a continuous, steady application of energy.

•The sections that show increasing temperature are the result of the particular phase being warmed.

q = cm(T)•The “flat” sections occur when all the applied energy is used to change the phase of the substance.

•Fusion – solid liquid•Vaporization – liquid gas

Chapter 6 17


•The energy required to change the phase of a substance is unique and is described in a physical constant.

•Solid Liquid•Heat of Fusion (water, 333J/g)

•Liquid Gas•Heat of Vaporization (water, 2256J/g)

•These constants can be used to determine the energy used in melting or vaporizing a substance.

q = (Heat of Fusion)(mass of sample)q = (Heat of Vapor.)(mass of sample)

Chapter 6 18


q = cm(T)

q = (Heat of Vapor.)(mass)

Chapter 6 19

First Law of ThermodynamicsFirst Law of ThermodynamicsInternal EnergyInternal Energy – sum of all kinetic and potential energy

in an object.• It is very hard to determine an objects internal energy,

but it is possible to determine the change in energy (E).

• Change in internal energy, E = Efinal - Einitial

– A positive E means Efinal > Einitial

or the system gained energy from the surroundings (endothermic)

– A negative E means Efinal < Einitial

or the system lost energy to the surroundings (exothermic)

Chapter 6 20

Relating E to Heat and Work

EE = = qq + + ww

q = heatq = heat w = workw = work

• Both heat energy and work can change a systems internal energy.

First Law of ThermodynamicsFirst Law of Thermodynamics

Chapter 6 21

State FunctionsState function – a process that is determined by its initial and final conditions.


Chapter 6 22

State FunctionsState function – a process that is determined by its initial and final conditions.

• “A process that is not path dependant.”• Work (w) and heat (q) are not state functions.• Energy change (E) is a state function.


Chapter 6 23

Enthalpy (H) - Heat transferred between the system and surroundings carried out under constant pressure.

EE = = qq + + ww

Most reactions occur under constant pressure, soMost reactions occur under constant pressure, soEE = = qq + (- + (-P(P(V))V))

If volume is also constant, V = 0V = 0EE = = qqpp

So, Energy change is due to heat transfer,EE = = HH = = qqpp


Chapter 6 24

Enthalpy Change (H) – The heat evolved or absorbed in a reaction at constant pressure

H = Hfinal - Hinitial = qP

EnthalpyEnthalpy

Chapter 6 25

Enthalpy Change (H) – The heat evolved or absorbed in a reaction at constant pressure

• H and H are state functions, depending only on the initial and final states.

EnthalpyEnthalpy

Chapter 6 26

Enthalpies of Reaction Enthalpies of Reaction

)reactants()products( HHH reaction

2 H2(g) + O2(g) 2 H2O(g) H = -483.6 J

Chapter 6 27

For a reaction

1. Enthalpy is an extensive property (magnitude H is

directly proportional to amount):

CH4(g) + 2O2(g) CO2(g) + 2H2O(g) H = -802 kJ

2CH4(g) + 4O2(g) 2CO2(g) + 4H2O(g) H = -1604 kJ


Chapter 6 28

For a reaction



2. When we reverse a reaction, we change the sign of

H:

CO2(g) + 2H2O(g) CH4(g) + 2O2(g) H = +802 kJ

CH4(g) + 2O2(g) CO2(g) + 2H2O(g) H = -802 kJ


Chapter 6 29

For a reaction



2. When we reverse a reaction, we change the sign of

H:

3. Change in enthalpy depends on state:

CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) H = -802 kJ

CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l) H = -890 kJ


Chapter 6 30

2 Mg(s) + O2(g) 2 MgO(s) H = -1205 kJ

a) Is this reaction endothermic or exothermic?


Chapter 6 31

2 Mg(s) + O2(g) 2 MgO(s) H = -1205 kJ

a) Is this reaction endothermic or exothermic?

Exothermic, this is indicated by the negative Exothermic, this is indicated by the negative

H.H.


Chapter 6 32

2 Mg(s) + O2(g) 2 MgO(s) H = -1205 kJ

b) Calculate the amount of heat transferred when 2.4g

of Mg reacts at constant pressure.


