Asked by Miaow

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

M2O3 (s) <=> 2M (s) + 3/2 O2 (g)

Given a delta Gf: -770 Kj for the M2O3
0 Kj for the M
0 Kj for the O2

What is the standard change in Gibbs energy for the reaction, as written, in the forward direction?

* I think -770 Kj ??

What is the equilibrium constant of this reaction, as written, in the forward direction at 298 K?

* K is products over reactants which leaves me with 0 ???

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

* I honestly have no idea and am not even sure how to approach this!

Any information is greatly appreciated, thank you!
Answered by DrBob222
Right digits; wrong sign.
dGrxn = (n*dG products) - (dG reactants) = (0) - (-770) = +770.

dG = -RT*lnK
Substitute and solve for Kp.

Kp = pO2<sup>3/2</sup>


Answered by Miaow
Thank you VERY much!
Answered by J
I still don't understand what to do after setting the K= pO2^(3/2). What do you substitute into the O2?
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