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Principles of Reactivity: Energy and Chemical Reactions. Chapter 6. Energy: 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 distance w = F  d

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energy some basics
Energy: 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 distance

w = Fd

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

Chapter 6

energy some basics1
Energy: Some Basics

Kinetic and Potential Energy

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

Chapter 6

energy some basics2
Energy: Some Basics

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

first law of thermodynamics

Energy: Some Basics

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

temperature and heat

Energy: Some Basics

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

slide7

Energy: Some Basics

Systems and Surroundings

System – portion of the universe we wish to study.

Surroundings – everything else.

Universe = System + Surroundings

Chapter 6

slide8

Energy: Some Basics

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

slide9

Energy: Some Basics

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

slide10

Energy: Some Basics

Energy Units

SI 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

Chapter 6

specific heat capacity
Specific 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

specific heat capacity1
Specific Heat Capacity

q = energy

c = specific heat capacity

DT = temperature change

Chapter 6

specific heat capacity2
Specific Heat Capacity

exothermic -DT -q

endothermic +DT +q

Chapter 6

specific heat capacity3
Specific Heat Capacity
  • 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

energy and changes of state1
Energy and Changes of State
  • 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(DT)
  • 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

energy and changes of state2
Energy and Changes of State
  • 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

energy and changes of state3
Energy and Changes of State

q = cm(DT)

q = (Heat of Vapor.)(mass)

Chapter 6

first law of thermodynamics1
First Law of Thermodynamics

Internal Energy

Internal 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 (DE).
  • Change in internal energy, DE = Efinal - Einitial
    • A positive DE means Efinal > Einitialor the system gained energy from the surroundings (endothermic)
    • A negative DE means Efinal < Einitialor the system lost energy to the surroundings (exothermic)

Chapter 6

slide20

First Law of Thermodynamics

  • Relating DE to Heat and Work
    • DE = q + w
    • q = heat w = work
    • Both heat energy and work can change a systems internal energy.

Chapter 6

slide21

First Law of Thermodynamics

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

Chapter 6

slide22

First Law of Thermodynamics

  • State Functions
  • State 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 (DE) is a state function.

Chapter 6

slide23

First Law of Thermodynamics

  • Enthalpy (H) - Heat transferred between the system and surroundings carried out under constant pressure.
  • DE = q + w
    • Most reactions occur under constant pressure, so
    • DE = q + (-P(DV))
    • If volume is also constant, DV = 0
    • DE = qp
    • So, Energy change is due to heat transfer,
    • DE = DH = qp

Chapter 6

slide24

Enthalpy

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

DH = Hfinal - Hinitial = qP

Chapter 6

slide25

Enthalpy

  • Enthalpy Change (DH) – The heat evolved or absorbed in a reaction at constant pressure
    • H and DH are state functions, depending only on the initial and final states.

Chapter 6

slide26

Enthalpies of Reaction

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

Chapter 6

slide27

Enthalpies of Reaction

  • For a reaction
  • Enthalpy is an extensive property (magnitude DH is directly proportional to amount):
  • CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) DH = -802 kJ
  • 2CH4(g) + 4O2(g)  2CO2(g) + 4H2O(g) DH = -1604 kJ

Chapter 6

slide28

Enthalpies of Reaction

  • For a reaction
  • Enthalpy is an extensive property (magnitude DH is directly proportional to amount):
  • When we reverse a reaction, we change the sign of DH:
  • CO2(g) + 2H2O(g)  CH4(g) + 2O2(g) DH = +802 kJ
  • CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) DH = -802 kJ

Chapter 6

slide29

Enthalpies of Reaction

  • For a reaction
  • Enthalpy is an extensive property (magnitude DH is directly proportional to amount):
  • When we reverse a reaction, we change the sign of DH:
  • Change in enthalpy depends on state:
  • CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)DH = -802 kJ
  • CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(l)DH = -890 kJ

Chapter 6

slide30

Enthalpies of Reaction

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

a) Is this reaction endothermic or exothermic?

Chapter 6

slide31

Enthalpies of Reaction

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

a) Is this reaction endothermic or exothermic?

Exothermic, this is indicated by the negative DH.

Chapter 6

slide32

Enthalpies of Reaction

  • 2 Mg(s) + O2(g)  2 MgO(s) DH = -1205 kJ
  • Calculate the amount of heat transferred when 2.4g of Mg reacts at constant pressure.

Chapter 6

slide33

Enthalpies of Reaction

  • 2 Mg(s) + O2(g)  2 MgO(s) DH = -1205 kJ
  • Calculate the amount of heat transferred when 2.4g of Mg reacts at constant pressure.

Chapter 6

slide34

Enthalpies of Reaction

  • 2 Mg(s) + O2(g)  2 MgO(s) DH = -1205 kJ
  • Calculate the amount of heat transferred when 2.4g of Mg reacts at constant pressure.

Chapter 6

slide35

Enthalpies of Reaction

  • 2 Mg(s) + O2(g)  2 MgO(s) DH = -1205 kJ
  • Calculate the amount of heat transferred when 2.4g of Mg reacts at constant pressure.

