Question

(a) Calculate the entropy and free energy change per mole when liquid water boils at 1 atm pressure. assume that for water ∆Hvap =2095.50 KJ/Kg
(b) State whether the process is spontaneous or not

Answer 1

(a).

Change in entropy of vaporization (ΔS_vap.) = ΔH_vap. (Enthalpy of vaporization) / T_vap.

ΔS_vap. = ΔH_vap. / T_vap.

Given, ΔH_vap. (Enthalpy of vaporization) = 2095.50 KJ/Kg

and

Temperature of vaporization T_vap. = 373.15 K

Conversion of 2095.50 KJ/Kg to KJ/mol :-

As,

For 1 Kg of water energy = 2095.50 KJ

Number of moles = Given mass in g / Gram molar mass

So,

Number of moles of water in 1 Kg = 1000 g / 18 g/mol

= 55.55 mol

i.e.

For 55.55 mol of water energy = 2095.50 KJ

So,

1 mol of water energy = 2095.50 KJ / 55.55 mol

= 37.72 KJ/mol

Now,

ΔS_vap. = ΔH_vap. / T_vap.

= 37.72 KJ/mol / 373.15 K

= 0.101085 KJ K^-1 mol^-1

Hence, Entropy of vaporization = ΔS_vap. = 101.085 J K^-1mol^-1

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From Gibb's Helmholtz equation is :

ΔG_vap. = ΔH_vap. - T.ΔS_vap.

= 37.72 KJ/mol - (373.15 K).(0.101085 KJ K^-1 mol^-1 )

= 37.72 KJ/mol - 37.72 KJ/mol

= 0.0 KJ/mol

i.e. at boiling point , ΔG_vap. = 0.0 KJ/mol

(b). Because, ΔG_vap. = 0, therefore reaction is not spontaneous, it is at equilibrium stage.