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Energy & Chemical Reactions

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					          Chapter 6
Energy & Chemical Reactions




       General Chemistry I
             T. Ara
                        A. Energy
• Thermodynamics: science of heat, work & the
  transformations of one to the other

• Thermochemistry: the study of energy changes that
  occur during physical processes & chemical reactions

• Energy: the capacity to do work
   – All energy can be classified as either kinetic or potential
   2. Kinetic & Potential Energy
• Kinetic Energy: energy that something has
  because it is moving

  – Macroscale: mechanical energy – objects in
    motion (people, cars, baseballs, etc.)

  – Nanoscale: thermal energy – nanoscale objects in
    motion (atoms, molecules, ions, etc.)
     2. Kinetic & Potential Energy
• Potential Energy: energy that something has as a
  result of its position and some force that is capable of
  changing that position

   – Macroscale: gravitational energy (E = mgh)

   – Nanoscale: electrostatic energy (coulombic
     attraction between anions & cations) & chemical
     potential energy (attractions among atomic nuclei
     and electrons in molecules)
        2. Kinetic & Potential Energy
Gravitational potential energy
 being converted into kinetic
           energy.




                                   The chemical potential energy stored in
                                 food molecules being released (converted
                                           into thermal energy).
              a) Units of Energy
• The SI Unit for energy is the joule (J).
• The joule is an SI derived unit.

                  1 joule =      1 kgm2
                                    s2
                  (Energy = Force x Distance)



• It is frequently more practical to use a larger unit –
  the kilojoule (kJ)
                      1 kJ = 103 J
              a) Units of Energy
• Another common unit of energy is the calorie.
                  1 cal = 4.184 J

• Energy changes related to most chemical processes are
  reported in kilocalories (kcal) or kilojoules (kJ).

          1 kcal = 1000 cal = 4.184 kJ = 4184 J
              a) Units of Energy
• The label on this Australian packet of Equal indicates
  that it supplies 16 kJ of nutritional energy. How many
  “American Calories” (kilocalories) are in this packet
  of Equal?
                 a) Units of Energy
• The label on this Australian packet of Equal indicates
  that it supplies 16 kJ of nutritional energy. How
  many “American Calories” (kilocalories) are in this
  packet of Equal?


    1 kcal = 4.184 kJ

16 kJ x      1 kcal = 3.82409
            4.184 kJ

          = 3.8 Calories
       3. Conservation of Energy
• The first law of thermodynamics states that energy
  can be neither created nor destroyed – the total energy
  of the universe is constant.
• We will be studying the transfer of energy from one
  form or one place to another.

• How is energy
  transferred?

  Work & Heat
                           a) Work
• Doing work (or working on
  an object) is one process
  that transfers energy to an
  object.
   – For example:
      • lifting an object against the
        force of gravity – increases the
        potential energy of the object
      • throwing a baseball – increases
        the kinetic energy of an object
                      b) Heat
• Transferring heat to an object is another way of
  transferring energy.
• Heating, or raising the temperature of, an object
  increases the thermal energy of that object.

• Thermal energy is kinetic energy on the nanoscale.
   – Matter consists of nanoscale particles that are
     constantly in motion.
   – Higher temperature  faster motion of particles 
     more thermal energy
                       b) Heat
• Heating refers to the energy transfer that occurs
  whenever two samples of matter at different
  temperatures are brought into contact.

• Energy always transfers from the hotter sample to the
  cooler sample until both are at the same temperature – a
  state called thermal equilibrium.

                                                The hot steel bar
                                               transfers energy to
                                              the cool water until
                                                 the two come to
                                                      thermal
                                                   equilibrium.
           c) Defining the System
• When analyzing the thermodynamics of a process, it is
  crucial to keep track of ALL energy transfers taking
  place – work and heat.

• In doing so, it is helpful to define the sample of matter
  you are investigating as the system (eg. a reaction, an
  object, etc.)
• Anything that can exchange energy with the system is
  defined as the surroundings.
          i) System vs. Surroundings
• The total amount of energy must remain constant, so any energy
  transferred out of the system must be transferred into the
  surroundings, and vice versa.
        ii) Thermodynamic Changes
• We will refer to E (change in energy) when keeping
  track of energy transfers.

