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# What is the lever rule in the Phase Diagram, Formula & Explanation

## What is the lever rule explain?

Lever rule: A phase diagram is a graphical representation of the system and relationships between different phases of matter coexisting within a specified range of temperature or pressure. They show the pressures, volumes, and temperatures at which different states may exist.

In chemistry, the lever rule is a formula used to determine the mole fraction (xi) or the mass fraction (wi) of each phase of a binary equilibrium phase

The lever rule is a mathematical procedure that allows us to calculate the fractions, percentages, or quantities of the phases present in equilibrium within a binary system. It is not only mathematical but also quite graphic and assertive, being very useful in physicochemical and engineering calculations.

This rule applies to phase diagrams for binary systems, regardless of the type of system itself. That is to say, the phases can be solid, as happens with alloys; or liquid and gaseous, as we see in systems in liquid-vapor equilibrium.

The lever rule can be applied directly, taking into account the values ​​plotted on the abscissa axis, where generally the global fractions or percentages of the most volatile component go, in the case of liquids; or refractory, in the case of metals in their alloys.

As will be seen below, its name is due to the immense similarity it has with the mathematical expressions that demonstrate the balance between two masses located at the ends of a seesaw with a fulcrum.

## Explanation

### Graphic aspects

In the middle of the diagram above we have a region where liquid and vapor coexist; that is, the region of liquid-vapor equilibrium. Above this region the mixture of A and B will be liquid, and below it will be gaseous because of the lower pressures.

Now, consider a mixture with a composition X B and whose pressure positions it at point D. From point D we draw a horizontal line that touches the line and the curve on the sides, originating points C and E, respectively. This line, which communicates the points C, D, and E, CDE, is what is known as the union line, and when projected towards the Y axis it must give us the pressure of the system.

Then, from these points, we draw other lines perpendicular to the junction line, which will touch the X axis. Since point E rests on the vapor curve, then we will have the mole fraction of B in the vapor phase (X V ) . Similarly, point C, on the straight line of liquid, will give us the mole fraction of B in the liquid phase (X L ).

The lever rule is based precisely on the connecting line, and the distances between X L , X B, and X V .

The global mole fraction of B is equal to:

B = B / ( L + V )

Where B is the total moles of B in both the liquid and vapor phases, and L and V are the respective moles for these phases. Solving for B we will have:

B = X L + X V (1)

On the other hand, B is also equal to:

B = L + V

= X L + X V  (2)

Now equating equations (1) and (2) will give us:

L + X V = X L + X V

And rearranging:

L (X B – X L ) = V (X V – X B ) (3)

L (CD) = V (SD)

These last two mathematical expressions are the lever rule. Note that X B – X L is the distance between points C and D; and X V – X B, is the distance between points DE: the two halves of the joint line (lever arms).

This equation is very similar to the one that describes the balance of the masses on a seesaw with a fulcrum:

1 = m 2

Thus, the lever rule will allow us to calculate the total moles L and V provided that the total moles of the mixture are known, T ( T = L + V ).

### Second way

The previous expression for the lever rule is used to calculate the quantities (masses, moles, etc.) of the phases in equilibrium at once. However, in the best-known version of the lever rule, we can calculate the fractions or percentages of each phase, taking only the distances between X B, X L, and X V.

Consider the same system as above, having another form of the lever rule below:

Where L and V are the mole fractions (or percentages, depending on the graph ) of the liquid and vapor phases, respectively. Note that obviously L and V have no units; while L and V do have units (moles, grams, etc.).

## Examples

#### Method 1

In a container, 28 moles of B and 12 moles of A are mixed. Determine the amounts and mole fractions for the phases that are formed.

