Question 7 of 17 incorrect Incorrect Consider the decomposition of a metal oxide

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Question 7 of 17 incorrect Incorrect Consider the decomposition of a metal oxide to its elements, where M represents a generic metal. Map Substance (kJ/mol) M203 10.10 M(s) O2(g) What is the standard change in Gibbs energy for the reaction, as written, in the forward direction? Number AG 10.10 kJI mol What is the equilibrium constant of this reaction, as written, in the forward direction at 298 K? Number What is the equilibrium pressure of O2(g) over M(s) at 298 K? Number bare A Previous Give Up & view solution Check Answer 0 Next a Exit

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dG° = G°products - G°rectants

dG° = 2*M + 3/2*O2 - (M2O3)

dG° = 2*0+ 3/2*0 - (-10.10)

dG° = 10.10 kJ/mol

b)

for K, we can relate, in equilibrium:

dG = dG° + RTln(Q)

Q in equilbirium, Q = Kc

dG° = we know it, dG = 0 in equilibrium, so

dG° = -RT*ln(K)

K = exp(-dG/(RT))

dG = 10.10 kJ = 10100 J/mol

K = exp(-10100 /(8.314*298))

K = 0.01696

then...

equilibrium pressure of O2 over M(s)

Relate:

Kc = Kp

Kp = Kc*(RT)

Kp = 0.01696*(0.082*298) = 0.41443

then...

Kp = (P-O2)^(3/2) (only for O2, since all other are solids, so activities = 1)

0.41443 = (P-O2)^(3/2)

(0.41443)^(2/3) = (P-O2)

P-O2 = 0.55586 atm

1 atm = 1.01325 bar

P-O2 = 0.55586 atm * 1.01325 bar/atm = 0.5632 bar

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