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Why and how are azeotropes formed?

I could not find a detailed explanation of how azeotropic mixtures are formed anywhere on the internet. Now, that I figured it out myself, I thought I'd share it.

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If we prepare a liquid-liquid solution of equimolar concentration, we know that the more volatile liquid component will have a greater composition in the vapour phase.

Let us keep increasing the concentration of the less volatile component in the solution. Initially, even though the solution is richer in the less volatile component, the vapour phase will be richer in the more volatile component but there will come a time when the vapour phase will be richer in the less volatile component. It may so happen that the composition of the vapour phase becomes equal to the composition in the solution phase. At this point, azeotrope is formed.

Note that if azeotrope is formed at all, both the solution and the vapour phase will be richer in the less volatile component.

Take the example of ethanol-water solution. Ethanol boils at 78.4 °C, water boils at 100 °C. Azeotropic mixture is obtained at about 95% percent of ethanol. Note that the less volatile component is present in greater amounts when the azeotropic mixture is formed.

I will keep adding more to it as it comes to me.

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  • $\begingroup$ Nice answer. But why these solutions are used? $\endgroup$ – Mockingbird Feb 14 '17 at 4:26
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    $\begingroup$ Let a mixture of alcohol & water be in equilibrium with their vapours. x is the mole fraction of A in one phase and y that in the second. The azeotrope occurs when $x=y$, which is that there is a particular composition of one phase in equilibrium with the other phase which has an equal mole fraction at a particular value of x . Then $(\partial T/\partial x)_P=0, (\partial P/\partial x)_T=0$ . $\endgroup$ – porphyrin Feb 14 '17 at 12:22
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In an Azeotrope or constant temperature boiling mixture, the vapour has the same composition as the liquid. This means that removing any distillate does not change the composition; the mixture thus acts as if it were a single phase.

To show this the Gibbs-Duhem (GD) equation is needed, which in its general form is $$ SdT - VdP + n_a d{\mu_a}+n_b d\mu_b =0 $$ where S, P and V are the total entropy, pressure and volume, $n_a, n_b$ are the number of moles. At constant temperature $dT=0$ then $$ n_ad\mu_a+n_bd\mu_b = VdP$$ It is easier to work with mole fractions $x_i$ and to express the chemical potential in terms of an ideal solution as partial pressures $p_i$ is

The definition of chemical potential allows us to replace $\mu$ with partial pressure $p_i$ : $$\mu_i^{soln}= \mu_i^{vap} = \mu_i^0 + RT\ln(p_i)$$ (The dependence of chemical potential $\mu$ on composition is known only for the special case of an ideal solution.) Differentiating gives $$d\ln(\mu_i) = RTd\ln(p_i)$$ and substituting into the GD equation produces $$RT \Sigma n_i d\ln(p_i)= VdP$$ Dividing both sides by $\Sigma n_i = N$ forms the mole fractions $x_i$ gives $$ \Sigma x_i d\ln(p_i) =\frac{VdP}{NRT}$$ Letting $v_l=V/N$ be the molar volume of the liquid and using $Pv_g=RT$ where $v_g $ is the mean molar volume of the vapour then $$ \Sigma x_i d\ln(p_i) =\frac{v_l}{v_g}d\ln(P)$$
(As $v_l << v_g$ the right hand side is very small and in some instances may be set to zero) In the case of two components $$ x_a d\ln(p_a) + x_b d\ln(p_b) =\frac{v_l}{v_g}d\ln(P) \tag 1 $$
Returning now to the Azeotrope calculation, let x be the mole fraction of component A in the condensed phase, and y the mole fraction in the vapour. The partial pressure of components A and B are $$\begin{array}{lcr} p_a = yP \\ p_b = (1-y)P\\ \end{array} $$ which can be substituted into equation $(1) $ to give $$ xd\ln(yP)+ (1-x)d\ln((1-y)P) = \frac{v_l}{v_g}d\ln(P)$$ which can be re-arranged using $d\ln(yP) = dP/(yP)$ and $d(yP)=ydP+Pdy$ to give $$\left({ \partial\ln(P)\over\partial y } \right)_T= {y-x\over y(1-y)(1-v_l/v_g)}$$ This equation describes how the total pressure varies with composition and as $v_l << v_g$ the right hand side is positive.

As the azeotrope occurs when the gradient of pressure vs composition is zero, the boiling point has a maximum or minimum value thus $$\left({ \partial\ln(P)\over\partial y } \right)_T= 0$$ and this can only be the case when $x=y$ which is when the composition of the condensed phase and vapour are equal to one another. In an alcohol/water mixture at $1 \pu{atm}$ the alcohol is present at a mole fraction of $\approx 0.46$.

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