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    Chapter 6 Thermochemistry


    (Bài giảng chưa được thẩm định)
    Nguồn:
    Người gửi: Trương Minh Chiến (trang riêng)
    Ngày gửi: 02h:03' 12-01-2011
    Dung lượng: 6.1 MB
    Số lượt tải: 11
    Số lượt thích: 0 người

    Mr. Truong Minh Chien ; losedtales@yahoo.com
    http://tailieu.vn/losedtales
    http://mba-programming.blogspot.com
    Chemistry, Julia Burdge, 2st Ed.
    McGraw Hill.
    Chapter 6
    Thermochemistry
    2011, NKMB Co., Ltd.
    2
    Heating Your Home
    most homes burn fossil fuels to generate heat
    the amount the temperature of your home increases depends on several factors
    how much fuel is burned
    the volume of the house
    the amount of heat loss
    the efficiency of the burning process
    can you think of any others?
    Chemistry, Julia Burdge, 2nd e., McGraw Hill.
    3
    Nature of Energy
    even though Chemistry is the study of matter, energy effects matter
    energy is anything that has the capacity to do work
    work is a force acting over a distance
    Energy = Work = Force x Distance
    energy can be exchanged between objects through contact
    collisions
    Chemistry, Julia Burdge, 2nd e., McGraw Hill.
    Tro, Chemistry: A Molecular Approach
    4
    Classification of
    Energy
    Kinetic energy is energy of motion or energy that is being transferred
    thermal energy is kinetic
    Tro, Chemistry: A Molecular Approach
    5
    Classification of Energy
    Potential energy is energy that is stored in an object, or energy associated with the composition and position of the object
    energy stored in the structure of a compound is potential
    Tro, Chemistry: A Molecular Approach
    6
    Law of Conservation of Energy
    energy cannot be created or destroyed
    First Law of Thermodynamics
    energy can be transferred between objects
    energy can be transformed from one form to another
    heat → light → sound
    Tro, Chemistry: A Molecular Approach
    7
    Some Forms of Energy
    Electrical
    kinetic energy associated with the flow of electrical charge
    Heat or Thermal Energy
    kinetic energy associated with molecular motion
    Light or Radiant Energy
    kinetic energy associated with energy transitions in an atom
    Nuclear
    potential energy in the nucleus of atoms
    Chemical
    potential energy in the attachment of atoms or because of their position
    8
    Units of Energy
    the amount of kinetic energy an object has is directly proportional to its mass and velocity
    KE = ½mv2
    Tro, Chemistry: A Molecular Approach
    9
    Units of Energy
    joule (J) is the amount of energy needed to move a 1 kg mass a distance of 1 meter
    1 J = 1 N∙m = 1 kg∙m2/s2
    calorie (cal) is the amount of energy needed to raise one gram of water by 1°C
    kcal = energy needed to raise 1000 g of water 1°C
    food Calories = kcals
    Tro, Chemistry: A Molecular Approach
    10
    Energy Use
    11
    Energy Flow and
    Conservation of Energy
    we define the system as the material or process we are studying the energy changes within
    we define the surroundings as everything else in the universe
    Conservation of Energy requires that the total energy change in the system and the surrounding must be zero
    DEnergyuniverse = 0 = DEnergysystem + DEnergysurroundings
    D is the symbol that is used to mean change
    final amount – initial amount
    Tro, Chemistry: A Molecular Approach
    12
    Internal Energy
    the internal energy is the total amount of kinetic and potential energy a system possesses
    the change in the internal energy of a system only depends on the amount of energy in the system at the beginning and end
    a state function is a mathematical function whose result only depends on the initial and final conditions, not on the process used
    DE = Efinal – Einitial
    DEreaction = Eproducts - Ereactants
    Tro, Chemistry: A Molecular Approach
    13
    State Function

