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Change in Entropy of Reaction

Discussion in 'MCAT Study Question Q&A' started by kkentm, Jul 27, 2011.

  1. kkentm

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    Kaplan tells me that this reaction demonstrates a decrease in entropy... I don't get it, unless im supposed to ignore the solid carbon for some reason?

    C(s)+2H2(g)→CH4(g)
     
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  3. Temperature101

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    Edit...
    Kaplan is right: 2 moles of gas (2H2) in the reactant side to one mole of gas (CH4) in the product side is a decrease in entropy. You dont even need to know the entropy values of these compounds.
     
    #2 Temperature101, Jul 27, 2011
    Last edited: Jul 27, 2011
  4. MD Odyssey

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    Kaplan is correct. I looked up the values for S at standard state for these compounds in the CRC. Here they are (all units in J / mol K)

    CH4 (g) = 186.3
    H2 (g) = 130.7
    C (s) = 158.1

    ΔS = ΔS(products) - ΔS(reactants)

    Thus,

    ΔS = 186.3 - (158.1 + 2(130.7)) = -233.2 J / mol K

    Just to be clear, there is more than one value for the entropy of the compound depending upon the phase and whether or not solid carbon is present as diamond or graphite. Naturally, these are the values for gases and graphite.
     
    #3 MD Odyssey, Jul 27, 2011
    Last edited: Jul 27, 2011
  5. paul411

    paul411 ANES
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    negative ΔS as in decrease in entropy

    Kaplan is right?
     
  6. MD Odyssey

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    I misspoke. The change in entropy is negative, which is precisely what the original poster was asking. Thanks.
     
  7. DrDotooMuch

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    I just think of it going from more moles of gas to less moles of g should be a decrease in entropy. 2 mole of gas is 2x(number of atoms in 1 mole of any substance). Twice as many atoms means more bumping around and more disorder. Sorry for the simple explanation. Its how I would have answered the question.
     

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