                Esystem = Efinal - Einitial

-The sign of E indicates the direction of transfer.
        ii) Thermodynamic Changes
• We will refer to E (change in energy) when keeping
  track of energy transfers.

                 Esystem = Efinal - Einitial

-The sign of E indicates the direction of transfer.

  Esystem > 0   Energy transferred into the system.
                    - Final E greater than initial E
  Esystem < 0   Energy transferred out of the system.
                     - Final E lower than initial E
    ii) Thermodynamic Changes

• In many simple cases:
     Esystem = qsystem + wsystem
          qsystem = heat transferred to the system
          wsystem = work done on the system
    ii) Thermodynamic Changes

• In many simple cases:
     Esystem = qsystem + wsystem
          qsystem = heat transferred to the system
          wsystem = work done on the system

   q>0    heat absorbed by the system
   q<0    heat given off by the system
    ii) Thermodynamic Changes
• In many simple cases:
     Esystem = qsystem + wsystem
          qsystem = heat transferred to the system
          wsystem = work done on the system

   q>0    heat absorbed by the system
   q<0    heat given off by the system

    w > 0 work done on the system
    w < 0 work done by the system
ii) Thermodynamic Changes
What is the change in internal energy of a system that does
        7.02 kJ of work and absorbs 888 J of heat?

                      E = q + w
What is the change in internal energy of a system that does
        7.02 kJ of work and absorbs 888 J of heat?

                      E = q + w

                 q = +888 J = 0.888 kJ
                      w = -7.02 kJ

                     E = q + w
                 E = 0.888 – 7.02
              E = -6.132 kJ = -6.13 kJ
           iii) Internal Energy
• The internal energy of a system is defined as
  the sum of the individual energies of all of the
  nanoscale particles in a sample of matter.

• Magnitude of internal energy depends on:
  – Temperature
  – Type of particles/molecules
  – Number of particles
                iii) Internal Energy
• As a hot cup of coffee cools, its
  internal energy decreases.
• The energy is transferred, in the
  form of heat, to the surrounding
  room.
• The room’s internal energy
  increases.

• If the coffee cools from 90 °C to 60
  °C (T = -30 °C), will the
  temperature in the room increase
  by 30 °C?
               B. Heat Transfer
1. Heat Capacity (C): quantity of energy required to
   increase the temperature of a sample by one degree

                       C = q/T

•    The magnitude of the heat capacity depends on:
    – Mass of the sample
    – Composition of the sample
              1. Heat Capacity
Calculate the heat capacity of an aluminum block that must
absorb 629 J of heat from its surroundings in order for its
temperature to rise from 22 C to 145 C.

    Heat Capacity (C):            C =q/T
              1. Heat Capacity
Calculate the heat capacity of an aluminum block that must
absorb 629 J of heat from its surroundings in order for its
temperature to rise from 22 C to 145 C.

    Heat Capacity (C):            C =q/T

            C = 629 J / (145 - 22 C)
              = 629 J / 123 C
              = 5.11 J/ C
               1. Heat Capacity
a) When comparing the heat capacities of different
   substances with different masses, it is more useful to
   compare specific heat capacities.

• Specific Heat (c): quantity of energy needed to
  increase the temperature of one gram of a substance
  by one degree Celsius

• Molar Heat Capacity (cm): related to specific heat,
  but for one mole of substance
1. Heat Capacity


                   Metals have low
                   specific heats.


                    Water has high
                     specific heat.
                    (4.184 J/gC or
                     1.000 cal/gC)
              1. Heat Capacity
When dealing with specific heat capacities (c):

    Given:         c = q/mT

    Derive:        q = cmT
                   T = Tfinal – Tinitial = q/cm
                   m = q/cT

                     q = thermal heat
                     c = specific heat
                    m = mass
                    T = change in temperature
If a 10.0 g sample of each substance below has 250 J
applied to it, which substance will have the greatest
increase in temperature?
     iron         (c = 0.46 J/gC)
     water (c = 4.184 J/gC)
     copper       (c = 0.39 J/gC)
     aluminum (c = 0.92 J/gC)
If a 10.0 g sample of each substance below has 250 J
applied to it, which substance will have the greatest
increase in temperature?
      iron        (c = 0.46 J/gC)
     water (c = 4.184 J/gC)
     copper       (c = 0.39 J/gC)
     aluminum (c = 0.92 J/gC)

