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Homework Help: Hess' Law, checking my answer

  1. Dec 18, 2005 #1


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    I don't have the answer to this, and it's a big question so I'd like to know if I got it right. Thanks

    Determine the [tex]\Delta H[/tex] value for reaction A:

    A)[tex]3CH_4 \rightarrow C_3H_8 + 2H_2[/tex]

    [tex]\Delta H=?[/tex]

    Okay so I was given a reference table and used these two equations for Hess' Law:

    B)[tex]C+2H_2 \rightarrow CH_4[/tex]

    [tex]\Delta H= -74.9[/tex]

    C)[tex]3C+4H_2 \rightarrow C_3H_8[/tex]

    [tex]\Delta H=-103.7[/tex]

    So I knew I the sum of B and C would have to give me A, so in order for the values to cancel properly I had to multiply B by three:

    B)[tex]3(C+2H_2 \rightarrow CH_4)[/tex]

    [tex]3(\Delta H=-74.9)[/tex]

    gives me:

    B')[tex]3C+6H_2 \rightarrow 3CH_4[/tex]

    [tex]\Delta H=-224.7[/tex]

    Then, since propane is on the left side of equation A, I would have to reverse B':

    [tex]3CH_4 \rightarrow 3C + 6H_2[/tex]

    [tex]\Delta H=+224.7[/tex]

    Now I can add them:

    [tex]3CH_4 +3C - 3C \rightarrow C_3H_8 + 6H_2 - 4H_2[/tex]


    [tex]3CH_4 \rightarrow C_3H_8 + 2H_2[/tex]

    which is equation A.

    So now I must add the change in enthalpy:

    [tex]\Delta H=+224.7+\Delta H=-103.7[/tex]

    [tex]\Delta H=121 kJ[/tex] for reaction A

    is that right?
  2. jcsd
  3. Dec 18, 2005 #2
    yes, it is.
  4. Dec 18, 2005 #3


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    thanks for all the help andrew
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