Posted by Kelsey 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):
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 - DrBob222, 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 - Kelsey, Thursday, March 15, 2012 at 12:56am
Yes it is correct! Thanks! :) I see my mistake!
- Chemistry - Bonjour, Tuesday, June 11, 2013 at 6:38pm
hello mon amie
- Chemistry - marie, 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 - Sophia, 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 - Lore, 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)
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)^(3/2)]^(2/3)= (0.0139)^(2/3)
-> pO2= 0.0578
- Chemistry - Lore, 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 - Lore, 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.
Answer This Question
More Related Questions
- Chemistry - Consider the decomposition of a metal oxide to its elements where m ...
- Chemistry - Consider the decomposition of a metal oxide to its elements, where M...
- chemistry pls help me asap - 2.08g of metal Y react with 0.96g of oxygen to form...
- Chemistry Problem - Consider an ionic compound, MX, composed of generic metal M ...
- Chemistry - Aluminum metal reacts with iron (III) oxide to produce aluminum ...
- Chemistry - In an experiment 1.015 g of a metal carbonate, containing an unknown...
- Chemistry - 0.356 grams of an unknown metal reacts with oxygen to form 452 ...
- chemistry - when aluminum metal reacts with iron (III) oxide to form aluminum ...
- Chemistry - I need help classifying these as physical and chemical properties! 1...
- Chemistry - When 1.187g of a metallic oxide is reduced with excess hydrogen, 1....