We calculate X B :

B = (28 moles B) / (28 moles B + 12 moles A)

= 0.7

This value corresponds to X B in the diagram above. The interceptions will give us, approximately, the following values ​​for X L and X V :

L = 0.41

V = 0.94

With the lever rule:

L (X B – X L ) = V (X V – X B )

And knowing that T = L + V, and that T = 40 moles, then we solve for L or V as a function of the other:

L (X B – X L ) = (40 moles – L ) (X V – X B )

Reordering and solving for L we will have:

L = (40 moles) (X V – X B ) / (X V – X L )

Isn’t this expression reminiscent of L? Now substituting we will have:

L = (40 moles) (0.94 – 0.70) / (0.94 – 0.41)

= 18.11 moles in the liquid phase

We can calculate V in two ways:

V = L (X B – X L ) / (X V – X B )

or

V = 40 moles – 18.11 moles

= 21.89 moles in the vapor phase

#### Method 2

What if we first calculate L and V?

L = (X V – X B ) / (X V – X L )

= (0.94 – 0.70) / (0.94 – 0.41)

= 0.4528 or 45.28%

That is, 45.2% of the moles are in the liquid phase, this amount being equal to:

L = T

= (0.4528) (40 moles)

= 18.11 moles

And V we can calculate it equally in two ways:

V = 1 – L

or

V = (X B – X L ) / (X V – X L )

Being its value:

V = 0.5472 or 54.72%

And therefore, V will be equal to:

V = T

= (0.5472) (40 moles)

= 21.89 moles

Note that by applying the two forms of the lever rule as alternative calculation methods, the same results can be reached. Method 2 seems more straightforward and simple; But if you look closely, after solving for L or V, you will see that both methods are actually just as easy.

## Solved exercises

Next, two other exercises will be solved, where now the systems considered will involve a liquid-solid and not liquid-vapor equilibrium. Also, the diagrams are plotted with respect to the temperature of the system and not its pressure.

### Exercise 1

We have above the phase diagram for an alloy between tantalum and tungsten, Ta-W. The X-axis represents the global mass percentages of tungsten, W% (m / m).

Within the region of liquid (Ta + W) and solid (alloy) equilibrium, there is a mixture at 3200 ºC. Determine the masses of each phase assuming 100 grams of the alloy was heated.

#### Process

This time we will proceed to solve the exercise using the second form of the lever rule. The junction line tells us that: in the solid phase, we have 63% tungsten, while in the liquid phase we have 37% tungsten. This is because tungsten melts at a higher temperature (3422 ºC) than tantalum (3020 ºC).

So, we have:

W% S or W S = 63%

W% L or W L = 37%

Plus:

0 = 50.1%

We apply the lever rule for L :

L = (63% – 50.1%) / (63% – 37%)

= 0.4961 or 49.61%

Note that the distance corresponding to the liquid phase is the arm of the lever close to the solid phase, the opposite side of the middle point.

The mass of the liquid phase is, therefore:

(0.4961) (100 grams) = 49.61 grams melted

And the solid phase will be equal to:

100 grams – 49.61 grams = 50.39 grams of high tungsten alloy

### Exercise 2

For the titanium-nickel alloy at 800 ° C, and with 70% nickel, determine how much of TiNi and TiNi 3 are present.

#### Process

This time they only ask us for the mass fractions of each phase. The red point is located in the equilibrium region between the TiNi and TiNi 3 phases, whose curves are where the junction line that leads to the values ​​of 58% Ni for the TiNi phase, and 77% Ni for the TiNi 3 phase touches.

Also, note that the red dot is closer to the TiNi 3 phase than to the TiNi phase. This means that there must be more TiNi 3 than TiNi; therefore, the distance or the arm of the lever that corresponds to TiNi 3 must be the longest, the opposite (70% -58%).

Knowing this, we proceed to calculate TiNi3 :

TiNi3 = (70% – 58%) / (77% – 58%)

= 0.6316 or 63.16%

Indeed, 63.16% of the alloy corresponds to the TiNi 3 phase. Meanwhile, the TiNi phase corresponds to the:

1 = TiNi3 + TiNi

TiNi = 1 – TiNi3

= 0.3684 or 36.84%

In conclusion to the exercises, we can say that the lever rule is very helpful to determine the fractions of each phase in equilibrium for a two-component system.

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