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Posted by on Thursday, March 15, 2012 at 12:18am.

Consider the decomposition of a metal oxide to its elements, where M represents a generic metal.

M203(s)---> 2M(s) + 3/2 O2(g)

info given for Gf(kJ/mol):
M203= -10.60
O2(g)= 0

1.) what is the standard change in Gibbs energy for rxn as written in forward direction? (kJ/mol)

2.) What is the equilibrium constant (K) of this rxn, as written in forward direction at 298K?

3.) What is the equilibrium pressure of O2(g) over M(s) at 298K? (atm)

For 1.) I got 10.6 kJ/mol and this is correct
For 2.) I got K= -4.28 and it marked me wrong. Now I got an answer of .01387. Is this correct???
For 3.) I got PO2= 0.0016 atm and it marked me wrong......I don't know how it is wrong.

Could you please check #2 and #3 and tell me what I did wrong and what are the answers thanks.

  • Chemistry - , Thursday, March 15, 2012 at 12:50am

    I agree with your answer to #2.
    For #3, if we do
    0.01378 = pO23/2
    and I evaluate that as 0.05748 atm but that has too many s.f. and I would round to 0.0575 atm. I tried 0.05748^1.5 and that = 0.01378. Check that carefully.

  • Chemistry - , Thursday, March 15, 2012 at 12:56am

    Yes it is correct! Thanks! :) I see my mistake!

  • Chemistry - , Tuesday, June 11, 2013 at 6:38pm

    hello mon amie

  • Chemistry - , Monday, June 2, 2014 at 12:38am

    Bonjour mon ami... Ca va? La chemie est un excellent science n'est-ce pas? Alors je vais partir maintenant, aie une bonne vie!

  • Chemistry - , Saturday, April 4, 2015 at 1:10pm

    @DrBob222 , can you explain how exactly you evaluated to get that number because im confused howd you got there

  • Chemistry - , Sunday, April 26, 2015 at 12:59am

    I see what asker did wrong too, because she left it incomplete at first.

    Since Kelsey found #1, 10.6, I'll leave it at that.

    Part 2: (Gibbs energy)= -RTln(K)
    -> 10.6 kJ/mol= -(8.314 J/molxK)(298K)ln(K)
    -> 10.6 x 10^3 (J/mol) = -(2477.6 J/mol)ln(K)
    -> Divide "-RT" (which is -2477.6 J/mol) from both sides
    -> -4.278 = ln(K)
    -> e^(-4.278) = K (because e and ln cancels out)
    -> 0.0139

    From what I see...
    Part 3: pO2 is (pO2)^(3/2) because of the 3/2 moles in M203(s)---> 2M(s) + 3/2 O2(g). So...
    ->(pO2)= 0.0139
    -> [(pO2)^(3/2)]^(2/3)= (0.0139)^(2/3)
    -> pO2= 0.0578

  • Chemistry - , Sunday, April 26, 2015 at 1:09am

    Sorry, for part three, it is not only because of 3/2 moles from O2(g). Part 3 is asking for "equilibrium pressure of O2(g) over M(s) at 298K? (atm)." Although it is 3/2 nevertheless, because M(s) also has 1 mole, it is due to O2(g)/M(s) so (3/2)/(1) =3/2. Please correct me if my thinking is wrong.

  • Chemistry - , Sunday, April 26, 2015 at 1:14am

    Oh, how embarrassing for the spam, but what I stated earlier, just above, was a work of negligence. Because I quickly looked at the problem, I mistook M203 for M(s). If only there was an editing button...I'll just leave my work up in the open, as I was trying to follow DrBob222's reasoning.

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