Question

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

M2O (s) ⇌ 2 M (s) + 1/2 O2 (g)

Substance: ----> ∆G°f (kg/mol): M2O(s) ----------->-7.00

M(s) --------------> 0

O2(g) --------------->0

1. What is the standard change in Gibbs energy for the reaction as written in the forward direction? ∆G°rxn = ? kj/mol

2. What is the equilibrium constant for this reaction as written in the forward direction at 298 K? K = ?

3. What is the equilibrium pressure of O2 (g) over M (s) at 298 K? PO2 = ? atm

Answer 1

1)

Lets calculate the standard change in Gibbs energy

G⁰ = G_f⁰_(products) - G_f⁰_(reactants)

= 0 - (-7)

= 7

2) to calculate equilibrium constant K, at 298K

lets use the formula

G = -RT ln(K)

R= 8.314 J/Kmol

T= 298K

so

ln (K) = G / (-RT)

= 7/ (-0.008314   298)

ln (K) = -2.83

so

e^{ln (K)} = K = e^-2.83

K = 0.06

3) At equilibrium

G⁰ = -RT Ln(K_p)

so from above (2)

K_p = 0.06 atm