Copper! Lower heat capacity means less energy
required to change the temperature – larger
temperature change.
              1. Heat Capacity
How much heat (in joules) does it take to raise the temperature
of 225 g of water from 25.0 to 100.0 C? (c = 4.184 J/gC)
              1. Heat Capacity
How much heat (in joules) does it take to raise the temperature
of 225 g of water from 25.0 to 100.0 C? (c = 4.184 J/gC)

                   q = cmT

    q = (4.184 J/gC)(225 g)(100.0 – 25.0 C)

                   q = 7.06x104 J
               1. Heat Capacity
What will be the final temperature of a 5.00 g silver ring at
37.0 C that gives off 25.0 J of heat to its surroundings
(c = 0.235 J/g C)?
               1. Heat Capacity
What will be the final temperature of a 5.00 g silver ring at
37.0 C that gives off 25.0 J of heat to its surroundings
(c = 0.235 J/g C)?

            T = Tfinal – Tinitial = q/cm

     Tfinal – 37.0 C = -25.0 J / (0.235 J/g C)(5.00 g)
     Tfinal – 37.0 C = -21.3 C
     Tfinal = 37.0 C - 21.3 C
     Tfinal = 15.7 C
              1. Heat Capacity
148 J of heat are transferred to a a piece of glass (c =
0.84 J/gC), raising the temperature from 25.0 C to
49.4 C. What is the mass of the glass?
              1. Heat Capacity
148 J of heat are transferred to a a piece of glass (c =
0.84 J/gC), raising the temperature from 25.0 C to
49.4 C. What is the mass of the glass?


           m = q/cT

           m = (148 J)/(0.84 J/gC)(24.4 C)
           m = 7.2 g
              2. Phase Changes
• We just saw that energy transfers ALWAYS
  accompany temperature changes.

• Energy transfers also accompany physical and
  chemical changes, even when there is no change in
  temperature.

• eg. Energy is always transferred into or out of a
  system during a phase change.
             a) Melting/Freezing
• During a phase change:
  – The temperature of the sample remains constant
  – Energy must be continually transferred into or out of
    the sample to facilitate the transformation

   – Melting: phase change from solid to liquid; energy is
     transferred into the sample
   – Freezing: phase change from liquid to solid; energy is
     transferred out of the sample
               a) Melting/Freezing
Heat of Fusion: quantity of thermal energy that must be
  transferred to a solid as it melts (qfusion = - qfreezing)



                                                               Water:

                                                        Heat of fusion =
                                                        +333 J/g at 0 C.

                                                       Specific Heat (l) =
                                                         1.00 cal/gC

                                                       Specific Heat (s)
                                                        depends on T:
                                                          0.5 cal/gC
                                                           near 0 C
           a) Melting/Freezing
• Adding energy to a sample can cause temperature
changes and/or phase changes.
                                • The energy required to
                                  melt a 1-gram ice cube
                                  at 0 °C would be
                                  enough to heat a 1-gram
                                  block of iron at 0 °C to
                                  738 °C (red hot) or to
                                  melt a 0.5-g ice cube
                                  and warm the resulting
                                  water to 80 °C.
    b) Vaporization/Condensation
• Vaporization: phase change from liquid to gas;
  energy is transferred into the sample
• Condensation: phase change from gas to liquid;
  energy transferred out of the sample

   Heat of Vaporization: energy that must be
    transferred to convert a liquid to a gas


             qvaporization = -qcondensation
     b) Vaporization/Condensation
• Heating the flask
  causes the water to
  boil – liquid turning
  to a vapor.
• When the heating is
  stopped, energy is
  transferred to the
  surroundings as the
  steam condenses back
  into liquid water.
How much heat is required to melt 125 g of ice at 0C?
How much heat is required to melt 125 g of ice at 0C?


 Water: Heat of fusion = 333 J/g at 0 C

         (125 g)(333 J/g) = 41625 J
                = 4.16x104 J
                 = 41.6 kJ
You have 1.00 g of ice at 0.0 C. How much energy would it take
 to melt all of the ice, warm the water from 0.0  C to 100.0  C,
          and boil the water to vapor (gas) at 100.0  C?
You have 1.00 g of ice at 0.0 C. How much energy would it take
 to melt all of the ice, warm the water from 0.0  C to 100.0  C,
          and boil the water to vapor (gas) at 100.0  C?

