Chapter 5 sem 2 by yk6k8j2

VIEWS: 49 PAGES: 46

									                    Chapter 5
• Brief review of
    – System, surroundings, universe
    – Energy, work, heat
    – Endothermic/exothermic


•   Enthalpies of formation
•   Heat capacity
•   Calorimetry
•   Foods and fuels
System and Surroundings

            • The system includes
              the molecules we want
              to study (here, the
              hydrogen and oxygen
              molecules).
            • The surroundings are
              everything else (here,
              the cylinder and
              piston).
         Heat, Work, Energy
• Heat (q) is a form of energy that moves
  when objects of different temperatures
  contact
• Work is energy expended in moving an
  object against an opposing force

                 DE = q + w

• E is a state function; q and w are not
Changes in Internal Energy
             • When energy is
               exchanged between
               the system and the
               surroundings, it is
               exchanged as either
               heat (q) or work (w).
             • That is, DE = q + w.
DE, q, w, and Their Signs
  First Law of Thermodynamics
• Energy is neither created nor destroyed.
• In other words, the total energy of the universe is
  a constant; if the system loses energy, it must be
  gained by the surroundings, and vice versa.



                               Use Fig. 5.5
   Exchange of Heat between
   System and Surroundings
• When heat is absorbed by the system from the
  surroundings, the process is endothermic.




                                   H, enthalpy, is qp
                                   (heat flow at constant
                                   pressure)
   Exchange of Heat between
   System and Surroundings
• When heat is absorbed by the system from the
  surroundings, the process is endothermic.
• When heat is released by the system to the
  surroundings, the process is exothermic.


                                   H, enthalpy, is qp
                                   (heat flow at constant
                                   pressure)
    Guidelines for Using DH Values
intensive property: value doesn’t depend         a property whose value depends
on the amount of material present               on the amount of material involved


     1.    Enthalpy is an extensive property. The size of DH will
           depend on the amount of matter involved in the
           chemical reaction

     For
                            2H2(g) + O2(g)  2H2O(g)      “enthalpy of reaction”


     If 2 mol H2(g) are combusted, DHrxn = -483.6 kJ
     If 4 mol H2(g) are combusted, DHrxn = -967.2 kJ
Guidelines for Using DH Values
2. The enthalpy change for a reaction is
   equal in magnitude, but opposite in sign,
   to DH for the reverse reaction

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

    CO2(g) + 2H2O(l)  CH4(g) + 2O2(g)   DH = 890 kJ
 Guidelines for Using DH Values

3. The enthalpy change for a reaction
   depends on the state of the reactants and
   products

 CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)   DH = -890 kJ
 CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)   DH = -802 kJ

(because 2H2O(l)  2H2O(g); DH = +88 kJ)
                 Hess’s Law
    Because DH is a state function, its value is independent
    of its history (i.e. it is a state function)


• Overall enthalpy change is
  independent of the number of steps or
  the particular nature of the path by
  which the reaction is carried out
• Useful for calculating enthalpy
  changes for reactions without actually
  carrying out the experiment
                             Hess’s Law
      • For example, let’s say we want to determine
        DH for the reaction
             C2H4(g) + 6F2(g)  2CF4(g) + 4HF(g)

      • And we know the following information:
         H2(g) + F2(g)  2HF(g)      DH = -537 kJ
         C(s) + 2F2(g)  CF4(g)      DH = -680 kJ
        2C(s) + 2H2(g)  C2H4(g)     DH = +52.3 kJ

Using the guidelines for DH values, we should be able to rearrange the
above equations, add them together to create the top (blue) equation,
and get DH for this reaction.
                        Hess’s Law
            C2H4(g) + 6F2(g)  2CF4(g) + 4HF(g)
 multiply by 2

        H2(g) + F2(g)  2HF(g)                 DH = -537 kJ
        C(s) + 2F2(g)  CF4(g)                 DH = -680 kJ
flip
        2C(s) + 2H2(g)  C2H4(g)               DH = +52.3 kJ
Each of the three lower equations involves at least one of the molecules
  in the top equation
Rearrange the three red equations (multiplying by coefficients when
  necessary so that they can be added together to yield the blue equation
             Hess’s Law
   C2H4(g) + 6F2(g)  2CF4(g) + 4HF(g)

2H2(g) + 2F2(g)  4HF(g)    DH = 2(-537 kJ)
2C(s) + 4F2(g)  2CF4(g)    DH = 2(-680 kJ)
 C2H4(g)  2C(s) + 2H2(g)    DH = -52.3 kJ
             Hess’s Law
   C2H4(g) + 6F2(g)  2CF4(g) + 4HF(g)

