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

M_{2}O_{3}(s) ⇌ 2M(s) + 3/2O_{2}(g)

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a) What is the standard change in the Gibbs energy for the reaction, as written, in the forward direction?

b) What is the equilibrium pressure of O_{₂}(g) over M(s) at 298K?

**Gibbs Energy Change**

The feasibility of a reaction is determined by an thermodynamic parameter, which is known as the Gibbs energy change (ΔG) . In order to make a feasible reaction, the it is necessary for the energy to lower than (negative). The equilibrium constant can be mathematically calculated through (ΔG).

Given Data:_{2}O_{3}(s) is −8.30kJ/mol._{2}(g) is 0.0kJ/molThe decomposition reaction in metal oxide is illustrated below.

M

_{2}O_{3}(s) ⇌ 2M(s) + 3/2O_{2}(g)a) A Gibbs-free energy change can be determined by the formula listed below.

ΔG

^{o}rxn = ∑m⋅ΔG^{o}_{f }(products) − ∑n⋅ΔG^{o}_{f }(reactants)Where,

^{o}_{f}(products) is the Gibbs energy formation of products.^{o}_{f}(reactants) is the Gibbs energy formation of reactants.For the specific reaction The Gibbs free energy is determined using the formula below.

ΔG

^{o}_{rxn }= [3/2.ΔG^{o}_{f}(O_{2}(g))+2ΔG^{o}_{f}(M(s))]−[1×ΔG^{o}_{f}(M_{2}O_{3}(s))]Substitute the required values in above formula.

ΔG

^{o}_{rxn }= [32×(0.00kJ/mol) + 2×(0.00kJ/mol)]−[1×(−8.30)kJ/mol]= 0kJ/mol−(−8.30kJ/mol)

= 8.30kJ/mol

Thus the change in energy free to trigger the reaction is 8.30kJ/mol

b) Since equilibrium constants are established in gaseous form, the term used to describe an equilibrium constant which is based on pressure works:

K=(PO

_{2})^{3/2}Where;

_{2}_{ }is the pressure of O_{2}.Substitute the values from the formula above.

P

_{O2 }= (0.035)^{3/2 }= 0.00655atmThus the pressure at equilibrium for O

_{2}is 0.00655atm