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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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Principles of reactivity energy and chemical reactions

Principles of Reactivity: Energy and Chemical Reactions

Chapter 6

Chapter 6


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


Energy: Some Basics

Systems and Surroundings

System – portion of the universe we wish to study.

Surroundings – everything else.

Universe = System + Surroundings

Chapter 6


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


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


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


    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


    First Law of Thermodynamics

    • State Functions

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

    Chapter 6


    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


    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


    Enthalpy

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

    DH = Hfinal - Hinitial = qP

    Chapter 6


    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


    Enthalpies of Reaction

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

    Chapter 6


    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


    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


    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


    Enthalpies of Reaction

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

    a) Is this reaction endothermic or exothermic?

    Chapter 6


    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


    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


    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


    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


    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


    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


    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


    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


    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


    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


    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


    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


    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


    Calorimetry

    Constant-Pressure Calorimetry

    Chapter 6


    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


    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


    Calorimetry

    Bomb Calorimetry (Constant-Volume Calorimetry)

    Chapter 6


    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


    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


    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


    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


    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


    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


    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


    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


    Homework Problems

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

    Chapter 6


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