MgmolmolggMgmoles 10.0/3.24

4.2

Chapter 6 33

2 Mg(s) + O2(g) 2 MgO(s) H = -1205 kJ




producedheat theand

used Mgof moles thebetween ratioa is there

10.0/3.244.2 Mgmolmolg

gMgmoles

Chapter 6 34

2 Mg(s) + O2(g) 2 MgO(s) H = -1205 kJ




mol

x

MgmolmolggMgmoles

10.0Mg2

kJ1205-

10.0/3.244.2

Chapter 6 35

2 Mg(s) + O2(g) 2 MgO(s) H = -1205 kJ




kJx

mol

x

MgmolmolggMgmoles

60

10.0Mg2

kJ1205-

10.0/3.244.2

Chapter 6 36

2 Mg(s) + O2(g) 2 MgO(s) H = -1205 kJ

c) How many grams of MgO are produced during an

enthalpy change of 96.0 kJ?


kJ

x

kJ

MgO

0.961205

2

Chapter 6 37

2 Mg(s) + O2(g) 2 MgO(s) H = -1205 kJ




MgOmolxkJ

x

kJ

MgO

16.00.961205

2

Chapter 6 38

2 Mg(s) + O2(g) 2 MgO(s) H = -1205 kJ




)/3.40(16.0

16.00.961205

2

molgmolMgOg

MgOmolxkJ

x

kJ

MgO

Chapter 6 39

2 Mg(s) + O2(g) 2 MgO(s) H = -1205 kJ




g

molgmolMgOg

MgOmolxkJ

x

kJ

MgO

42.6

)/3.40(16.0

16.00.961205

2

Chapter 6 40

2 Mg(s) + O2(g) 2 MgO(s) H = -1205 kJ

d) How many kilojoules of heat are absorbed when

7.50g of MgO is decomposed into Mg and O2 at

constant pressure?


Chapter 6 41

2 MgO(s) 2 Mg(s) + O2(g) H = 1205 kJ



constant pressure?


mol

molggMgOmoles

186.0

/3.4050.7

Chapter 6 42

2 MgO(s) 2 Mg(s) + O2(g) H = 1205 kJ



constant pressure?


mol

x

MgO

kJ

mol

molggMgOmoles

186.02

1205

186.0

/3.4050.7

Chapter 6 43

2 MgO(s) 2 Mg(s) + O2(g) H = 1205 kJ



constant pressure?


kJx

mol

x

MgO

kJ

mol

molggMgOmoles

112

186.02

1205

186.0

/3.4050.7

Chapter 6 44

Constant-Pressure CalorimetryConstant-Pressure Calorimetry

Calorimetry Calorimetry

Chapter 6 45

Constant-Pressure CalorimetryConstant-Pressure CalorimetryAtmospheric pressure is constant!

H = qP

qsystem = -qsurroundings

- The surroundings are composed of the water in the calorimeter and the calorimeter.

qsystem = -(qwater + qcalorimeter)


Chapter 6 46

Constant-Pressure CalorimetryConstant-Pressure CalorimetryAtmospheric pressure is constant!

H = qP

qsystem = -qsurroundings

- The surroundings are composed of the water in the calorimeter and the calorimeter.

- For most calculations, the qcalorimeter can be ignored.qsystem = - qwater

csystemmsystem Tsystem = - cwatermwater Twater


Chapter 6 47

Bomb Calorimetry Bomb Calorimetry (Constant-Volume Calorimetry)(Constant-Volume Calorimetry)


Chapter 6 48

Bomb Calorimetry Bomb Calorimetry (Constant-Volume Calorimetry)(Constant-Volume Calorimetry)

- Special calorimetry for combustion reactionsSpecial calorimetry for combustion reactions- Substance of interest is placed in a “bomb” and filled Substance of interest is placed in a “bomb” and filled

to a high pressure of oxygento a high pressure of oxygen- The sealed bomb is ignited and the heat from the The sealed bomb is ignited and the heat from the

reaction is transferred to the waterreaction is transferred to the water- This calculation must take into account the heat This calculation must take into account the heat

capacity of the calorimeter (this is grouped together capacity of the calorimeter (this is grouped together with the heat capacity of water).with the heat capacity of water).