2H2(g) + 2F2(g)  4HF(g)      DH = 2(-537 kJ)
2C(s) + 4F2(g)  2CF4(g)      DH = 2(-680 kJ)
 C2H4(g)  2C(s) + 2H2(g)      DH = -52.3 kJ

     C2H4(g) + 6F2(g)  2CF4(g) + 4HF(g)
         Adding, get DH = -2.49x103 kJ
          Enthalpies of Formation
   An enthalpy of formation, DHf, is defined as the
   enthalpy change for the reaction in which one
   mole of compound is made from its constituent
   elements in their elemental forms.
these are the elemental (natural) forms of carbon and oxygen


C(graphite) + O2(g)  CO2(g)
2C(graphite) + 2H2(g) + O2(g)  HC2H3O2(l)                       qP = DHf

Mg(s) + ½O2(g)  MgO(s)
                                      DHf defined per one mole of product
Standard Enthalpies of Formation
  Standard enthalpies of formation, DHof, are enthalpies of
  formation measured when all reactants and products are
  present in their standard states at (25°C (298 K) and 1.00 atm
  pressure (for any gases present).
   The symbol “o” indicates standard conditions
          Enthalpies of formation
• Thus, the enthalpy change that accompanies the
   formation of one mole of a compound from its constituent
   elements, with all substances in their standard states, is
   DHof
e.g.
       2C(graphite) + 3H2(g) + ½O2(g)  C2H5OH(l)   DHof= -277.7 kJ

This reaction represents DHof[C2H5OH(l)]

Notice that all the reactants are pure elements (this reaction shows the
  formation of 1 mol of ethanol from the elements that make up an
  ethanol molecule, C, H, and O. These elements are written as they
  appear in nature)
  Elements in their standard states
• Some examples:
   – Carbon: C(s) (graphite)
   – Oxygen: O2(g)
   – Hydrogen: H2(g)
   – Sodium: Na(s)
   – Chlorine: Cl2(g)
   – Bromine: Br2(l)
   – Iodine: I2(s)
   – Phosphorus: P4(s)
   – Iron: Fe(s)

   Trés important: Enthalpy of formation of any element in
     its standard state is zero
     e.g. DHof of H2(g) would correspond to the reaction:

                                 H2(g)  H2(g)              DHof = 0kJ/mol
          Enthalpies of formation
• Which of the following represent standard
  enthalpy of formation reactions?
a)2K(l) + Cl2(g)  2KCl(s)
b)C6H12O6(s)  6C(diamond) + 6H2(g) + 3O2(g)
c)2Na(s) + ½O2(g)  Na2O(s)


(remember: DHf: enthalpy change that accompanies the formation of one mole
of a compound from its constituent elements)
  Enthalpy Changes for Chemical
            Reactions
We calculated enthalpy changes for chemical
 reactions using Hess’s Law in the first semester.
Using enthalpies of formation, enthalpy changes
 for chemical reactions may also be calculated


       DH rxn    DH f products   DH f reactants
                                       Example
   • For the question we solved earlier using Hess’s
     Law
                              C2H4(g) + 6F2(g)  2CF4(g) + 4HF(g)

   If we knew the formation enthalpies for the reactants and
      products involved, we could get DHrxn as follows*:
   DHof C2H4(g) = 52.30 kJ/mol
   DHof F2(g) = 0 kJ/mol
   DHof CF4(g) = -679.9 kJ/mol
   DHof HF(g) = -268.61 kJ/mol
DH rxn    DH f products   DH f reactants
DH rxn   2DH f CF4 ( g )  4DH f  HF( g )   DH f C2 H 4( g )  6DH f  F2( g ) 
DH rxn   2(679.9kJ / mol )  4(268.61kJ / mol )   (52.30kJ / mol )  6(0kJ / mol ) 
DH rxn   2486.54kJ / mol  2.49 x103 kJ / mol
                                                *Technically only valid under standard conditions
       Heat Capacity and Specific Heat
• The amount of energy required                  Calorimetry:
                                          Heat evolved/consumed by
  to raise the temperature of a           eaction causes heat change
  substance by 1 K (1C) is its         in surroundings (measurable) –
                                        observe a temperature change
  heat capacity
    Note:  100oC  101oC; DT = 1oC
           373.15K  374.15K; DT = 1K
    DT same for celsius and Kelvin
     Specific Heat and Molar Heat
               Capacity
• We define specific heat capacity (or simply
  specific heat) as the amount of energy
  required to raise the temperature of 1 g of
  a substance by 1 K (or 1oC)
                          (quantity _ of _ heat _ transferred )
 specific _ heat 
                   ( grams _ of _ subs tan ce)(temperature _ change)
         q
 s
     (m)(DT )

