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)
 

Temperature101

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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)
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.
 
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Jun 26, 2010
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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)
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.
 
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paul411

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Kaplan tells me that this reaction demonstrates a decrease in entropy
Temperature101 said:
This must be a mistake...Two moles of H2(g) as a reactant and one mole of CH4(g) as product should be a decrease in entropy.
MD Odyssey said:
Kaplan is wrong... ΔS = -233.2 J / mol K
negative ΔS as in decrease in entropy

Kaplan is right?
 
Jul 27, 2011
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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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