Lecture 4: Hess� Law by 0K3gdl

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									       Lecture 4: Hess’ Law
• Reading: Zumdahl 9.5

• Outline
  – Definition of Hess’ Law
  – Using Hess’ Law (many examples)
           Hess’ Law Defined
• From lecture 3: Enthalpy is a state function. As
  such, DH for going from some initial state to some
  final state is pathway independent.

• Hess’ Law: DH for a process involving the
  transformation of reactants into products is not
  dependent on pathway. Therefore, we can pick a
  single-step pathway to calculate DH for a reaction.
Hess’ Law: An Example
                   Using Hess’ Law
                                 • When calculating DH for
N2 (g) + 2O2 (g)     2NO 2 (g)     a chemical reaction as a
                                   single step, we can use
      2NO 2 (g)
                                   combinations of
                                   reactions as “pathways”
                            q      to determine DH for our
  N2 (g) + 2O2 (g)                 “single step” reaction.
             Example (cont.)
• Our reaction of interest is:
  N2(g) + 2O2(g)       2NO2(g)     DH = 68 kJ


• This reaction can also be carried out in two
  steps:

 N2 (g) + O2 (g)     2NO(g) DH = 180 kJ
2NO (g) + O2 (g)     2NO2(g) DH = -112 kJ
            Example (cont.)
• If we take the previous two reactions and add
  them, we get the original reaction of interest:

 N2 (g) + O2 (g)      2NO(g) DH = 180 kJ
2NO (g) + O2 (g)      2NO2(g) DH = -112 kJ

 N2 (g) + 2O2 (g)      2NO2(g) DH =       68 kJ
             Example (cont.)
• Note the important things about this example, the
  sum of DH for the two reaction steps is equal to
  the DH for the reaction of interest.

• Big point: We can combine reactions of known
  DH to determine the DH for the “combined”
  reaction.
         Hess’ Law: Details
• One can always reverse the direction of a
  reaction when making a combined reaction.
  When you do this, the sign of DH changes.

   N2(g) + 2O2(g)      2NO2(g)    DH = 68 kJ

    2NO2(g)         N2(g) + 2O2(g) DH = -68 kJ
               Details (cont.)
• The magnitude of DH is directly proportional to
  the quantities involved (it is an “extensive”
  quantity).
• As such, if the coefficients of a reaction are
  multiplied by a constant, the value of DH is also
  multiplied by the same integer.


  N2(g) + 2O2(g)          2NO2(g)        DH = 68 kJ

 2N2(g) + 4O2(g)          4NO2(g)       DH = 136 kJ
           Using Hess’ Law
• When trying to combine reactions to form a
  reaction of interest, one usually works
  backwards from the reaction of interest.

• Example:
  What is DH for the following reaction?
    3C (gr) + 4H2 (g)       C3H8 (g)
                   Example (cont.)
       3C (gr) + 4H2 (g)         C3H8 (g) DH = ?

   •   You’re given the following reactions:
       C (gr) + O2 (g)      CO2 (g)    DH = -394 kJ

C3H8 (g) + 5O2 (g)       3CO2 (g) + 4H2O (l)   DH = -2220 kJ

  H2 (g) + 1/2O2 (g)       H2O (l)    DH = -286 kJ
             Example (cont.)
• Step 1. Only reaction 1 has C (gr).
  Therefore, we will multiply by 3 to get the
  correct amount of C (gr) with respect to our
  final equation.
                 Initial:
   C (gr) + O2 (g)            CO2 (g)   DH = -394 kJ
                     Final:
   3C (gr) + 3O2 (g)            3CO2 (g)   DH = -1182 kJ
               Example (cont.)
  • Step 2. To get C3H8 on the product side of
    the reaction, we need to reverse reaction 2.
                      Initial:
C3H8 (g) + 5O2 (g)      3CO2 (g) + 4H2O (l)   DH = -2220 kJ

