Answer : The value of
for this reaction is, 
Explanation :
The given chemical reaction is:

Now we have to calculate value of
.

![\Delta G^o=[n_{HCH_3CO_2(g)}\times \Delta G^0_{(HCH_3CO_2(g))}]-[n_{CH_3OH(g)}\times \Delta G^0_{(CH_3OH(g))}+n_{CO(g)}\times \Delta G^0_{(CO(g))}]](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo%3D%5Bn_%7BHCH_3CO_2%28g%29%7D%5Ctimes%20%5CDelta%20G%5E0_%7B%28HCH_3CO_2%28g%29%29%7D%5D-%5Bn_%7BCH_3OH%28g%29%7D%5Ctimes%20%5CDelta%20G%5E0_%7B%28CH_3OH%28g%29%29%7D%2Bn_%7BCO%28g%29%7D%5Ctimes%20%5CDelta%20G%5E0_%7B%28CO%28g%29%29%7D%5D)
where,
= Gibbs free energy of reaction = ?
n = number of moles
= -389.8 kJ/mol
= -161.96 kJ/mol
= -137.2 kJ/mol
Now put all the given values in this expression, we get:
![\Delta G^o=[1mole\times (-389.8kJ/mol)]-[1mole\times (-163.2kJ/mol)+1mole\times (-137.2kJ/mol)]](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo%3D%5B1mole%5Ctimes%20%28-389.8kJ%2Fmol%29%5D-%5B1mole%5Ctimes%20%28-163.2kJ%2Fmol%29%2B1mole%5Ctimes%20%28-137.2kJ%2Fmol%29%5D)

The relation between the equilibrium constant and standard Gibbs, free energy is:

where,
= standard Gibbs, free energy = -89.4 kJ/mol = -89400 J/mol
R = gas constant = 8.314 J/L.atm
T = temperature = 
= equilibrium constant = ?
Now put all the given values in this expression, we get:


Thus, the value of
for this reaction is, 