Phase 1 (Melting Ice):             Heat of Fusion = 333 J/g at 0 C
                 q = (1.00 g)(333 J/g) = 333 J

Phase 2 (Warming Water): c = 4.184 J/g C
                q = (4.184 J/gC)(1.00 g)(100.0 C)
                q = 418.4 J

Phase 3 (Vaporizing Water):        Heat of Vap. = 2260 J/g at 100 C

                 q = (1.00 g)(2260 J/g) = 2260 J

                 q = 333 + 418.4 + 2260 = 3011 J = 3.011 kJ
   3. Endothermic vs. Exothermic
• Processes involving energy transfers can be described
  as endothermic or exothermic.

endothermic: energy must be transferred into the
  system to maintain a constant temperature

exothermic: energy must be transferred out of the
  system to maintain a constant temperature
   3. Endothermic vs. Exothermic
• Melting and Vaporization are both endothermic
  processes (q > 0)
   – Heat must be transferred into the system to facilitate these
     phase changes
   – Heat is absorbed by the system

• Freezing and Condensation are both exothermic
  processes (q < 0)
   – Heat must be transferred out of the system to facilitate
     these phase changes
   – Heat is given off by the system
                C. Enthalpy (H)
• The quantity of thermal energy transferred into a system
  at constant pressure (qp) is called the enthalpy change
  (H).

   – Enthalpy is an extensive property used to quantify the
     heat evolved or absorbed during a physical or chemical
     change.
   – Extensive Property: dependent on the quantity of
     substance in the sample (the enthalpy of 2 mol of A is
     exactly twice the enthalpy of 1 mol of A)
                          C. Enthalpy
• Enthalpy is a state function – a property whose value is always
  the same if a system is in the same state.
• A change in a state function does not depend on the path by
  which the system changes from one state to another.

                                           This allows us to apply
                                          laboratory measurements
                                            to real-life situations.

                                          eg. The enthalpy change
                                          of a chemical reaction is
                                           the same – whether it
                                          occurs in a lab or in your
                                                    body.
      eg. Altitude is a state function.
     1. Enthalpy of Chemical Reactions
• We have already investigated energy transfers during
  physical transformations, what about chemical
  transformations – reactions?

                H = Hproducts – Hreactants

• An endothermic reaction (H > 0) is a reaction in
  which heat is absorbed from the surroundings.

• An exothermic reaction (H < 0) is a reaction in which
  heat is given off to the surroundings.
     1. Enthalpy of Chemical Reactions
• The combustion of hydrogen in the presence of oxygen
  is highly exothermic. This means that the energy of
  the reactants is higher than the energy of the products.
  The extra energy is released during the reaction.


2 H2 (g) + O2 (g) 

 2 H2O (g)
    2. Thermochemical Equations

• Thermochemical Equation: a balanced chemical
  equation (including phase labels) with the molar
  enthalpy of reaction written directly after the equation

 N2 (g) + 3 H2 (g)  2 NH3 (g); H = -91.8 kJ

a) Molar Interpretation: When 1 mol of nitrogen gas
   reacts with 3 mol of hydrogen gas to form 2 mol of
   ammonia gas, 91.8 kJ of energy is given off.
        b) Physical States of Matter
• Physical states are important in thermochemical equations.
• Phase changes are accompanied by enthalpy changes, so
  H depends on the phases of the reactants and products.


  H2 (g) + ½ O2 (g)  H2O (g) ; H = - 241.8 kJ

  H2 (g) + ½ O2 (g)  H2O (l) ; H = - 285.8 kJ
              c) Quantity of Matter
• Thermochemical equations obey the rules of stoichiometry
  – the larger the magnitude of the reaction, the more energy
  is transferred.


  H2 (g) + ½ O2 (g)  H2O (g) ; H = - 241.8 kJ

 2 H2 (g) + O2 (g)  2 H2O (g) ; H = - 483.6 kJ
   d) Manipulating Thermochemical
              Equations

• When a thermochemical equation is multiplied by any
  factor, the value of H for the new equation is
  obtained by multiplying the H in the original
  equation by that same factor.

