Solids and Phase Changes. How is Kinetic Theory Applied to Solids? As liquid cools, particles move more slowly. Finally particles take fixed positions.

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Presentation transcript:

Solids and Phase Changes

How is Kinetic Theory Applied to Solids? As liquid cools, particles move more slowly. Finally particles take fixed positions in a regular geometric pattern – crystallization, freezing. Strong intermolecular forces make this happen.

Properties of Solids Are rigid, less compressible than liquids Solids have definite shapes and volumes Are much denser than gases, but usually only slightly denser than liquids. Solids do not flow – they are rigid bodies.

Freezing and Melting The temperature at which solidification takes place is the freezing point. Going the other way, the melting point. The heat energy required to melt a sample of a substance is called the heat of fusion. In the reverse process, it is called the heat of crystallization.

Crystalline Solids Most solid substances are crystalline. In a crystal, atoms, ions, or molecules are arranged in an orderly, repeating 3D pattern – the crystal lattice. All crystals have a regular shape, which reflects the arrangement of particles within the solid. The angles of the faces of the crystal always are the same for a substance.

Examples of Crystalline Solids

Seven Crystal Systems Crystal SystemAnglesFaces CubicAll = 90ºa=b=c TetragonalAll = 90ºa=b OrthorhombicAll = 90ºNo faces equal Monoclinic2 angles = 90ºNo faces equal TriclinicNo angles equal No faces equal Hexagonal2 angles = 90º, 3 rd = 120º a=b RhombohedralNo angles = 90ºa=b=c

Sublimation The change of a solid to a gas or vapor without passing through the liquid state is called sublimation. What causes this? Solids also have vapor pressures – some are fairly high. Example: Iodine, dry ice (carbon dioxide)Iodinedry ice I 2 (s)  I 2 (g)CO 2 (s)  CO 2 (g)

Why Doesn’t CO 2 Form a Liquid?

Antarctic Meteorites and Sublimation

Gas  Solid (Deposition) The opposite of sublimation is deposition: H 2 O(g)  H 2 O(s) + energy

Heating Curves When a solid is heated and the transitions to liquids and gases occur, the following curve is obtained:

Heating Curves for Two Substances

Why are there “plateaus” where the temperature does not change? Temperature is a measure of average kinetic energy - energy of motion. At the melting point, molecules separate from each other somewhat increasing potential energy, not kinetic energy. At the boiling point, the separation becomes more dramatic – a greater increase in potential energy.

Cooling Curves When a gas is allowed to cool down, the opposite curve is produced:

Heat of Vaporization (H v ) When the temperature of a liquid is at its boiling point, the distance between molecules (potential energy) increases, while kinetic energy remains the same. The amount of heat required to change liquid to gas is called the heat of vaporization, expressed in J/gram.

How much heat, in joules, is needed to vaporize a 18.0 gram sample of water at 100°C? A coffee cup filled with boiling water loses 10,000 J at 100°C. How much water is in the cup?

Heat of Fusion (H f ) When a substance melts, there is no increase in molecular motion (KE), but a significant increase in potential energy. The energy associated with melting is called the heat of fusion. For water, H f = 334 J/g, considerably less than the heat of vaporization.

A 50.0 g sample of ice at 0°C is melted. How much heat was absorbed by the ice? 6000 J of heat is absorbed by an ice cube at 0°C, completely melting it. What is the mass of the cube?

Specific Heat When atoms absorb energy, they begin to move and/or vibrate faster – their kinetic energy increases. The amount of movement varies depending on the substance. Specific heat is the amount of energy required to raise 1 g of a substance 1°C.

The Specific Heat Equation It takes very little heat to raise the temperature of metals – they have very low specific heats. Covalent compounds, like water, have much higher specific heats.

The slopes of A-B, C-D, and E-F are the specific heats for solid, liquid, and gas, respectively.

A gram sample of water is heated from 20.5°C to 88.0°C. How much heat did it absorb? When 3000 J is added to a sample of water at 15°C, the temperature rises to 18°C. What is the mass of the sample?

A student melts 230 g of ice by placing the ice in a beaker in a cold room at 0˚C. How much heat did the ice absorb to melt completely at 0˚C? That same student boils the 230 g of water at 100˚C. How much heat did she add? A sample of water required removing 1000 J for it to melt. What was the mass of the water?

500 g of water was heated from 10˚C to 45˚C using a hot plate. How much heat was absorbed by the water? 10,500 J of heat were added to a sample of water. The temperature changed from 25˚C to 45˚C. What was the mass of the water? An ice cube was placed in 100 g of water at 20˚C. The temperature of the water dropped to 4˚C. when the ice completely melted. How much did the ice cube weigh?

Heat of Fusion Lab

Procedure Heat approximately 120 ml of water to about 45ºC in a 400 ml beaker. Measure the mass of a styrofoam calorimeter. Measure 100 ml of warm water into the calorimeter; measure the mass. Determine the temperature of the water. Holding an ice cube with a pair of tongs, shake off any excess water and add several cubes to the warm water in the calorimeter. Stir until the temperature is less than 1ºC. Record this final temperature. Using tongs remove the unmelted ice. Allow any water on the ice to drip back into the cup. Measure the mass of the water and the cup.

Results and Calculations Prepare a data table with all of your measurements. To calculate the heat of fusion, use the specific heat equation: Heat lost by the water = q = m water cΔT where q=heat (in Joules) c = specific heat of water(J/ºC) m=mass of warm water, and ΔT= change of temperature What is the sign of q in this equation?

Calculations Assuming that no heat is lost to the air (a dubious assumption), we can say that - Heat lost by the water = Heat gained by the ice = q So q=m ice ΔH fusion. So the heat of fusion = heat lost by water/mass of ice!

Using the data on your reference tables, calculate your % error.

Data Table 1. Mass of calorimeter __________________ 2. Mass of calorimeter + water __________________ 3. Mass of warm water __________________ (2-1) 4. Mass of calorimeter, water, ice __________________ 5. Mass of ice ___________________(4-2)