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.
[tex]M_3O_4(s)\rightleftharpoons 3M(s)+2O_2(g)[/tex]
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:
[tex]PCl_3(l)\rightarrow PCl_3(g)[/tex]
First we have to calculate the Gibbs energy of reaction [tex](\Delta G^o)[/tex].
[tex]\Delta G^o=G_f_{product}-G_f_{reactant}[/tex]
[tex]\Delta G^o=[n_{M(s)}\times \Delta G^0_{(M(s))}+n_{O_2(g)}\times \Delta G^0_{(O_2(g))}]-[n_{M_3O_4(s)}\times \Delta G^0_{(M_3O_4(s))}][/tex]
where,
[tex]\Delta G^o[/tex] = Gibbs energy of reaction = ?
n = number of moles
Now put all the given values in this expression, we get:
[tex]\Delta G^o=[3mole\times (0kJ/mol)+2mole\times (0kJ/mol)]-[1mole\times (-9.50kJ/K.mol)][/tex]
[tex]\Delta G^o=9.50kJ/mol[/tex]
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:
[tex]\Delta G^o=-RT\times \ln K[/tex]
where,
[tex]\Delta G^o[/tex] = 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 = ?
[tex]9500J/mol=-(8.314J/K.mol)\times (298K)\times \ln (K)[/tex]
[tex]K=0.0216[/tex]
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:
[tex]K=(P_{O_2})^2[/tex]
[tex]0.0216=(P_{O_2})^2[/tex]
[tex]P_{O_2}=0.147atm[/tex]
The equilibrium pressure of O₂(g) is, 0.147 atm
The free energy change of the reaction is 9.40 kJ/mol. The equilibrium constant is 0.0225. The partial pressure of oxygen is 0.15 atm.
The equation is represented as follows;
M3O4 ( s ) ⇄ 3M ( s ) + 2O2 ( g )
The enthalpy change of the reaction is obtained using the relation;
ΔG∘rxn = ΔGf∘products - ΔGf∘reactants
ΔG∘rxn =[0 kJ/mol- 0kJ/mol] - (-9.40 kJ/mol)
ΔG∘rxn = 9.40 kJ/mol
Now;
ΔG∘ = -RTlnK
lnK = -ΔG∘/RT
K = e^-ΔG∘/RT
K = e^-(9.40 × 10^3J/mol/8.314 J/Kmol × 298 K)
K = 0.0225
Now;
K =(pO2)^2
pO2 = √K
pO2 =√ 0.0225
pO2 = 0.15 atm
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