Answer : The equilibrium
pressure is, 
Explanation :
The given balanced chemical reaction is,

First we have to calculate the standard free energy of reaction
.

![\Delta G^o=[n_{S(s)}\times \Delta G_f^0_{(S(s))}+n_{H_2O(g)}\times \Delta G_f^0_{(H_2O(g))}]-[n_{SO_2(g)}\times \Delta G_f^0_{(SO_2(g))}+n_{H_2S(g)}\times \Delta G_f^0_{(H_2S(g))}]](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo%3D%5Bn_%7BS%28s%29%7D%5Ctimes%20%5CDelta%20G_f%5E0_%7B%28S%28s%29%29%7D%2Bn_%7BH_2O%28g%29%7D%5Ctimes%20%5CDelta%20G_f%5E0_%7B%28H_2O%28g%29%29%7D%5D-%5Bn_%7BSO_2%28g%29%7D%5Ctimes%20%5CDelta%20G_f%5E0_%7B%28SO_2%28g%29%29%7D%2Bn_%7BH_2S%28g%29%7D%5Ctimes%20%5CDelta%20G_f%5E0_%7B%28H_2S%28g%29%29%7D%5D)
where,
= standard free energy of reaction = ?
n = number of moles
Now put all the given values in this expression, we get:
![\Delta G^o=[3mole\times (0kJ/mol)+2mole\times (-228.57kJ/mol)]-[1mole\times (-300.4kJ/mol)+2mole\times (-33.01kJ/mol)]](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo%3D%5B3mole%5Ctimes%20%280kJ%2Fmol%29%2B2mole%5Ctimes%20%28-228.57kJ%2Fmol%29%5D-%5B1mole%5Ctimes%20%28-300.4kJ%2Fmol%29%2B2mole%5Ctimes%20%28-33.01kJ%2Fmol%29%5D)

Now we have to calculate the value of 

where,
= standard Gibbs free energy = -90.72 kJ/mol
R = gas constant = 8.314 J/mole.K
T = temperature = 298 K
= equilibrium constant = ?
Now put all the given values in this expression, we get:


Now we have to calculate the value of
.
The given balanced chemical reaction is,

The expression for equilibrium constant will be :

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.
Let the equilibrium
pressure be, x
Pressure of
= Pressure of
= x
Now put all the given values in this expression, we get


Thus, the equilibrium
pressure is, 