Iron (iii) oxide can be reduced to metallic iron by beating it in the presence of carbon. Balance the following equation, then d
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This is an incomplete question, here is a complete question.
Consider the decomposition of a metal oxide to its elements, where M represents a generic metal.
Substance ΔG°f (kJ/mol)
M₃O₄ -9.50
M(s) 0
O₂(g) 0
What is the standard change in Gibbs energy for the reaction, as written, in the forward direction? delta G°rxn = kJ / mol.
What is the equilibrium constant of this reaction, as written, in the forward direction at 298 K?
What is the equilibrium pressure of O₂(g) over M(s) at 298 K?
Answer :
The Gibbs energy of reaction is, 9.50 kJ/mol
The equilibrium constant of this reaction is, 0.0216
The equilibrium pressure of O₂(g) is, 0.147 atm
Explanation :
The given chemical reaction is:
First we have to calculate the Gibbs energy of reaction .
where,
= Gibbs energy of reaction = ?
n = number of moles
Now put all the given values in this expression, we get:
The Gibbs energy of reaction is, 9.50 kJ/mol
Now we have to calculate the equilibrium constant of this reaction.
The relation between the equilibrium constant and standard Gibbs free energy is:
where,
= standard Gibbs free energy = 9.50kJ/mol = 9500 J/mol
R = gas constant = 8.314 J/K.mol
T = temperature = 298 K
K = equilibrium constant = ?
The equilibrium constant of this reaction is, 0.0216
Now we have to calculate the equilibrium pressure of O₂(g).
The expression of equilibrium constant is:
The equilibrium pressure of O₂(g) is, 0.147 atm