Calorimetry Example If 10.0 g of Al metal is heated to 75.00 C and transferred to a coffee cup calor

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Thermochemistry
Calorimetry Example If 10.0 g of Al metal is
heated to 75.00 ?C and transferred to a coffee
cup calorimeter containing 95.0 g H2O at 25.00
?C , what is the specific heat of Al if the
contents of the calorimeter end up at 26.11
?C? Calorimetry Example try this one on
your own! If 25.0 g Fe at 95.00 ?C were added to
a coffee cup calorimeter containing 75.0 g H2O
at 28.00 ?C, what would be the final temperature
of the contents of the calorimeter when
equilibrium is established?
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Energy

• Enthalpy (H) is the heat energy associated with
  systems at constant pressure
• Most systems studied by chemists are at constant
  pressure
• ?H Hfinal-Hinitial qP the enthalpy change
  accompanying a change in a system is the
  amount of heat gained or lost by a system
  when the change occurs at constant pressure
• Enthalpy is a state function
• ?H 0 when enthalpy (qP) is gained by the system
• This is an endothermic process
• ?H
• This is an exothermic process
• Relationship between E and H
• E H - PV
• at constant pressure
• ?E ?H - P?V
• for many processes at constant
  pressure, ?V is small

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Thermochemistry

• Enthalpies of Reaction are the changes in
  enthalpy accompanying a chemical reaction
• Thermochemical equations include the usual
  balanced equation with the enthalpy change
  appended to the right of the products
• Example 2H2(g) O2(g) 2H2O(g) DH
  -483 kJ
• DHrxn H(products) - H(reactants)
• if DHrxn enthalpy is lost from the system to the
  surroundings
• DHrxn
• if DHrxn 0, H(products) H(reactants) and
  enthalpy is added to the system from the
  surroundings
• DHrxn 0 means the reaction is endothermic

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Thermochemistry

• Thermochemical Equations
• Enthalpy is an extensive property and it is
  understood that the equation is written on a
  mole basis
• 483 kJ are produced for 2 mol H2(g), 1 mol O2(g),
  2 mol H2O(g)
• 2H2(g) O2(g) 2H2O(g) DH -483
  kJ
• 4H2(g) 2O2(g) 4H2O(g) DH (-483
  kJ) x 2 -966 kJ
• If the equation is reversed, the enthalpy change
  will be the same size but opposite sign
• 2H2O(g) 2H2(g) O2(g) DH
  483 kJ
• The enthalpy change for a reaction depends on the
  state of the reactants and products and states
  must be included in the equation
• 2H2(g) O2(g) 2H2O(g)
  DH -483 kJ
• 2H2(g) O2(g) 2H2O(l )
  DH -571 kJ
• 2H2O(g) 2H2O(l ) DH - 88
  kJ

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Thermochemistry

• Hesss Law
• If one considers a reaction taking place in a
  series of steps, the enthalpy change for the
  overall reaction will be the sum of the enthalpy
  changes of the individual steps.
• This results from the fact that H is a state
  function.
• One must consider the physical state of all
  reactants and products.
• Example
• 2H2(g) O2(g) 2H2O(g)
  DH -483 kJ
• 2H2O(g) 2H2O(l )
  DH - 88 kJ
• (sum) 2H2(g) O2(g) 2H2O(l )
  DH -571 kJ
• Example From the following thermochemical
  equations, find the enthalpy change for NO(g)
  O2 (g) NO2 (g).
• N2 (g) O2 (g) NO (g) DH 90.37
  kJ
• N2 (g) O2 (g) NO2 (g) DH 33.8
  kJ

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Thermochemistry

• Step 1 write the equation for which the DH is
  sought
• NO(g) O2 (g) NO2 (g) DH ?
• Step 2 rearrange the equations for the steps so
  the correct reactants are on the left and the
  correct products are on the right.
• Some equations may be reversed this changes the
  sign on DH
• Some equations may be multiplied by an integer or
  fraction
• N2 (g) O2 (g) NO2 (g)
  DH 33.8 kJ
• NO (g) N2 (g) O2 (g)
  DH -90.37 kJ
• NO(g) O2 (g) NO2 (g)
  DH ?
• Step 3 Cancel substances in the equations above
  the line.
• Substances not in the final equation should all
  cancel
• The coefficients for some substances should
  cancel to give the coefficient for those
  substances in the final equation.
• O2(g) - O2 (g) O2 (g) on the left above
• N2 (g) - N2 (g) 0N2 (g) above N2(g) is not
  in the final equation

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Thermochemistry

• Step 4 add the DHs to get the DHrxn for the
  reaction.
• DHrxn33.8 kJ - 90.7 kJ -56.6 kJ
• Another Example From the following equations,
  calculate the DH for
• S(s) O2(g) SO2(g)
• 2SO2 (g) O2 (g) 2SO3 (g) DH
  -196 kJ
• 2S(s) 3O2 (g) 2SO3 (g) DH
  -790 kJ
• (2SO3 (g) 2SO2 (g) O2
  (g) DH 196 kJ) x
• (2S 3O2 (g) 2SO3 (g)
  DH -790 kJ) x
• S(s) O2(g) SO2(g)
  DH ?