qrxn = -Ccalorimeter(T)


Chapter 6 49

NH4NO3(s) NH4+(aq) + NO3

-(aq) Twater = 16.9oC – 22.0oC = -5.1oC

mwater = 60.0gcwater = 4.184J/goCmsample = 4.25g

qsample = -qwater

qsample = -cwatermwater Twater

qsample = -(4.184J/goC)(60.0g)(-5.1oC)qsample = 1280.3J

- Now calculate H in kJ/mol


Chapter 6 50

NH4NO3(s) NH4+(aq) + NO3

-(aq) Twater = 16.9oC – 22.0oC = -5.1oC

mwater = 60.0gcwater = 4.184J/goCmsample = 4.25g

qsample = 1280.3Jmoles NH4NO3 = 4.25g/80.032g/mol = 0.0529 mol

H = qsample/moles H = 1280.3J/0.0529mol H = 24.2 kJ/mol


Chapter 6 51

2 C8H18 + 25O2 16 CO2 + 18 H2O Twater = 28.78oC – 21.36oC = 7.42oC

Ccal = 11.66kJ/oCmsample = 1.80g

qrxn = -Ccal (Twater)qrxn = -11.66kJ/oC(7.42oC)qrxn = -86.52kJ


Chapter 6 52



qrxn = -86.52kJHcombustion(in kJ/g)

Hcombustion = -86.52kJ/1.80g =


Chapter 6 53




Hcombustion = -86.52kJ/1.80g = -48.1 kJ/gHcombustion(in kJ/mol)

Hcombustion = -86.52kJ/0.01577mol =


Chapter 6 54




Hcombustion = -86.52kJ/1.80g = -48.1 kJ/gHcombustion(in kJ/mol)

Hcombustion = -86.52kJ/0.01577mol = -5485 kJ/mol


Chapter 6 55

Hess’s law - if a reaction is carried out in a series of steps, H for the overall reaction is the sum of H’s for each individual step.

For example:

CH4(g) + 2O2(g) CO2(g) + 2H2O(g) H = -802 kJ

2H2O(g) 2H2O(l) H = -88 kJ

CH4(g) + 2O2(g) CO2(g) + 2H2O(l)H = -890 kJ

Hess’s Law Hess’s Law

Chapter 6 56

Enthalpies of Formation Enthalpies of Formation (Heat of Formation)(Heat of Formation)- There are many type of H, depending on what you

want to know

Hvapor – enthalpy of vaporization (liquid gas)

Hfusion – enthalpy of fusion (solid liquid)

Hcombustion – enthalpy of combustion

(energy from burning a substance)

Chapter 6 57

Enthalpies of Formation Enthalpies of Formation (Heat of Formation)(Heat of Formation)- A fundamental H is the Standard Enthalpy of

Formation ( )

Standard Enthalpy of Formation ( ) – The enthalpy change that accompanies the formation of one mole of a substance from the most stable forms of its component elements at 298 Kelvin and 1 atmosphere pressure.

“The standard enthalpy of formation of the most stable form on any element is zero”

ofH

ofH

Chapter 6 58

Enthalpies of FormationEnthalpies of Formation

Chapter 6 59

Enthalpies of FormationEnthalpies of FormationUsing Enthalpies of Formation to Calculate Using Enthalpies of Formation to Calculate Enthalpies of ReactionEnthalpies of Reaction

For a reaction:

reactantsproducts of

of

orxn HmHnH

Chapter 6 60

Homework ProblemsHomework Problems

4, 14, 20, 24, 28, 36, 40, 44, 46, 52, 54, 56a

Chapter 61 Principles of Reactivity: Energy and Chemical Reactions Chapter 6.

Documents

surroundings energy

energy change e

measure of heat energy

j energy

applied energy

heat q

q slide

f d heat energy