    Tro, Chemistry: A Molecular Approach
    14
    Energy Diagrams
    energy diagrams are a “graphical” way of showing the direction of energy flow during a process
    if the final condition has a
    larger amount of internal
    energy than the initial
    condition, the change in the
    internal energy will be +
    if the final condition has a
    smaller amount of internal
    energy than the initial
    condition, the change in the
    internal energy will be ─
    Tro, Chemistry: A Molecular Approach
    15
    Energy Flow
    when energy flows out of a system, it must all flow into the surroundings
    when energy flows out of a system, DEsystem is ─
    when energy flows into the surroundings, DEsurroundings is +
    therefore:
    ─ DEsystem= DEsurroundings

    Tro, Chemistry: A Molecular Approach
    16
    Energy Flow
    when energy flows into a system, it must all come from the surroundings
    when energy flows into a system, DEsystem is +
    when energy flows out of the surroundings, DEsurroundings is ─
    therefore:
    DEsystem= ─ DEsurroundings

    Tro, Chemistry: A Molecular Approach
    17
    How Is Energy Exchanged?
    energy is exchanged between the system and surroundings through heat and work
    q = heat (thermal) energy
    w = work energy
    q and w are NOT state functions, their value depends on the process
    DE = q + w
    Tro, Chemistry: A Molecular Approach
    18
    Energy Exchange
    energy is exchanged between the system and surroundings through either heat exchange or work being done
    Tro, Chemistry: A Molecular Approach
    19
    Heat & Work
    on a smooth table, most of the kinetic energy is transferred from the first ball to the second – with a small amount lost through friction
    Tro, Chemistry: A Molecular Approach
    20
    Heat & Work
    on a rough table, most of the kinetic energy of the first ball is lost through friction – less than half is transferred to the second
    Tro, Chemistry: A Molecular Approach
    21
    Heat Exchange
    heat is the exchange of thermal energy between the system and surroundings
    occurs when system and surroundings have a difference in temperature
    heat flows from matter with high temperature to matter with low temperature until both objects reach the same temperature
    thermal equilibrium
    Tro, Chemistry: A Molecular Approach
    22
    Quantity of Heat Energy Absorbed
    Heat Capacity
    when a system absorbs heat, its temperature increases
    the increase in temperature is directly proportional to the amount of heat absorbed
    the proportionality constant is called the heat capacity, C
    units of C are J/°C or J/K
    q = C x DT
    the heat capacity of an object depends on its mass
    200 g of water requires twice as much heat to raise its temperature by 1°C than 100 g of water
    the heat capacity of an object depends on the type of material
    1000 J of heat energy will raise the temperature of 100 g of sand 12°C, but only raise the temperature of 100 g of water by 2.4°C
    Tro, Chemistry: A Molecular Approach
    23
    Specific Heat Capacity
    measure of a substance’s intrinsic ability to absorb heat
    the specific heat capacity is the amount of heat energy required to raise the temperature of one gram of a substance 1°C
    Cs
    units are J/(g∙°C)
    the molar heat capacity is the amount of heat energy required to raise the temperature of one mole of a substance 1°C
    the rather high specific heat of water allows it to absorb a lot of heat energy without large increases in temperature
    keeping ocean shore communities and beaches cool in the summer
    allows it to be used as an effective coolant to absorb heat
    Tro, Chemistry: A Molecular Approach
    24
    Quantifying Heat Energy
    the heat capacity of an object is proportional to its mass and the specific heat of the material
    so we can calculate the quantity of heat absorbed by an object if we know the mass, the specific heat, and the temperature change of the object
    Heat = (mass) x (specific heat capacity) x (temp. change)
    q = (m) x (Cs) x (DT)
    Example 6.2 – How much heat is absorbed by a copper penny with mass 3.10 g whose temperature rises from
    -8.0°C to 37.0°C?



    q = m ∙ Cs ∙ DT
    Cs = 0.385 J/g (Table 6.4)
    the unit and sign are correct
    T1= -8.0°C, T2= 37.0°C, m=3.10 g
    q, J
    Check:
    Check
    Solution:
    Follow the Concept Plan to Solve the problem
    Concept Plan:



