Question

The partial molar enthalpies of a species in a simple binary mixture can sometimes be approximated by the following expressions

a) For these expressions, show that b₁ must equal b₂ .

b) For these expressions, show that the enthalpy of mixing per mole of solution at constant temperature and pressure is given by

c)  If N₁ moles of species 1 and N₂ moles of species 2, both initially at a temperature of T_i , are adiabatically mixed at constant pressure, show that the final temperature is equal to

where C_P,1 and C_P,2 are the constant molar isobaric heat capacities of species 1 and 2, respectively.

Answer 1

The enthalpy of mixing (also called heat of mixing) is the heat that is taken up or released upon mixing of two (non-reacting) chemical substances. When the enthalpy of mixing is positive, mixing is endothermic while negative enthalpy of mixing signifies exothermicmixing.

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Enthalpy of mixing

We know that in an ideal system ΔG=ΔH−TΔSΔG=ΔH−TΔS, but this equation can also be applied to the thermodynamics of mixing and solved for the enthalpy of mixing so that it reads

ΔmixH=ΔmixG+TΔmixS(1.15)(1.15)ΔmixH=ΔmixG+TΔmixS

Plugging in our expressions for ΔmixGΔmixG (Equation 1.101.10) and ΔmixSΔmixS (Equation 1.141.14) , we get

ΔmixH=nRT(x1lnx1+x2lnx2)+T[−nR(x1lnx1+x2lnx2)]=0(1.16)(1.16)ΔmixH=nRT(x1ln⁡x1+x2ln⁡x2)+T[−nR(x1ln⁡x1+x2ln⁡x2)]=0

This result makes sense when considering the system. The molecules of ideal gas are spread out enough that they do not interact with one another when mixed, which implies that no heat is absorbed or produced and results in a ΔmixHΔmixH of zero. Figure 2 illustrates how TΔmixSTΔmixS and ΔmixGΔmixG change as a function of the mole fraction so that ΔmixHΔmixH of a solution will always be equal to zero (this is for the mixing of two ideal gasses).

TAmxS AmixH 0 mix 0.0 0.2 0.4 0.6 0.8 1.0 1

I can go on but the details require statistical thermodymics modelling of enthalpy of mixing.