Respuesta :
Answer:
For 1: The standard Gibbs free energy change of the reaction is 10.60 kJ/mol
For 2: The equilibrium constant for the given reaction at 298 K is [tex]1.386\times 10^{-2}[/tex]
For 3: The equilibrium pressure of oxygen gas is 0.0577 atm
Explanation:
- For 1:
The equation used to calculate standard Gibbs free energy change of a reaction is:
[tex]\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f_{(product)}]-\sum [n\times \Delta G^o_f_{(reactant)}][/tex]
For the given chemical reaction:
[tex]M_2O_3(s)\rightarrow 2M(s)+\frac{3}{2}O_2(g)[/tex]
The equation for the standard Gibbs free energy change of the above reaction is:
[tex]\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(M(s))})+(\frac{3}{2}\times \Delta G^o_f_{(O_2(g))})]-[(1\times \Delta G^o_f_{(M_2O_3(s))})][/tex]
We are given:
[tex]\Delta G^o_f_{(M_2O_3(s))}=-10.60kJ/mol\\\Delta G^o_f_{(M(s))}=0kJ/mol\\\Delta G^o_f_{(O_2(g))}=0kJ/mol[/tex]
Putting values in above equation, we get:
[tex]\Delta G^o_{rxn}=[(2\times (0))+(\frac{3}{2}\times (0))]-[(1\times (-10.60))]\\\\\Delta G^o_{rxn}=10.60kJ/mol[/tex]
Hence, the standard Gibbs free energy change of the reaction is 10.60 kJ/mol
- For 2:
Relation between standard Gibbs free energy and equilibrium constant follows:
[tex]\Delta G^o=-RT\ln K_{eq}[/tex]
where,
[tex]\Delta G^o[/tex] = 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
[tex]K_{eq}[/tex] = equilibrium constant = ?
Putting values in above equation, we get:
[tex]10600J/mol=-(8.314J/Kmol)\times 298K\times \ln (K_{eq})\\\\K_{eq}=1.386\times 10^{-2}[/tex]
Hence, the equilibrium constant for the given reaction at 298 K is [tex]1.386\times 10^{-2}[/tex]
- For 3:
The expression of [tex]K_{eq}[/tex] for above equation follows:
[tex]K_{eq}=p_{O_2}^{3/2}[/tex]
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:
[tex]1.386\times 10^{-2}=p_{O_2}^{3/2}\\\\p_{O_2}=0.0577atm[/tex]
Hence, the equilibrium pressure of oxygen gas is 0.0577 atm