    Relationships:
    Strategize
    Given:

    Find:
    Sort Information
    Tro, Chemistry: A Molecular Approach
    26
    Pressure -Volume Work
    PV work is work that is the result of a volume change against an external pressure
    when gases expand, DV is +, but the system is doing work on the surroundings so w is ─
    as long as the external pressure is kept constant
    ─Work = External Pressure x Change in Volume
    w = ─PDV
    to convert the units to joules use 101.3 J = 1 atm∙L
    Example 6.3 – If a balloon is inflated from 0.100 L to 1.85 L against an external pressure of 1.00 atm, how much work is done?



    the unit and sign are correct
    V1=0.100 L, V2=1.85 L, P=1.00 atm
    w, J
    Check:
    Solution:
    Concept Plan:



    Relationships:
    Given:

    Find:
    101.3 J = 1 atm L
    Tro, Chemistry: A Molecular Approach
    28
    Exchanging Energy Between
    System and Surroundings
    exchange of heat energy
    q = mass x specific heat x DTemperature
    exchange of work
    w = −Pressure x DVolume
    Tro, Chemistry: A Molecular Approach
    29
    Measuring DE,
    Calorimetry at Constant Volume
    since DE = q + w, we can determine DE by measuring q and w
    in practice, it is easiest to do a process in such a way that there is no change in volume, w = 0
    at constant volume, DEsystem = qsystem
    in practice, it is not possible to observe the temperature changes of the individual chemicals involved in a reaction – so instead, we use an insulated, controlled surroundings and measure the temperature change in it
    the surroundings is called a bomb calorimeter and is usually made of a sealed, insulated container filled with water
    qsurroundings = qcalorimeter = ─qsystem
    ─DEreaction = qcal = Ccal x DT
    Tro, Chemistry: A Molecular Approach
    30
    Bomb Calorimeter
    used to measure DE because it is a constant volume system
    31
    Example 6.4 – When 1.010 g of sugar is burned in a bomb calorimeter, the temperature rises from 24.92°C to 28.33°C. If Ccal = 4.90 kJ/°C, find DE for burning 1 mole
    qcal = Ccal x DT = -qrxn
    MM C12H22O11 = 342.3 g/mol
    the units and sign are correct
    1.010 g C12H22O11, T1 = 24.92°C, T2 = 28.33°C, Ccal = 4.90 kJ/°C
    DErxn, kJ/mol
    Check:
    Solution:
    Concept Plan:



    Relationships:
    Given:

    Find:
    Tro, Chemistry: A Molecular Approach
    32
    Enthalpy
    the enthalpy, H, of a system is the sum of the internal energy of the system and the product of pressure and volume
    H is a state function
    H = E + PV
    the enthalpy change, DH, of a reaction is the heat evolved in a reaction at constant pressure
    DHreaction = qreaction at constant pressure
    usually DH and DE are similar in value, the difference is largest for reactions that produce or use large quantities of gas
    33
    Endothermic and Exothermic Reactions
    when DH is ─, heat is being released by the system
    reactions that release heat are called exothermic reactions
    when DH is +, heat is being absorbed by the system
    reactions that absorb heat are called endothermic reactions
    chemical heat packs contain iron filings that are oxidized in an exothermic reaction ─ your hands get warm because the released heat of the reaction is absorbed by your hands
    chemical cold packs contain NH4NO3 that dissolves in water in an endothermic process ─ your hands get cold because they are giving away your heat to the reaction
    34
    Molecular View of
    Exothermic Reactions
    in an exothermic reaction, the temperature rises due to release of thermal energy
    this extra thermal energy comes from the conversion of some of the chemical potential energy in the reactants into kinetic energy in the form of heat
    during the course of a reaction, old bonds are broken and new bonds made
    the products of the reaction have less chemical potential energy than the reactants
    the difference in energy is released as heat
    Tro, Chemistry: A Molecular Approach
    35
    Molecular View of
    Endothermic Reactions
    in an endothermic reaction, the temperature drops due to absorption of thermal energy
    the required thermal energy comes from the surroundings
    during the course of a reaction, old bonds are broken and new bonds made
    the products of the reaction have more chemical potential energy than the reactants
    to acquire this extra energy, some of the thermal energy of the surroundings is converted into chemical potential energy stored in the products
    Tro, Chemistry: A Molecular Approach
    36
    Enthalpy of Reaction
    the enthalpy change in a chemical reaction is an extensive property
    the more reactants you use, the larger the enthalpy change
    by convention, we calculate the enthalpy change for the number of moles of reactants in the reaction as written
    C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g) DH = -2044 kJ

