<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 ![1.386\times 10^{-2}](https://tex.z-dn.net/?f=1.386%5Ctimes%2010%5E%7B-2%7D)
<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:
![\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f_{(product)}]-\sum [n\times \Delta G^o_f_{(reactant)}]](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20G%5Eo_f_%7B%28product%29%7D%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20G%5Eo_f_%7B%28reactant%29%7D%5D)
For the given chemical reaction:
![M_2O_3(s)\rightarrow 2M(s)+\frac{3}{2}O_2(g)](https://tex.z-dn.net/?f=M_2O_3%28s%29%5Crightarrow%202M%28s%29%2B%5Cfrac%7B3%7D%7B2%7DO_2%28g%29)
The equation for the standard Gibbs free energy change of the above reaction is:
![\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))})]](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_%7Brxn%7D%3D%5B%282%5Ctimes%20%5CDelta%20G%5Eo_f_%7B%28M%28s%29%29%7D%29%2B%28%5Cfrac%7B3%7D%7B2%7D%5Ctimes%20%5CDelta%20G%5Eo_f_%7B%28O_2%28g%29%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20G%5Eo_f_%7B%28M_2O_3%28s%29%29%7D%29%5D)
We are given:
![\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](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_f_%7B%28M_2O_3%28s%29%29%7D%3D-10.60kJ%2Fmol%5C%5C%5CDelta%20G%5Eo_f_%7B%28M%28s%29%29%7D%3D0kJ%2Fmol%5C%5C%5CDelta%20G%5Eo_f_%7B%28O_2%28g%29%29%7D%3D0kJ%2Fmol)
Putting values in above equation, we get:
![\Delta G^o_{rxn}=[(2\times (0))+(\frac{3}{2}\times (0))]-[(1\times (-10.60))]\\\\\Delta G^o_{rxn}=10.60kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_%7Brxn%7D%3D%5B%282%5Ctimes%20%280%29%29%2B%28%5Cfrac%7B3%7D%7B2%7D%5Ctimes%20%280%29%29%5D-%5B%281%5Ctimes%20%28-10.60%29%29%5D%5C%5C%5C%5C%5CDelta%20G%5Eo_%7Brxn%7D%3D10.60kJ%2Fmol)
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:
![\Delta G^o=-RT\ln K_{eq}](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo%3D-RT%5Cln%20K_%7Beq%7D)
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:
![10600J/mol=-(8.314J/Kmol)\times 298K\times \ln (K_{eq})\\\\K_{eq}=1.386\times 10^{-2}](https://tex.z-dn.net/?f=10600J%2Fmol%3D-%288.314J%2FKmol%29%5Ctimes%20298K%5Ctimes%20%5Cln%20%28K_%7Beq%7D%29%5C%5C%5C%5CK_%7Beq%7D%3D1.386%5Ctimes%2010%5E%7B-2%7D)
Hence, the equilibrium constant for the given reaction at 298 K is ![1.386\times 10^{-2}](https://tex.z-dn.net/?f=1.386%5Ctimes%2010%5E%7B-2%7D)
The expression of
for above equation follows:
![K_{eq}=p_{O_2}^{3/2}](https://tex.z-dn.net/?f=K_%7Beq%7D%3Dp_%7BO_2%7D%5E%7B3%2F2%7D)
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:
![1.386\times 10^{-2}=p_{O_2}^{3/2}\\\\p_{O_2}=0.0577atm](https://tex.z-dn.net/?f=1.386%5Ctimes%2010%5E%7B-2%7D%3Dp_%7BO_2%7D%5E%7B3%2F2%7D%5C%5C%5C%5Cp_%7BO_2%7D%3D0.0577atm)
Hence, the equilibrium pressure of oxygen gas is 0.0577 atm