Intermolecular Forces

Changes of State

The three states of matter include

• solid
• liquid
• gas

In general, matter in one state can be changed into either of the other two states.

• Whenever the change involves the disruption of intermolecular forces, energy must be supplied
• The disruption of intermolecular forces accompanies the state going towards a less ordered state
• As the strengths of the intermolecular forces increase, greater amounts of energy are required to overcome them during a change in state

The melting process for a solid is also referred to as fusion

• The enthalpy change associated with melting a solid is often called the heat of fusion (DH_fus)
• Ice DH_fus = 6.01 kJ/mol

The heat needed for the vaporization of a liquid is called the heat of vaporization (DH_vap)

• Water DH_vap = 40.67 kJ/mol

Vaporization requires the input of heat energy.

• Our bodies use this as a mechanism to remove excess heat from ourselves. We sweat, and its evaporation requires heat input (the excess heat from ourselves).
• Refrigerators use the evaporation of Freon (CCl₂F₂) to remove heat inside the fridge. The Freon is condensed outside the cabinet (usually in coils at the back) in a process which releases heat energy (the coils will be warm)

The heating of ice at -25 °C to +125 °C at constant pressure (1 atm) will exhibit the following characteristics

• Initially, the heat input is used to increase the temperature of the ice, but the ice does not change phase (remains a solid)
• As the temperature approaches some critical point (i.e. the melting temperature of ice), the kinetic energy of the molecules of water is sufficient to allow the molecules to begin sliding past one another.
• As the ice begins to melt, additional input of heat energy does not raise the temperature of the water, rather it is used to overcome the intermolecular attraction during the phase change from solid to liquid
• Once the water is in a liquid phase, increasing the amount of heat input raises the temperature of the liquid water
• As the temperature approaches another critical point (the vaporization, or boiling, temperature of water) the kinetic energy of the molecules is sufficient to allow the separation of molecules into the gas phase
• As the liquid begins to boil. Additional input of heat energy does not raise the temperature of the water, rather it is used to overcome the intermolecular attractions during the phase change from liquid to gas
• Once the water is in the gas phase, additional heat input raises the temperature of the water vapor

Note: greater energy is needed to vaporize water than to melt it

In the region of the curve where we are not undergoing a phase transition, we are simply changing the temperature of one particular phase of water (either solid, liquid or gas) as a function of heat input

• The slope of the lines relates temperature to heat input
• The greater the slope, the greater the temperature change for a given unit of heat input
• The amount of heat needed to change the temperature of a substance is given by the specific heat or molar heat capacity
• Specific heat of ice = 2.09 J/g °K
• Specific heat of water = 4.18 J/g °K
• Specific heat of water vapor = 1.84 J/g °K

In the regions of the curve where we are undergoing a phase transition, the heat energy input is not raising the temperature of the sample, rather it is being used to disrupt the intermolecular forces

• DH_fus = 6.01 kJ/mol
• DH_vap = 40.67 kJ/mol

Calculate the enthalpy change for converting 2 moles of ice at -25°C to +125°C.

• Converting to grams: (2 mol)*(18 g/mol) = 36 g
• Heating ice from -25 to 0°C: (25°C)*(2.09 J/g °K)*(36 g) = 1.88 kJ
• Fusion of ice to liquid water: (6.01 kJ/mol)*(2 mol) = 12.02 kJ
• Heating of water from 0 to 100°C: (100°C)*(4.18 J/g °K)*(36 g) = 15.05 kJ
• Vaporization of water to water vapor: (40.67 kJ/mol)*(2 mol) = 81.34 kJ
• Heating of water vapor from 100 to 125°C: (1.84 J/g °K)*(25°C)*(36 g) = 1.66 kJ
• Grand total: 1.88 + 12.02 + 15.05 + 81.34 + 1.66 = 111.95 kJ

Gases can be liquified by either decreasing the temperature or increasing the pressure

• As long as the temperature is not too high, we can use pressure to liquefy a gas
• As temperatures increase it becomes more difficult to use pressure to liquefy a gas (due to the increasing kinetic energy)
• For every substance there is a temperature above which it is impossible to liquefy the gas regardless of the increase in pressure

For example, oxygen has a critical temperature of 154.4 °K. It cannot be liquefied until the temperature is reduced to this point. At this temperature, the pressure needed to liquefy oxygen is 49.7 atm.

1996 Michael Blaber