• When a thermochemical equation is reversed, the
  value of H is reversed in sign.
   d) Manipulating Thermochemical
              Equations
Given the equation:
    3 O2 (g)  2 O3 (g) ; H = +285.4 kJ

Calculate H for the following equation:
         3/2 O2 (g)  O3 (g) ; H = ?
   d) Manipulating Thermochemical
              Equations
Given the equation:
    3 O2 (g)  2 O3 (g) ; H = +285.4 kJ

Calculate H for the following equation:
         3/2 O2 (g)  O3 (g) ; H = ?

          H = 285.4 kJ / 2 = +142.7 kJ
    d) Manipulating Thermochemical
               Equations
Given the equation:
 H2 (g) + Cl2 (g)  2 HCl (g) ; H = -184.6 kJ

Calculate H for the following equation:
      2 HCl (g)  H2 (g) + Cl2 (g) ; H = ?
    d) Manipulating Thermochemical
               Equations
Given the equation:
 H2 (g) + Cl2 (g)  2 HCl (g) ; H = -184.6 kJ

Calculate H for the following equation:
      2 HCl (g)  H2 (g) + Cl2 (g) ; H = ?

           H = - (-184.6) = + 184.6 kJ
   d) Manipulating Thermochemical
              Equations
• Given the following equation:
N2 (g) + 3 H2 (g)  2 NH3 (g); H = -91.80 kJ

Write the thermochemical equation for:
a) Formation of one mole of ammonia gas.
   d) Manipulating Thermochemical
              Equations
• Given the following equation:
N2 (g) + 3 H2 (g)  2 NH3 (g); H = -91.80 kJ

Write the thermochemical equation for:
a) Formation of one mole of ammonia gas.

  1/2 N2 (g) + 3/2 H2 (g)  NH3 (g); H = -45.90 kJ
   d) Manipulating Thermochemical
              Equations
• Given the following equation:
  N2 (g) + 3 H2 (g)  2 NH3 (g); H = -91.80 kJ

Write the thermochemical equation for:
b) Decomposition of 6 mol of ammonia gas.
   d) Manipulating Thermochemical
              Equations
• Given the following equation:
  N2 (g) + 3 H2 (g)  2 NH3 (g); H = -91.80 kJ

Write the thermochemical equation for:
b) Decomposition of 6 mol of ammonia gas.

    2 NH3 (g)  3 H2 (g) + N2 (g); H = +91.80 kJ

   6 NH3 (g)  9 H2 (g) + 3 N2 (g); H = +275.4 kJ
When 0.250 mol of solid CaO is mixed with 0.250 mol of liquid
water, solid Ca(OH)2 is formed and 16.3 kJ of heat is released.
 Write a thermochemical equation for the reaction producing 1
                      mol of Ca(OH)2.
When 0.250 mol of solid CaO is mixed with 0.250 mol of liquid
water, solid Ca(OH)2 is formed and 16.3 kJ of heat is released.
 Write a thermochemical equation for the reaction producing 1
                      mol of Ca(OH)2.

     0.250 CaO (s) + 0.250 H2O (l)  _____ Ca(OH)2 (s)

     0.250 CaO (s) + 0.250 H2O (l)  0.250 Ca(OH)2 (s)
                        H = -16.3 kJ

  [ 0.250 CaO (s) + 0.250 H2O (l)  0.250 Ca(OH)2 (s) ] x 4
                     [ H = -16.3 kJ ] x 4

           CaO (s) + H2O (l)  Ca(OH)2 (s)
                    H = -65.2 kJ
What is the enthalpy change when 22.5 g of CH4 are burned in
                         excess O2?
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(l); H= -890 kJ
 What is the enthalpy change when 22.5 g of CH4 are burned in
                          excess O2?
 CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(l); H= -890 kJ


  CH4:          12.011 + 4(1.008) = 16.043 g/mol

        22.5 g CH4 = 1.402 mol CH4
       16.043 g/mol

(1.402 mol CH4)(-890 kJ/mol) = -1247.78 kJ
                = -1.25x103 kJ
             D. Chemical Energy
• In an exothermic chemical reaction (H < 0), where
  does the energy come from?
• In an endothermic reaction (H > 0), where does the
  energy go?