Chapter 6

slide36

Enthalpies of Reaction

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

c) How many grams of MgO are produced during an enthalpy change of 96.0 kJ?

Chapter 6

slide37

Enthalpies of Reaction

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

c) How many grams of MgO are produced during an enthalpy change of 96.0 kJ?

Chapter 6

slide38

Enthalpies of Reaction

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

c) How many grams of MgO are produced during an enthalpy change of 96.0 kJ?

Chapter 6

slide39

Enthalpies of Reaction

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

c) How many grams of MgO are produced during an enthalpy change of 96.0 kJ?

Chapter 6

slide40

Enthalpies of Reaction

2 Mg(s) + O2(g)  2 MgO(s) DH = -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

slide41

Enthalpies of Reaction

2 MgO(s)  2 Mg(s) + O2(g) DH = 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

slide42

Enthalpies of Reaction

2 MgO(s)  2 Mg(s) + O2(g) DH = 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

slide43

Enthalpies of Reaction

2 MgO(s)  2 Mg(s) + O2(g) DH = 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

slide44

Calorimetry

Constant-Pressure Calorimetry

Chapter 6

slide45

Calorimetry

  • Constant-Pressure Calorimetry
  • Atmospheric pressure is constant!
  • DH = qP
  • qsystem = -qsurroundings
  • The surroundings are composed of the water in the calorimeter and the calorimeter.
  • qsystem = -(qwater + qcalorimeter)

Chapter 6

slide46

Calorimetry

  • Constant-Pressure Calorimetry
  • Atmospheric pressure is constant!
  • DH = 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 DTsystem = - cwatermwater DTwater

Chapter 6

slide47

Calorimetry

Bomb Calorimetry (Constant-Volume Calorimetry)

Chapter 6

slide48

Calorimetry

  • Bomb Calorimetry
  • (Constant-Volume Calorimetry)
  • Special calorimetry for combustion reactions
  • Substance of interest is placed in a “bomb” and filled to a high pressure of oxygen
  • The sealed bomb is ignited and the heat from the reaction is transferred to the water
  • This calculation must take into account the heat capacity of the calorimeter (this is grouped together with the heat capacity of water).
  • qrxn = -Ccalorimeter(DT)

Chapter 6

slide49

Calorimetry

NH4NO3(s)  NH4+(aq) + NO3-(aq)

DTwater = 16.9oC – 22.0oC = -5.1oC

mwater = 60.0g

cwater = 4.184J/goC

msample = 4.25g

qsample = -qwater

qsample = -cwatermwater DTwater

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

qsample = 1280.3J

- Now calculate DH in kJ/mol

Chapter 6

slide50

Calorimetry

NH4NO3(s)  NH4+(aq) + NO3-(aq)

DTwater = 16.9oC – 22.0oC = -5.1oC

mwater = 60.0g

cwater = 4.184J/goC

msample = 4.25g

qsample = 1280.3J

moles NH4NO3 = 4.25g/80.032g/mol = 0.0529 mol

DH = qsample/moles

DH = 1280.3J/0.0529mol

DH = 24.2 kJ/mol

Chapter 6

slide51

Calorimetry

2 C8H18 + 25O2 16 CO2 + 18 H2O

DTwater = 28.78oC – 21.36oC = 7.42oC

Ccal = 11.66kJ/oC

msample = 1.80g

qrxn = -Ccal (DTwater)

qrxn = -11.66kJ/oC(7.42oC)

qrxn = -86.52kJ

Chapter 6

slide52

Calorimetry

2 C8H18 + 25O2 16 CO2 + 18 H2O

DTwater = 28.78oC – 21.36oC = 7.42oC

Ccal = 11.66kJ/oC

msample = 1.80g

qrxn = -86.52kJ

DHcombustion(in kJ/g)

DHcombustion = -86.52kJ/1.80g =

Chapter 6

slide53

Calorimetry

2 C8H18 + 25O2 16 CO2 + 18 H2O

DTwater = 28.78oC – 21.36oC = 7.42oC

Ccal = 11.66kJ/oC

msample = 1.80g

qrxn = -86.52kJ

DHcombustion(in kJ/g)

DHcombustion = -86.52kJ/1.80g = -48.1 kJ/g

DHcombustion(in kJ/mol)

DHcombustion = -86.52kJ/0.01577mol =

Chapter 6

slide54

Calorimetry

2 C8H18 + 25O2 16 CO2 + 18 H2O

DTwater = 28.78oC – 21.36oC = 7.42oC

Ccal = 11.66kJ/oC

msample = 1.80g

qrxn = -86.52kJ

DHcombustion(in kJ/g)

DHcombustion = -86.52kJ/1.80g = -48.1 kJ/g

DHcombustion(in kJ/mol)

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

Chapter 6

slide55

Hess’s Law

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

Chapter 6

enthalpies of formation heat of formation
Enthalpies 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

enthalpies of formation heat of formation1
Enthalpies 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”

Chapter 6

enthalpies of formation1
Enthalpies of Formation

Using Enthalpies of Formation to Calculate Enthalpies of Reaction

For a reaction:

Chapter 6

slide60

Homework Problems

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

Chapter 6

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