Question

What proportions of benzene and ethylbenzene should be mixed (a) by mole fraction, (b) by mass in order to reach the lowest Gibbs free energy of mixing. Repeat the same question for a mixture of (c) methanol and ethanol, and for a mixture of ethanol and 1-propanol. The above mixtures are considered ideal.

Answer 1

Gmix = -nRT X_i ln (X_i)  . Let mole fractionof benzene is x₁ & ethyl benzene x₂ . Therefore,

Gmix = -nRT[X₁ln (X₁)+ X₂ln (X₂) ] . Now for lowest free energy of mixing, ( Gmix ) =0 , X₁ +X₂= 1

On solving we get, X₁ 1/X₁ + ln (X₁) + (1- X₁ ) -1/ (1- X₁ ) + ln  (1- X₁ ) (-1) = 0

or, ln ( X₁ / 1- X₁ ) = ln1 ; 2 X₁ = 1; or X₁ = 0.5. Therefore  X₂ = 1- X₁ = 1- 0.5 = 0.5. 1

Thus equal molefraction of ethyl benzene & benzene to be mixed to reach lowest Gibbs free energy of mixing.

m _benzene = 0.5 mole = 0.5 78 gm = 39 gm ; m ethylbenzene = 0.5 mole = 0.5 106 gm= 53 gm

Part -B: For an ideal gas mixture there is no enthalpy of mixing. Hence the gibbs free energy of mixing is given by entropy term only. The lowest value is when the mole fraction is 0.5 for a mixture of two components.