• In a chemical reaction, chemical bonds are broken and
  new chemical bonds are formed.
• As nuclei and electrons move closer together or further
  apart, their energies change.
           1. Bond Enthalpies
• Bond Enthalpy (Bond Energy): the enthalpy
  change that occurs when two bonded atoms in a
  gas-phase molecule are separated completely at
  constant pressure
   – Bond breaking is always an endothermic
     process – energy must be put in to break a
     chemical bond.
   – Bond enthalpies are always positive.
   – The potential energy of atoms are lower when
     they are bonded together.
1. Bond Enthalpies

                 • Atoms that are
                   chemically
                   bonded
                   together are
                   thought of as
                   being in a
                   stable “energy
                   well.” (More
                   on this later.)
            1. Bond Enthalpies
• Bond breaking is always endothermic:
        Cl2 (g)  2 Cl (g) ; H = + 242 kJ

• Bond formation (the reverse reaction) is always
  exothermic:
         2 Cl (g)  Cl2 (g); H = - 242 kJ

• Bond enthalpies are usually expressed per mole of
  bonds.
     eg. The bond enthalpy (bond strength) of Cl2 is
     +242 kJ/mol.
                1. Bond Enthalpies
• Knowing the bond enthalpies for the bonds broken & formed in a
  chemical reaction allows you to predict whether the reaction will
  be endothermic or exothermic.
                 1. Bond Enthalpies
• Use the table of bond enthalpies to answer the
  questions.
                              a) Which of the molecules in the table
  Bond      Bond Enthalpy     has the strongest bond? The weakest
                              bond?
               (kJ/mol)   _
    H-H            436        b) Calculate the enthalpy change for the
    H-Br           366        following reactions:

    H-F            566
                              H-H + Br-Br  2 H-Br
    Br-Br          193        H-H + F-F  2 H-F
    F-F            158
                              c) Which is the most exothermic?
          1. Bond Enthalpies
• Strongest Bond?

        Highest Bond Enthalpy  HF

• Weakest Bond?

         Lowest Bond Enthalpy  F2
    1. Bond Enthalpies
H-H (g) + Br-Br (g)  2 H-Br (g) ; H = ?
            1. Bond Enthalpies
        H-H (g) + Br-Br (g)  2 H-Br (g) ; H = ?


Bonds Broken:     H-H           H = + 436 kJ/mol
                  Br-Br         H = + 193 kJ/mol
Bonds Formed:     H-Br          H = - 366 kJ/mol


     H = 436 + 193 – (2 x 366)
     H = - 103 kJ/mol
    1. Bond Enthalpies
H-H (g) + F-F (g)  2 H-F (g) ; H = ?
           1. Bond Enthalpies
      H-H (g) + F-F (g)  2 H-F (g) ; H = ?

Bonds Broken:   H-H         H = + 436 kJ/mol
                F-F         H = + 158 kJ/mol
Bonds Formed:   H-F         H = - 566 kJ/mol


     H = 436 + 158 –(2 x 566)
     H = - 538 kJ/mol
        1. Bond Enthalpies
H-H + Cl-Cl  2 H-Cl ; H = - 184 kJ/mol
 H-H + F-F  2 H-F ; H = - 538 kJ/mol

     Which is the most exothermic?
        1. Bond Enthalpies
H-H + Cl-Cl  2 H-Cl ; H = - 184 kJ/mol
 H-H + F-F  2 H-F ; H = - 538 kJ/mol

     Which is the most exothermic?

           Formation of H-F
  2. Calorimetry: Measuring Enthalpy
               Changes
• Thermochemical equations and bond enthalpies can
  be used to calculate H for a chemical reaction.

• Thermal energy transfers can also be measured
  directly using a calorimeter.

• Calorimeter: device for measuring the quantity of
  thermal energy transferred during a chemical reaction
  or some other process
2. Calorimetry: Measuring Enthalpy Changes
• When at least one of the reactants or products is a gas, a bomb
  calorimeter is used because it is closed to the atmosphere – at
  constant volume.

• System = burning sample (reaction)
• Surroundings = water & bomb

  qreaction = -(qbomb + qwater)

       qreaction = Ereaction
A bomb calorimeter has a heat capacity of 783 J/C and contains
254 g of water. How much energy is evolved by a reaction when
    the temperature of the calorimeter goes from 23.73C to
                            26.01C?
  A bomb calorimeter has a heat capacity of 783 J/C and contains
  254 g of water. How much energy is evolved by a reaction when
      the temperature of the calorimeter goes from 23.73C to
                              26.01C?
T = 2.28 C

qbomb = (heat capacity)(T) = (783 J/C)(2.28 C)
qbomb = 1785 J

qwater = cmT =(4.184 J/gC)(254 g)(2.28 C)
qwater = 2423 J

qreaction = -(qbomb + qwater) = - (1785 + 2423) = -4208 J
qreaction = -4.21x103 J = 4.21 kJ
  2. Calorimetry: Measuring Enthalpy
               Changes
• When a reaction takes place in solution, it is easier to
  use a calorimeter that is open to the atmosphere.