    37
    Example 6.6 – How much heat is evolved in the complete combustion of 13.2 kg of C3H8(g)?




    1 kg = 1000 g, 1 mol C3H8 = -2044 kJ, Molar Mass = 44.09 g/mol
    the sign is correct and the value is reasonable
    13.2 kg C3H8,
    q, kJ/mol
    Check:
    Solution:
    Concept Plan:



    Relationships:
    Given:

    Find:
    Calorimetry animation
    Tro, Chemistry: A Molecular Approach
    38
    Tro, Chemistry: A Molecular Approach
    39
    Measuring DH
    Calorimetry at Constant Pressure
    reactions done in aqueous solution are at constant pressure
    open to the atmosphere
    the calorimeter is often nested foam cups containing the solution
    qreaction = ─ qsolution = ─(masssolution x Cs, solution x DT)
    DHreaction = qconstant pressure = qreaction
    to get DHreaction per mol, divide by the number of moles
    40
    Example 6.7 – What is DHrxn/mol Mg for the reaction
    Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g) if 0.158 g Mg reacts in
    100.0 mL of solution changes the temperature from 25.6°C to 32.8°C?




    1 kg = 1000 g, 1 mol C3H8 = -2044 kJ, Molar Mass = 44.09 g/mol
    the sign is correct and the value is reasonable
    0.158 g Mg, 100.0 mL,
    q, kJ/mol
    Check:
    Solution:
    Concept Plan:



    Relationships:
    Given:

    Find:
    41
    Example 6.7 – What is DHrxn/mol Mg for the reaction
    Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g) if 0.158 g Mg reacts in
    100.0 mL of solution to change the temperature from 25.6°C to 32.8°C?



    qsoln = m x Cs x DT = -qrxn
    the units and sign are correct
    0.158 g Mg, 100.0 mL sol’n, T1 = 25.6°C, T2 = 32.8°C, Cs = 4.18 J/°C, dsoln = 1.00 g/mL
    DHrxn, J/mol Mg
    Check:
    Solution:
    Concept Plan:


    Relationships:
    Given:

