What is the equilibrium pressure of o2(g) over m(s) at 298 k?

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

M203(s)—> 2M(s) + 3/2 O2(g)

info given for Gf(kJ/mol):
M203= -6.70
M(s)=0
O2(g)= 0

what is the standard change in Gibbs energy for rxn as written in forward direction? (kJ/mol)
What is the equilibrium constant (K) of this rxn, as written in forward direction at 298K?

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

Answer

General guidance

Concepts and reason

• The difference between free energy of products and free energy of reactants is represented as standard free energy change of the reaction. • The ratio of concentration of a product to the ratio of concentration of a reactant is represented as equilibrium constant. The equilibrium constant is expressed as. • The equilibrium constant, in terms of partial pressure, is represented as equilibrium pressure.

Fundamentals

The standard Gibb’s free energy change can be calculated by: AG.com = {4G products - E46 Coretants
AG = Sum of Gibbs free energy change of products.
AGuna = Sum of Gibbs free energy ch The equilibrium constant in terms of Gibb’s free energy can be calculated by: AG= -RTIK R= gas constant (8.314J/mol.K)
T =Temperature
K=equilibrium constant. If the reaction contains gases, their concentrations are generally reported in terms of partial pressure. For a simple equilibrium reaction, aA +bB
cC+dD The equilibrium constant in terms of partial pressure can be represented as. K,= (C)*(D)
(A)(B)

Step-by-step

Step 1 of 3

Given reaction: M20,(3) —>2M(s)+20,(8) Given:
M20,(s)= -6.70 kJ/mol
M(s)=0
02 (8)= 0
Calculation:
AGC = 4G produto - 46. ctenos
= (2x0)+(x0)]-[1*-6.70]
AG = 6.70 kJ

The Gibb’s free energy change of products and reactants were given. Gibb’s free change for reaction in forward direction can be calculated by substituting the given values in Gibb’s free energy formula.

Step 2 of 3

Given:
T = 298 K
R=gas constant (8.314J/mol.K)
AGº=6.7 kJ/mol
= 6.7kJ x 1000)
= 6.7kJ * kJ
AGº=6700J/mol
Calculation:
Ink - 4

From the given temperature, the equilibrium constant K is calculated by substituting the given values in Gibb’s free energy formula with respect to equilibrium constant.

Step 3 of 3

M20,69) —>2M()+ 20:(8)
K = pẢ0, (8)
pO20)=K
K = 0.0669
PO26) = (0.0669)
p026) = 0.164 atm

The standard change in Gibb’s free energy for the reaction in forward direction is6.7 kJ/mol. The equilibrium constant K of this reaction is 0.0669. The equilibrium pressure of 02(g)over M(s) at 298K is 0.164 atm.


From the given reaction, only oxygen is gas. Therefore, the equilibrium pressure of oxygen is calculated in terms of partial pressure using equilibrium constant value.

Answer

The standard change in Gibb’s free energy for the reaction in forward direction is6.7 kJ/mol. The equilibrium constant K of this reaction is 0.0669. The equilibrium pressure of 02(g)over M(s) at 298K is 0.164 atm.

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