                      Final:
3CO2 (g) + 4H2O (l)       C3H8 (g) + 5O2 (g) DH = +2220 kJ
               Example (cont.)
  • Step 3: Add two “new” reactions together
    to see what is left:
      3C (gr) + 3O2 (g)        3CO2 (g)   DH = -1182 kJ
3CO2 (g) + 4H2O (l)       C3H8 (g) + 5O2 (g) DH = +2220 kJ
                                  2
 3C (gr) + 4H2O (l)        C3H8 (g) + 2O2 DH = +1038 kJ
                Example (cont.)
• Step 4: Compare previous reaction to final
  reaction, and determine how to reach final
  reaction:
3C (gr) + 4H2O (l)    C3H8 (g) + 2O2 DH = +1038 kJ
 H2 (g) + 1/2O2 (g)    H2O (l)       DH = -286 kJ

  3C (gr) + 4H2 (g)      C3H8 (g)

       Need to multiply second reaction by 4
                Example (cont.)
• Step 4: Compare previous reaction to final
  reaction, and determine how to reach final
  reaction:
3C (gr) + 4H2O (l)    C3H8 (g) + 2O2 DH = +1038 kJ
 4H2 (g) + 2O2 (g)     4H2O (l)       DH = -1144 kJ

  3C (gr) + 4H2 (g)       C3H8 (g)
                Example (cont.)
• Step 4 (cont.):

 3C (gr) + 4H2O (l)   C3H8 (g) + 2O2 DH = +1038 kJ
 4H2 (g) + 2O2 (g)    4H2O (l)       DH = -1144 kJ


  3C (gr) + 4H2 (g)     C3H8 (g)    DH = -106 kJ
             Another Example
• Calculate DH for the following reaction:
      H2(g) + Cl2(g)     2HCl(g)
 Given the following:
  NH3 (g) + HCl (g)      NH4Cl(s) DH = -176 kJ
  N2 (g) + 3H2 (g) 2NH3 (g) DH = -92 kJ
  N2 (g) + 4H2 (g) + Cl2 (g) 2NH4Cl(s) DH = -629 kJ
      Another Example (cont.)
• Step 1: Only the first reaction contains the
  product of interest (HCl). Therefore,
  reverse the reaction and multiply by 2 to get
  stoichiometry correct.
   NH3 (g) + HCl (g)   NH4Cl(s) DH = -176 kJ

  2NH4Cl(s)     2NH3 (g) + 2HCl (g) DH = 352 kJ
      Another Example (cont.)
• Step 2. Need Cl2 as a reactant, therefore,
  add reaction 3 to result from step 1 and see
  what is left.
   2NH4Cl(s)      2NH3 (g) + 2HCl (g) DH = 352 kJ
 N2 (g) + 4H2 (g) + Cl2 (g)    2NH4Cl(s) DH = -629 kJ

  N2 (g) + 4H2 (g) + Cl2 (g)    2NH3(g) + 2HCl(g)
                   DH = -277 kJ
          Another Example (cont.)
  • Step 3. Use remaining known reaction in
    combination with the result from Step 2 to
    get final reaction.
N2 (g) + 4H2 (g) + Cl2 (g)   2NH3(g) + 2HCl(g)       DH = -277 kJ

       ( N2 (g) + 3H2(g)      2NH3(g)              DH = -92 kJ)


                 H2(g) + Cl2(g)      2HCl(g)         DH = ?

           Need to take middle reaction and reverse it
          Another Example (cont.)
  • Step 3. Use remaining known reaction in
    combination with the result from Step 2 to
    get final reaction.
N2 (g) + 4H2 (g) + Cl2 (g)    2NH3(g) + 2HCl(g)   DH = -277 kJ
        1
         2NH3(g)         3H2 (g) + N2 (g)         DH = +92 kJ


       H2(g) + Cl2(g)      2HCl(g)                DH = -185 kJ

								
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