Predict whether the change in entropy, for the following reaction will be positive or negative H2(g) + (1/2) O2(g) → H2O(g) . Why is this positive, even though there are less mols in the product?

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Predict whether the change in entropy, for the following reaction will be positive or negative H2(g) + (1/2) O2(g) → H2O(g) . Why is this positive, even though there are less mols in the product?

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Answer 1 · 2018-03-31T21:05:35

Because you got lied to? #DeltaS_(rxn)^@ = -"44.37 J/K"#...

By pretending that these are ideal gases (in addition, that they are spherical particles), we would predict #Deltan_"gas" = -1/2# so that we might think #DeltaS < 0#... and it is...

#"H"_2(g) + 1/2"O"_2(g) -> "H"_2"O"(g)#

Here are the entropies:

And thus,

#color(blue)(DeltaS_(rxn)^@) = sum_P S_P^@ - sum_R S_R^@#

#= "1 mol" cdot "188.72 J/mol"cdot"K" - ["1 mol" cdot "130.57 J/mol"cdot"K" + 1/2 "mols" cdot "205.03 J/mol"cdot"K"]#

#= color(blue)(-"44.37 J/K")#

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Predict whether the change in entropy, for the following reaction will be positive or negative H2(g) + (1/2) O2(g) → H2O(g) . Why is this positive, even though there are less mols in the product?

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