Raoult's law

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Raoult's law states that the vapor pressure of each component in an ideal solution is dependent on the vapor pressure of the individual component and the mole fraction of the component present in the solution.

Once the components in the solution have reached chemical equilibrium, the total vapor pressure of the solution is:

$\ P_{solution}= (P_{1})_{pure} X_1 + (P_{2})_{pure} X_2 \cdots$

and the individual vapor pressure for each component is

$\ P_{i}=(P_{i})_{pure} X_i$

where

(P_i)_pure is the vapor pressure of the pure component
X_i is the mole fraction of the component in solution

Consequently, as the number of components in a solution increases, the individual vapor pressures decrease, since the mole fraction of each component decreases with each additional component. If a pure solute which has zero vapor pressure (it will not evaporate) is dissolved in a solvent, the vapor pressure of the final solution will be lower than that of the pure solvent.

This law is strictly valid only under the assumption that the chemical bond between the two liquids is equal to the bonding within the liquids, the conditions of an ideal solution. Therefore, comparing actual measured vapor pressures to predicted values from Raoult's law allows information about the relative strength of bonding between liquids to be obtained. If the measured value of vapor pressure is less than the predicted value, fewer molecules have left the solution than expected. This is put down to the strength of bonding between the liquids being greater than the bonding within the individual liquids, so fewer molecules have enough energy to leave the solution. Conversely, if the vapor pressure is greater than the predicted value more molecules have left the solution than expected, due to the bonding between the liquids being less strong than the bonding within each.

[edit] Deduction of Raoult’s Law (or Raoult’s Equation)

We define an ideal solution, as the solution for which the chemical potential of the component $i$ is

$\mu _i = \mu _i ^\circ + RT\ln x_i$

Where the reference state is the pure substance at work $P$ and $T$ .

If the system is at equilibrium, then the chemical potential of the component $i$ must be the same in the liquid solution and in the vapor above it. That is,

$\mu _{i,L} = \mu _{i,V}\,$

If the liquid is an ideal solution, and using the formula for a gas’ chemical potential

$\mu _{i,L,P} ^\circ + RT\ln x_i = \mu _{i,1 \mathrm{bar}} ^\circ + RT\ln \frac{{f_i }} {{P^\circ }}$ (1)

(where $f$ is the fugacity of the vapor of $i$ )

If we study the component i in its pure state, we would have

$\mu _{i,L}^* = \mu _{i,V}^*\,$

Where * indicates that we study a pure component.

$\mu _{i,L,P_i^*} ^\circ + RT\ln x_i = \mu _{i,1{\mathrm{ bar}}} + RT\ln \frac{{f_i^*}} {{P^\circ }}$

But now, $x i = 1$ , so

$\mu _{i,L,P_i^*} ^\circ = \mu _{i,1{\mathrm{ bar}}} + RT\ln \frac{{f_i^*}} {{P^\circ }}$ (2)

Substracting (1)-(2) gives us

$(\mu _{i,L,P_i}^\circ - {\mu _{i,L,P_i^*} ^\circ }) + RT\ln x_i = RT\ln \frac{{f_i }} {{f_i^* }}$

which can be written as

$\frac{{\Delta \mu }} {{RT}} + \ln x_i = \ln \frac{{f_i }} {{f_i^*}}$

$e^{\frac{{\Delta \mu }} {{RT}}}x_i = \frac{{f_i }} {{f_i^*}}$

This equation is valid for the ideal solution.

Now, let’s suppose the vapor of the solution behaves as an ideal gas. In this case, fugacity and pressure are identical, and we get

$e^{\frac{{\Delta \mu }} {{RT}}} x_i = \frac{{P_i }} {{P_i^*}}$

At equilibrium we have $Δμ = 0$ , and then

$x_i \approx \frac{{P_i }} {{P_i^*}}$

Finally,

$P_i \approx x_i P_i^*$

This last equality is what is known as Raoult’s Law.

[edit] Ideal mixing

An ideal solution can be said to follow Raoult's Law but it must be kept in mind that in the strict sense ideal solutions do not exist. The fact that the vapor is taken to be ideal is the least of our worries. Interactions between gas molecules are typically quite small especially if the vapor pressures are low. The interactions in a liquid however are very strong. For a solution to be ideal we must assume that it does not matter whether a molecule A has another A as neighbor or a B molecule. This is only approximately true if the two species are almost identical chemically. We can see that from considering the Gibbs free energy change of mixing:

When we ad n₁ moles of component 1 and n₂ moles of component 2 to form an ideal liquid solution this is generally a spontaneous process. Let us consider the Gibbs free energy change of that process:

ΔG_mix= G(T,P,n₁,n₂)^sln - G*₁(T,P,n₁)* - G*₂(T,P,n₂)

G^sln = n₁μ^sln₁ + n₂μ^sln₂

G*₁ =n₁μ*₁

G*₂ =n₂μ*₂

ΔG_mix= n₁μ^sln₁ + n₂μ^sln₂-n₁μ*₁-n₂μ*₂

Using μ^sln = μ* + RTlnx we find:

ΔG_mix= = RT(n₁ln[x₁]+n₂ln[x₂])

This is always negative, so mixing is spontaneous. However the expression is -apart from a factor -T- equal to the entropy of mixing.

ΔG_mix= = -T.{-R(n₁lnx₁+n₂lnx₂)}= -T{ΔS_mix}

This leaves no room at all for an enthalpy effect:

ΔG_mix =ΔH_mix-TΔS_mix

ΔG_mix^ideal = 0 -TΔS_mix^ideal

This implies that ΔH_mix must be equal to zero and this can only be if the interactions U between the molecules are indifferent. On the average we must have:

2U_AB=U_AA+U_BB

This is seldom fulfilled even approximately.

It can be shown using the Gibbs-Duhem equation that if Raoult's law holds over the entire concentration range x=0 to 1 in abinary solution that for the second component the same must hold.

If the deviations from ideality are not too strong, Raoult's law will still be valid in a narrow concentration range when approaching x=1 for the majority phase (the solvent). The solute will also show a lineair limiting law but with a different coefficient. This law is known as Henry's law.

The presence of these limited lineair regimes has been experimentally verified in a great number of cases.

Categories: Articles lacking sources | Solutions | Distillation | Thermodynamics | Eponymous laws

Raoult's law

From Wikipedia, the free encyclopedia

[edit] Deduction of Raoult’s Law (or Raoult’s Equation)

[edit] Ideal mixing

[edit] See also

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