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Where possible, classify these systems as reactant-favored or product-favored at 298 K. If the direction cannot be determined from the information given, classify the reaction as "Insufficient information".

a)2A(g) + 3B(g) -->4C(g) delta H = -95kJ
b)2A(g) + 2B(g) -->3C(g) delta H = 254kJ
c)2A(g) + B(g) --> 4C(g) delta H =322 kJ
d)2A(g) + B(g) --> 3C(g) delta H -193kJ

I thought that a)reactant favored, b) reactant favored, c) product favored, and d) insufficient information because that in a and b there are more reactant than products and in c there are more products than reactant and d is equal amount. But I got the answer wrong. Could someone explain to me?

  • chemistry -

    I think D and B have definite answers.
    D. 3 mols left; 3 right. dS zero so TdS zero. that + a - dH gives dG - so we go with products favored.

    B. 4 mols left; 3 right. dS is - which makes -TdS + and that added to + dH means +dG so reactants favored.

    A and B are up for grabs. Here is what I would do.
    A. 5 mols left; 4 right so dS is - which makes -TdS +. So the question is can the -dH given overcome the + TdS term? If we guess that dS might be 50 and that x 298, then TdS might be in the neighborhood of 15,000 J. Even at 100 for dS, then TdS is in the neighborhood of 30,000. When added to -95,000 it still carries a - sign and products are favored. All of that depends upon how reasonable we are at guessing for the value of dS. Even at 300 for dS it still favors the products. So I would go with favored; however, since you don't really know you may feel better choosing that you can't tell.

    C. 3 mols left; 4 right so dS is + which makes TdS -. Add that to a large (322,000 J) means the TdS term has a lot to overcome to get a - number out of it. Go through the same reasoning as in A above. Even at 400 for dS, then TdS is about 120,000 and that isn't even close to making it -. I would go with reactants favored. I don't have enough experience in thermodynamics to know what reasonable values are for dH and dS.

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