# Boiling points and intermolecular forces relationship goals

### Intermolecular forces (video) | Khan Academy

LEARNING GOALS. , Deduce the relative boiling points and melting points of substances when given their structures or formulas. , Deduce the High melting points indicate STRONG intermolecular forces. Molecules are difficult to. In this lesson we will review what intermolecular forces are and how they will affect physical properties such as boiling point, freezing point. When you apply heat to a collection of molecules the different intermolecular forces are as follows in strength from weakest to greatest. 1. London Dispersion.

At that point, the vapor became stable. For this to be true, the pressure created by the vapor which is of course the vapor pressure must at least be equal to the pressure applied externally to the piston.

If the vapor pressure is less than the applied pressure, the vapor cannot resist the applied pressure, the piston moves in, and all of the vapor condenses into the liquid. Therefore, for the liquid to boil, the temperature must be high enough for the vapor pressure to equal the applied pressure.

Only at this temperature or above will the rate of evaporation be great enough to offset the rate of condensation created by the externally applied pressure. To find the boiling point temperature at 1 atm pressure, we need to find the temperature at which the vapor pressure is 1 atm.

Intermolecular Forces - Hydrogen Bonding, Dipole Dipole Interactions - Boiling Point & Solubility

To do so, we find the point on the graph where the vapor pressure is 1 atm and read off the corresponding temperature, which must be the boiling point.

Of course, this will work at any given pressure. We just read off of Figure 1 the temperature at which the vapor pressure equals the applied pressure, and that will be the temperature at which the liquid boils at that pressure. This means that Figure.

## Intermolecular forces

They are the same graph! Remember that in the experiment, at the boiling point we observed that both liquid and gas are at equilibrium with one another. Both phases are present at the boiling point. This is true at every combination of applied pressure and boiling point temperature. Therefore, for every combination of temperature and pressure along the curve on the graph in Figure 1, we observe liquid-gas equilibrium. What happens at combinations of temperature and pressure which are not on the line drawn in Figure 1?

We first start at any temperature-pressure combination on the curve and elevate the temperature while holding the applied pressure constant. In Figure 1, this moves us to the right of the curve. We observe that all of the liquid vaporizes, and there is only gas in the container. What happened to the equilibrium?

At higher temperature, the vapor pressure of the liquid rises, but if the applied pressure does not also increase, then the vapor pressure will be greater than the applied pressure.

### The Four Intermolecular Forces and How They Affect Boiling Points — Master Organic Chemistry

The vapor pushes back the piston and the liquid evaporates. We must therefore not be at equilibrium anymore.

For all temperature and pressure combinations to the right of the curve, only vapor exists. We observe that all of the gas condenses into the liquid.

This is because the vapor pressure is below the applied pressure, and the piston moves in against the gas until it all condenses into the liquid.

For all temperature and pressure combinations to the left of the curve, only liquid exists. What if we start at a temperature-pressure combination on the curve and elevate the applied pressure without raising the temperature? The applied pressure is now greater than the vapor pressure, and as before all of the gas will condense into the liquid.

Just as before, for all points to the left of or above the curve, only liquid exists. The opposite reasoning applies if we decrease the applied pressure. Figure 1 thus actually reveals to us what phase or phases are present at each combination of temperature and pressure: We know that, if the temperature is low enough, we expect that the water will freeze into solid. To complete the phase diagram, we need additional observations. We go back to our apparatus we used before, with a piston in a cylinder trapping liquid water and vapor in phase equilibrium.

If we slowly lower the temperature, the vapor pressure decreases slowly as well, as shown in Figure 1. However, if we continue to lower the temperature, we observe an interesting transition, as shown in the more detailed Figure 2.

Below this temperature, the pressure continues to vary smoothly, but along a slightly different curve. To understand what we have observed, we examine the contents of the container. As with the liquid-vapor curve, we can interpret this new curve in two ways.

The solid-gas curve gives the vapor pressure of the solid water as a function of temperature, and also gives the sublimation temperature as a function of applied pressure. Figure 2 is still not a complete phase diagram, because we have not included the combinations of temperature and pressure at which solid and liquid are at equilibrium.

Branching creates more spherical shapes noting that the sphere allows the maximum volume with the least surface area. The H-bonding of ethanol results in a liquid for cocktails at room temperature, while the weaker dipole-dipole of the dimethylether results in a gas a room temperature. In the last example, we see the three IMFs compared directly to illustrate the relative strength IMFs to boiling points. Boiling points and melting points The observable melting and boiling points of different organic molecules provides an additional illustration of the effects of noncovalent interactions.

The overarching principle involved is simple: Higher melting and boiling points signify stronger noncovalent intermolecular forces. Consider the boiling points of increasingly larger hydrocarbons.

More carbons means a greater surface area possible for hydrophobic interaction, and thus higher boiling points. As you would expect, the strength of intermolecular hydrogen bonding and dipole-dipole interactions is reflected in higher boiling points. Just look at the trend for hexane nonpolar London dispersion interactions only3-hexanone dipole-dipole interactionsand 3-hexanol hydrogen bonding.

Of particular interest to biologists and pretty much anything else that is alive in the universe is the effect of hydrogen bonding in water.

Because it is able to form tight networks of intermolecular hydrogen bonds, water remains in the liquid phase at temperatures up to OC, slightly lower at high altitude.

The world would obviously be a very different place if water boiled at 30 OC. Based on their structures, rank phenol, benzene, benzaldehyde, and benzoic acid in terms of lowest to highest boiling point. Solution By thinking about noncovalent intermolecular interactions, we can also predict relative melting points.

• 2.11: Intermolecular Forces & Relative Boiling Points (bp)
• Intermolecular Forces
• The Four Intermolecular Forces and How They Affect Boiling Points

All of the same principles apply: Ionic compounds, as expected, usually have very high melting points due to the strength of ion-ion interactions there are some ionic compounds, however, that are liquids at room temperature.