• The molar heat capacity is the heat
  required to raise the temperature of one
  mole of substance by 1 K
                      Problem
• The specific heat of iron metal is 0.450
  J/g.K. How many J of heat are required to
  raise the temperature of a 1.05-kg block of
  iron from 25.0oC to 88.5oC?

      q                               again, just to show…
s           q  msDT
   (m)(DT )                       DT  Tf  Ti
                                  DT  88.5o C  25.0o C  63.5o C
                   J 
q  1050g  0.450 . 63.5K 
            
                  gK  
                                  DT  361.65K  298.15K  63.5K

q  30003J  3.00 x10 4 J
                           Problem
• What is the heat capacity of 185 g of liquid
  water? (s = 4.184 J/g.K)
                  J 
            4.184 . 185g   774
                                    J
                 gK
                                  K

• What is the molar heat capacity of water?
 Have the specific heat of water (4.184 J/g.K). Want to know how much heat
 is needed to raise the temperature of 1 mol of water (18.0152g) by 1K

                     J  18.0152g _ H 2O             J
               4.184 . 
                                            75.375
                    g K  1mol _ H 2O 
                                                   mol. K
Calorimetry
       Since we cannot
       know the exact
       enthalpy of the
       reactants and
       products, we measure
       DH through
       calorimetry, the
       measurement of heat
       flow.
Constant Pressure Calorimetry
                           By carrying out a
                           reaction in aqueous
                           solution in a simple
                           calorimeter such as this
                           one, one can indirectly
                           measure the heat
      this is an example
     of an “open system”   change for the system
                           by measuring the heat
                           change for the water in
                           the calorimeter.
       Constant Pressure Calorimetry

                                               Because the specific heat
                                               for water is well known
                                               (4.184 J/g.K), we can
                                               measure DH for the reaction
                                               with this equation:

                                               Heat that flows into solution
                                               causes a temperature
                                               increase: qrxn=-qsol’n     “m” in this
                                                                              Equation is
                                                                              the mass of
                                                                              the solution
                                                        qrxn = -msDT

For an exothermic reaction, qrxn will be negative                      rxn = system
For an endothermic reaction, qrxn will be positive                 sol’n = surroundings
                  Problem
• In a coffee cup calorimetry experiment, two
  aqueous solutions are combined (50.0 mL of
  0.100 M AgNO3 and 50.0 mL of 0.100 M HCl).
  The following reaction occurs:

   AgNO3(aq) + HCl(aq)  AgCl(s) + HNO3(aq)

• The temperature is observed to increase from
  22.20oC to 23.11oC in the experiment. Calculate
  DH for this reaction in kJ/mol AgNO3 (assuming
  a total solution mass of 100.0g and a specific
  heat of 4.184 J/g.K)
                                qrxn  msDT
                                  Problem
    • Use the equation for the constant pressure
      calorimeter to solve for qrxn:           DT = 0.91oC
                 q rxn  msDT
                                          J 
                 q rxn
                                   
                                               
                          100g  4.184 o  23.11o C  22.2 o C   
                                         g. C 
                                               
                 q rxn    380.744J  0.380 kJ

    • …and then divide by the number of moles of
      AgNO3 (what the question asks you to do)

                      0.38 kJ
          DH                      76 .1 kJ / mol _ AgNO3
               0.0050 mol _ AgNO3

Remember C = n/V  n = CV
For AgNO3 solution: n = (0.100M)(0.05000L) = 0.00500 mol AgNO3
            Bomb Calorimetry
                 Constant Volume Calorimetry

Reactions can be
carried out in a sealed
“bomb,” such as this
one, and measure the
heat absorbed by the
water.     This is an example of
               a “closed system”

   q  C cal DT

Ccal is the heat capacity of the calorimeter
(often given or needs to be determined separately)
           Bomb Calorimetry

• Because the volume
  in the bomb
  calorimeter is
  constant, what is
  measured is really the
  change in internal
  energy, DE, not DH.
• For most reactions,
  the difference is very
  small.
        A bomb calorimeter problem
                                                In this problem, you are given Ccal

