<u>Answer:</u>
<u>For 1:</u> The standard Gibbs free energy change of the reaction is 10.60 kJ/mol
<u>For 2:</u> The equilibrium constant for the given reaction at 298 K is
<u>For 3:</u> The equilibrium pressure of oxygen gas is 0.0577 atm
<u>Explanation:</u>
The equation used to calculate standard Gibbs free energy change of a reaction is:
For the given chemical reaction:
The equation for the standard Gibbs free energy change of the above reaction is:
We are given:
Putting values in above equation, we get:
Hence, the standard Gibbs free energy change of the reaction is 10.60 kJ/mol
Relation between standard Gibbs free energy and equilibrium constant follows:
where,
= Standard Gibbs free energy = 10.60 kJ/mol = 10600 J/mol (Conversion factor: 1 kJ = 1000 J )
R = Gas constant = 8.314 J/K mol
T = temperature = 298 K
= equilibrium constant = ?
Putting values in above equation, we get:
Hence, the equilibrium constant for the given reaction at 298 K is
The expression of for above equation follows:
The concentration of pure solids and pure liquids are taken as 1 in the expression. That is why, the concentration of metal and metal oxide is taken as 1 in the expression.
Putting values in above expression, we get:
Hence, the equilibrium pressure of oxygen gas is 0.0577 atm