Answer : The values of
are
respectively.
Explanation :
The given balanced chemical reaction is,

First we have to calculate the enthalpy of reaction
.

![\Delta H^o=[n_{C_6H_6(g)}\times \Delta H_f^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta H_f^0_{(C_6H_6(l))}]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo%3D%5Bn_%7BC_6H_6%28g%29%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28C_6H_6%28g%29%29%7D%5D-%5Bn_%7BC_6H_6%28l%29%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28C_6H_6%28l%29%29%7D%5D)
where,
= enthalpy of reaction = ?
n = number of moles
= standard enthalpy of formation of gaseous benzene = 82.93 kJ/mol
= standard enthalpy of formation of liquid benzene = 49.04 kJ/mol
Now put all the given values in this expression, we get:
![\Delta H^o=[1mole\times (82.93kJ/mol)]-[1mole\times (49.04J/mol)]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo%3D%5B1mole%5Ctimes%20%2882.93kJ%2Fmol%29%5D-%5B1mole%5Ctimes%20%2849.04J%2Fmol%29%5D)

Now we have to calculate the entropy of reaction
.

![\Delta S^o=[n_{C_6H_6(g)}\times \Delta S^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta S^0_{(C_6H_6(l))}]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo%3D%5Bn_%7BC_6H_6%28g%29%7D%5Ctimes%20%5CDelta%20S%5E0_%7B%28C_6H_6%28g%29%29%7D%5D-%5Bn_%7BC_6H_6%28l%29%7D%5Ctimes%20%5CDelta%20S%5E0_%7B%28C_6H_6%28l%29%29%7D%5D)
where,
= entropy of reaction = ?
n = number of moles
= standard entropy of formation of gaseous benzene = 269.2 J/K.mol
= standard entropy of formation of liquid benzene = 173.26 J/K.mol
Now put all the given values in this expression, we get:
![\Delta S^o=[1mole\times (269.2J/K.mol)]-[1mole\times (173.26J/K.mol)]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo%3D%5B1mole%5Ctimes%20%28269.2J%2FK.mol%29%5D-%5B1mole%5Ctimes%20%28173.26J%2FK.mol%29%5D)

Now we have to calculate the Gibbs free energy of reaction
.
As we know that,

At room temperature, the temperature is
.


Therefore, the values of
are
respectively.