    Find:
    Tro, Chemistry: A Molecular Approach
    42
    Relationships Involving DHrxn
    when reaction is multiplied by a factor, DHrxn is multiplied by that factor
    because DHrxn is extensive
    C(s) + O2(g) → CO2(g) DH = -393.5 kJ
    2 C(s) + 2 O2(g) → 2 CO2(g) DH = 2(-393.5 kJ) = -787.0 kJ
    if a reaction is reversed, then the sign of DH is reversed
    CO2(g) → C(s) + O2(g) DH = +393.5 kJ
    Hess’s Law animation
    Tro, Chemistry: A Molecular Approach
    43
    Tro, Chemistry: A Molecular Approach
    44
    Relationships Involving DHrxn
    Hess’s Law
    if a reaction can be expressed as a series of steps, then the DHrxn for the overall reaction is the sum of the heats of reaction for each step
    Tro, Chemistry: A Molecular Approach
    45
    Sample – Hess’s Law
    Given the following information:
    2 NO(g) + O2(g)  2 NO2(g) DH° = -173 kJ
    2 N2(g) + 5 O2(g) + 2 H2O(l)  4 HNO3(aq) DH° = -255 kJ
    N2(g) + O2(g)  2 NO(g) DH° = +181 kJ
    Calculate the DH° for the reaction below:
    3 NO2(g) + H2O(l)  2 HNO3(aq) + NO(g) DH° = ?
    [2 NO2(g)  2 NO(g) + O2(g)] x 1.5 DH° = 1.5(+173 kJ)
    [2 N2(g) + 5 O2(g) + 2 H2O(l)  4 HNO3(aq)] x 0.5 DH° = 0.5(-255 kJ)
    [2 NO(g)  N2(g) + O2(g)] DH° = -181 kJ
    [3 NO2(g)  3 NO(g) + 1.5 O2(g)] DH° = (+259.5 kJ)
    [1 N2(g) + 2.5 O2(g) + 1 H2O(l)  2 HNO3(aq)] DH° = (-128 kJ)
    [2 NO(g)  N2(g) + O2(g)] DH° = -181 kJ
    3 NO2(g) + H2O(l)  2 HNO3(aq) + NO(g) DH° = - 49 kJ
    Tro, Chemistry: A Molecular Approach
    46
    Standard Conditions
    the standard state is the state of a material at a defined set of conditions
    pure gas at exactly 1 atm pressure
    pure solid or liquid in its most stable form at exactly 1 atm pressure and temperature of interest
    usually 25°C
    substance in a solution with concentration 1 M
    the standard enthalpy change, DH°, is the enthalpy change when all reactants and products are in their standard states
    the standard enthalpy of formation, DHf°, is the enthalpy change for the reaction forming 1 mole of a pure compound from its constituent elements
    the elements must be in their standard states
    the DHf° for a pure element in its standard state = 0 kJ/mol
    by definition
    Tro, Chemistry: A Molecular Approach
    47
    Formation Reactions
    reactions of elements in their standard state to form 1 mole of a pure compound
    if you are not sure what the standard state of an element is, find the form in Appendix IIB that has a DHf° = 0
    since the definition requires 1 mole of compound be made, the coefficients of the reactants may be fractions
    48
    Writing Formation Reactions
    Write the formation reaction for CO(g)
    the formation reaction is the reaction between the elements in the compound, which are C and O
    C + O → CO(g)
    the elements must be in their standard state
    there are several forms of solid C, but the one with DHf° = 0 is graphite
    oxygen’s standard state is the diatomic gas
    C(s, graphite) + O2(g) → CO(g)
    the equation must be balanced, but the coefficient of the product compound must be 1
    use whatever coefficient in front of the reactants is necessary to make the atoms on both sides equal without changing the product coefficient
    C(s, graphite) + ½ O2(g) → CO(g)
    Tro, Chemistry: A Molecular Approach
    49
    Calculating Standard Enthalpy Change for a Reaction
    any reaction can be written as the sum of formation reactions (or the reverse of formation reactions) for the reactants and products
    the DH° for the reaction is then the sum of the DHf° for the component reactions
    DH°reaction = S n DHf°(products) - S n DHf°(reactants)
    S means sum
    n is the coefficient of the reaction
    Tro, Chemistry: A Molecular Approach
    50
    The Combustion of CH4

    Tro, Chemistry: A Molecular Approach
    51
    Sample - Calculate the Enthalpy Change in the Reaction
    2 C2H2(g) + 5 O2(g) ® 4 CO2(g) + 2 H2O(l)
    1. Write formation reactions for each compound and
    determine the DHf° for each
    2 C(s, gr) + H2(g) ® C2H2(g) DHf° = +227.4 kJ/mol
    C(s, gr) + O2(g) ® CO2(g) DHf° = -393.5 kJ/mol
    H2(g) + ½ O2(g) ® H2O(l) DHf° = -285.8 kJ/mol
    Tro, Chemistry: A Molecular Approach
    52
    Sample - Calculate the Enthalpy Change in the Reaction
    2 C2H2(g) + 5 O2(g) ® 4 CO2(g) + 2 H2O(l)
    2 C2H2(g) ® 4 C(s) + 2 H2(g) DH° = 2(-227.4) kJ
    4 C(s) + 4 O2(g) ® 4CO2(g) DH° = 4(-393.5) kJ
    2 H2(g) + O2(g) ® 2 H2O(l) DH° = 2(-285.8) kJ
    2. Arrange equations so they add up to desired reaction
    2 C2H2(g) + 5 O2(g) ® 4 CO2(g) + 2 H2O(l) DH = -2600.4 kJ
    Tro, Chemistry: A Molecular Approach
    53
    Sample - Calculate the Enthalpy Change in the Reaction
    2 C2H2(g) + 5 O2(g) ® 4 CO2(g) + 2 H2O(l)
    DH°reaction = S n DHf°(products) - S n DHf°(reactants)