• This allows for the direct measurement of the
  enthalpy change (H – q at constant pressure).

• A cheap and easy way to do this is with a coffee-cup
  calorimeter.
            Coffee Cup Calorimeter
 Hreaction = -qsolution

- The energy transferred to
  the solution comes from
  the reaction.
- The masses of other
  substances in the solution
  are so small compared to
  the mass of the solvent
  (water) that their heat
  capacities can usually be
  ignored (c  4.184 J/gC)
          Coffee Cup Calorimeter
When a 13.0 g sample of NaOH (s) dissolves in 400.0 g water in
   a coffee cup calorimeter, the temperature of the water
   changes from 22.6 to 30.7 C. Calculate:
a) The heat transfer from “system” to “surroundings.”
b) H for the process: NaOH (s)  Na+ (aq) + OH- (aq)
           Coffee Cup Calorimeter
When a 13.0 g sample of NaOH (s) dissolves in 400.0 g water in
   a coffee cup calorimeter, the temperature of the water
   changes from 22.6 to 30.7 C. Calculate:
a) The heat transfer from “system” to “surroundings.”
b) H for the process: NaOH (s)  Na+ (aq) + OH- (aq)

qsolution = cmT
          = (4.184 J/g C)(400.0 + 13.0 g)(30.7 – 22.6 C)
          = (4.184 J/g C)(413.0 g)(8.1 C)
          = 13996.735 = 1.4x104 J
          Coffee Cup Calorimeter
When a 13.0 g sample of NaOH (s) dissolves in 400.0 g water in
   a coffee cup calorimeter, the temperature of the water
   changes from 22.6 to 30.7 C. Calculate:
a) qsolution = 13996.735 = 1.4x104 J
b) H for the process: NaOH (s)  Na+ (aq) + OH- (aq)

               H = qreaction / mols NaOH

              qreaction = - qsolution = - 1.4x104 J

    13.0 g NaOH x       1 mol    = 0.325 mol NaOH
                        39.9971 g
          Coffee Cup Calorimeter
When a 13.0 g sample of NaOH (s) dissolves in 400.0 g water in
   a coffee cup calorimeter, the temperature of the water
   changes from 22.6 to 30.7 C. Calculate:
a) qsolution = 13996.735 = 1.4x104 J
b) H for the process: NaOH (s)  Na+ (aq) + OH- (aq)

               H = (-13996.735)/(0.325 mol)
               H = - 4.3x104 J
               H = - 43 kJ
          Coffee Cup Calorimeter
A 19.6-g sample of metal was heated to 61.67 C and placed in
  26.70 g of water in a coffee cup calorimeter. The temperature
  of the water increased from 25.00 C to 30.00 C. What is the
  specific heat of the metal?

  - First, calculate the magnitude of the heat transferred to the
  water. Then, use qwater to calculate c for the metal.
          Coffee Cup Calorimeter
A 19.6-g sample of metal was heated to 61.67 C and placed in
  26.70 g of water in a coffee cup calorimeter. The temperature
  of the water increased from 25.00 C to 30.00 C. What is the
  specific heat of the metal?

  - First, calculate the magnitude of the heat transferred to the
  water. Then, use qwater to calculate c for the metal.

  qwater = cmT = (4.184 J/g C)(26.7 g)(5.00 C)
                = 558.56 J

                       qmetal = - qwater
          Coffee Cup Calorimeter
A 19.6-g sample of metal was heated to 61.67 C and placed in
  26.70 g of water in a coffee cup calorimeter. The temperature
  of the water increased from 25.00 C to 30.00 C. What is the
  specific heat of the metal?

       qwater = 558.56 J             qmetal = - qwater

qmetal = - 558.56 J = c(19.6 g)(30.00 – 61.67 C)

              c = (- 558.56 J) / (19.6 g)(- 31.67 C)

              c = 0.900 J/gC

				
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