Question

Consider the decomposition of a metal oxide to its elements, where M represents a generic metal. M 3 O 4 ( s ) − ⇀ ↽ − 3 M ( s ) + 2 O 2 ( g ) What is the standard change in Gibbs energy for the reaction, as written, in the forward direction? Δ G ∘ rxn = kJ/mol What is the equilibrium constant of this reaction, as written, in the forward direction at 298 K? K = What is the equilibrium pressure of O2(g) over M(s) at 298 K? P O 2 = atm

Answer 1

a) ΔGrxn = 6.7 kJ/mol

b) K = 0.066

c) PO2 = 0.16 atm

Step-by-step explanation:

a) The reaction is:

M₂O₃ = 2M + 3/2O₂

The expression for Gibbs energy is:

ΔGrxn = ∑Gproducts - ∑Greactants

Where

M₂O₃ = -6.7 kJ/mol

M = 0

O₂ = 0

`deltaG_(rxn) =((2*0)+(3/2*0))-(1*(-6.7))=6.7kJ/mol`

b) To calculate the constant we have the following expression:

`lnK=-(deltaG_(rxn) )/(RT)`

Where

ΔGrxn = 6.7 kJ/mol = 6700 J/mol

T = 298 K

R = 8.314 J/mol K

`lnK=-(6700)/(8.314*298) =-2.704\\K=0.066`

c) The equilibrium pressure of O₂ over M is:

`K=P_(O2) ^(3/2) \\P_(O2)=K^(2/3) =0.066^(2/3) =0.16atm`