    DHrxn = [(4•DHCO2 + 2•DHH2O) – (2•DHC2H2 + 5•DHO2)]

    DHrxn = [(4•(-393.5) + 2•(-285.8)) – (2•(+227.4) + 5•(0))]

    DHrxn = -2600.4 kJ
    54
    Example 6.11 – How many kg of octane must be combusted to supply 1.0 x 1011 kJ of energy?




    MMoctane = 114.2 g/mol, 1 kg = 1000 g
    the units and sign are correct
    the large value is expected
    1.0 x 1011 kJ
    mass octane, kg
    Check:
    Solution:
    Concept Plan:





    Relationships:
    Given:
    Find:
    Write the balanced equation per mole of octane
    from
    above
    C8H18(l) + 25/2 O2(g) → 8 CO2(g) + 9 H2O(g)
    Look up the DHf°
    for each material
    in Appendix IIB
    Tro, Chemistry: A Molecular Approach
    55
    Energy Use and the Environment
    in the U.S., each person uses over 105 kWh of energy per year
    most comes from the combustion of fossil fuels
    combustible materials that originate from ancient life
    C(s) + O2(g) → CO2(g) DH°rxn = -393.5 kJ
    CH4(g) +2 O2(g) → CO2(g) + 2 H2O(g) DH°rxn = -802.3 kJ
    C8H18(g) +12.5 O2(g) → 8 CO2(g) + 9 H2O(g) DH°rxn = -5074.1 kJ
    fossil fuels cannot be replenished
    at current rates of consumption, oil and natural gas supplies will be depleted in 50 – 100 yrs.
    Tro, Chemistry: A Molecular Approach
    56
    Energy Consumption
    the distribution of energy consumption in the US
    the increase in energy consumption in the US
    Tro, Chemistry: A Molecular Approach
    57
    The Effect of Combustion Products
    on Our Environment
    because of additives and impurities in the fossil fuel, incomplete combustion and side reactions, harmful materials are added to the atmosphere when fossil fuels are burned for energy
    therefore fossil fuel emissions contribute to air pollution, acid rain, and global warming
    Tro, Chemistry: A Molecular Approach
    58
    Global Warming
    CO2 is a greenhouse gas
    it allows light from the sun to reach the earth, but does not allow the heat (infrared light) reflected off the earth to escape into outer space
    it acts like a blanket
    CO2 levels in the atmosphere have been steadily increasing
    current observations suggest that the average global air temperature has risen 0.6°C in the past 100 yrs.
    atmospheric models suggest that the warming effect could worsen if CO2 levels are not curbed
    some models predict that the result will be more severe storms, more floods and droughts, shifts in agricultural zones, rising sea levels, and changes in habitats
    Tro, Chemistry: A Molecular Approach
    59
    CO2 Levels

    Tro, Chemistry: A Molecular Approach
    60
    Renewable Energy
    our greatest unlimited supply of energy is the sun
    new technologies are being developed to capture the energy of sunlight
    parabolic troughs, solar power towers, and dish engines concentrate the sun’s light to generate electricity
    solar energy used to decompose water into H2(g) and O2(g); the H2 can then be used by fuel cells to generate electricity
    H2(g) + ½ O2(g) → H2O(l) DH°rxn = -285.8 kJ
    hydroelectric power
    wind power
     
     
     
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