# Partial Molar Properties: Binary Solutions

In this example, I’d like to demonstrate how

we can get partial molar properties when we’re looking at a binary mixture from just a graphical

plot of the molar property of the mixture versus the mole fraction of one of the components.

So I’ve plotted on the black line some partial molar properties. Could be enthalpy, entropy,

Gibbs free energy, volume. And then I’ve indicated M1 which is the pure component. So this is

the pure component value for component 1, the enthalpy for example of component 1, and

likewise over here for component 2, this is the value for the solution. So at any mole

fraction, we are given a value which is not the same as the pure component values, is

not the same as an average. And then these are the partial molar quantities. So again,

partial molar enthalpy of component 1 and over here component 2. So what I’d like to

demonstrate by calculations is that when I draw a tangent line. If I pick this point,

this molar property at this mole fraction, then I can draw a tangent line to that point

and where it intersects the axes gives me the partial molar properties. And we’re going

to start off with basically how we define these partial molar quantities and we’re going

to take advantage of some equations and some manipulations to get to a relation that demonstrates

that indeed these intersections of this tangent are the partial molar properties. So I’m starting

with the definition so the property is some state function. It’s a property of the mixture

is equal to the mole fraction of component 1 times the partial molar property of component

1 plus the mole fraction of component 2 and the partial molar properties. So we’re saying

we take partial molar properties and weight them by the mole fraction to get the property

of the mixture. So what we’re going to do next is take a differential of both sides

of this equation. So here I’ve used the product rule to get two terms. First I hold x1 constant

and take the differential of the partial molar property then I hold the partial molar property

and take the differential of x1. So it’s going to turn out this is going to simplify because

this term plus this term when added together equal to zero. And this sit eh result of something

called the Gibbs-Duhem equation. So the Gibbs-Duhem equation is at constant temperature and pressure,

x1 times the differential of the partial molar quantity for component 1 plus x2 the differential

of the partial molar quantity for component 2, that equals zero. So the two terms circled

in red add to zero and as a result, the equation reducers to just 2 terms. Now we’re going

to simplify this because we know x1+x2 adds to 1. Binary systems, the two mole fractions

add to 1. If we take the differentials of both sides, right sides a constant. So this

means dx1 is equal to -dx2. So we can simplify this equation and replace dx2 by minus dx1.

And so I make that assumption if I then divide through by that differential. So my objective

here is to get an equation to solve for the partial molar quantity for M1. That partial

molar quantity in terms of M. So I want to replace this, and what I’m going to do is

go back to our original starting equation where M, the overall property for the mixture

is x1 times partial molar quantity and x2 times the partial molar quantity. And I can

use this to substitute for M2. So here I’ve solved for partial molar quantity of component

2 in the equation on the right. I’m going to substitute that back into here and I end

up then with this equation. And so now I’ve made the substitution and multiplied through

by x2 and then i’m going to rearrange by bringing the solution property to the other side. So

I have the equation in terms of the solution property and mole fractions on one side. The

other side I have x1+x2 times partial molar quantity for property 1 which of course since

x1 plus x2 adds to 1 is equal to partial molar quantity. So keep in mind this equation, we

need a bar over this symbol here to represent the partial molar quantity. And if I rearrange

this equation to just solve for the derivative which would be the same as the slope at that

point. If I pick a point here. So the property M and I’ll just draw a line so that’s M at

this mole fraction. So this is a value for x1 corresponding to M. So this would be x2

at that point. So I can see the difference. M1, partial molar quantity, minus M that’s

right here, this difference, divided by x2, that’s the slope at this point, and therefore,

that’s equal to the derivative of M with respect to x1 which means this intercept here must

be the partial molar quantity because that’s what this equation indicates. And the same

thing for partial molar quantity M2 at the other end. So we can look at one of these

graphical representations and see how the partial molar quantity changes as we get closer

to component 1, for example a point here the partial molar quantity is going to get close

to the molar quantity as we expect but the partial molar quantity for a component that’s

very dilute can be quite different from the molar quantity. Partial molar quantity if

I take a point here and draw a straight line, partial molar quantity is way down here for

component 2. And so for binary solutions, we can just look at the graph and very quickly

see how the partial molar quantity changes as we change the mole fraction.

Ternary system formules ?

Thank You!

Is the graphic still convex if you have volume instead of M?

thankyouuuuuuuu

Thank you, that was very helpful.