1SO3 (g) 1SO2 (g) O2 (g) DH
(196 kJ) x 1S O2 (g) 1SO3
(g) DH (-790 kJ) x S(s)
O2(g) SO2(g) DH
? DHrxn (196 kJ) (-790 kJ) -297 kJ
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Thermochemistry

• Standard Enthalpy of Formation, ?Hf, is the
  enthalpy change associated with the formation of
  one mole of a compound from its elements in their
  standard states.
• The standard state is the most stable physical
  state the compound or element has at 298.15 K
  (25 C) and 1 atmosphere pressure
• Oxygen exists as O2 gas at 25 C
• Carbon exists as solid graphite at 25 C.
• Water is H2O(l ) in its standard state.
• ?Hf for elements is defined as zero when the
  elements are in their standard states.
• Examples
• H2(g) O2(g) H2O(l ) ?Hf
  -285.8 kJ/mol
• 3C(s) 4H2(g) C3H8(g) ?Hf -103.85
  kJ/mol
• Table 6.2, page 270, as does Appendix L in Kotz
  Treichel gives many ?Hfs without balanced
  chemical equations.

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Thermochemistry

• Other kinds of enthalpies
• Enthalpy of vaporization the enthalpy change
  accompanying the conversion of one mol of a
  substance from liquid to gas
• H2O(l ) H2O(g) ?Hvap 44 kJ/mol
• Enthalpy of fusion the enthalpy change
  accompanying the conversion of one mol of
  substance from solid to liquid.
• H2O(s) H2O(l ) ?Hfusion 6.008
  kJ/mol
• Enthalpy of combustion the enthalpy change
  accompanying the complete combustion of one mole
  of a substance.
• C7H8(l ) 9O2(g) 7CO2(g) 4H2O(l )
  ?Hcomb -3,910 kJ

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Thermochemistry

• Applications of enthalpies of formation
• Calculate enthalpy changes for other reactions
• Example calculate the ?Hrxn for the combustion
  of benzene, C6H6(l )
• 6C(s) 6O2(g) 6CO2(g)
  ?Hf 6(-393.5 kJ)
• 3H2(g) O2(g) 3H2O(l )
  ?Hf 3(-285.8 kJ)
• C6H6(l ) 6C(s) 3H2(g)
  ?Hf (-49.04 kJ)
• C6H6(l ) O2(g) 6CO2(g)
  3H2O(l ) ?Hcomb ?
• ?Hcomb 6(-393.5 kJ) 3(-285.8 kJ) (-49.04
  kJ) -3,267.4 kJ
• Notice we have added the heats of formation of
  CO2(g) and H2O(l ), weighted for their number of
  moles, and subtracted the heat of formation of
  C6H6(l ).
• This amounts to adding the enthalpies
  of formation of the products and subtracting the
  sum of the enthalpies of formation of reactants
  to get the heat of reaction. These enthalpies
  are weighted for their respective numbers of
  moles.

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Thermochemistry

• Calculate enthalpy changes for other reactions
  (cont.)
• This can be summarized
• ?Hrxn ?n ?Hf(products) - ?m ?Hf(reactants)
• where ? means the sum of
• n is the respective molar coefficient
  for each product
• m is the respective molar coefficient
  for each reactant
• Example find ?Hrxn for NH4NO3(s)
  N2(g) 2H2O(g) O2(g)
• ?Hrxn ?Hf(N2(g) ) 2?Hfo(H2O(g))
  ?Hf(O2(g)) -?Hf(NH4NO3(s))
• ?Hrxn 0 2 x (-241.8 kJ)
  0 - -365.6
• ?Hrxn -118 kJ

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Thermochemistry

• Constant-volume Calorimetry - the use of bombs!
• A bomb in this case is a sealed container,
  usually made from steel, in which a reaction
  can be carried out.
• The major kinds of reactions carried out in bombs
  are combustion reactions.
• It turns out that qV DE
• DH can be calculated from DH DE PDV, where DV
  is the change in volume that would occur if the
  process were carried out at constant pressure,
  P.
• Normally, the heat capacity, CcalorimeterCwaterC
  bomb, of the calorimeter must be known.
• Combust a quantity of a substance having a known
  qV.
• From Ccalorimeter and DT for a mass of another
  substance, qV DE can be calculated.

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Thermochemistry

• Constant-volume Calorimetry - the use of bombs!
• Example When benzoic acid, C7H6O2(s) is
  combusted in a bomb calorimeter, it has
  qv-26.38 kJ/g. The heat capacity of a bomb is
  determined by sealing 0.3684 g of benzoic acid
  in a bomb with excess O2 , immersing the bomb in
  855.0 g H2O. The temperature is allowed to come
  to equilibrium at 22.350 oC and the acid is
  combusted. After thermal equilibrium, the
  temperature is measured to be 24.500 oC.

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Thermochemistry

• Constant-volume Calorimetry - the use of bombs!
• Example The bomb from the previous example is
  used to determine the heat of combustion of
  naphthalene, C10H8(s). 0.1447 g of naphthalene is
  sealed in the bomb with excess O2 and the bomb
  is immersed in 825.3 g H2O. Combustion causes
  the temperature of the bomb and water to
  increase by 1.324 oC. What is the molar heat of
  combustion of naphthalene?

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Thermochemistry

• Constant pressure calorimetry
• Appropriate for measuring heat changes of
  reactions in solution
• Heat measurements give DH qP
• Heat gained by the solution (specific heat of
  soln) x (soln mass) x DT
• When a reaction occurs qrxn -qsoln
• If solution is a dilute aqueous solution,
  specific heat 4.18 J/g
• Use an adiabatic calorimeter all heat is
  contained within the calorimeter

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Thermochemistry
Calorimetry Example A student mixes 50.0 mL of
1.00 M HCl and 50.0 mL of NaOH in a coffee cup
calorimeter and finds the mixture goes from 21.00
?C to 27.50 ?C. Calculate the change in
enthalpy for the reaction, assuming that the
calorimeter loses only a negligible quantity of
heat, that the total volume of the solution is
100.0 mL, that its density is 1.000g/mL and its
specific heat is 4.18 J/g K. HCl(aq)
NaOH(aq) H2O(l ) NaCl(aq) DH
-5.4 kJ