ttp/www saplingleaming.com/ibiscmu/mod/bswiew.ho CConvervations Fort Hays State

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ttp/www saplingleaming.com/ibiscmu/mod/bswiew.ho CConvervations Fort Hays State : Print Calculator hd Periodic Table estion 17 of 20 sapin9,learning AG Substance (kJ/mol) M20(s)-6.90 M(S) OAg) Consider the decomposition of a metal oxide to its elements, where M represents a generic metal 0 0 M,O What is the standard change in Gibbs energy for the reaction, as written, in the forward direction? Number AG kJ/mol What is the equilibrium constant of this reaction, as written, in the forward direction at 298 K? Number K- What is the equilibrium pressure of OAg) over M(s) at 298 K? Number atm o, Hint Previous Give up & View Solution 9 Check Answer 0 Next DOLL

Explanation / Answer

This reaction is the reverse of the formation of M2O(s), so the DG0rxn is going to be 6.9 kJ/mol.

The equilibrium constant can be calculated with this equation:

DG0= -RTLnK

DG0= 6.9= -8.314x10-3kJ/mol.K x 298K LnK

K= 0.062

Now let´s write the equation for K, you must know that liquids and solids are not included into the equilibrium constant:

K= PO2= 0.0621/2 = 0.248 atm

Let me know if you have any doubt

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