 • A 2.200-g sample of quinone was burned in a
   bomb calorimeter, whose heat capacity is 7.854
   kJ/oC. The temperature of the calorimeter
   changes from 23.44oC to 30.57oC. What is the
   heat of combustion per gram (and mole) of                                        O


   quinone?
                                    Per gram
                                                                                    O
                                                                    Per mole
q rxn  C cal DT                  56.0kJ
                                            25.4 54
                                                      kJ
                                   2.200g             g
                  kJ
q rxn    (7.854 o )(7.13o C )
                   C                         kJ  108.098 g                       kJ
                                    25.4 54                        2.751 x103
q rxn    56.0kJ                 
                                             g  1mol _ C 6 H 4 O2 
                                                                                  mol
                Calibration of the Bomb
    • Bomb calorimeter equations involve the Ccal
      term, which is determined through an
      experiment in which a quantity of a substance
      that releases a known amount of heat per gram
      is combusted in the calorimeter.
    • Example, combustion of 1.000g benzoic acid is
      known to release 26.38 kJ of heat. If
      combustion of 1.000g benzoic acid causes the
      temperature in a bomb calorimeter to rise by
      5.75oC,
                                    q rxn  C cal DT
Ccal is always positive (it’s the
                                               q rxn    (26 .38 kJ )          kJ
amount of heat energy needed        C cal                          4.58 7 o
to raise the calorimeter’s                     DT         5.75 o C               C
temperature by 1oC)
       2-Part Bomb Calorimeter Problem

 Use this        2.50g of benzoic acid is combusted in a
  info to        bomb calorimeter, causing the temperature
determine
    Ccal
                 to increase by 8.35oC. (benzoic acid is
                 known to release 26.38 kJ/g).
                 In a separate experiment, 2.00g of glucose
After finding
 Ccal, use it    (C6H12O6) is combusted using the same
with this info   calorimeter, producing a temperature
 to find qrxn    increase of 3.95oC. What is qrxn for the
  in kJ/mol
                 combustion of glucose, in kJ/mol?


                           q  Ccal DT
                             Solution
 • Must solve the problem in two basic parts
     – Ccal determination
     – qrxn for glucose combustion


Remember that combustion of 1.000g benzoic acid releases 26.38 kJ of heat…
how much heat does combustion of 2.50g benzoic acid create?



                          kJ 
                    26.38 2.50g   65.95 kJ
                  
                          g 
    First Part of Solution (Ccal)
• Using the heat evolved from the
  combustion of 2.50g benzoic acid, we can
  determine Ccal:

                          qrxn
               Ccal  
                          DT
                          (65.95 kJ )   Ccal is always going
               Ccal                       to be positive
                            8.35o C
                               kJ
               Ccal    7.898 o
                                C
                          Second Part
• Now that the calibration constant is known,
  can determine the heat evolved from the
  combustion of 2.00g glucose
                 q rxn  C cal DT                       Part II: “2.00g of glucose
                                                         is combusted using the
                                    kJ
                 q rxn  (7.898   o
                                       )(3.95o C )       same calorimeter,
                                     C                   producing a temperature
                                                         increase of 3.95oC”
                 q rxn    31.19 kJ

To find the heat in kJ/mol glucose:

                      31 .19 kJ  180 .0 g                  kJ
            qrxn                            2.80 7 x10 3
                        2g        1mol                      mol
           Food and Fuels
• Our body requires energy for
  – maintenance of body temperature
  – muscle contraction
  – tissue construction/repair


• We are able to get energy from the body’s
  breakdown of carbohydrates, fats, and
  proteins
             Carbohydrates
• Carbohydrates (e.g. starch) are transformed into
  sugars in the intestines (e.g. glucose, C6H12O6),
  which is absorbed into the blood and transported
  to the cells

         C6H12O6 + 6O2  6CO2 + 6H2O

• The breakdown of carbohydrates is fast, so this
  energy gets used quickly (carbohydrates tend
  not to get stored by the body)
• Average fuel value of carbohydrates is 17kJ/g
                     Fats
• Fats also break down into CO2 and H2O. For
  example, the combustion of C57H110O6:

    2C57H110O6 + 163O2  114CO2 + 110H2O

• Fats are water-insoluble, and so they are easily
  stored.
• Also, they have a higher average fuel value than
  carbohydrates (38kJ/g), and so are well-suited
  as a body’s energy reserve.
                   Proteins
• The combustion of proteins in a bomb
  calorimeter yields N-containing products
  (released as NO2)
• In the body, nitrogen is released mainly as urea
  (NH2)2CO
• Proteins are mainly used as building materials
  for organ walls, skin, hair, muscle, etc.
• The average fuel value for proteins is 17kJ/g
Energy in Foods
         Most of the fuel in the
         food we eat comes
         from carbohydrates
         and fats.
                                Fuels

   The vast majority
   of the energy
   consumed in this
   country comes
   from fossil fuels.

The greater the percentage of C and H
in a fuel, the higher its fuel value

								
To top